Mole Concept and Stoichiometry

October 17, 2025

1. The Mole

A mole is a counting unit equal to 6.022 × 10²³ particles (atoms, molecules, ions). This is Avogadro’s number.

Molar Mass: The mass of 1 mole of a substance (element or compound) expressed in grams. It is numerically equal to its atomic/molecular mass.

2. Gay-Lussac’s Law

Gases react in simple volume ratios which are also simple whole-number ratios of their molecules, provided temperature and pressure are constant.

3. Avogadro’s Law

Equal volumes of all gases, under the same temperature and pressure, contain an equal number of molecules.

4. Important Formulae

Number of Moles (n) = Mass of substance (g) / Molar Mass (g/mol)

Number of Moles (n) = Volume of gas at STP (L) / 22.4 L/mol

Number of Particles = Number of Moles × 6.022 × 10²³

5. Stoichiometry

Uses the mole concept and balanced chemical equations to calculate:

Masses of reactants and products.

Volumes of gases involved.

Calculations are based on the law of conservation of mass.

6. Key Terms

Vapour Density: Ratio of the mass of a certain volume of a gas to the mass of an equal volume of hydrogen under the same conditions.

Vapour Density (V.D.) = Molar Mass / 2 (approximately)

Percentage Composition: The percentage by mass of each element present in a compound.

Exercise 5(A)

1. State :

(a) Gay-Lussac’s Law of combining volumes, (2011)

(b) Avogadro’s law.

Ans: (a) Gay-Lussac’s Law of Combining Volumes

Gay-Lussac’s Law of Combining Volumes states that when gases react with each other, they do so in volumes which bear a simple whole number ratio to one another and to the volumes of the gaseous products, provided all volumes are measured at the same temperature and pressure.

In simpler terms: If you measure the volumes of reacting gases and the volumes of any gaseous products, you will find that these volumes can be expressed as small whole numbers. For example, in the reaction where hydrogen combines with oxygen to form water vapour, two volumes of hydrogen always combine with one volume of oxygen to produce two volumes of steam.

(b) Avogadro’s Law

Avogadro’s Law states that equal volumes of all gases, under the same conditions of temperature and pressure, contain an equal number of molecules.

In simpler terms: It doesn’t matter what type of gas it is—one litre of hydrogen, one litre of oxygen, or one litre of any other gas—if they are at the same temperature and pressure, they will all have the exact same number of molecules. This law was crucial in distinguishing between atoms and molecules and provided the foundation for understanding gaseous reactions.

2.(a) Define atomicity of a gas. State the atomicity of Hydrogen, Phosphorus and Sulphur.

(b) Differentiate between N2 and 2N.

Ans: (a) Atomicity refers to the number of atoms present in a single molecule of an element. It classifies elements based on whether their molecules are composed of single atoms, two atoms, or more. For the given elements, Hydrogen exists as a diatomic molecule (H₂), so its atomicity is 2. Phosphorus commonly exists as a tetra-atomic molecule (P₄) in its white form, so its atomicity is 4. Sulphur, in its most stable form at room temperature, exists as an octa-atomic molecule (S₈), which means its atomicity is 8.

(b) The difference between N₂ and 2N lies in their fundamental chemical meaning. N₂ represents a molecule of nitrogen gas, which is the stable form of the element in nature. This molecule is composed of two nitrogen atoms chemically bonded together. In contrast, 2N does not represent a molecule; it signifies two separate, individual atoms of nitrogen that are not bonded to each other. It is often used in chemical equations to denote the atomic state or to show that two nitrogen atoms are required for a reaction.

3. Explain Why ?

(a) “The number of atoms in a certain volume of hydrogen is twice the number of atoms in the same volume of helium at the same temperature and pressure.”

(b) “When stating the volume of a gas, the pressure and temperature should also be given.”

(c) Inflating a balloon seems to violate Boyle’s law.

Ans:(a) This occurs because hydrogen (H₂) and helium (He) are both diatomic and monatomic gases respectively under standard conditions. According to Avogadro’s law, equal volumes of all gases at the same temperature and pressure contain an equal number of molecules. Therefore, a certain volume will contain the same number of molecules for both gases. However, each hydrogen molecule (H₂) is composed of two atoms, while each helium molecule is a single atom. Consequently, the total number of atoms in the hydrogen gas is twice that in the same volume of helium.

(b) The volume of a gas is highly dependent on its pressure and temperature because gases are compressible and expand when heated. Stating a volume without specifying the pressure and temperature is ambiguous, as that same amount of gas can occupy vastly different volumes under different conditions. Therefore, providing the pressure and temperature is essential to define the state of the gas and make the volume measurement meaningful and reproducible.

(c) Inflating a balloon seems to violate Boyle’s law because the law states that at a constant temperature, the volume of a gas is inversely proportional to its pressure. When you inflate a balloon, you are adding more air molecules, increasing the internal pressure, yet the volume also increases. This seems contradictory. The resolution is that Boyle’s law applies to a fixed mass of gas. When inflating, the mass of gas inside the balloon is not constant; it is increasing. The increase in volume is due to this addition of more air, not because the existing air is expanding against its pressure.

NUMERICAL PROBLEMS

4.(a) Calculate the volume of oxygen at STP required for the complete combustion of 100 litres of carbon monoxide at the same temperature and pressure.

2CO+O2→2CO2

Ans:Based on the balanced chemical equation 2CO + O₂ → 2CO₂, the reaction shows that 2 volumes of carbon monoxide need 1 volume of oxygen. Therefore, for 100 litres of carbon monoxide, the required oxygen volume is 50 litres at standard temperature and pressure.

(b) 200 cm3 of hydrogen and 150 cm3 of oxygen are mixed and ignited, as per following reaction,

2H2+O2→2H2O

What volume of oxygen remains unreacted ?

Ans: In the reaction 2H₂ + O₂ → 2H₂O, the gases react in a 2:1 volume ratio.

We have 200 cm³ of H₂, which would require 100 cm³ of O₂ to react completely (since 200:100 = 2:1).

We started with 150 cm³ of O₂, so after 100 cm³ reacts, the remaining oxygen is:

150 cm³ – 100 cm³ = 50 cm³.

Thus, 50 cm³ of oxygen remains unreacted.

5. 24 cc Marsh gas (CH4) was mixed with 106 cc oxygen and then exploded. On cooling, the volume of the mixture became 82 cc, of which, 58 cc was unchanged oxygen. Which law does this experiment support? Explain with calculations.

Ans: This experiment supports the Law of Combining Volumes (or Gay-Lussac’s Law of Gaseous Volumes).

The initial volumes are:

Marsh gas (CH₄) = 24 cc

Oxygen = 106 cc

After explosion and cooling, the mixture volume is 82 cc, which includes 58 cc of unreacted oxygen.

Volume of oxygen used in reaction = Initial oxygen − Unchanged oxygen

= 106 cc − 58 cc = 48 cc oxygen used.

Let the reaction be:

CH₄ + 2O₂ → CO₂ + 2H₂O (liquid after cooling)

From stoichiometry: 1 vol CH₄ reacts with 2 vol O₂.

So, 24 cc CH₄ would require 2 × 24 = 48 cc O₂, which matches exactly the oxygen consumed.

The products after cooling:

CO₂ volume produced = volume of CH₄ used (from stoichiometry) = 24 cc

Water is liquid, so negligible volume.

Final gaseous mixture = CO₂ (24 cc) + Unused O₂ (58 cc) = 82 cc, matching the observed volume.

Since the volume ratio CH₄ : O₂ : CO₂ = 24 : 48 : 24 = 1 : 2 : 1 follows small whole numbers, it confirms the Law of Combining Volumes.

6. What volume of oxygen would be required to burn 400 mL of acetylene [C2H2] ? Also, calculate the volume of carbon dioxide formed.

2C2H2+5O2→4CO2+2H2O(l)

Ans: According to the balanced equation 2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O(l), 2 volumes of acetylene require 5 volumes of oxygen.

For 400 mL of acetylene, the required oxygen volume is calculated as (5/2) × 400 mL = 1000 mL.

Similarly, 2 volumes of acetylene produce 4 volumes of carbon dioxide, so the volume of CO₂ formed is (4/2) × 400 mL = 800 mL.Thus, burning 400 mL of acetylene requires 1000 mL of oxygen and produces 800 mL of carbon dioxide.

7. 112 cm3 of H2S(g) is mixed with 120 cm3 of Cl2(g) at STP to produce HCl(g) and sulphur(s). Write a balanced equation for this reaction and calculate

(i) the volume of gaseous product formed

(ii) composition of the resulting mixture.

Ans: The balanced chemical equation for the reaction is:

H2S(g)+Cl2(g)→2HCl(g)+S(s)

At STP, 112 cm³ of H₂S is 0.005 mol (112/22400) and 120 cm³ of Cl₂ is approximately 0.00536 mol. From the stoichiometry, 1 mole of H₂S reacts with 1 mole of Cl₂. Since H₂S has the lesser amount, it is the limiting reactant.

(i) Volume of gaseous product formed:

The only gaseous product is HCl. 0.005 mol of H₂S produces 0.01 mol of HCl. At STP, this is

0.01×22400=224 cm3

(ii) Composition of the resulting mixture:

The reaction consumes all 0.005 mol of H₂S and an equal amount of Cl₂. The leftover Cl₂ is

0.00536−0.005=0.00036 mol, which is 0.00036×22400≈8 cm3

The solid sulfur formed is 0.005 mol and is part of the mixture.

Thus, the final mixture consists of 224 cm³ of HCl gas, 8 cm³ of unreacted Cl₂ gas, and 0.005 mol of solid sulfur.

8. 2500 cc of oxygen was burnt with 600 cc of ethane [C2H6]. Calculate the volume of unused oxygen and the volume of carbon dioxide formed, after writing a balanced equation :

Ethane + Oxygen → Carbon dioxide + Water vapour

Ans: In the balanced equation for the combustion of ethane, 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O, 2 volumes of ethane require 7 volumes of oxygen.

Given 600 cc of ethane, the required oxygen is (7/2) × 600 = 2100 cc.

Since 2500 cc of oxygen was supplied, the unused oxygen is 2500 – 2100 = 400 cc.

From the equation, 2 volumes of ethane produce 4 volumes of carbon dioxide.

So, 600 cc of ethane will produce (4/2) × 600 = 1200 cc of carbon dioxide.

Final Answer: Unused oxygen = 400 cc, Carbon dioxide formed = 1200 cc.

9. What volume of oxygen at STP is required to affect the combustion of 11 litres of ethylene [C2H4] at 273°C and 380 mm of Hg pressure?

C2H4+3O2→2CO2+2H2O

Ans: In the balanced equation for the combustion of ethane, 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O, 2 volumes of ethane require 7 volumes of oxygen.

Given 600 cc of ethane, the required oxygen is (7/2) × 600 = 2100 cc.

Since 2500 cc of oxygen was supplied, the unused oxygen is 2500 – 2100 = 400 cc.

From the equation, 2 volumes of ethane produce 4 volumes of carbon dioxide.

So, 600 cc of ethane will produce (4/2) × 600 = 1200 cc of carbon dioxide.

Final Answer: Unused oxygen = 400 cc, Carbon dioxide formed = 1200 cc.

10. Calculate the volume of HCl gas formed and chlorine gas required when 40 mL of methane reacts completely with chlorine at STP.

CH4+2Cl2→CH2Cl2+2HCl

Ans: When 40 mL of methane (CH₄) reacts completely with chlorine, the balanced equation is:

CH₄ + 2Cl₂ → CH₂Cl₂ + 2HCl

According to Gay-Lussac’s law of combining volumes, the mole ratio is the same as the volume ratio for gases at STP.

From the equation:

1 volume CH₄ produces 2 volumes HCl

So, 40 mL CH₄ produces:

Volume of HCl = 40 × 2 = 80 mL

Also, 1 volume CH₄ reacts with 2 volumes Cl₂

So, 40 mL CH₄ requires:

Volume of Cl₂ = 40 × 2 = 80 mL

Final answer:

Volume of HCl gas formed = 80 mL

Volume of chlorine gas required = 80 mL

11. What volume of propane is burnt for every 500 cm3 of air used in the reaction under the same conditions ? (Assuming oxygen is 1/5th of air)

C3H8+5O2→3CO2+4H2O

Ans: Based on the balanced chemical equation, 1 volume of propane requires 5 volumes of oxygen for complete combustion. The 500 cm³ sample of air contains 100 cm³ of oxygen, since oxygen constitutes one-fifth of the air’s volume. Applying the 1:5 ratio, the volume of propane that can be burnt is calculated as 100 cm³ divided by 5, which equals 20 cm³.

12. 450 cm3 of nitrogen monoxide and 200 cm3 of oxygen are mixed together and ignited. Calculate the composition of the resulting mixture.

2NO+O2→2NO2

Ans: When 450 cm³ of nitrogen monoxide (NO) and 200 cm³ of oxygen (O₂) are mixed, the reaction 2NO + O₂ → 2NO₂ occurs.From the stoichiometry, 2 volumes of NO react with 1 volume of O₂.

Thus, 450 cm³ of NO would require 225 cm³ of O₂.

However, only 200 cm³ of O₂ is available, so O₂ is the limiting reactant.

200 cm³ of O₂ will react with 400 cm³ of NO (since 2NO : 1O₂).

This produces 400 cm³ of NO₂ (since 2NO₂ : 2NO).

After reaction, NO left = 450 − 400 = 50 cm³, and O₂ left = 0 cm³.

Resulting mixture: 400 cm³ NO₂ and 50 cm³ NO.

13. If 6 litres of hydrogen and 4 litres of chlorine are mixed and exploded and if water is added to the gases formed, find the volume of the residual gas. (2015)

Ans: When 6 litres of hydrogen and 4 litres of chlorine are mixed, they react to form hydrogen chloride gas according to the reaction:

H₂ + Cl₂ → 2HCl.

Here, hydrogen and chlorine react in a 1:1 volume ratio. Since only 4 litres of chlorine is available, it is the limiting reactant.

Thus, 4 litres of hydrogen reacts with 4 litres of chlorine to produce 8 litres of HCl.

This leaves (6 − 4) = 2 litres of hydrogen unreacted.

When water is added, HCl gas dissolves completely in water, but hydrogen gas does not.

Therefore, the residual gas is the unreacted hydrogen, and its volume is 2 litres.

14. Ammonia may be oxidised to nitrogen monoxide in the presence of a catalyst according to the following equation.

4NH3+5O2→4NO+6H2O

If 27 litres of reactants are consumed, what volume of nitrogen monoxide is produced at the same temperature and pressure ?

Ans: When 27 litres of the reactants (ammonia and oxygen) are consumed in the given reaction:

4NH3+5O2→4NO+6H2O

the total mole ratio of gaseous reactants is

4+5=9 moles.

Under the same temperature and pressure, volume is proportional to moles.

The products include nitrogen monoxide (NO) and water, but since water is liquid under standard conditions, only NO is considered as a gas here.

From the equation, 9 moles of reactants produce 4 moles of NO.

Thus, volume of NO produced =

49×2794 ×27 litres = 12 litres.

Final answer: 12 litres of nitrogen monoxide is produced.

15. A mixture of hydrogen and chlorine occupying 36 cm3 exploded. On shaking it with water, 4 cm3 of hydrogen was left behind. Find the composition of the mixture.

Ans: When 36 cm³ of hydrogen and chlorine mixture exploded, the reaction was:

H₂ + Cl₂ → 2HCl.

The product HCl dissolved in water on shaking, leaving 4 cm³ of unreacted hydrogen.

This means the hydrogen was in excess.

Let the volume of chlorine be x cm³.

Then hydrogen used in the reaction = x cm³ (since 1:1 mole ratio).

Total hydrogen present = hydrogen used + leftover hydrogen = x+4 cm³.

Total mixture volume = hydrogen present + chlorine present

(x+4)+x=36

2x+4=36

2x=32

x=16

So, chlorine = 16 cm³, hydrogen = 16+4=20 cm³.

Composition: Hydrogen = 20 cm³, Chlorine = 16 cm³.

16. What volume of air (containing 20% O2by volume) will be required to burn completely 10 cm3 each of methane and acetylene.

CH4+2O2→CO2+2H2O

2C2H2+5O2→4CO2+2H2O

Ans: To burn 10 cm³ of methane (CH₄), the reaction CH₄ + 2O₂ → CO₂ + 2H₂O shows that 1 volume of CH₄ requires 2 volumes of O₂. Thus, 10 cm³ CH₄ requires 20 cm³ of O₂.

For 10 cm³ of acetylene (C₂H₂), the reaction 2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O indicates that 2 volumes of C₂H₂ require 5 volumes of O₂. So, 10 cm³ C₂H₂ requires (5/2) × 10 = 25 cm³ of O₂.

Total O₂ needed = 20 + 25 = 45 cm³.

Given that air contains 20% O₂ by volume, the volume of air required = (100 / 20) × 45 = 225 cm³.

17. LPG has 60% propane and 40% butane : 10 litres of this mixture is burnt. Calculate the volume of carbon dioxide added to atmosphere.

Ans: C3H8+5O2→3CO2+4H2O

2C4H10+13O2→8CO2+10H2O

Ans: We are given that LPG contains 60% propane (C₃H₈) and 40% butane (C₄H₁₀) by volume, and 10 litres of the mixture is burnt.

From the combustion equations:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

1 volume of propane produces 3 volumes of CO₂.

2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

2 volumes of butane produce 8 volumes of CO₂, so 1 volume of butane produces 4 volumes of CO₂.

Step 1: Volumes of each gas in 10 litres

Propane = 60% of 10 L = 6 L

Butane = 40% of 10 L = 4 L

Step 2: CO₂ from propane

6 L C₃H₈ → 6 × 3 = 18 L CO₂

Step 3: CO₂ from butane

4 L C₄H₁₀ → 4 × 4 = 16 L CO₂

Step 4: Total CO₂

18 L + 16 L = 34 L

Thus, the volume of carbon dioxide added to the atmosphere is 34 litres.

18.200cm3 of CO2 is collected at STP when a mixture of acetylene and oxygen is ignited. Calculate the volume of acetylene and oxygen at STP in the original mixture.

2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)

Ans: In the given reaction, 2 volumes of C₂H₂ react with 5 volumes of O₂ to produce 4 volumes of CO₂.

We know 18.200 cm³ of CO₂ is collected at STP.

From the stoichiometry, 4 volumes of CO₂ are produced from 2 volumes of C₂H₂.

So, volume of C₂H₂ used = (2/4) × 18.200 = 9.100 cm³.

Similarly, 4 volumes of CO₂ come from 5 volumes of O₂.

So, volume of O₂ used = (5/4) × 18.200 = 22.750 cm³.

Thus, the original mixture at STP contained 9.100 cm³ of acetylene and 22.750 cm³ of oxygen.

19. You have collected (a) 2 litres of CO2

(b) 3 litres of chlorine (c) 5 litres of hydrogen (d) 4 litres of nitrogen and (e) 1 litre of SO2, under similar conditions of temperature and pressure. Which gas sample will have :

(a) the greatest number of molecules, and

(b) the least number of molecules ?

Justify your answers.

Ans: Under the same temperature and pressure, equal volumes of all gases contain an equal number of molecules (Avogadro’s law). Therefore, the number of molecules in each gas sample depends only on its volume.

The gas with the largest volume of 5 litres (hydrogen) will have the greatest number of molecules. Conversely, the gas with the smallest volume of 1 litre (sulfur dioxide, SO₂) will have the least number of molecules. The chemical identity of the gas does not affect the number of molecules, only the volume does under these conditions.

20. The gases chlorine, nitrogen, ammonia and sulphur dioxide are collected under the same conditions of temperature and pressure. The following table gives the volumes of gases collected and the number of molecules (x) in 20 litres of nitrogen. You are to complete the table giving the number of molecules in the other gases in terms of x.

Ans: Under the same temperature and pressure, equal volumes of gases contain an equal number of molecules (Avogadro’s law).

The volume of nitrogen is 20 L and it contains *x* molecules.

Chlorine (Volume: 10 L):

Since 20 L contains *x* molecules, 10 L will contain half the number of molecules.

→ Number of molecules = x/2

Ammonia (Volume: 40 L):

40 L is double the volume of nitrogen, so it will contain double the molecules.

→ Number of molecules = 2x

Sulphur dioxide (Volume: 5 L):

5 L is one-fourth of 20 L, so it will contain one-fourth the molecules.

→ Number of molecules = x/4

21. If 100 cm3 of oxygen contains Y molecules, how many molecules of nitrogen will be present in 50 cm3 of nitrogen under the same conditions of temperature and pressure?

Ans: Under the same temperature and pressure, an equal volume of any gas contains the same number of molecules.Since 100 cm³ of oxygen contains Y molecules, the number of molecules in 50 cm³ of any gas (including nitrogen) will be half of Y.Therefore, the number of nitrogen molecules present in 50 cm³ is Y/2.

Exercise 5(B)

1.(a) The relative atomic mass of Cl atom is 35.5 a.m.u. Explain this statement.

(b) What is the value of Avogadro’s number?

(c) What is the value of molar volume of a gas at S.T.P.?

Ans: (a) The relative atomic mass of chlorine being 35.5 a.m.u. indicates that a natural sample of chlorine atoms is a mixture of isotopes, primarily chlorine-35 and chlorine-37. The value 35.5 is not the mass of a single atom, but rather the average mass of all the atoms in the sample, calculated by considering the proportional abundance of each isotope.

(b) Avogadro’s number is defined as the number of constituent particles (atoms, molecules, or ions) present in one mole of a substance. Its fixed value is 6.022 × 10²³ mol⁻¹.

(c) The molar volume of any gas at Standard Temperature and Pressure (S.T.P.), which is defined as 0°C (273 K) and 1 atmosphere of pressure, occupies a volume of 22.4 litres or 22.4 dm³.

2. Define or explain the terms :

(a) vapour density,

(b) molar volume,

(c) relative atomic mass,

(d) relative molecular mass,

(e) Avogadro’s number,

(f) Gram atom,

(g) Mole.

Ans: (a) Vapour Density

Vapour density is defined as the mass of a certain volume of a gas or vapour compared to the mass of an equal volume of hydrogen, measured under the same conditions of temperature and pressure. It is a unitless quantity. Practically, it provides a simple way to determine the molecular mass of a gas, as the molecular mass is approximately twice its vapour density.

(b) Molar Volume

Molar volume is the volume occupied by one mole of any substance in the gaseous state at a specific temperature and pressure. At standard temperature and pressure (STP, 0°C and 1 atm), one mole of any ideal gas occupies a volume of approximately 22.4 litres. This concept is fundamental in relating the volume of a gas to the amount of substance present.

Relative atomic mass is the average mass of an atom of a chemical element, compared to one-twelfth the mass of a carbon-12 atom.It is the standard atomic weight value found on the periodic table and represents the average mass of all the naturally occurring isotopes of an element.

(d) Relative Molecular Mass

Relative molecular mass is the sum of the relative atomic masses of all the atoms that constitute a molecule of a substance. Like relative atomic mass, it is a dimensionless number. It was formerly known as molecular weight and is calculated directly from the chemical formula and the standard atomic weights.

(e) Avogadro’s Number

Avogadro’s number is the fundamental constant that represents the number of constituent particles (atoms, molecules, or ions) present in one mole of a substance. Its value is approximately 6.022 × 10²³. This number provides a crucial link between the macroscopic world we can measure and the microscopic world of atoms and molecules.

(f) Gram Atom

A gram atom is a term, now less commonly used, that refers to the quantity of a chemical element whose mass in grams is numerically equal to its relative atomic mass. In modern terminology, it is simply equivalent to one mole of atoms of that element. For example, one gram atom of carbon-12 is exactly 12 grams of carbon-12.

(g) Mole

The mole is the SI unit for the amount of substance. One mole contains exactly 6.02214076 × 10²³ (Avogadro’s number) elementary entities, which may be atoms, molecules, ions, or other particles. It is a counting unit that allows chemists to work with a manageable number of particles when dealing with macroscopic quantities in the laboratory.

3.(a) What are the main applications of Avogadro’s Law?

(b) How does Avogadro’s Law explain Gay-Lussac’s Law of combining volumes?

Ans: (a) Avogadro’s Law has several key applications. It provides the foundation for determining the molar volume of gases, stating that one mole of any gas occupies the same volume at a given temperature and pressure (approximately 22.4 litres at STP). This principle is crucial in determining the molecular masses of unknown gases by comparing their densities. Furthermore, it helps in establishing the molecular formulae of gaseous compounds by relating the volumes of reacting gases to the number of molecules involved, making it indispensable for stoichiometric calculations in chemical reactions involving gases.

(b) Avogadro’s Law offers a clear particle-based explanation for Gay-Lussac’s Law of combining volumes. Gay-Lussac observed that gases react in simple whole-number volume ratios. Avogadro proposed that equal volumes of all gases, under the same conditions, contain an equal number of molecules. Therefore, the simple volume ratios observed in reactions directly correspond to simple ratios in the number of reacting molecules. For instance, the reaction where two volumes of hydrogen combine with one volume of oxygen to form two volumes of water vapour implies that two molecules of hydrogen combine with one molecule of oxygen to yield two molecules of water, perfectly aligning the macroscopic volume ratios with the microscopic molecular ratios.

4. Calculate the relative molecular masses of :

(a) Ammonium chloroplatinate

(NH4)2PtCl6

(b) Potassium chlorate,

(c)CuSO4⋅5H2O

(d)(NH4)2SO4

(e)CH3COONa

(f)CHCl3

(g)(NH4)2Cr2O7

Ans: The relative molecular mass (also known as molecular weight) is calculated by summing the atomic masses of all the atoms present in the chemical formula. The atomic masses used are from the periodic table (C=12, H=1, N=14, O=16, Na=23, S=32, Cl=35.5, K=39, Cr=52, Pt=195, Cu=63.5).

(a) Ammonium chloroplatinate, (NH₄)₂PtCl₆

Nitrogen (N): 2 atoms × 14 = 28

Hydrogen (H): 8 atoms × 1 = 8

Platinum (Pt): 1 atom × 195 = 195

Chlorine (Cl): 6 atoms × 35.5 = 213

Relative Molecular Mass = 28 + 8 + 195 + 213 = 444

(b) Potassium chlorate, KClO₃

Potassium (K): 1 atom × 39 = 39

Chlorine (Cl): 1 atom × 35.5 = 35.5

Oxygen (O): 3 atoms × 16 = 48

Relative Molecular Mass = 39 + 35.5 + 48 = 122.5

This formula includes 5 water molecules. We calculate the mass of CuSO₄ and the mass of 5H₂O separately and then add them.

Copper Sulphate (CuSO₄):

Copper (Cu): 1 atom × 63.5 = 63.5

Sulphur (S): 1 atom × 32 = 32

Oxygen (O): 4 atoms × 16 = 64

Mass of CuSO₄ = 63.5 + 32 + 64 = 159.5

Water of Crystallisation (5H₂O):

Hydrogen (H): 10 atoms × 1 = 10

Oxygen (O): 5 atoms × 16 = 80

Mass of 5H₂O = 10 + 80 = 90

Relative Molecular Mass = Mass of CuSO₄ + Mass of 5H₂O = 159.5 + 90 = 249.5

(d) Ammonium Sulphate, (NH₄)₂SO₄

Nitrogen (N): 2 atoms × 14 = 28

Hydrogen (H): 8 atoms × 1 = 8

Sulphur (S): 1 atom × 32 = 32

Oxygen (O): 4 atoms × 16 = 64

Relative Molecular Mass = 28 + 8 + 32 + 64 = 132

(e) Sodium Acetate, CH₃COONa

Carbon (C): 2 atoms × 12 = 24

Hydrogen (H): 3 atoms × 1 = 3

Oxygen (O): 2 atoms × 16 = 32

Sodium (Na): 1 atom × 23 = 23

Relative Molecular Mass = 24 + 3 + 32 + 23 = 82

(f) Chloroform, CHCl₃

Carbon (C): 1 atom × 12 = 12

Hydrogen (H): 1 atom × 1 = 1

Chlorine (Cl): 3 atoms × 35.5 = 106.5

Relative Molecular Mass = 12 + 1 + 106.5 = 119.5

(g) Ammonium Dichromate, (NH₄)₂Cr₂O₇

Nitrogen (N): 2 atoms × 14 = 28

Hydrogen (H): 8 atoms × 1 = 8

Chromium (Cr): 2 atoms × 52 = 104

Oxygen (O): 7 atoms × 16 = 112

Relative Molecular Mass = 28 + 8 + 104 + 112 = 252

Summary of Answers:

(a) (NH₄)₂PtCl₆ : 444

(b) KClO₃ : 122.5

(d) (NH₄)₂SO₄ : 132

(e) CH₃COONa : 82

(f) CHCl₃ : 119.5

(g) (NH₄)₂Cr₂O₇ : 252

5. Find the :

(a) number of molecules in 73 g of HCl,

(b) weight of 0.5 mole of O2,

(c) number of molecules in 1.8 g of H2O,

(d) number of moles in 10 g of CaCO3,

(e) weight of 0.2 mole of H2gas,

(f) number of molecules in 3.2 g of SO2.

Ans: (a) 73 g of HCl

Molar mass of HCl = 36.5 g/mol

Moles = 73 ÷ 36.5 = 2 moles

Molecules = 2 × 6.022 × 10²³ = 1.2044 × 10²⁴ molecules

(b) 0.5 mole of O₂

Molar mass of O₂ = 32 g/mol

Mass = 0.5 × 32 = 16 g

Molar mass of H₂O = 18 g/mol

Moles = 1.8 ÷ 18 = 0.1 mole

Molecules = 0.1 × 6.022 × 10²³ = 6.022 × 10²² molecules

(d) 10 g of CaCO₃

Molar mass of CaCO₃ = 100 g/mol

Moles = 10 ÷ 100 = 0.1 mole

(e) 0.2 mole of H₂ gas

Molar mass of H₂ = 2 g/mol

Mass = 0.2 × 2 = 0.4 g

(f) 3.2 g of SO₂

Molar mass of SO₂ = 64 g/mol

Moles = 3.2 ÷ 64 = 0.05 mole

Molecules = 0.05 × 6.022 × 10²³ = 3.011 × 10²² molecules

6. Which of the following would weigh most :

(a) 1 mole of H2O,

(b) 1 mole of CO2,

(c) 1 mole of NH3,

(d) 1 mole of CO ?

Ans: (b) 1 mole of CO2,

7. Which of the following contains maximum number of molecules :

(a) 4 g of O2,

(b) 4 g of NH3,

(c) 4 g of CO2,

(d) 4 g of SO2?

Ans: Step 1: Understanding the Concept

The number of molecules in a given mass of a substance is directly proportional to the number of moles it contains. This is because one mole of any substance contains the same number of molecules, which is Avogadro’s number (6.022×1023 ).

Therefore, to find which sample has the most molecules, we simply need to find which one has the highest number of moles.

The number of moles (n) is calculated as:

n= Given Mass (g) / Molar Mass (g/mol)

Since the given mass is the same (4 g) for all the substances, the substance with the smallest molar mass will have the largest number of moles, and hence, the maximum number of molecules.

Step 2: Calculate the Molar Mass of Each Substance

(a) Oxygen gas (O₂):

Molar mass = 2×16=32 g/mol

(b) Ammonia (NH₃):

Molar mass =

14+(3×1)=17 g/mol

Molar mass =

12+(2×16)=44 g/mol

(d) Sulfur dioxide (SO₂):

Molar mass =

32+(2×16)=64 g/mol

Step 3: Calculate the Number of Moles in 4 g of Each Substance

(a) Moles of O₂:

n= 432= 0.125 mol

(b) Moles of NH₃:

n= 174 ≈0.235 mol

n= 444 ≈0.091 mol

(d) Moles of SO₂:

n= 64 =0.0625 mol

Step 4: Compare the Number of Moles

Comparing the calculated moles:

NH3(0.235 mol)>O2(0.125 mol)>CO2(0.091 mol)>SO2(0.0625 mol)

Ammonia (NH₃) has the highest number of moles.

Step 5: Final Conclusion

Since 4 g of NH₃ has the largest number of moles, it contains the maximum number of molecules.

(b) 4 g of NH3

8. Calculate the number of :

(a) particles in 0.1 mole of any substance,

(b) hydrogen atoms in 0.1 mole of H2SO4,

(c) molecules in one kg of calcium chloride.

Ans: (a) Particles in 0.1 mole of any substance

In chemistry, the mole is a fundamental unit for counting particles. One mole of any substance, whether it’s atoms, molecules, or ions, always contains the same number of particles, known as Avogadro’s constant. This value is approximately 6.022 × 10²³.Therefore, for 0.1 mole of a substance, the number of particles is calculated by multiplying Avogadro’s constant by 0.1.

Number of particles = 0.1 × (6.022 × 10²³) = 6.022 × 10²²

Conclusion: There are 6.022 × 10²² particles in 0.1 mole of any substance.

(b) Hydrogen atoms in 0.1 mole of H₂SO₄

To find the number of hydrogen atoms, we first need to find the number of sulfuric acid (H₂SO₄) molecules. As established, 0.1 mole of any substance contains 6.022 × 10²² fundamental units. In this case, the units are H₂SO₄ molecules.

Number of H₂SO₄ molecules = 0.1 × (6.022 × 10²³) = 6.022 × 10²² molecules

Now, looking at the chemical formula H₂SO₄, we see that each molecule contains 2 hydrogen (H) atoms. To find the total number of hydrogen atoms, we multiply the total number of molecules by 2.

Total H atoms = (6.022 × 10²² molecules) × 2 = 1.2044 × 10²³ atoms

Conclusion: There are approximately 1.204 × 10²³ hydrogen atoms in 0.1 mole of H₂SO₄.

First, we determine the molar mass of calcium chloride (CaCl₂).

Atomic mass of Calcium (Ca) = 40 g/mol

Atomic mass of Chlorine (Cl) = 35.5 g/mol

Molar mass of CaCl₂ = 40 + 2(35.5) = 40 + 71 = 111 g/mol.

This means that one mole of CaCl₂ has a mass of 111 grams.

The given mass is 1 kilogram, which is 1000 grams. To find out how many moles this represents, we use the formula:

Number of moles = Given mass / Molar mass

Number of moles = 1000 g / 111 g/mol ≈ 9.009 moles

Finally, we calculate the number of molecules by multiplying the number of moles by Avogadro’s constant.

Number of molecules = 9.009 moles × (6.022 × 10²³ molecules/mole)

Number of molecules ≈ 5.426 × 10²⁴ molecules

Conclusion: There are approximately 5.426 × 10²⁴ molecules in one kilogram of calcium chloride.

9. How many grams of :

(a) Al are present in 0.2 mole of it,

(b) HCl are present in 0.1 mole of it ?

(c) H2O are present in 0.2 mole of it,

(d) CO2is present in 0.1 mole of it ?

Ans: The mass of a substance is calculated using the formula:

Mass = Number of moles × Molar mass

(a) Mass of Aluminium (Al)

Number of moles = 0.2 mol

Molar mass of Al = 27 g/mol

Mass = 0.2 mol × 27 g/mol = 5.4 grams

(b) Mass of Hydrochloric Acid (HCl)

Number of moles = 0.1 mol

Molar mass of HCl = Atomic mass of H + Atomic mass of Cl = 1 + 35.5 = 36.5 g/mol

Mass = 0.1 mol × 36.5 g/mol = 3.65 grams

Number of moles = 0.2 mol

Molar mass of H₂O = (2 × Atomic mass of H) + Atomic mass of O = (2 × 1) + 16 = 18 g/mol

Mass = 0.2 mol × 18 g/mol = 3.6 grams

(d) Mass of Carbon Dioxide (CO₂)

Number of moles = 0.1 mol

Molar mass of CO₂ = Atomic mass of C + (2 × Atomic mass of O) = 12 + (2 × 16) = 44 g/mol

Mass = 0.1 mol × 44 g/mol = 4.4 grams

10.(a) The mass of 5.6 litres of a certain gas at STP is 12 g. What is the relative molecular mass or molar mass of the gas ?

(b) Calculate the volume occupied at S.T.P. by 2 moles of SO2.

Ans: (a)Volume at STP = 5.6 L

Mass = 12 g

Moles of gas =

5.6/22.4=0.25mol

Molar mass =

12/0.25=48g/mol

Relative molecular mass = 48

(b)1 mole of gas at STP occupies 22.4 L

2 moles of SO2

occupy =2×22.4=44.8 L

Volume = 44.8 L

11. Calculate the number of moles of :

(a) CO2which contain 8.00 g of O2,

(b) methane in 0.80 g of methane.

Ans: (a) Moles of CO₂ containing 8.00 g of O₂

Molar mass of O₂ = 32 g/mol

Moles of O₂ = 8.00 / 32 = 0.25 mol

Each mole of O₂ has 2 moles of O atoms → 0.25 × 2 = 0.5 mol O atoms

Each CO₂ has 2 O atoms → Moles of CO₂ = 0.5 / 2 = 0.25 mol

(b) Moles of methane in 0.80 g

Molar mass of CH₄ = 16 g/mol

Moles = 0.80 / 16 = 0.05 mol

12. Calculate the actual mass of :

(a) an atom of oxygen,

(b) an atom of hydrogen,

(c) a molecule of NH3,

(d) the atom of silver,

(e) the molecule of oxygen,

(f) 0.25 gram atom of calcium.

Ans: (a) Actual mass of an atom of oxygen

= Atomic mass / Avogadro’s number

= 16 g/mol / (6.022 × 10²³ atoms/mol)

≈ 2.656 × 10⁻²³ g

(b) Actual mass of an atom of hydrogen

= 1 g/mol / (6.022 × 10²³ atoms/mol)

≈ 1.660 × 10⁻²⁴ g

Molecular mass of NH₃ = 14 + 3(1) = 17 g/mol

= 17 g/mol / (6.022 × 10²³ molecules/mol)

≈ 2.823 × 10⁻²³ g

(d) Actual mass of an atom of silver

= 108 g/mol / (6.022 × 10²³ atoms/mol)

≈ 1.793 × 10⁻²² g

(e) Actual mass of a molecule of oxygen (O₂)

Molecular mass of O₂ = 32 g/mol

= 32 g/mol / (6.022 × 10²³ molecules/mol)

≈ 5.313 × 10⁻²³ g

(f) Mass of 0.25 gram atom of calcium

Gram atomic mass of Calcium (Ca) = 40 g

Mass = 0.25 × 40 g

= 10 g

13. Calculate the mass of 0.1 mole of each of the following:

(a) CaCO3,

(b) Na2SO4· 10H2O,

(c) CaCl2,

(d) Mg.

(Ca = 40, Na = 23, Mg = 24, S = 32, C = 12, Cl = 35.5, O = 16, H = 1)

Ans: (a) 10 g (b) 32.2 g (c) 11.1 g (d) 2.4 g

14. Calculate the number of oxygen atoms in 0.10 mole of Na2CO3· 10H2O.

Ans: Step 1: Determine number of oxygen atoms in one formula unit

Formula: Na₂CO₃·10H₂O

Na₂CO₃ contains 3 O atoms

10H₂O contains 10 × 1 = 10 O atoms? Wait — each H₂O has 1 O atom, so 10 H₂O has 10 O atoms.

Total O atoms per formula unit = 3 (from CO₃) + 10 (from 10H₂O) = 13 O atoms.

Step 2: Calculate total O atoms in 0.10 mol

Number of formula units in 0.10 mol

= 0.10 × (6.022 × 10²³)

= 6.022 × 10²² formula units.

Each formula unit has 13 O atoms, so:

O atoms = 6.022 × 10²² × 13

= 78.286 × 10²²

= 7.8 × 10²³ O atoms.

Final Answer:7.8×10 *23

15. What mass of Ca will contain the same number of atoms as are present in 3.2 g of S ?

Ans: Step 1: Find moles of sulfur (S).

Molar mass of S = 32 g/mol

Given mass = 3.2 g

Moles of S = 3.2/32=0.1mol

Step 2: Number of atoms same → moles same.

So, moles of Ca needed = 0.1 mol

Step 3: Mass of Ca

Molar mass of Ca = 40 g/mol

Mass = 0.1×40=4.0 g

Final answer: 4.0 g

16. Calculate the number of atoms in each of the following :

(a) 52 moles of He,

(b) 52 amu of He,

(c) 52 g of He.

Ans: Step 1 – Recall constants

1 mole of atoms = 6.022×1023 atoms

Molar mass of He = 4 g/mol

Mass of 1 He atom ≈ 4 amu

(a) 52 moles of He

Number of atoms=52×6.022×1023=3.131×1025 atoms

(b) 52 amu of He

Number of atoms= 4 amu/atom/52 amu =13 atoms

Moles=52/4=13 moles

Atoms=13×6.022×1023=7.829×10/24 atoms

Final answers:(a) 3.131×10/25 atoms

(b) 13 atoms

17. Calculate the number of atoms of each kind in 5.3 grams of sodium carbonate.

Ans: Step 1: Write the chemical formula of sodium carbonate.

The formula is Na₂CO₃.

Step 2: Calculate the molecular mass of Na₂CO₃.

Mass of 2 Sodium (Na) atoms = 2 × 23 = 46 u

Mass of 1 Carbon (C) atom = 1 × 12 = 12 u

Mass of 3 Oxygen (O) atoms = 3 × 16 = 48 u

Molecular mass of Na₂CO₃ = 46 + 12 + 48 = 106 u

So, the gram molecular mass of Na₂CO₃ is 106 grams/mol. This means 106 grams of sodium carbonate contain 1 mole of Na₂CO₃ molecules.

Step 3: Find the number of moles in 5.3 grams of Na₂CO₃.

Number of moles (n) = (Given Mass) / (Molar Mass)

n = 5.3 g / 106 g/mol

n = 0.05 moles

Step 4: Calculate the number of molecules of Na₂CO₃.

1 mole of any substance contains 6.022 × 10²³ molecules (Avogadro’s number).

Number of molecules = Number of moles × Avogadro’s number

= 0.05 mol × 6.022 × 10²³ molecules/mol

= 3.011 × 10²² molecules

Step 5: Determine the number of atoms of each kind.

Now, we analyze one molecule of Na₂CO₃:

One molecule of Na₂CO₃ contains:

2 atoms of Sodium (Na)

1 atom of Carbon (C)

3 atoms of Oxygen (O)

Therefore, the total number of atoms of each element is:

Number of Sodium (Na) atoms = (Number of molecules) × 2

= (3.011 × 10²²) × 2

= 6.022 × 10²² atoms

Number of Carbon (C) atoms = (Number of molecules) × 1

= (3.011 × 10²²) × 1

= 3.011 × 10²² atoms

Number of Oxygen (O) atoms = (Number of molecules) × 3

= (3.011 × 10²²) × 3

= 9.033 × 10²² atoms

18.(a) Calculate the mass of nitrogen supplied to soil by 5 kg of urea [CO(NH2)

2].[O = 16; N = 14; C = 12; H = 1]

(b) Calculate the volume occupied by 320 g of sulphur dioxide at STP [S = 32; O = 16]

Ans: (a) Mass of nitrogen in 5 kg of urea

Step 1: Molar mass of Urea, CO(NH₂)₂

C = 12 g/mol

O = 16 g/mol

N = 14 × 2 = 28 g/mol

H = 1 × 4 = 4 g/mol

Molar mass = 12 + 16 + 28 + 4 = 60 g/mol

Step 2: Mass of N in 1 mol urea

1 mol urea (60 g) contains 28 g of N.

Step 3: Mass of N in 5 kg (5000 g) urea

Mass of N = (28 / 60) × 5000 g

= (28 × 5000) / 60

= 140000 / 60

= 2333.33 g or 2.33 kg

Answer (a): 2.33 kg

(b) Volume of 320 g of SO₂ at STP

Step 1: Molar mass of SO₂

S = 32 g/mol

O = 16 × 2 = 32 g/mol

Molar mass = 32 + 32 = 64 g/mol

Step 2: Moles in 320 g of SO₂

Moles = Given mass / Molar mass

= 320 g / 64 g mol⁻¹

= 5 moles

Step 3: Volume at STP

1 mole at STP = 22.4 L

5 moles = 5 × 22.4 L

= 112 L

Answer (b): 112 L

19.(a) What do you understand by the statement that ‘vapour density of carbon dioxide is 22’ ?

(b) Atomic mass of chlorine is 35.5. What is its vapour density ?

Ans:(a)Vapour density of carbon dioxide is 22 means that the mass of a certain volume of carbon dioxide gas is 22 times the mass of an equal volume of hydrogen gas measured under the same conditions of temperature and pressure.

(b)Vapour density (VD) of a gas = Molecular mass/2

For chlorine (Cl₂), molecular mass = 2×35.5=71

So, VD = 71/2=35.52

Ans:

(a) Carbon dioxide is 22 times heavier than hydrogen.

(b) Vapour density of chlorine is 35.5.

20. What is the mass of 56 cm3of carbon monoxide at STP?

[C = 12; O = 16]

Ans: Step 1: Moles of gas

Volume = 56 cm³

Molar volume = 22400 cm³/mol

Moles = 56/22400=0.0025

Step 2: Molar mass of CO

Molar mass = 12 + 16 = 28 g/mol

Step 3: Mass

Mass = 0.0025×28=0.07g

Final Answer: 0.07 g

21. Determine the no. of molecules in a drop of water which weighs 0.9 g.

Ans: Step 1: Understand what we are given and what we need.

We know:

Mass of the water drop = 0.9 grams

We need to find the number of molecules.

We also need to recall a fundamental concept from chemistry:

The molar mass of water (H₂O) is 18 g/mol. This means that 18 grams of water contain one mole of water molecules.

Avogadro’s number (Nₐ) is 6.022 × 10²³.

So, our plan is to find out how many moles are in 0.9 grams of water, and then multiply that by Avogadro’s number to find the number of molecules.

Step 2: Calculate the number of moles.

The formula for the number of moles (n) is:

n=Given Mass/Molar Mass

Plugging in our values:

N =0.9g/18g/mol=0.05moles

So, the drop of water contains 0.05 moles of H₂O molecules.

Step 3: Calculate the number of molecules.

Now, we use Avogadro’s number to convert moles to molecules.

Number of Molecules=0.05mol×(6.022×1023molecules/mol)

Number of Molecules=0.05mol×(6.022×1023 molecules/mol)

Let’s do the multiplication:

0.05×6.022=0.3011

So,Number of Molecules=0.3011×10/23

To write this in proper scientific notation, we can express it as:

Number of Molecules=3.011×1022

Final Answer:

3.011×1022

This is the number of individual water molecules in a 0.9-gram drop of water. It’s a staggeringly large number, which shows just how tiny a single molecule is

22. The molecular formula for elemental sulphur is S8. In a sample of 5.12 g of sulphur :

(a) How many moles of sulphur are present,

(b) How many molecules and atoms are present ?

Ans: Step 1: Calculate the molar mass of S₈

Atomic mass of S = 32 g/mol

Molar mass of S₈ = 8 × 32 = 256 g/mol

(a) Moles of sulphur

Moles=Mass5.12/256=0.02 mol

Moles= 256/5.12 =0.02 mol

(b) Number of molecules of S₈

Molecules=Moles×N /A

=0.02×6.022×10 23=1.2044×1022 molecules

Number of sulphur atoms

Each S₈ molecule has 8 atoms, so:

Atoms=1.2044×10 22 ×8=9.635×10 *22 atoms

23. If phosphorus is considered to contain P4 molecules, then calculate the number of moles in 100 g of phosphorus?

Ans: Step 1: Understand the given information

We are told that phosphorus exists as P₄ molecules. This means the basic unit for our calculation is a molecule made of 4 phosphorus atoms.

The mass of the phosphorus sample given is 100 g.

Step 2: Determine the molar mass of P₄

The atomic mass of phosphorus (P) is 31 g/mol.

Since one P₄ molecule contains 4 phosphorus atoms, its molar mass is:

Molar mass of P₄ = 4 × 31 g/mol = 124 g/mol.

This means that one mole of P₄ molecules has a mass of 124 grams.

Step 3: Apply the mole formula

Number of moles=Given mass/Molar mass

Number of moles= Molar mass/Given mass

Step 4: Substitute the values and calculate

Number of moles of P₄=100g/124 g/mol

Number of moles of P₄= 0.80645mol

Number of moles of P₄=0.80645mol

Step 5: State the final answer

Rounding off to an appropriate number of significant figures, the number of moles in 100 g of phosphorus (P₄) is:0.806

24. Calculate :

(a) the gram molecular mass of chlorine if 308 cm3 of it at STP weighs 0.979 g,

(b) the volume of 4 g of H2 at 4 atmosphere,

(c) the mass of oxygen in 2.2 litres of CO2 at STP.

Ans: 24. (a) Gram Molecular Mass of Chlorine

Given:Volume of chlorine gas at STP = 308 cm³

Mass of this volume = 0.979 g

Step 1: Recall the principle.

The gram molecular mass of any gas occupies 22.4 litres at STP.

Step 2: Convert the given volume to litres.

We know that 1000 cm³ = 1 litre.

So, 308 cm³ = 308 / 1000 = 0.308 litres.

Step 3: Find the mass of 22.4 litres.

If 0.308 litres of chlorine weighs 0.979 g,

Then, 1 litre of chlorine would weigh 0.979 g.

Therefore, 22.4 litres (the molar volume) would weigh

0.979/0.308×22.4g.

Step 4: Perform the calculation.

First,

0.979/0.308=3.17857

Then,

3.17857×22.4=71.2 g (approximately)

Final Answer for (a):

The gram molecular mass of chlorine is 71.2 g.

(b) Volume of 4 g of H₂ at 4 atmosphere

Given:

Mass of Hydrogen gas (H₂) = 4 g

Pressure = 4 atm

Temperature is assumed to be constant (not specified, so we assume it’s the same as STP for the application of Boyle’s Law).

Step 1: Find the number of moles of H₂.

Molecular mass of H₂ = 2 g/mol.

Number of moles (n) =

Given Mass

Molar Mass=42=2

Molar Mass

Given Mass = 24 =2 moles.

Step 2: Find the volume at STP (1 atm).

At STP, 1 mole of any gas occupies 22.4 litres.

So, volume of 2 moles at STP (V₁) = 2×22.4=44.8 litres.

Pressure at STP (P₁) = 1 atm.

Step 3: Find the volume at the new pressure (4 atm) using Boyle’s Law.

Boyle’s Law states: P₁V₁ = P₂V₂ (at constant temperature).

We have:

P₁ = 1 atm

V₁ = 44.8 litres

P₂ = 4 atm

V₂ = ? (This is what we need to find)

Step 4: Apply Boyle’s Law.

P1V1=P2V2

1×44.8=4×V 2

V2=44.8/4 =11.2 litres.

Final Answer for (b):

The volume of 4 g of H₂ at 4 atmosphere is 11.2 litres.

Given:Volume of CO₂ at STP = 2.2 litres

Step 1: Find the number of moles of CO₂.

At STP, 22.4 litres of any gas = 1 mole.

So, number of moles in 2.2 litres = 22.4=22/

224=11/112 moles.

Step 2: Analyze the composition of CO₂.

One molecule of CO₂ contains one atom of Carbon and two atoms of Oxygen.

Therefore, 1 mole of CO₂ contains 2 moles of Oxygen atoms.

Step 3: Find the number of moles of Oxygen.

Moles of O = 2×

moles of CO₂=2×moles of CO₂=2× 112/11 = 112/22= 56/11moles.

Step 4: Calculate the mass of Oxygen.

Atomic mass of Oxygen (O) = 16 g/mol.

Since oxygen exists as O₂ in this context, but the mass calculation remains the same for the atoms, the mass of oxygen is:

Mass = Number of moles of O atoms × Molar mass of O

Mass = 11/56×16 g

Step 5: Perform the calculation.

11/56×16=11×16/56=176/56=3.142=3.142 g (approximately)

Final Answer for (c):

The mass of oxygen in 2.2 litres of CO₂ at STP is 3.14 g.

25. A student puts his signature with graphite pencil. If the mass of carbon in the signature is 10−12g, calculate the number of carbon atoms in the signature.

Ans: Calculating the Number of Atoms in a Microscopic Carbon Sample

Let’s determine how many atoms are in an incredibly small speck of carbon, weighing just one picogram (which is 0.000000000001 grams or 10⁻¹² g).

Step 1: Find the Number of Moles of Carbon

First, we need to understand that a “mole” is a chemist’s counting unit, similar to how a “dozen” means 12 of something. One mole of any substance contains exactly Avogadro’s number of particles (atoms, molecules, etc.), which is approximately 6.022 × 10²³.

The number of moles (*n*) is calculated by dividing the given mass of a substance by its molar mass (the mass of one mole of that substance).

Given Mass of Carbon: 10⁻¹² grams

Molar Mass of Carbon: 12 grams per mole (from the periodic table)

The calculation is:

*n* = (10⁻¹² g) / (12 g/mol) = 8.33 × 10⁻¹⁴ moles

So, this tiny carbon sample amounts to 0.0000000000000833 moles.

Step 2: Find the Number of Atoms

Now that we know how many moles we have, we can find the actual number of atoms. We simply multiply the number of moles by Avogadro’s number, which tells us how many atoms are in each mole.

Number of Moles (*n*): 8.33 × 10⁻¹⁴ mol

Avogadro’s Number (Nₐ): 6.022 × 10²³ atoms/mol

The calculation is:

N = (8.33 × 10⁻¹⁴ mol) × (6.022 × 10²³ atoms/mol)

Let’s break that multiplication down:

Multiply the numbers: 8.33 × 6.022 ≈ 50.15

Add the exponents: 10⁻¹⁴ × 10²³ = 10⁽⁻¹⁴ ⁺ ²³⁾ = 10⁹

This gives us a preliminary result of 50.15 × 10⁹

To express this properly in scientific notation, we adjust it to 5.015 × 10¹⁰.

Final Answer: Therefore, a picogram (10⁻¹² g) sample of carbon contains approximately 5.02 × 10¹⁰ atoms.

To put this enormous number into perspective, even in such an unimaginably small sample, there are still over 50 billion individual carbon atoms.

26. An unknown gas shows a density of 3 g per litre at 273°C and 1140 mm Hg pressure. What is the gram molecular mass of this gas ?

Ans: Step 1: Given data

Density d=3 g/L

Temperature

T=273

C=273+273=546 K

Pressure

P=1140 mm Hg

Gas constant

Step 2: Convert pressure to atm

P=1140/760 atm=1.5 atm

Step 3: Use the relationship

For an ideal gas:

PM=dRT

where

M = molar mass (g/mol).

M=dRTP

Step 4: Substitute values

M=3×0.0821×546/1.5

M= 1.5

M=134.4798/1.5≈89.65 g/mol

Final Answer:89.65

27. Cost of Sugar (C12H22O11) is ₹ 40 per kg; calculate its cost per mole.

Ans: Step 1: Find the molar mass of sugar (C₁₂H₂₂O₁₁)

Carbon (C): 12×12=144 g/mol

Hydrogen (H): 22×1=22 g/mol

Oxygen (O): 11×16=176 g/mol

Molar mass = 144+22+176=342 g/mol

Step 2: Convert price per kg to price per mole

1 kg = 1000 g

Cost per gram = 1000/40 =₹0.04 per gram

Cost of 342 g (1 mole) = 342×0.04=₹13.68

Final Answer:13.68

28. Which of the following weighs the least ?

(a) 2 g atom of N

(b) 3 ×10251 atoms of carbon

(c) 1 mole of sulphur

(d) 7 g silver

Ans: (d) 7 g silver

29. Four grams of caustic soda contains :

(a) 6.02 ×1023 atoms of it,

(b) 4 g atom of sodium,

(c)6.02×1022 molecules

(d) 4 moles of NaOH.

Ans: The correct answer is (c) 6.02 × 10²² molecules.

4 grams of NaOH corresponds to 0.1 moles.

Multiplying 0.1 moles by Avogadro’s number (6.02 × 10²³ molecules/mol) gives 6.02 × 10²² molecules.

The other options are incorrect due to miscalculations in the number of atoms, moles, or gram-atoms.

30. The number of molecules in 425 g of ammonia is :

(a)1.0×1023

(b)1.5×1023

(c)2.0×1023

(d)3.5×1023

Ans: (b)1.5×1023

31. Correct the statements, if required.

(a) One mole of chlorine contains 6.023×1022atoms of chlorine,

(b) Under similar conditions of temperature and pressure, two volumes of hydrogen combined with two volumes of oxygen will give two volumes of water vapour,

(c) Relative atomic mass of an element is the number of times one molecule of an element is heavier than 1/12 the mass of an atom of carbon [C12],

(d) Under the same conditions of the temperature and pressure, equal volumes of all gases contain the same number of atoms.

Ans:

(a) Incorrect.

Corrected statement: One mole of chlorine gas (Cl₂) contains

6.023×1023

molecules, and each molecule has 2 atoms, so total atoms = 1.2046×1024

(b) Incorrect.

Corrected statement: Under similar conditions, two volumes of hydrogen combine with one volume of oxygen to give two volumes of water vapour.

Corrected statement: Relative atomic mass of an element is the number of times one atom of the element is heavier than

1/12

the mass of a carbon-12 atom.

(d) Incorrect.

Corrected statement: Under the same conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules (not atoms).

Exercise 5(C)

1. Give three kinds of information conveyed by the formula H₂O.

Ans:Substance Identity: It identifies the substance as water.

Elemental Composition: It shows water is composed of hydrogen (H) and oxygen (O) elements.

Atomic Ratio: It indicates two hydrogen atoms are bonded to one oxygen atom in a 2:1 ratio.

2. Explain the terms, empirical formula and molecular formula.

Ans: Empirical Formula: The Simplified Ratio

This shows the simplest whole-number ratio of the atoms in a compound. It is like a recipe reduced to its most basic proportions.

Example: Hydrogen peroxide’s molecule is H₂O₂. The ratio of H to O is 2:2, which simplifies to 1:1. So, its empirical formula is HO.

Molecular Formula: The Actual Count

This shows the real number of each type of atom in a single molecule. It is always a multiple of the empirical formula.

Example: For hydrogen peroxide, the molecular formula is H₂O₂. This tells you one molecule actually contains 2 hydrogen atoms and 2 oxygen atoms.

The Key Difference

Think of it this way: the empirical formula gives you the simplified ratio of ingredients, while the molecular formula gives you the exact count used in one full serving.

3. Give the empirical formula of :

(a) C₃H₆

(b) C₄H₄O₆

(c) C₂H₂

(d) CH₃COOH

Ans:

(a) C₃H₆ → CH₂

(b) C₄H₄O₆ → C₂H₂O₃

(d) CH₃COOH → CH₂O

4. Find the percentage mass of water in the Epsom salt MgSO₄·7H₂O.

Ans: Step 1: Molar mass of MgSO₄

Mg = 24.3 g/mol

S = 32.1 g/mol

O₄ = 4 × 16.0 = 64.0 g/mol

MgSO₄ = 24.3 + 32.1 + 64.0 = 120.4 g/mol

Step 2: Molar mass of 7H₂O

H₂O = (2 × 1.0) + 16.0 = 18.0 g/mol

7H₂O = 7 × 18.0 = 126.0 g/mol

Step 3: Molar mass of MgSO₄·7H₂O

120.4 + 126.0 = 246.4 g/mol

Step 4: Mass percentage of water

% mass of water = (mass of water / total mass) × 100

= (126.0 / 246.4) × 100

≈ 51.14%

Final Answer: 51.1% (to 3 significant figures)

5. Calculate the percentage of phosphorus in :

(a) Calcium hydrogen phosphate Ca(H₂PO₄)₂

(b) Calcium phosphate Ca₃(PO₄)₂

Ans: The percentage composition of an element in a compound is calculated using the formula:

Percentage of an element = (Total mass of the element in the compound / Molar mass of the compound) × 100

(a) Calcium Hydrogen Phosphate, Ca(H₂PO₄)₂

Step 1: Calculate the molar mass of Ca(H₂PO₄)₂.

First, find the atomic masses of the constituent elements:

Calcium (Ca) = 40 u

Hydrogen (H) = 1 u

Phosphorus (P) = 31 u

Oxygen (O) = 16 u

Now, calculate the total mass:

Mass of Ca = 40

Mass of H₂PO₄ unit = (2×1) + 31 + (4×16) = 2 + 31 + 64 = 97

Since there are two H₂PO₄ groups, their total mass = 2 × 97 = 194

Molar Mass of Ca(H₂PO₄)₂ = 40 + 194 = 234 g/mol

Step 2: Calculate the total mass of Phosphorus (P) in the compound.

Each H₂PO₄ unit contains 1 P atom.

So, Ca(H₂PO₄)₂ contains 2 P atoms.

Total mass of P = 2 × 31 = 62 g

Step 3: Calculate the percentage of Phosphorus.

Percentage of P = (Total mass of P / Molar mass of compound) × 100

Percentage of P = (62 / 234) × 100

Percentage of P = 26.5%

Answer for (a): The percentage of phosphorus in Ca(H₂PO₄)₂ is 26.5%.

(b) Calcium Phosphate, Ca₃(PO₄)₂

Step 1: Calculate the molar mass of Ca₃(PO₄)₂.

Atomic masses:

Calcium (Ca) = 40 u

Phosphorus (P) = 31 u

Oxygen (O) = 16 u

Now, calculate the total mass:

Mass from Ca atoms = 3 × 40 = 120

Mass from P atoms = 2 × 31 = 62

Mass from O atoms: Each PO₄ group has 4 O atoms, so two groups have 8 O atoms.

Mass = 8 × 16 = 128

Molar Mass of Ca₃(PO₄)₂ = 120 + 62 + 128 = 310 g/mol

Step 2: Calculate the total mass of Phosphorus (P) in the compound.

The compound has 2 atoms of P.

Total mass of P = 2 × 31 = 62 g

Step 3: Calculate the percentage of Phosphorus.

Percentage of P = (Total mass of P / Molar mass of compound) × 100

Percentage of P = (62 / 310) × 100

Percentage of P = 20.0%

Answer for (b): The percentage of phosphorus in Ca₃(PO₄)₂ is 20.0%.

6. Calculate the percentage composition of Potassium chlorate, KClO₃.

Ans: Step 1: Molar Mass of KClO₃

K = 39 g/mol, Cl = 35.5 g/mol, O = 48 g/mol

Total = 39 + 35.5 + 48 = 122.5 g/mol

Step 2: Percentage Composition

Potassium (K): (39 ÷ 122.5) × 100 ≈ 31.84%

Chlorine (Cl): (35.5 ÷ 122.5) × 100 ≈ 28.98%

Oxygen (O): (48 ÷ 122.5) × 100 ≈ 39.18%

Final Answer:

Potassium ≈ 31.84%, Chlorine ≈ 28.98%, Oxygen ≈ 39.18%

7. Find the empirical formula of the compounds with the following percentage composition :

Pb = 62.5%; N = 8.5%; O = 29.0%.

Ans: Step 1: Assume 100 g of the compound

Mass of Pb = 62.5 g

Mass of N = 8.5 g

Mass of O = 29.0 g

Step 2: Convert mass to moles

Molar mass of Pb = 207 g/mol

Moles of Pb = 62.5/207≈0.302

Molar mass of N = 14 g/mol

Moles of N = 8.5/14≈0.607

Molar mass of O = 16 g/mol

Moles of O = 29.0/16≈1.812

Step 3: Divide by the smallest number of moles

Smallest moles = 0.302

Pb: 0.302/0.302=1

N: 0.607/0.302≈2.01≈2

O: 1.812/0.302≈6.00≈6

Step 4: Write the empirical formula

Pb : N : O = 1 : 2 : 6

Empirical formula: PbN2O6

This is often written as

Pb(NO3)2 (lead nitrate) in compound form.

8. Calculate the mass of iron in 10 kg of iron ore which contains 80% of pure ferric oxide.

Ans: Step 1: Find the mass of pure Fe₂O₃ in the ore.

The ore is 80% pure ferric oxide (Fe₂O₃).

Mass of Fe₂O₃ = 80% of 10 kg

0.80×10

= 8 kg.

Step 2: Find the mass of iron in Fe₂O₃.

Molar mass of Fe₂O₃ =2×56+3×16

= 112+48

= 160 g/mol.

Mass of Fe in Fe₂O₃ = 112/160 of Fe₂O₃ mass.

So, mass of Fe = 112/160×8 kg

=0.7×8

= 5.6 kg.

Final Answer:

The mass of iron is 5.6 kg.

9. If the empirical formula of two compounds is CH and their vapour densities are 13 and 39 respectively. Find their molecular formula. (2015)

Ans:Step 1: Calculate molecular mass

Molecular mass = 2 × Vapour Density

First compound: 2 × 13 = 26 g/mol

Second compound: 2 × 39 = 78 g/mol

Step 2: Empirical formula mass

Empirical formula = CH

Empirical mass = 12 + 1 = 13 g/mol

Step 3: Determine molecular formula

=Molecular mass/Empirical mass

First compound:

n=26/13=2→ C₂H₂

Second compound:

n=78/13=6→ C₆H₆

Final Answer:

C₂H₂ and C₆H₆

10. Find the empirical formula of a compound containing 17.7% hydrogen and 82.3% nitrogen.

Ans: To determine the empirical formula from 17.7% hydrogen and 82.3% nitrogen, we analyze a 100 g sample.

Mass to moles:

Hydrogen: 17.7 g ÷ 1.008 g/mol ≈ 17.56 mol

Nitrogen: 82.3 g ÷ 14.01 g/mol ≈ 5.875 mol

Mole ratio:

Divide by the smallest mole value (5.875):

H: 17.56 ÷ 5.875 ≈ 3

N: 5.875 ÷ 5.875 = 1

Empirical formula:

The ratio of N to H is 1:3, giving the formula NH₃.

11. On analysis, a substance was found to contain :

C = 54.54%, H = 9.09%, O = 36.36%.

The vapour density of the substance is 44, calculate :

(a) its empirical formula,

(b) its molecular formula.

Ans: Step 1 – Empirical Formula

We take a 100 g sample:

C = 54.54 g → 54.54 ÷ 12 ≈ 4.545 mol

H = 9.09 g → 9.09 ÷ 1 = 9.09 mol

O = 36.36 g → 36.36 ÷ 16 ≈ 2.2725 mol

Divide by the smallest (2.2725):

C: 4.545 ÷ 2.2725 ≈ 2

H: 9.09 ÷ 2.2725 ≈ 4

O: 1

Empirical formula = C₂H₄O

Empirical mass = 44 g/mol

Step 2 – Molecular Formula

Vapour density = 44

Molecular mass = 2 × 44 = 88 g/mol

n = Molecular mass ÷ Empirical mass = 88 ÷ 44 = 2

Molecular formula = (C₂H₄O)₂ = C₄H₈O₂

Final answers:

(a) Empirical formula: C₂H₄O

(b) Molecular formula: C₄H₈O₂

12. An organic compound, whose vapour density is 45, has the following percentage composition,

H = 2.22%; O = 71.19%; and remaining carbon.

Calculate :

(a) its empirical formula,

(b) its molecular formula.

Ans: Step 1: Percentage composition

Hydrogen (H) = 2.22%

Oxygen (O) = 71.19%

Carbon (C) = 100 − (2.22 + 71.19) = 26.59%

Step 2: Moles of each element

H = 2.22 ÷ 1 = 2.22 mol

C = 26.59 ÷ 12 ≈ 2.216 mol

O = 71.19 ÷ 16 ≈ 4.449 mol

Step 3: Simplest ratio

Divide by smallest (2.216):

H ≈ 1.00

C = 1.00

O ≈ 2.00

Empirical formula = CHO₂

Step 4: Empirical formula mass

= 12 + 1 + (16 × 2) = 45 g/mol

Step 5: Molecular formula

Vapour density = 45

Molecular mass = 2 × 45 = 90

n = 90 ÷ 45 = 2

Molecular formula = (CHO₂)₂ = C₂H₂O₄

Final answers:

(a) Empirical formula: CHO₂

(b) Molecular formula: C₂H₂O₄

13. An organic compound contains 4.07% hydrogen, 71.65% chlorine and remaining carbon. Its molar mass is 98.96. Find its,

(a) Empirical formula (b) Molecular formula

Ans: Step 1: Determine the percentage composition by mass

Given:

Hydrogen, H=4.07%

Chlorine, Cl=71.65%

The rest is carbon.

Carbon percentage = 100−(4.07+71.65)

C=100−75.72=24.28%

Step 2: Convert percentages to moles

Using atomic masses:

H=1.008,

Cl=35.45,

C=12.01.

Moles of C= 12.01/24.28 ≈2.022

Moles of H=4.071.008≈4.038

Moles of H= 1.008/4.07 ≈4.038

Moles of Cl= 35.45/71.65 ≈2.021

Step 3: Find the simplest ratio

Divide each by the smallest number of moles (≈ 2.021):

C: 2.021/2.022 ≈1.000 (1)

H:4.038/2.021≈1.998(2)

Cl: 2.021/2.021=1

So the empirical formula is CH2Cl

Step 4: Calculate empirical formula mass

C=12.01,

H2=2.016,

Cl=35.45

E.F. mass=12.01+2.016+35.45=49.476 g/mol

Step 5: Find molecular formula

Molar mass given = 98.96 g/mol.

n=98.96/49.476≈2.00

So molecular formula =

(CH2Cl)2=C2H4Cl2

Final Answer:

(a) Empirical formula: CH 2Cl

(b) Molecular formula: C 2 H4 Cl 2

14. A hydrocarbon contains 4.8 g of carbon per gram of hydrogen. Calculate :

(a) the gram atom of each,

(b) find the empirical formula,

(c) find molecular formula, if its vapour density is 29.

Ans: Mass ratio: Carbon : Hydrogen = 4.8 g : 1 g

(a) Gram atoms of each

Gram atoms = mass (g) / atomic mass

Carbon:

Atomic mass = 12 g/mol

Mass of C = 4.8 g

Gram atoms of C = 4.8/12=0.4

Hydrogen:

Atomic mass = 1 g/mol

Mass of H = 1 g

Gram atoms of H = 1/1=1

So:

Gram atoms of C=0.4

Gram atoms of H=1

(b) Empirical formula

Divide by smallest (0.4):

C:0.4/0.4=1

H:1/0.4=2.5

Multiply by 2 to get whole numbers:

C2H5

Empirical formula = C2H5

Vapour density (VD) = 29

Molecular mass = 2×VD=2×29=58 g/mol

Empirical formula mass of

C2H5 : (2×12)+(5×1)=24+5=29

n=Molecular mass/Empirical mass

=58/29=2

Molecular formula = (C2H5)2=C4H10

Final answers:

(a) C: 0.4 gram atoms, H: 1 gram atom

(b) Empirical formula: C2H5

15. Combine 0.2 g atom of silicon with 21.3 g of chlorine. Find the empirical formula of the compound formed.

Ans: Step 1: Interpret “0.2 g atom of silicon”

“g atom” means moles of atoms.

So, moles of silicon = 0.2 mol.

Mass of silicon = 0.2×28.086 g/mol≈5.617 g.

Step 2: Moles of chlorine

Mass of chlorine given = 21.3 g.

Molar mass of Cl = 35.45 g/mol.

Moles of Cl = 21.3/35.45 ≈0.601 mol.

Step 3: Mole ratio

Moles Si : Moles Cl = 0.200:0.601.

Divide by smallest (0.200):1:3.005≈1:3.

Step 4: Empirical formula

Si : Cl = 1 : 3 → SiCl3

16. A gaseous hydrocarbon contains 82.76% of carbon. Given that its vapour density is 29, find its molecular formula. (2016)

Ans: Step 1: Find the empirical formula

Given:

Carbon = 82.76%

Hydrogen = 100 – 82.76 = 17.24%

Atomic masses: C = 12, H = 1

Moles ratio:

C:82.76/12=6.8967

H:17.24/1=17.24

Divide by smallest (6.8967):C:1

H:17.24/6.8967≈2.5

Multiply by 2 to get whole numbers:

C:2,H:5

So empirical formula = C2H5

Empirical mass = 2×12+5×1=29

Step 2: Find molecular mass

Vapour density (VD) = 29

Molecular mass=2×VD=2×29=58

Step 3: Find molecular formula

n=Molecular mass/Empirical mass

=58/29

Molecular formula = (C2H5)2=C4H10

Final answer:C4H10

17. In a compound of magnesium (Mg = 24) and nitrogen (N = 14), 18 g of magnesium combines with 7g of nitrogen. Deduce the simplest formula by answering the following questions :

(a) How many gram-atoms of magnesium are equal to 18g ?

(b) How many gram-atoms of nitrogen are equal to 7g of nitrogen ?

(c) Calculate simple ratio of gram-atoms of magnesium to gram-atoms of nitrogen and hence the simplest formula of the compound formed.

Ans: (a) Gram-atoms of magnesium in 18 g:

Mass of Mg = 18 g

Atomic mass of Mg = 24 g

Gram-atoms = 18/24=0.75

Answer: 0.75 gram-atoms of Mg.

(b) Gram-atoms of nitrogen in 7 g:

Mass of N = 7 g

Atomic mass of N = 14 g

Gram-atoms = 7/14=0.5

Answer: 0.5 gram-atoms of N.

Ratio = 0.75 : 0.5 = 1.5 : 1

Multiply by 2 → 3 : 2

Empirical formula: Mg₃N₂

18. Barium chloride crystals contain 14.8% water of crystallisation. Find the number of molecules of water of crystallisation per molecule.

Ans: Step 1: Problem setup

Barium chloride hydrate contains 14.8% water by mass.

We need to find BaCl2⋅xH2O

Step 2: Molar mass calculation

Molar mass of BaCl2=137.3+2×35.5=208.3 g/mol

Molar mass of xH2O = 18x g/mol

Total molar mass = 208.3+18x

Step 3: Water percentage equation

Mass of water in 1 mole = 18x

Given:

18x/208.3+18x×100=14.8

18×208.3+18x=0.148

Step 4: Solve for

18x=0.148×(208.3+18x)

18x=30.8284+2.664x

18x−2.664x=30.8284

15.336x=30.8284

≈30.8284/15.336≈2.01

Step 5: Conclusion

Since

x≈2, the formula is

BaCl2⋅2H2O

19. Urea is a very important nitrogenous fertilizer. Its formula is CON₂H₄. Calculate the percentage of nitrogen in urea (C = 12, O = 16, N = 14 and H = 1).

Ans: The percentage of nitrogen in urea (CON₂H₄) is calculated as follows:

Molar mass of urea = 60 g/mol.

Mass of nitrogen in one mole of urea = 28 g.

Percentage of nitrogen = (28 / 60) × 100 = 46.67%.

Final Answer: 46.67%

20. Determine the formula of the organic compound if its molecule contains 12 atoms of carbon. The percentage compositions of hydrogen and oxygen are 6.48 and 51.42 respectively.

Ans:

Step 1: Given data

Carbon atoms in one molecule = 12

Percentage composition:

Hydrogen = 6.48%

Oxygen = 51.42%

Carbon = 100−6.48−51.42=42.10%

Step 2: Find mole ratio in compound

Assume 100 g sample:

Moles of C = 42.10/12.01≈3.505

Moles of H = 6.48/1.008≈6.429

Moles of O = 51.42/16.00≈3.214

Step 3: Simplify mole ratio

Divide by smallest (3.214):

C : H : O=3.505/3.214:6.429/3.214:1

≈1.09:2.00:1

Multiply by 9 to get near whole numbers:

9.81:18.0:9=10:18:9

So empirical formula: C10H18O9

Empirical mass = 10×12+18×1+9×16=120+18+144=282.

Step 4: Use given C atoms = 12 per molecule

Empirical formula has 10 C atoms → multiply by 12/10=1.2

=1.2 → not integer.

Let’s check our ratio again.

From moles: C = 3.505, H = 6.429, O = 3.214.

Divide by 3.214:

C = 1.090, H = 2.000, O = 1.000.

Multiply by 11: C = 11.99 (~12), H = 22, O = 11.

So molecular formula: C12H22O11

Step 5: Verify percentages

Molar mass = 12×12+22×1+11×16=144+22+176=342.

%C = 144/342 ×100≈42.11% ✓

%H = 22/342 ×100≈6.43% ✓ (slight rounding diff from 6.48)

%O = 176/342 ×100≈51.46% ✓

Small discrepancies due to rounding atomic masses.

C 12 H 22 O 11

This is the formula of the organic compound (likely sucrose or another disaccharide).

21.(a) A compound with empirical formula AB₂, has the vapour density equal to its empirical formula weight. Find its molecular formula.

(b) A compound with empirical formula AB has vapour density three times its empirical formula weight. Find the molecular formula.

Ans: (a) Molecular Formula: A₂B₄

Vapour density (VD) equals the empirical formula weight (EFW).

Molecular weight (MW) = 2 × VD.

So, MW = 2 × EFW.

But MW = n × EFW.

Thus, n = 2.

Molecular formula = (AB₂)₂ = A₂B₄.

(b) Molecular Formula: A₆B₆

Vapour density (VD) is three times the empirical formula weight (EFW).

So, VD = 3 × EFW.

Molecular weight (MW) = 2 × VD = 2 × 3 × EFW = 6 × EFW.

But MW = n × EFW.

Thus, n = 6.

Molecular formula = (AB)₆ = A₆B₆.

22. A hydride of nitrogen contains 87.5 per cent by mass of nitrogen. Determine the empirical formula of this compound.

Ans:

Step 1: Assume a 100 g sample.

Mass of Nitrogen = 87.5 g

Mass of Hydrogen = 100 − 87.5 = 12.5 g

Step 2: Convert masses to moles.

Moles of N = 87.5 ÷ 14 = 6.25 mol

Moles of H = 12.5 ÷ 1 = 12.5 mol

Step 3: Determine the simplest mole ratio.

Divide both by 6.25:

N = 6.25 ÷ 6.25 = 1

H = 12.5 ÷ 6.25 = 2

Step 4: Write the empirical formula.

Empirical formula = NH₂

23. A compound has O = 61.32%, S = 11.15%, H = 4.88% and Zn = 22.65%. The relative molecular mass of the compound is 287 a.m.u. Find the molecular formula of the compound, assuming that all the hydrogen is present as water of crystallisation.

Ans: Step 1: Calculate moles in 100 g of compound

Using atomic masses: Zn = 65.4, S = 32.07, O = 16.00, H = 1.008

Zn: 22.65/65.4≈0.346mol

S: 11.15/32.07≈0.347mol

O: 61.32/16.00≈3.832 mol

H: 4.88/1.008≈4.841 mol

Step 2: Determine empirical formula

Divide by smallest (0.346):

Zn : 1, S : 1, O : 11.07 ≈ 11, H : 13.99 ≈ 14

Empirical formula: ZnS O11H14

Step 3: Assign water of crystallisation

H14O7 → 7 H₂O molecules.

Remaining O = 4 → SO₄ group.

Thus: ZnSO₄·7H₂O

Step 4: Verify molar mass

ZnSO₄·7H₂O = 65.4 + 32.07 + 64 + 7(18.016)

≈ 287.58 g/mol (matches given 287 g/mol)

Final Answer:

ZnSO4⋅7H2O

Exercise 5(D)

2. The reaction between 15 g of marble and nitric acid is given by the following equation :

CaCO3+2HNO3→Ca(NO3)2+H2O+CO2

Calculate :

(a) the mass of anhydrous calcium nitrate formed,

(b) the volume of carbon dioxide evolved at STP.

Ans: Step 1: Write balanced equation & molar masses

CaCO₃ + 2HNO₃ → Ca(NO₃)₂ + H₂O + CO₂

Molar masses:

CaCO₃ = 40 + 12 + 48 = 100 g/mol

Ca(NO₃)₂ = 40 + 2×(14 + 48) = 164 g/mol

Step 2: (a) Mass of Ca(NO₃)₂ formed

From equation: 100 g CaCO₃ → 164 g Ca(NO₃)₂

So 15 g CaCO₃ → (164 / 100) × 15 = 24.6 g

Step 3: (b) Volume of CO₂ at STP

From equation: 100 g CaCO₃ → 22.4 L CO₂ at STP

So 15 g CaCO₃ → (22.4 / 100) × 15 = 3.36 L

Final Answer:

(a) 24.6 g of calcium nitrate

(b) 3.36 L of carbon dioxide at STP

3. 66 g ammonium sulphate is produced by the action of ammonia on sulphuric acid.

Write a balanced equation and calculate :

(a) mass of ammonia required,

(b) the volume of the gas used at STP,

(c) the mass of acid required.

Ans: Step 1: Balanced Chemical Equation

The balanced equation for the reaction is:

2NH3+H2SO4→(NH4)2SO4

Step 2: Calculate Molar Masses

(Using atomic masses: N=14, H=1, S=32, O=16)

Ammonia (NH3): 14+(3×1)=17 g/mol

Sulphuric Acid (H2SO4 ): (2×1)+32+(4×16)=98 g/mol

Ammonium Sulphate 2[14+(4×1)]+32+(4×16)=132 g/mol

Step 3: (a) Mass of Ammonia Required

From the equation: 132 g of ammonium sulphate is produced from

2×17=34 g of ammonia.

Mass of ammonia required to produce 66 g of ammonium sulphate is:

34/132×66=17 g

Step 4: (b) Volume of Ammonia Gas at STP

From the equation: 132 g of ammonium sulphate requires 2×22.4=44.8 L of ammonia at STP. Volume of ammonia required to produce 66 g of ammonium sulphate is:

44.8/132 ×66=22.4 L

Step 5: (c) Mass of Acid Required

From the equation: 132 g of ammonium sulphate is produced from 98 g of sulphuric acid.Mass of acid required to produce 66 g of ammonium sulphate is:

132/98 ×66=49 g

Final Answers:

(a) Mass of ammonia = 17 g

(b) Volume of ammonia at STP = 22.4 L

4. The reaction between red lead and hydrochloric acid is given below :

Pb3O4+8HCl→3PbCl2+4H2O+Cl2

Calculate :

(a) the mass of lead chloride formed by the action of 6.85 g of red lead,

(b) the mass of chlorine and

(c) the volume of chlorine evolved at STP.

Ans: Step 1: Write down the balanced equation and atomic masses.

Balanced Equation:

Pb3O4+8HCl→3PbCl2+4H2O+Cl2

Atomic Masses (g/mol): Pb = 207, O = 16, Cl = 35.5, H = 1.

Step 2: Calculate molar masses of key compounds.

Molar mass of Pb₃O₄ (Red lead):

(3×207)+(4×16)=621+64=685 g/mol

Molar mass of PbCl₂ (Lead chloride):

207+(2×35.5)=207+71=278 g/mol

Molar mass of Cl₂ (Chlorine gas):

2×35.5=71 g/mol

Step 3: (a) Calculate the mass of lead chloride (PbCl₂) formed.

From the equation:

685 g of Pb₃O₄ produces

3×278=834 g of PbCl₂.

Therefore, 6.85 g of Pb₃O₄ produces:

Mass of PbCl₂=( 685/834 )×6.85=8.34 g

Step 4: (b) Calculate the mass of chlorine (Cl₂) evolved.

From the equation:

685 g of Pb₃O₄ produces 71 g of Cl₂.

Therefore, 6.85 g of Pb₃O₄ produces:

Mass of Cl₂=( 685/71 )×6.85=0.71 g

Step 5: (c) Calculate the volume of chlorine at STP.

From the equation:

685 g of Pb₃O₄ produces 22.4 L of Cl₂ at STP.

Therefore, 6.85 g of Pb₃O₄ produces:

Volume of Cl₂=( 685/22.4 )×6.85=0.224 L

Final Answers:

(a) Mass of lead chloride = 8.34 g

(b) Mass of chlorine = 0.71 g

5. Find the mass of KNO3 required to produce 126 kg of nitric acid. Find whether a larger or smaller mass of NaNO3 is required for the same purpose.

KNO3+H2SO4→KHSO4+HNO3

NaNO3+H2SO4→NaHSO4+HNO3

Ans: Step 1: Mass of KNO₃ required

The balanced equation is:

KNO3+H2SO4→KHSO4+HNO3

Molar mass of KNO₃ = 39 + 14 + 48 = 101 g/mol

Molar mass of HNO₃ = 1 + 14 + 48 = 63 g/mol

From the equation: 101 g of KNO₃ produces 63 g of HNO₃.

Let the required mass of KNO₃ be x kg to produce 126 kg of HNO₃.

x= 63/101 ×126=202 kg

Step 2: Mass of NaNO₃ required

The balanced equation is:

NaNO3+H2SO4→NaHSO4+HNO3

Molar mass of NaNO₃ = 23 + 14 + 48 = 85 g/mol

Molar mass of HNO₃ = 63 g/mol

From the equation: 85 g of NaNO₃ produces 63 g of HNO₃.

Let the required mass of NaNO₃ be y kg to produce 126 kg of HNO₃.

y= 63/85 ×126=170 kg

Step 3: Comparison

Mass of KNO₃ required = 202 kg

Mass of NaNO₃ required = 170 kg

Since 170 kg < 202 kg, a smaller mass of NaNO₃ is required.

Final Answer:

Mass of KNO₃ required = 202 kg

A smaller mass of NaNO₃ is required for the same purpose.

Exercise 5(D)

1. Complete the following blanks in the equation as indicated.

CaH2(s)+2H2O(aq)→Ca(OH)2(s)+2H2(g)

(a) Moles: 1 mole + …… → …… + ……

(b) Grams: 42 g + …… → …… + ……

(c) Molecules:6.02×1023+ …… → …… + ……

Ans:

(a) Moles:1 mole + 2 moles → 1 mole + 2 moles.

(b) Grams:42 g + 36 g → 74 g + 4 g. Total mass is 78 g on both sides.

2. The reaction between 15 g of marble and nitric acid is given by the following equation :

CaCO3+2HNO3→Ca(NO3)2+H2O+CO2

Calculate :

(a) the mass of anhydrous calcium nitrate formed,

(b) the volume of carbon dioxide evolved at STP.

Ans: Step 1: Write balanced equation & molar masses

CaCO₃ + 2HNO₃ → Ca(NO₃)₂ + H₂O + CO₂

Molar masses:

CaCO₃ = 40 + 12 + 48 = 100 g/mol

Ca(NO₃)₂ = 40 + 2×(14 + 48) = 164 g/mol

Step 2: (a) Mass of Ca(NO₃)₂ formed

From equation: 100 g CaCO₃ → 164 g Ca(NO₃)₂

So 15 g CaCO₃ → (164 / 100) × 15 = 24.6 g

Step 3: (b) Volume of CO₂ at STP

From equation: 100 g CaCO₃ → 22.4 L CO₂ at STP

So 15 g CaCO₃ → (22.4 / 100) × 15 = 3.36 L

Final Answer:

(a) 24.6 g of calcium nitrate

(b) 3.36 L of carbon dioxide at STP

3. 66 g ammonium sulphate is produced by the action of ammonia on sulphuric acid.

Write a balanced equation and calculate :

(a) mass of ammonia required,

(b) the volume of the gas used at STP,

(c) the mass of acid required.

Ans: Step 1: Balanced Chemical Equation

The balanced equation for the reaction is:

2NH3+H2SO4→(NH4)2SO4

Step 2: Calculate Molar Masses

(Using atomic masses: N=14, H=1, S=32, O=16)

Ammonia (NH3): 14+(3×1)=17 g/mol

Sulphuric Acid (H2SO4 ): (2×1)+32+(4×16)=98 g/mol

Ammonium Sulphate 2[14+(4×1)]+32+(4×16)=132 g/mol

Step 3: (a) Mass of Ammonia Required

From the equation: 132 g of ammonium sulphate is produced from

2×17=34 g of ammonia.

Mass of ammonia required to produce 66 g of ammonium sulphate is:

34/132×66=17 g

Step 4: (b) Volume of Ammonia Gas at STP

From the equation: 132 g of ammonium sulphate requires 2×22.4=44.8 L of ammonia at STP. Volume of ammonia required to produce 66 g of ammonium sulphate is:

44.8/132 ×66=22.4 L

Step 5: (c) Mass of Acid Required

From the equation: 132 g of ammonium sulphate is produced from 98 g of sulphuric acid.Mass of acid required to produce 66 g of ammonium sulphate is:

132/98 ×66=49 g

Final Answers:

(a) Mass of ammonia = 17 g

(b) Volume of ammonia at STP = 22.4 L

4. The reaction between red lead and hydrochloric acid is given below :

Pb3O4+8HCl→3PbCl2+4H2O+Cl2

Calculate :

(a) the mass of lead chloride formed by the action of 6.85 g of red lead,

(b) the mass of chlorine and

(c) the volume of chlorine evolved at STP.

Ans: Step 1: Write down the balanced equation and atomic masses.

Balanced Equation:

Pb3O4+8HCl→3PbCl2+4H2O+Cl2

Atomic Masses (g/mol): Pb = 207, O = 16, Cl = 35.5, H = 1.

Step 2: Calculate molar masses of key compounds.

Molar mass of Pb₃O₄ (Red lead):

(3×207)+(4×16)=621+64=685 g/mol

Molar mass of PbCl₂ (Lead chloride):

207+(2×35.5)=207+71=278 g/mol

Molar mass of Cl₂ (Chlorine gas):

2×35.5=71 g/mol

Step 3: (a) Calculate the mass of lead chloride (PbCl₂) formed.

From the equation:

685 g of Pb₃O₄ produces

3×278=834 g of PbCl₂.

Therefore, 6.85 g of Pb₃O₄ produces:

Mass of PbCl₂=( 685/834 )×6.85=8.34 g

Step 4: (b) Calculate the mass of chlorine (Cl₂) evolved.

From the equation:

685 g of Pb₃O₄ produces 71 g of Cl₂.

Therefore, 6.85 g of Pb₃O₄ produces:

Mass of Cl₂=( 685/71 )×6.85=0.71 g

Step 5: (c) Calculate the volume of chlorine at STP.

From the equation:

685 g of Pb₃O₄ produces 22.4 L of Cl₂ at STP.

Therefore, 6.85 g of Pb₃O₄ produces:

Volume of Cl₂=( 685/22.4 )×6.85=0.224 L

Final Answers:

(a) Mass of lead chloride = 8.34 g

(b) Mass of chlorine = 0.71 g

5. Find the mass of KNO3 required to produce 126 kg of nitric acid. Find whether a larger or smaller mass of NaNO3 is required for the same purpose.

KNO3+H2SO4→KHSO4+HNO3

NaNO3+H2SO4→NaHSO4+HNO3

Ans: Step 1: Mass of KNO₃ required

The balanced equation is:

KNO3+H2SO4→KHSO4+HNO3

Molar mass of KNO₃ = 39 + 14 + 48 = 101 g/mol

Molar mass of HNO₃ = 1 + 14 + 48 = 63 g/mol

From the equation: 101 g of KNO₃ produces 63 g of HNO₃.

Let the required mass of KNO₃ be x kg to produce 126 kg of HNO₃.

x= 63/101 ×126=202 kg

Step 2: Mass of NaNO₃ required

The balanced equation is:

NaNO3+H2SO4→NaHSO4+HNO3

Molar mass of NaNO₃ = 23 + 14 + 48 = 85 g/mol

Molar mass of HNO₃ = 63 g/mol

From the equation: 85 g of NaNO₃ produces 63 g of HNO₃.

Let the required mass of NaNO₃ be y kg to produce 126 kg of HNO₃.

y= 63/85 ×126=170 kg

Step 3: Comparison

Mass of KNO₃ required = 202 kg

Mass of NaNO₃ required = 170 kg

Since 170 kg < 202 kg, a smaller mass of NaNO₃ is required.

Final Answer:

Mass of KNO₃ required = 202 kg

A smaller mass of NaNO₃ is required for the same purpose.

6. Pure calcium carbonate and dilute hydrochloric acid are reacted and 2 litres of carbon dioxide was collected at 27°C and normal pressure.

CaCO3+2HCl→CaCl2+H2O+CO2

Calculate :

(a) the mass of salt required,

(b) the mass of the acid required.

Ans: Given:

Volume of CO2 collected = 2L

Temperature T=27∘C=27+273=300

Normal pressure P=1atm

Reaction:

CaCO3+2HCl→CaCl2+H2O+CO2

Step 1: Moles of CO2 produced

Using ideal gas equation:

PV=nRT

n=PV/RT=1×2/0.0821×300

n=2/24.63≈0.0812mol

So, moles of CO2=0.0812mol.

Step 2: Moles of CaCO3 and CaCl2

From reaction:

1 mole CaCO3→ 1 mole CO2→ 1 mole CaCl2

So moles of CaCO3 reacted = 0.0812 mol

Moles of CaCl2 produced = 0.0812 mol

(a) Mass of salt (CaCl2) required?

They ask for mass of salt required — here “salt” means CaCl2 produced.

Molar mass of CaCl2 = 40+2×35.5=111g/mol

Mass=0.0812×111≈9.01g

(b) Mass of acid (HCl) required

From reaction: 2 moles HCl → 1 mole CO2

So moles of HCl = 2×0.0812=0.1624 mol

Molar mass of HCl = 1+35.5=36.5g/mol

Mass=0.1624×36.5≈5.93g

Final Answers:

(a) 9.01g(b) 5.93g

7. Calculate the mass and volume of oxygen at STP, which will be evolved on electrolysis of 1 mole (18 g) of water.

Ans: From 1 mole of water (H₂O) during electrolysis:

Moles of oxygen gas (O₂) produced = 0.5 moles

Mass of oxygen gas = 16 grams

Volume at STP = 11.2 litres

8. 1.56 g of sodium peroxide reacts with water according to the following equation :

2Na2O2+2H2O→4NaOH+O2

Calculate :

(a) mass of sodium hydroxide formed,

(b) volume of oxygen liberated at STP,

(c) mass of oxygen liberated.

Ans:

Step 1: Write the balanced equation

2Na2O2+2H2O→4NaOH+O2

Step 2: Molar masses

2×23+2×16=46+32=78 g/mol

NaOH

23+16+1=40 g/mol

32 g/mol

Step 3: Moles of Na₂O₂ given

Mass=1.56 g

nNa2O2 = 78/1.56

=0.02 mol

Step 4: Moles of NaOH formed

From equation:

2mol Na2O2→4 mol NaOH

So 1mol Na2O2→2 mol NaOH

n NaOH =0.02×2=0.04 mol

Mass of NaOH:

m=0.04×40=1.6 g

(a) Answer:1.6 g

Step 5: Moles of O₂ liberated

From equation:

2mol Na2O2→1mol O2

1mol Na2O2→0.5mol O2

nO2=0.02×0.5=0.01 mol

Step 6: Volume of O₂ at STP

At STP, 1 mol=22.4 L

V=0.01×22.4=0.224 L=224 mL

(b) Answer:0.224 L

Step 7: Mass of O₂ liberated

mO2 =0.01×32=0.32 g

Final answers:

(a) 1.6 g NaOH

(b) 0.224 L O₂

9.(a) Calculate the mass of ammonia that can be obtained from 21.4 g of NH4Cl by the reaction :

2NH4Cl+Ca(OH)2→CaCl2+2H2O+2NH3

(b) What will be the volume of ammonia when measured at STP ?

Ans: (a) Mass of ammonia from 21.4 g of NH₄Cl

1.Molar mass of NH₄Cl

N = 14, H = 4×1 = 4, Cl = 35.5

Molar mass = 14 + 4 + 35.5 = 53.5 g/mol

2.Moles of NH₄Cl used

n=21.4/53.5

n≈0.4 mol

3.Reaction stoichiometry

2NH4Cl→2NH3

So, moles of NH₃ = moles of NH₄Cl = 0.4 mol

4.Molar mass of NH₃

N = 14, H = 3×1 = 3 → 17 g/mol

5.Mass of NH₃

m=0.4×17=6.8 g

(b) Volume of ammonia at STP

At STP, 1 mole of gas = 22.4 L

Volume = 0.4×22.4=8.96 L

Final Answer:

(a) 6.8 g NH₃

(b) 8.96 L at STP

10. The reaction between aluminium carbide and water takes place according to the following equation :

Al4C3+12H2O→3CH4+4Al(OH)3

Calculate the volume of methane measured at STP released from 14.4 g of aluminium carbide by excess of water.

Ans: Step 1: Balanced chemical equation:

Al4C3+12H2O→3CH4+4Al(OH)3

Step 2: Molar mass of Al4C3 :

(4×27)+(3×12)=144 g/mol

Step 3: Moles of Al4C3 in 14.4 g:

14.4/144=0.1 mol

Step 4: Moles of CH4 produced:

From equation, 1 mol Al4C3 gives 3 mol CH4

0.1×3=0.3 mol

Step 5: Volume at STP (1 mole = 22.4 L):

0.3×22.4=6.72 L

Final answer:6.72

11.MnO2+4HCl→MnCl2+2H2O+Cl2 0.02 moles of pure MnO2 is heated strongly with conc. HCl. Calculate :

(a) mass of MnO2 used,

(b) moles of salt formed,

(c) mass of salt formed,

(d) moles of chlorine gas formed,

(e) mass of chlorine gas formed,

(f) volume of chlorine gas formed at STP,

(g) moles of acid required,

(h) mass of acid required.

Ans: Reaction:

MnO2+4HCl→MnCl2+2H2O+Cl2

Given:

nMnO2=0.02 mol

(a) Mass of MnO₂ used

Molar mass of MnO₂ = 55+32=87 g/mol

m=0.02×87=1.74 g

Answer: 1.74 g

(b) Moles of salt formed

Salt is MnCl2

From reaction: 1 mol MnO₂ → 1 mol MnCl₂

nMnCl2=0.02 mol

Answer: 0.02 mol

Molar mass of MnCl₂ = 55+2×35.5=55+71=126 g/mol

m=0.02×126=2.52 g

Answer: 2.52 g

(d) Moles of chlorine gas formed

From reaction: 1 mol MnO₂ → 1 mol Cl₂

nCl2=0.02 mol

Answer: 0.02 mol

(e) Mass of chlorine gas formed

Molar mass of Cl₂ = 71 g/mol

m=0.02×71=1.42 g

Answer: 1.42 g

(f) Volume of chlorine gas at STP

At STP, 1 mol gas = 22.4 L

V=0.02×22.4=0.448 L=448 mL

Answer: 0.448 L

(g) Moles of acid required

From reaction: 1 mol MnO₂ → 4 mol HCl

n HCl =0.02×4=0.08 mol

Answer: 0.08 mol

(h) Mass of acid required

Molar mass HCl = 1+35.5=36.5 g/mol

m=0.08×36.5=2.92 g

Answer: 2.92 g

Final answers:

(a) 1.74 g

(b) 0.02 mol

(d) 0.02 mol

(e) 1.42 g

(f) 0.448 L

(g) 0.08 mol

(h) 2.92 g

12. Nitrogen and hydrogen react to form ammonia.

N2(g)+3H2(g)→2NH3(g) If 1000 g of H2 react with 2000 g of N2

(a) Will any of the two reactants remain unreacted ? If yes, which one and what will be its mass ?

(b) Calculate the mass of ammonia (NH3) that will be formed.

Ans:

Step 1: Write the balanced equation

N2+3H2→2NH3N2

Step 2: Molar masses

N2: 2×14=28 g/mol

H2:2×1=2 g/mol

NH3: 14+3=17 g/mol

Step 3: Calculate moles available

nN2=2000/28=71.4286 mol

nH2=1000/2=500 mol

Step 4: Find limiting reagent

From the balanced equation:

1 mol N₂ requires 3 mol H₂.

So for 71.4286 mol N₂, H₂ required = 71.4286×3=214.2857 mol.

We have 500 mol H₂ available, which is more than 214.2857 mol.

So N₂ is limiting reagent, H₂ is in excess.

Step 5: Mass of H₂ reacted

Moles H₂ reacted = 214.2857 mol

Mass H₂ reacted = 214.2857×2=428.5714 g

Step 6: Mass of H₂ left unreacted

Initial H₂ = 1000 g

Mass left =1000−428.5714=571.4286 g

Step 7: Mass of NH₃ formed

From stoichiometry: 1 mol N₂ → 2 mol NH₃

Moles NH₃ = 71.4286×2=142.8572 mol

Mass NH₃ = 142.8572×17=2428.5714 g

(a) Yes, H₂ will remain unreacted. Mass left = 571.43 g

(b) Mass of NH₃ formed = 2428.57 g

571.43 g, 2428.57 g

MISCELLANEOUS EXERCISE

1. From the equation for burning of hydrogen and oxygen

2H2+O2→2H2O(steam)

Write down the number of mole (or moles) of steam obtained from 0.5 moles of oxygen.

Ans: From the balanced equation:

2H₂ + O₂ → 2H₂O (steam)

1 mole of O₂ produces 2 moles of steam.

So, 0.5 moles of O₂ will produce:

0.5×2=1 mole of steam

Answer: 1 mole of steam

2. From the equation,

3Cu+8HNO3→3Cu(NO3)2+4H2O+2NO

Calculate :

(a) mass of copper needed to react with 63 g of HNO3,

(b) volume of nitric oxide at STP that can be collected.

Ans: Given reaction:

3Cu+8HNO3→3Cu(NO3)2+4H2O+2NO

(a) Mass of copper needed to react with 63 g of HNO₃

1.Molar masses:

HNO3 =1+14+48=63 g/mol

Cu=63.5 g/mol

2.Moles of HNO₃ in 63 g:

63/63 =1 mol

3.From reaction:

8 mol HNO₃ react with 3 mol Cu

1 mol HNO₃→ 8/3 mol Cu

Moles of Cu=0.375 mol

4.Mass of Cu:

0.375×63.5=23.81 g

(a) Answer: 23.81 g Cu

(b) Volume of NO at STP from 63 g HNO₃

From reaction:

8 mol HNO₃ produce 2 mol NO

1 mol HNO₃→2/8=0.25 mol NO

Moles of NO from 1 mol HNO₃: 0.25 mol

Volume at STP: 0.25×22.4=5.6 L

(b) Answer: 5.6 L NO

Final answers:

(a) 23.81 g

(b) 5.6 L

3.(a) Calculate the number of moles in 7 g of nitrogen.

(b) What is the volume at STP of 7.1 g of chlorine?

(c) What is the mass of 56 cm3of carbon monoxide at STP?

Ans: (a) Calculate the number of moles in 7 g of nitrogen.

Molar mass of Nitrogen gas (N₂) = 2 × 14 = 28 g/mol

Number of moles = Mass / Molar mass

Number of moles = 7 g / 28 g/mol = 0.25 mol

(b) What is the volume at STP of 7.1 g of chlorine?

Molar mass of Chlorine gas (Cl₂) = 2 × 35.5 = 71 g/mol

Number of moles = 7.1 g / 71 g/mol = 0.1 mol

At STP, 1 mole of any gas occupies 22.4 litres.

Volume at STP = 0.1 mol × 22.4 L/mol = 2.24 L

First, convert the volume to litres: 56 cm³ = 0.056 L

Number of moles at STP = Volume (L) / 22.4 L/mol

Number of moles = 0.056 L / 22.4 L/mol = 0.0025 mol

Molar mass of Carbon monoxide (CO) = 12 + 16 = 28 g/mol

Mass = Moles × Molar mass = 0.0025 mol × 28 g/mol = 0.07 g

Final Answers:

(a) 0.25 mol

(b) 2.24 L

4. Some of the fertilizers are sodium nitrate NaNO3, ammonium sulphate (NH4)2SO4 and urea CO(NH2)2. Which of these contains the highest percentage of nitrogen?

Ans: Sodium Nitrate (NaNO₃)

Its formula has 1 nitrogen atom (Atomic mass N = 14).

The total molar mass is: Sodium (23) + Nitrogen (14) + three Oxygens (3 × 16 = 48) = 85 g/mol.

So, the percentage of nitrogen is (14 / 85) × 100, which works out to approximately 16.47%.

Ammonium Sulphate ((NH₄)₂SO₄)

Its formula has two nitrogen atoms (2 × 14 = 28).

The total molar mass is: two Nitrogen atoms (28) + eight Hydrogen atoms (8 × 1 = 8) + Sulphur (32) + four Oxygens (4 × 16 = 64) = 132 g/mol.

The percentage of nitrogen is (28 / 132) × 100, which is approximately 21.21%.

Urea (CO(NH₂)₂)

Its formula also has two nitrogen atoms (2 × 14 = 28).

The total molar mass is: Carbon (12) + Oxygen (16) + two Nitrogen atoms (28) + four Hydrogen atoms (4 × 1 = 4) = 60 g/mol.

The percentage of nitrogen is (28 / 60) × 100, which is approximately 46.67%.

5. Water decomposes to O2 and H2 under suitable conditions as represented by the equation below:

2H2O→2H2+O2

(a) If 2500 cm3 of H2 is produced, what volume of O2 is liberated at the same time and under the same conditions of temperature and pressure?

(b) The 2500 cm3 of H2 is subjected to 1.5 times increase in pressure (temp. remaining constant). What volume of H2will now occupy?

(c) Taking the volume of H2 calculated in 5(b), what changes must be made in kelvin (absolute) temperature to return the volume to 2500 cm3 pressure remaining constant.

Ans: Step 1 – Volume of O₂ from 2500 cm³ of H₂

From the reaction:

2H2O→2H2+O2

2 volumes of H₂ → 1 volume of O₂

So,

Volume of O2=2500/2=1250cm3

Step 2 – New H₂ volume when pressure increases 1.5 times

Boyle’s Law:

P1V1=P2V2

Let

P1=P

V1=2500cm³,

P2=1.5P

P×2500=1.5P×V 2

V2=2500/1.5=1666.67cm3

Step 3 – Temperature change to return to 2500 cm³ at constant pressure

Charles’s Law:

V1/T1=V2/T2

Here

V1=1666.67cm³,

V2=2500 cm³,

T1=T

1666.67/T=2500T2

T2=T×2500/1666.67=1.5T

So temperature must be increased to 1.5 times the original Kelvin temperature.

Final Answers:

(a) 1250 cm 3

(b) 1666.67 cm 3

6. Urea [CO(NH2)2] is an important nitrogenous fertilizer, and is sold in 50 kg sacks. What mass of nitrogen is in one sack of urea?

Ans: The molar mass of urea, CO(NH₂)₂, is 60 g/mol.

It contains 28 g of nitrogen per mole, since nitrogen’s atomic mass is 14 and there are two nitrogen atoms.

Thus, the fraction of nitrogen in urea is 28/60.

For a 50 kg sack:

Nitrogen mass = (28/60) × 50 kg

= 1400/60 ≈ 23.33 kg

Final Answer:

A 50 kg sack of urea contains about 23.33 kg of nitrogen.

7. Find the molecular formula of a hydrocarbon having vapour density 15, which contains 20% of Hydrogen.

Ans: Step 1: Find empirical formula

Carbon = 80%, Hydrogen = 20%

Moles: C = 80 ÷ 12 = 6.667

H = 20 ÷ 1 = 20

Divide by smallest (6.667):

C = 6.667 ÷ 6.667 = 1

H = 20 ÷ 6.667 ≈ 3

Empirical formula = CH₃

Empirical mass = 12 + 3 = 15

Step 2: Find molecular formula

Vapour density = 15

Molecular mass = 2 × 15 = 30

n = 30 ÷ 15 = 2

Molecular formula = C₂H₆

Final Answer: C₂H₆

8. The following experiment was performed in order to determine the formula of a hydrocarbon. The hydrocarbon X is purified by fractional distillation.0.145 g of X were heated with dry copper (II) oxide and 224 cm3 of carbon dioxide was collected at STP.

(a) Which elements does X contain?

(b) What was the purpose of Copper (II) oxide?

(c) Calculate the empirical formula of X by the following steps:

(i) Calculate the number of moles of carbon dioxide gas.

(ii) Calculate the mass of carbon contained in this quantity of carbon dioxide and thus the mass of carbon in sample X.

(iii) Calculate the mass of hydrogen in sample X.

(iv) Deduce the ratio of atoms of each element in X (empirical formula).

Ans: (a) Hydrocarbon X contains carbon and hydrogen only.

(b) Copper(II) oxide acts as an oxidizing agent, supplying oxygen to convert:

Carbon in X → CO₂

Hydrogen in X → H₂O

(i) Volume of CO₂ = 224 cm³ = 0.224 dm³

Molar volume at STP = 22.4 dm³/mol

Moles of CO₂ = 0.224/22.4=0.01mol

(ii) Each mole of CO₂ contains 1 mole of C atoms.

Moles of C = 0.01 mol

Mass of C = 0.01×12=0.12 g

(iii) Mass of hydrogen in sample = Total mass – mass of C

= 0.145−0.12=0.025 g

(iv) Moles of H = =0.025 mol

Mole ratio C : H = 0.01:0.025=1:2.5=2:5

Empirical formula = C₂H₅

9. A compound is formed by 24 g of X and 64 g of oxygen. If atomic mass of X = 12 and O = 16, calculate the simplest formula of compound.

Ans: Step 1: Find moles of each element

Moles of X = 24/12=2

Moles of O = 64/16=4

Step 2: Find simplest mole ratio

Mole ratio X : O = 2:4=1:2

Step 3: Write empirical formula

Empirical formula = XO2

Final Answer:XO2

10. A gas cylinder filled with hydrogen holds 5 g of the gas. The same cylinder holds 85 g of gas X under same temperature and pressure. Calculate:

(a) vapour density of gas X,

(b) molecular weight of gas X.

Ans: (a) Vapour density of gas X

Vapour density =Mass of certain volume of gas / Mass of same volume of hydrogen

= 85 g / 5 g

= 17

(b) Molecular weight of gas X

Molecular weight = 2 × Vapour density

= 2 × 17

= 34

Final Answer:

(a) Vapour density of gas X = 17

(b) Molecular weight of gas X = 34

11.(a) When carbon dioxide is passed over red hot carbon, carbon monoxide is produced according to the equation:

CO2+C→2CO What volume of carbon monoxide at STP can be obtained from 3 g of carbon?

(b) 60 cm3 of oxygen was added to 24 cm of carbon monoxide and the mixture ignited. Calculate:

(i) volume of oxygen used up, and

(ii) volume of carbon dioxide formed.

Ans: (a)Mass of carbon = 3 g

Molar mass of carbon = 12 g/mol

Moles of carbon = 3/12 =0.25 mol

Reaction:

C+CO2→2CO

From the equation, 1 mole C gives 2 moles CO.

So, moles of CO = 0.25×2=0.5 mol

Volume at STP = 0.5×22.4=11.2 dm³

(b)Reaction:

2CO+O2→2CO2

Given: CO = 24 cm³, O₂ = 60 cm³

From stoichiometry: 2 volumes CO react with 1 volume O₂

So, 24 cm³ CO requires

24/2=12 cm³ O₂

(i) Oxygen used = 12 cm³

(ii) CO₂ formed = 24 cm³ (since 2 volumes CO produce 2 volumes CO₂)

12. How much calcium oxide is obtained by heating 82 g of calcium nitrate? Also find the volume of NO2 evolved:

2Ca(NO3)2→2CaO+4NO2+O2

Ans: Step 1: Write the balanced equation

2Ca(NO3)2→2CaO+4NO2+O2

Step 2: Molar masses

Ca(NO3)2 :Ca = 40, N = 14, O = 16

40+2×[14+3×16]=40+2×62=164 g/mol

CaO:40+16=56 g/mol

NO2 :14+32=46 g/mol

Step 3: Moles of calcium nitrate heated

Moles of Ca(NO3)2=82/164=0.5 mol

Step 4: Find CaO formed

From equation:

2mol Ca(NO3)2 → 2 mol CaO

0.5 mol Ca(NO 3 ) 2 → 0.5 mol CaO

Mass of CaO = 0.5×56=28 g

Step 5: Find NO₂ evolved

From equation:

2mol Ca(NO3)2 → 4mol NO2

0.5mol Ca(NO3)2 → 1.0mol NO2

Volume at STP = 1.0×22.4=22.4 L

Final Answer:

Calcium oxide obtained = 28 g

Volume of NO₂ evolved at STP = 22.4 L

13. The equation for the burning of octane is:

2C8H18+25O2→16CO2+18H2O

(i) How many moles of carbon dioxide are produced when one mole of octane burns?

(ii) What volume, at STP, is occupied by the number of moles determined in (i)?

(iii) If the relative molecular mass of carbon dioxide is 44, what is the mass of carbon dioxide produced by burning two moles of octane?

(iv) What is the empirical formula of octane?

Ans: (i) Moles of CO₂ from 1 mole of octane:

From the equation: 2 C₈H₁₈ → 16 CO₂

So, 1 mole C₈H₁₈ gives

16/2=8 moles of CO₂

Ans: 8 moles

(ii) Volume at STP of this CO₂:

1 mole gas at STP = 22.4 dm³

8 moles = 8×22.4=179.2 dm³

Ans: 179.2 dm³

(iii) Mass of CO₂ from burning 2 moles of octane:

2 moles C₈H₁₈ → 16 moles CO₂

Molar mass of CO₂ = 44 g/mol

Mass = 16×44=704 g

Ans: 704 g

(iv) Empirical formula of octane:

Molecular formula = C₈H₁₈

Divide by 2 → C₄H₉

Ans: C₄H₉

14. Ordinary chlorine gas has two isotopes

35Cl and 37Cl in the ratio of 3 : 1. Calculate the relative atomic mass of chlorine.

Ans: The relative atomic mass of chlorine is calculated from its two main isotopes:

Isotope 35Clwith abundance 3 parts

Isotope 37 Cl with abundance 1 part

Total parts = 3+1=4

Relative atomic mass=

4/(35×3)+(37×1) = 4/105+37= 4/142 =35.5

Thus, the relative atomic mass of chlorine is 35.5.

15. Silicon (Si = 28) forms a compound with chlorine (Cl = 35.5) in which 5.6 g of silicon combines with 21.3 g of chlorine. Calculate the empirical formula of the compound.

Ans: Step 1: Moles of silicon

Mass of Si = 5.6 g

Atomic mass of Si = 28 g/mol

Moles of Si = 5.6/28=0.2

Step 2: Moles of chlorine

Mass of Cl = 21.3 g

Atomic mass of Cl = 35.5 g/mol

Moles of Cl = 21.3/35.5=0.6

Step 3: Simplest ratio

Si : Cl = 0.2 : 0.6

Divide both by 0.2 → 1 : 3

Step 4: Empirical formula

SiCl₃

16. An acid of phosphorus has the following percentage composition: Phosphorus = 38.27%; hydrogen = 2.47%; oxygen = 59.26%. Find the empirical formula of the acid and its molecular formula, given that its relative molecular mass is 162.

Ans:Step 1: Find the empirical formula

We start by assuming we have a 100 g sample of the acid. This means:

Mass of Phosphorus = 38.27 g

Mass of Hydrogen = 2.47 g

Mass of Oxygen = 59.26 g

Next, we convert these masses into moles using atomic masses (P = 31, H = 1, O = 16):

Moles of P = 38.27/31≈1.235

Moles of H = 2.47/1=2.47

Moles of O = 59.26/16≈3.704

Now, we divide each mole value by the smallest number of moles (which is 1.235) to get the simplest ratio:

P: 1.235/1.235

H: 2.47/1.235≈2

O: 3.704/1.235≈3

So, the empirical formula is H₂PO₃.

Step 2: Calculate the empirical formula mass

Using atomic masses:

H = 2 × 1 = 2

P = 1 × 31 = 31

O = 3 × 16 = 48

Empirical mass = 2 + 31 + 48 = 81 g/mol

Step 3: Find the molecular formula

We are given the molecular mass = 162 g/mol.

n=Molecular mass/Empirical mass

=162/81

So, the molecular formula is twice the empirical formula:

Molecular formula=(H2PO3)2=H4P2O6

Final Answer:

Empirical formula: H₂PO₃

Molecular formula: H₄P₂O₆

17. Calculate the mass of substance ‘A’ which in gaseous form occupies 10 litres at 27°C and 700 mm pressure. The molecular mass of ‘A’ is 60.

Ans: Step 1: Known values

Volume V=10L

Temperature

T=27∘ C=27+273=300K

Pressure P=700mmHg

Molecular mass M=60g/mol

Gas constant

Step 2: Use ideal gas law

PV=nRT

n=PV/RT

n= 62.3637×300/700×10

n=7000/18709.11

≈0.374mol

Step 3: Find mass

mass=n×M

mass=0.374×60≈22.44g

Final Answer:22.44g

18. A gas cylinder can hold 1 kg of hydrogen at room temperature and pressure.

(a) What mass of carbon dioxide can it hold under similar conditions of temperature and pressure?

(b) If the number of molecules of hydrogen in the cylinder is X, calculate the number of carbon dioxide molecules in the cylinder. Give reasons for your answer.

Ans: (a) Mass of carbon dioxide the cylinder can hold

The cylinder has a fixed volume and holds gases at the same temperature and pressure.

According to Avogadro’s law, equal volumes of gases under the same conditions contain equal number of moles.

Molar mass of H₂ = 2 g/mol, Molar mass of CO₂ = 44 g/mol.

Number of moles of H₂ in 1 kg (1000 g) = 1000/2=500moles.

The cylinder can hold 500 moles of any gas under the same conditions.

Mass of CO₂ = 500×44=22000 g=22 kg.

(b) Number of CO₂ molecules

If number of H₂ molecules = X then number of CO₂ molecules is also X

Reason: Equal number of moles of gases contain equal number of molecules (Avogadro’s law). Since the cylinder holds the same number of moles of any gas, the number of molecules remains the same.

19. Following questions refer to one mole of chlorine gas.

(a) What is the volume occupied by this gas at STP?

(b) What will happen to volume of gas, if pressure is doubled?

(c) What volume will it occupy at 273°C?

(d) If the relative atomic mass of chlorine is 35.5, what will be the mass of 1 mole of chlorine gas?

Ans: (a) Volume at STP

One mole of any gas at STP occupies 22.4 dm³. So, the volume of one mole of chlorine gas is 22.4 dm³.

(b) Effect of doubling pressure

According to Boyle’s law, volume is inversely proportional to pressure at constant temperature. If pressure is doubled, the volume becomes half of the original volume.

At STP, temperature = 273 K.

Given temperature = 273 °C = 273 + 273 = 546 K.

Volume is directly proportional to absolute temperature (Charles’s law).

Since temperature doubles, volume also doubles.

New volume = 22.4 dm³ × 2 = 44.8 dm³.

(d) Mass of 1 mole of chlorine gas

Chlorine is diatomic (Cl₂).

Relative atomic mass of Cl = 35.5.

Molecular mass of Cl₂ = 2 × 35.5 = 71 u.

Mass of 1 mole of Cl₂ gas = 71 g.

20.(a) A hydrate of calcium sulphate CaSO4⋅xH2O contains 21% water of crystallisation. Find the value of x.

(b) What volume of hydrogen and oxygen measured at STP will be required to prepare 1.8 g of water.

(c) How much volume will be occupied by 2 g of dry oxygen at 27°C and 740 mm pressure?

(d) What would be the mass of CO2 occupying a volume of 44 litres at 25°C and 750 mm pressure.

(e) 1 g of a mixture of sodium chloride and sodium nitrate is dissolved in water. On adding silver nitrate solution, 1.435 g of AgCl is precipitated.

AgNO3(aq) + NaCl (aq)→ AgCl (s) + NaNO3.

Calculate the percentage of NaCl in the mixture.

Ans: (a) Hydrate is CaSO₄·xH₂O; water of crystallisation = 21%.

Molar mass of CaSO₄ = 40 + 32 + 64 = 136 g/mol

Mass of water in 1 mole of hydrate = 18x g

Total molar mass of hydrate = 136 + 18x

% water = 18×136+18x ×100=21

136+18x/18x=0.21

18x=0.21(136+18x)

18x=28.56+3.78x

14.22x=28.56

x≈2

Answer: x=2 (CaSO₄·2H₂O)

(b) Prepare 1.8 g of water.

Molar mass of H₂O = 18 g/mol

Moles of water =

1.8/18=0.1mol

Reaction:

2H2+O2→2H2O

From stoichiometry: 2 mol H₂O → 2 mol H₂ and 1 mol O₂

So 0.1 mol H₂O → 0.1 mol H₂ and 0.05 mol O₂

Volume at STP:

H₂: 0.1×22.4=2.24 L

O₂: 0.05×22.4=1.12 L

Answer: H₂ = 2.24 L, O₂ = 1.12 L at STP

Moles of O₂ = 2/32=0.0625mol

T = 27 + 273 = 300 K, P = 740/760 atm

V= PnRT = 740/760/0.0625×0.0821×300

V≈ 0.9737/1.539

≈1.58 L

Answer: 1.58 L

(d) Mass of CO₂ occupying 44 L at 25°C and 750 mm pressure.

T = 25 + 273 = 298 K, P = 750/760 atm

n=PV/RT=(750/760)×440.0821×298

n≈43.421/24.4658

≈1.775 mol

Mass = 1.775×44≈78.1 g

Answer: 78.1 g

(e) 1 g mixture of NaCl and NaNO₃, AgNO₃ gives 1.435 g AgCl.

Molar mass AgCl = 108 + 35.5 = 143.5 g/mol

Moles AgCl = 1.435/143.5=0.01mol

1 mol AgCl ↔ 1 mol NaCl

Moles NaCl in mixture = 0.01 mol

Mass NaCl = 0.01×58.5=0.585 g

% NaCl = 0.585/1×100=58.5

Answer: 58.5%

21.(a) From the equation:

C+2H2SO4→CO2+2H2O+2SO2

Calculate:

(i) the mass of carbon oxidized by 49 g of sulphuric acid.

(ii) The volume of sulphur dioxide measured at STP, liberated at the same time.

(b) (i) A compound has the following percentage composition by mass: carbon 14.4%, hydrogen 1.2% and chlorine 84.5%. Determine the empirical formula of this compound. Work correct to 1 decimal place. (H = 1; C = 12; Cl = 35.5)

(ii) The relative molecular mass of this compound is 168, so what is its molecular formula?

Ans: (a) From the equation: C + 2H₂SO₄ → CO₂ + 2H₂O + 2SO₂

(i) Mass of carbon oxidized by 49 g of sulphuric acid.

The molar mass of H₂SO₄ is 98 g/mol. From the balanced equation, 2 moles (196 g) of H₂SO₄ oxidize 1 mole (12 g) of carbon. This gives a mass ratio of 196 g H₂SO₄ : 12 g C.

Therefore, for 49 g of H₂SO₄, the mass of carbon (C) oxidized is = (12 / 196) × 49 = 3 g.

(ii) Volume of SO₂ liberated at STP.

From the equation, 2 moles (196 g) of H₂SO₄ produce 2 moles of SO₂ gas. So, 196 g of H₂SO₄ produce 44.8 liters of SO₂.Therefore, 49 g of H₂SO₄ will produce = (44.8 / 196) × 49 = 11.2 liters of SO₂ at STP.

(b) (i) Determine the empirical formula.

Consider a 100 g sample of the compound:

Mass of Carbon (C) = 14.4 g

Mass of Hydrogen (H) = 1.2 g

Mass of Chlorine (Cl) = 84.5 g

Now, find the number of moles of each:

Moles of C = 14.4 / 12 = 1.2

Moles of H = 1.2 / 1 = 1.2

Moles of Cl = 84.5 / 35.5 ≈ 2.38

Find the simplest mole ratio by dividing by the smallest number of moles (1.2):

C : H : Cl = (1.2/1.2) : (1.2/1.2) : (2.38/1.2) ≈ 1 : 1 : 2

Thus, the empirical formula is CHCl₂.

(ii) Determine the molecular formula.

The empirical formula mass of CHCl₂ = 12 + 1 + (2 × 35.5) = 84 g/mol.

The given relative molecular mass is 168.

The multiplier (n) = Molecular Mass / Empirical Formula Mass = 168 / 84 = 2.

Therefore, the molecular formula is (CHCl₂)₂ = C₂H₂Cl₄.

Final Answers:

(a)(i) 3 g of Carbon

(a)(ii) 11.2 liters of SO₂ at STP

(b)(i) Empirical Formula: CHCl₂

(b)(ii) Molecular Formula: C₂H₂Cl₄

22. Find the percentage of

(a) oxygen in magnesium nitrate crystals [Mg(NO3)2⋅6H2O].

(b) boron in Na2B2O7⋅10H2O, [H = 1, B = 11, O = 16, Na = 23].

(c) phosphorus in the fertilizer superphosphate Ca(H2PO4)2

Ans:

(a) Percentage of oxygen in Mg(NO₃)₂·6H₂O

First, calculate the molar mass of the crystal.

Mg = 24 g/mol

N = 14 g/mol, so two NO₃ groups: 2 × (14 + 48) = 124 g/mol

Six H₂O molecules: 6 × (2 + 16) = 108 g/mol

Total Molar Mass = 24 + 124 + 108 = 256 g/mol

The total mass of oxygen (O) comes from:

The six oxygen atoms in two NO₃ groups: 6 × 16 = 96 g/mol

The six oxygen atoms in six H₂O molecules: 6 × 16 = 96 g/mol

Total Oxygen Mass = 96 + 96 = 192 g/mol

Percentage of oxygen = (192 / 256) × 100 = 75%

(b) Percentage of boron in Na₂B₂O₇·10H₂O

First, calculate the molar mass.

Na₂: 2 × 23 = 46 g/mol

B₂: 2 × 11 = 22 g/mol

O₇: 7 × 16 = 112 g/mol

10H₂O: 10 × (2 + 16) = 180 g/mol

Total Molar Mass = 46 + 22 + 112 + 180 = 360 g/mol

The mass of boron (B) is 22 g/mol.

Percentage of boron = (22 / 360) × 100 = 6.11%

First, calculate the molar mass.

Ca = 40 g/mol

Two H₂PO₄ groups:

H₄: 4 × 1 = 4 g/mol

P₂: 2 × 31 = 62 g/mol

O₈: 8 × 16 = 128 g/mol

Total Molar Mass = 40 + 4 + 62 + 128 = 234 g/mol

The mass of phosphorus (P) is 62 g/mol.

Percentage of phosphorus = (62 / 234) × 100 = 26.5%

23. A gas occupied 360 cm3 at 87°C and 380 mm Hg pressure. If the mass of gas is 0.546 g, find its relative molecular mass.

Ans: The gas occupies 360 cm³ (0.360 L) at 87°C (360 K) and 380 mm Hg pressure. To find the relative molecular mass, we use the ideal gas equation, PV = nRT.

First, convert pressure to atm: 380 mm Hg / 760 = 0.5 atm. Using R = 0.0821 L·atm·mol⁻¹·K⁻¹, the number of moles (n) is calculated as n = PV / RT = (0.5 atm × 0.360 L) / (0.0821 × 360 K) = 0.00609 mol.

The mass of the gas is given as 0.546 g. The molar mass is mass divided by moles: 0.546 g / 0.00609 mol ≈ 89.6 g/mol. Therefore, the relative molecular mass of the gas is approximately 90.

24. Solid ammonium dichromate decomposes as under:

(NH4)2Cr2O7→N2+Cr2O3+4H2O

If 63 g of ammonium dichromate decomposes. Calculate:

(a) the quantity in moles of (NH4)2Cr2O7

(b) the quantity in moles of nitrogen formed

(c) the volume of N2 evolved at STP.

(d) what will be the loss of mass?

(e) calculate the mass of chromium (III) oxide formed at the same time.

Ans: We are given the decomposition reaction:

(NH₄)₂Cr₂O₇ → N₂ + Cr₂O₃ + 4H₂O

Mass of ammonium dichromate = 63 g

Molar mass of (NH₄)₂Cr₂O₇ = (2×18) + (2×52) + (7×16) = 36 + 104 + 112 = 252 g/mol

(a) Moles of (NH₄)₂Cr₂O₇

Moles = Mass / Molar mass = 63 g / 252 g/mol = 0.25 mol

(b) Moles of nitrogen (N₂) formed

From the balanced equation:

1 mole of (NH₄)₂Cr₂O₇ produces → 1 mole of N₂

So, 0.25 moles of (NH₄)₂Cr₂O₇ will produce → 0.25 moles of N₂

Volume of N₂ = 0.25 mol × 22.4 L/mol = 5.6 L

(d) Loss of mass

The solid reactants produce solid Cr₂O₃ and gaseous N₂ & H₂O vapor. The mass loss is due to gases (N₂ and H₂O) escaping.

From the equation:

1 mole (NH₄)₂Cr₂O₇ (252 g) → 1 mole N₂ (28 g) + 4 moles H₂O (72 g)

So mass of gaseous products per mole of reactant = 28 + 72 = 100 g

For 0.25 moles of reactant:

Mass of gases = 0.25 × 100 g = 25 g

So, loss of mass = 25 g

(e) Mass of chromium(III) oxide (Cr₂O₃) formed

From the equation:

1 mole (NH₄)₂Cr₂O₇ → 1 mole Cr₂O₃

Molar mass of Cr₂O₃ = (2×52) + (3×16) = 104 + 48 = 152 g/mol

Moles of Cr₂O₃ formed = 0.25 mol

Mass of Cr₂O₃ = 0.25 × 152 = 38 g

25. Hydrogen sulphide gas burns in oxygen to yield 12.8 g of sulphur dioxide gas as under :

2H2S+3O2→2H2O+2SO2

Calculate the volume of hydrogen sulphide at STP.Also, calculate the volume of oxygen required at STP, which will complete the combustion of hydrogen sulphide determined in (litres).

Ans: Based on the reaction 2H₂S + 3O₂ → 2H₂O + 2SO₂, the production of 12.8 grams of SO₂ is analyzed. Since the molar mass of SO₂ is 64 g/mol, 12.8 grams corresponds to 0.2 moles.

From the reaction’s stoichiometry, the mole ratio of H₂S to SO₂ is 1:1. Therefore, 0.2 moles of SO₂ were produced by 0.2 moles of H₂S. At STP, this equals a volume of 0.2 × 22.4 = 4.48 litres of H₂S.

Similarly, the mole ratio of O₂ to SO₂ is 3:2. Thus, the amount of O₂ required is (3/2) × 0.2 = 0.3 moles. At STP, this equals a volume of 0.3 × 22.4 = 6.72 litres of O₂.

Final Answer:

Volume of H₂S = 4.48 L

Volume of O₂ = 6.72 L

26. Ammonia burns in oxygen and the combustion, in the presence of a catalyst, may be represented by:

2NH3+212O2→2NO+3H2O[H=1,N=14,O=16].

What mass of steam is produced when 1.5 g of nitrogen monoxide is formed?

Ans: When ammonia burns with oxygen in the presence of a catalyst, the balanced equation given is:

2NH3+21/2O2→2NO+3H2O

From the equation,

2 moles of NO are produced along with 3 moles of H₂O.

Molar mass of NO = 14 + 16 = 30 g/mol.

Molar mass of H₂O = 18 g/mol.

So,

2 moles of NO ≡ 3 moles of H₂O

Mass of 2 moles of NO = 2×30=60 g

Mass of 3 moles of H₂O = 3×18=54 g

Thus, 60 g NO → 54 g H₂O.

Given 1.5 g of NO is formed:

Mass of steam (H₂O)=54/60×1.5=1.35 g

Final answer: 1.35 g of steam is produced.

27. If a crop of wheat removes 20 kg of nitrogen per hectare of soil, what mass of the fertilizer, calcium nitrate Ca(NO3)2 would be required to replace the nitrogen in a 10 hectare field?

Ans: To replenish 20 kg of nitrogen per hectare over a 10-hectare field, 200 kg of nitrogen is required in total. Using calcium nitrate, Ca(NO₃)₂, which supplies 28 grams of nitrogen per 164 grams of fertilizer, the nitrogen content ratio is 28/164. Dividing the total nitrogen needed by this fraction, 200 kg ÷ (28/164), gives approximately 1171 kg. Thus, about 1171 kilograms of calcium nitrate is needed to meet the nitrogen requirement for the 10-hectare area.

28. Concentrated nitric acid oxidises phosphorus to phosphoric acid according to the following equation:

P+5HNO3→H3PO4+5NO2+H2O

If 6.2 g of phosphorus was used in the reaction calculate:

(a) Number of moles of phosphorus taken and mass of phosphoric acid formed.

(b) What mass of nitric acid will be consumed at the same time?

(c) What would be the volume of steam produced at the same time if measured at 760 mm Hg pressure and 273∘C?

Ans: The balanced equation is:

P + 5HNO₃ → H₃PO₄ + 5NO₂ + H₂O

Mass of Phosphorus (P) used = 6.2 g

Molar mass of P = 31 g/mol

(a) Moles of Phosphorus and Mass of Phosphoric Acid

Moles of P taken = Mass / Molar mass = 6.2 g / 31 g/mol = 0.2 mol

From the equation, 1 mole of P produces 1 mole of H₃PO₄.

So, 0.2 mol of P will produce 0.2 mol of H₃PO₄.

Molar mass of H₃PO₄ = 3(1) + 31 + 4(16) = 98 g/mol

Mass of H₃PO₄ formed = Moles × Molar mass = 0.2 mol × 98 g/mol = 19.6 g

(b) Mass of Nitric Acid Consumed

From the equation, 1 mole of P reacts with 5 moles of HNO₃.

So, 0.2 mol of P will react with 0.2 × 5 = 1.0 mol of HNO₃.

Molar mass of HNO₃ = 1 + 14 + 48 = 63 g/mol

Mass of HNO₃ consumed = Moles × Molar mass = 1.0 mol × 63 g/mol = 63 g

From the equation, 1 mole of P produces 1 mole of H₂O (steam).

So, 0.2 mol of P will produce 0.2 mol of H₂O vapor.

The gas is measured at 760 mm Hg and 273°C.

Pressure (P) = 760 mm Hg = 1 atm

Temperature (T) = 273°C + 273 = 546 K

Moles of steam (n) = 0.2 mol

Gas constant (R) = 0.0821 L·atm·mol⁻¹·K⁻¹

Using the Ideal Gas Equation: PV = nRT

Volume of steam (V) = (nRT) / P = (0.2 mol × 0.0821 L·atm·mol⁻¹·K⁻¹ × 546 K) / 1 atm

V = 8.96 L

29. 112 cm of a gaseous fluoride of phosphorus has a mass of 0.63 g. Calculate the relative molecular mass of the fluoride. If the molecule of the fluoride contains only one atom of phosphorus, then determine the formula of the phosphorus fluoride. [F = 19; P = 18.]

Ans:Volume of gaseous phosphorus fluoride = 112 cm³ = 0.112 L

Mass of gas = 0.63 g

To find the relative molecular mass, we use the fact that 1 mole of any gas at STP (0°C and 1 atm) occupies 22.4 L. However, the given volume is not at STP, so we must find the volume at STP first.

Let’s find the volume at STP using the gas law proportion (since pressure is constant, we use Charles’s Law):

V₁ / T₁ = V₂ / T₂

V₁ = 0.112 L

T₁ = 273°C + 273 = 546 K (Given temperature, assumed for calculation)

T₂ = 273 K (STP temperature)

V₂ (Volume at STP) = (V₁ × T₂) / T₁ = (0.112 L × 273 K) / 546 K = 0.056 L

Now, 0.056 L of the gas has a mass of 0.63 g.

22.4 L of the gas (1 mole) would have a mass of:

Mass = (0.63 g / 0.056 L) × 22.4 L = 252 g

Relative Molecular Mass = 252 g/mol

Determining the Formula:

The molecule contains one atom of phosphorus (P).

Given: Atomic mass of P = 18 (as per question), F = 19.

Let the formula be PFₓ.

Molecular Mass = Mass of P + (x × Mass of F)

252 = 18 + (x × 19)

252 – 18 = 19x

234 = 19x

x = 234 / 19 ≈ 12.3

This value is not a whole number, which suggests a possible discrepancy with the given atomic mass of P (18). Let’s recalculate using the standard atomic mass of P = 31.

Molecular Mass = Mass of P + (x × Mass of F)

252 = 31 + (x × 19)

252 – 31 = 19x

221 = 19x

x = 221 / 19 = 11.63 ≈ 12

Again, we do not get a perfect integer. However, a known stable phosphorus fluoride is Phosphorus Pentafluoride (PF₅, Molar mass = 126). The calculated mass (252) is exactly double of 126. Therefore, the gas is likely dimerized, meaning two PF₅ molecules join to form a larger molecule.

The formula of the dimer of phosphorus pentafluoride is P₂F₁₀.

Let’s verify:

Molar mass of P₂F₁₀ = (2 × 31) + (10 × 19) = 62 + 190 = 252 g/mol. This matches perfectly.

Final Answer: The relative molecular mass is 252, and the formula of the phosphorus fluoride is P₂F₁₀.

30. Washing soda has the formula Na2CO3⋅10H2O. What mass of anhydrous sodium carbonate is left when all the water of crystallization is expelled by heating 57.2 g of washing soda?

Ans: When 57.2 g of washing soda (Na₂CO₃·10H₂O) is heated, the hydrate loses its water of crystallization, leaving behind anhydrous sodium carbonate (Na₂CO₃).

The molar mass of Na₂CO₃·10H₂O is:

Na₂CO₃ = 106 g/mol

10H₂O = 180 g/mol

Total = 106 + 180 = 286 g/mol

The mass of Na₂CO₃ in the hydrate is:

(106 g Na₂CO₃ / 286 g hydrate) × 57.2 g = 21.2 g

Thus, 21.2 g of anhydrous sodium carbonate remains after heating.

31. A metal M forms a volatile chloride containing 65.5% chlorine. If the density of the chloride relative to hydrogen is 162.5, find the molecular formula of the chloride (M = 56).

Ans: The metal M has an atomic mass of 56 g/mol.

The chloride contains 65.5% chlorine by mass, so the mass of chlorine in the compound is 65.5 g per 100 g of the compound.

The mass of metal M per 100 g of compound = 100 − 65.5 = 34.5 g.

Let the molecular formula be MCln

Mass ratio: Mass of Cl/Mass of M

=n×35.5/56

Given:

n×35.5/56=65.5/34.5

n×35.5×34.5=65.5×56

n=65.5×56/35.5×34.5≈2.98≈3

So the empirical formula is MCl3

Vapor density relative to hydrogen = 162.5

Molecular mass = 2×162.5=325 g/mol.

Empirical formula mass of MCl3=56+3×35.5=162.5 g/mol.

Number of empirical units in molecule = 325/162.5=2

Thus, molecular formula = M2Cl6

Final answer: M2Cl6

32. A compound X consists of 4.8% carbon and 95.2% bromine by mass:

(i) Determine the empirical formula of this compound working correct to one decimal place (C = 12; Br = 80).

(ii) If the vapour density of the compound is 252, what is the molecular formula of the compound?

Ans: (i) Determination of the Empirical Formula

To find the empirical formula, we calculate the number of moles of each element in a 100 g sample.

Mass of Carbon = 4.8 g → Moles of C = 4.8 / 12 = 0.4 mol

Mass of Bromine = 95.2 g → Moles of Br = 95.2 / 80 = 1.19 mol

Next, we find the simplest mole ratio by dividing both values by the smallest number of moles (0.4).

C : 0.4 / 0.4 = 1

Br : 1.19 / 0.4 ≈ 2.975

The ratio is approximately 1 : 3. Rounding 2.975 to one decimal place gives 3.0. Therefore, the empirical formula, based on a 1:3 ratio of C to Br, is CBr₃.

(ii) Determination of the Molecular Formula

The vapour density (VD) is given as 252. The molecular mass (M) is twice the vapour density.

Molecular Mass (M) = 2 × VD = 2 × 252 = 504 g/mol

The empirical formula is CBr₃. Its empirical mass is (12 + 3×80) = 252 g/mol.

To find the molecular formula, we determine how many empirical units fit into the molecular mass.

n = (Molecular Mass) / (Empirical Mass) = 504 / 252 = 2

The molecular formula is (CBr₃)₂, which is C₂Br₆.

Final Answers:

(i) The empirical formula is CBr₃.

(ii) The molecular formula is C₂Br₆.

33. The reaction: 4N2O+CH4→CO2+2H2O+4N2 takes place in the gaseous state. If all volumes are measured at the same temperature and pressure, calculate the volume of dinitrogen oxide (N2O) required to give 150 cm of steam.

Ans: Based on the balanced chemical equation, 4 volumes of N₂O produce 2 volumes of steam (H₂O gas). This simplifies to a 2:1 volume ratio between N₂O and steam.Therefore, for every 1 cm³ of steam produced, 2 cm³ of N₂O is required. To produce 150 cm³ of steam, the volume of N₂O needed is 2 × 150 = 300 cm³.

34. Samples of the gases O2, N2, CO2 and CO under the same conditions of temperature and pressure contain the same number of molecules x.The molecules of oxygen occupy V litres and have a mass of 8 g under the same conditions of temperature and pressure.

What is the volume occupied by:

(a) x molecules of N2

(b) 3x molecules of CO,

(c) What is the mass of CO2 in grams?

(d) In answering the above questions, which law have you used?

Ans: (a) Volume of x molecules of N₂

Under the same temperature and pressure, equal number of molecules occupy equal volume (Avogadro’s law).

Since x molecules of O₂ occupy V litres, x molecules of N₂ will also occupy V litres.

(b) Volume of 3x molecules of CO

Number of molecules is proportional to volume.

If x molecules occupy V litres, then 3x molecules will occupy 3V litres.

x molecules of O₂ have mass = 8 g (given).

Molar mass of O₂ = 32 g/mol, so x molecules correspond to 8/32=0.25moles.

Thus x molecules = 0.25 × NA molecules.

For CO₂ (molar mass = 44 g/mol),

Mass of x molecules = 0.25×44=11 g.

So mass of CO₂ = 11 grams.

(d) Law used

Avogadro’s law — equal volumes of gases at the same temperature and pressure contain equal number of molecules.

35. The percentage composition of sodium phosphate as determined by analysis is 42.1% sodium, 18.9% phosphorus and 39% oxygen. Find the empirical formula of the compound.

Ans: Based on the percentage composition, the mass of each element in a 100 g sample is 42.1 g of sodium (Na), 18.9 g of phosphorus (P), and 39.0 g of oxygen (O). Using the atomic masses, the mole ratio is calculated as approximately 3 atoms of sodium to 1 atom of phosphorus to 4 atoms of oxygen. Therefore, the empirical formula of the compound is Na₃PO₄.

36. What volume of oxygen is required to burn completely a mixture of 22.4 dm3 of methane and 11.2 dm3 of hydrogen into carbon dioxide and steam?

CH4+2O2→CO2+2H2O

Ans: To solve this, we first determine the oxygen needed for each gas in the mixture, using the fact that 1 mole of any gas occupies 22.4 dm³ at standard conditions.

For Methane (CH₄):

The reaction is CH₄ + 2O₂ → CO₂ + 2H₂O. This shows 1 volume of CH₄ requires 2 volumes of O₂.

We have 22.4 dm³ of CH₄, which is 1 mole.

Thus, the oxygen required is 1 × 2 = 2 moles.

Volume of O₂ = 2 × 22.4 dm³ = 44.8 dm³.

For Hydrogen (H₂):

The reaction is 2H₂ + O₂ → 2H₂O. This shows 2 volumes of H₂ require 1 volume of O₂.

We have 11.2 dm³ of H₂.

Volume of O₂ required = (11.2 / 2) = 5.6 dm³.

Total Oxygen Required:

Adding both amounts gives the total volume of oxygen needed: 44.8 dm³ + 5.6 dm³ = 50.4 dm³.Therefore, 50.4 dm³ of oxygen is required to burn the mixture completely.

37. The gases hydrogen, oxygen, carbon dioxide, sulphur dioxide and chlorine are arranged in order of their increasing relative molecular masses. Given 8 g of each gas at STP, which gas will contain the least number of molecules and which gas the most?

Ans: Among the given gases—hydrogen (H₂), oxygen (O₂), carbon dioxide (CO₂), sulphur dioxide (SO₂), and chlorine (Cl₂)—each has a mass of 8 g at STP.

The number of molecules depends on the number of moles, which is calculated as mass divided by molecular mass.

Since hydrogen has the smallest molecular mass (2 g/mol), it will have the largest number of molecules.

Chlorine, with the largest molecular mass (71 g/mol), will have the smallest number of molecules.

Thus, hydrogen contains the most molecules, and chlorine contains the least.

38. 10 g of a mixture of sodium chloride and anhydrous sodium sulphate is dissolved in water. An excess of barium chloride solution is added and 6.99 g of barium sulphate is precipitated according to the equation given below:

Na2SO4+BaCl2→BaSO4+2NaCl Calculate the percentage of sodium sulphate in the original mixture.

Ans: Of the 10 g mixture, only sodium sulfate reacts with barium chloride to form barium sulfate precipitate.

From the reaction:

Na2SO4+BaCl2→BaSO4+2NaCl

Molar mass of BaSO4=137+32+64=233 g/mol

Molar mass of Na2 SO4=46+32+64=142 g/mol

Mass of BaSO4 precipitated = 6.99 g

Moles of BaSO4=233/6.99 ≈0.03 mol

From stoichiometry, 1 mole of Na2 SO4 gives 1 mole of BaSO4

So, moles of Na2 SO4 in mixture = 0.03 mol

Mass of Na2 SO4=0.03×142=4.26 g

Percentage of Na2SO4 in mixture =4.26/10×100=42.6%

Answer: The original mixture contains 42.6% sodium sulfate.

39. When heated, potassium permanganate decomposes according to the following equation:

2KMnO4→K2MnO4+MnO2+O2solid residue

(a) Some potassium permanganate was heated in the test tube. After collecting one litre of oxygen at room temperature, it was found that the test tube had undergone a loss in mass of 1.32 g. If one litre of hydrogen under the same conditions of temperature and pressure has a mass of 0.0825 g, calculate the relative molecular mass of oxygen

(b) Given that the molecular mass of potassium permanganate is 158, what volume of oxygen (measured at room temperature) would be obtained by the complete decomposition of 15.8 g of potassium permanganate? (Molar volume at room temperature is 24 litres).

Ans: (a)In the experiment, the relative molecular mass of oxygen was found using Avogadro’s Law. Since one litre of oxygen had a mass of 1.32 g and one litre of hydrogen had a mass of 0.0825 g under the same conditions, the mass ratio is 1.32 / 0.0825 = 16. Taking hydrogen’s molecular mass as 2, the molecular mass of oxygen is 16 × 2 = 32.

(b)For the decomposition of 15.8 g of KMnO₄, the moles used are 0.1. According to the reaction 2KMnO₄ → K₂MnO₄ + MnO₂ + O₂, 0.1 moles of KMnO₄ produce 0.05 moles of oxygen gas. Using the molar volume of 24 litres at room temperature, the oxygen volume obtained is 0.05 × 24 = 1.2 litres.

40. (a) A flask contains 3.2 g of sulphur dioxide. Calculate the following:

(i) The moles of sulphur dioxide present in the flask.

(ii) The number of molecules of sulphur dioxide present in the flask.

(iii) The volume occupied by 3.2 g of sulphur dioxide at STP.

(S = 32, O = 16)

(b) An experiment showed that in a lead chloride solution, 6.21 g of lead is combined with 4.26 g of chlorine. What is the empirical formula of this chloride? (Pb = 207; Cl = 35.5).

Ans: (a) Sulphur Dioxide Calculations

The molar mass of sulphur dioxide (SO₂) is calculated as S (32) + 2×O (16) = 64 g/mol.

(i) Moles of SO₂:

Number of moles = Mass / Molar mass = 3.2 g / 64 g/mol = 0.05 mol.

(ii) Number of Molecules:

Number of molecules = Moles × Avogadro’s number = 0.05 mol × (6.022 × 10²³ molecules/mol) = 3.011 × 10²² molecules.

(iii) Volume at STP:

At STP, 1 mole of any gas occupies 22.4 litres.

Volume = Moles × 22.4 L/mol = 0.05 mol × 22.4 L/mol = 1.12 L.

(b) Empirical Formula of Lead Chloride

To find the empirical formula, we first determine the number of moles of each element in the sample.

Moles of Lead (Pb) = 6.21 g / 207 g/mol ≈ 0.03 mol

Moles of Chlorine (Cl) = 4.26 g / 35.5 g/mol ≈ 0.12 mol

Next, we find the simplest mole ratio by dividing both values by the smallest number of moles (0.03).

Pb : Cl = 0.03 / 0.03 : 0.12 / 0.03 = 1 : 4

Therefore, the empirical formula of the lead chloride is PbCl₄.

41. The volumes of gases A, B, C and D are in the ratio 1 : 2 : 2 : 4 under the same conditions of temperature and pressure.

(i) Which sample of gas contains the maximum number of molecules?

(ii) If the temperature and the pressure of gas A are kept constant, then what will happen to the volume of A when the number of molecules is doubled?

(iii) If this ratio of gas volumes refers to the reactants and products of a reaction, which gas law is being observed?

(iv) If the volume of A is actually 5.6 dm3 at STP, calculate the number of molecules in the actual volume of D at STP (Avogadro’s Number is 6×1023).

(v) Using your answer from (iv), state the mass of D if the gas is dinitrogen oxide (N2O).

Ans: (i) Which sample of gas contains the maximum number of molecules?

According to Avogadro’s law, under the same conditions of temperature and pressure, equal volumes of gases contain an equal number of molecules. Since gas D has the largest volume (a ratio of 4), it contains the maximum number of molecules.

(ii) What will happen to the volume of A when the number of molecules is doubled?

Avogadro’s law states that the volume of a gas is directly proportional to the number of molecules at constant temperature and pressure. Therefore, if the number of molecules of gas A is doubled while its temperature and pressure are held constant, its volume will also double.

(iii) Which gas law is being observed?

If the ratio of gas volumes refers to the reactants and products of a reaction, the law being observed is Gay-Lussac’s Law of Gaseous Volumes. This law states that gases combine or are produced in simple volume ratios when they react with one another, provided all measurements are made at the same temperature and pressure.

(iv) Calculate the number of molecules in the actual volume of D at STP.

The volume ratio of A : D is 1 : 4.

The actual volume of A at STP is 5.6 dm³.

Therefore, the actual volume of D = 4 × 5.6 dm³ = 22.4 dm³.

At STP, 22.4 dm³ (or 22.4 liters) of any gas contains one mole of molecules, which is Avogadro’s number (6 × 10²³).

Thus, the number of molecules in 22.4 dm³ of gas D is 6 × 10²³.

(v) State the mass of D if the gas is dinitrogen oxide (N₂O).

The molar mass of dinitrogen oxide (N₂O) is (2×14) + 16 = 44 g/mol.

We have one mole of N₂O gas (as calculated from its volume of 22.4 dm³ at STP).

Therefore, the mass of gas D is 44 grams.

42. The equations given below relate to the manufacture of sodium carbonate (molecular weight of Na2CO3= 106).

NaCl + NH3+ CO2 + H2O → NaHCO3+ NH4Cl

2NaHCO3 → Na2CO3 + H2O + CO2

Equations (1) and (2) are based on the production of 21.2 g of sodium carbonate.

(a) What mass of sodium hydrogen carbonate must be heated to give 21.2 g of sodium carbonate?

(b) To produce the mass of sodium hydrogen carbonate calculated in (a), what volume of carbon dioxide, measured at STP, would be required?

Ans:(a) Mass of Sodium Hydrogen Carbonate

The second reaction shows that 2 moles of sodium hydrogen carbonate (NaHCO₃) decompose to produce 1 mole of sodium carbonate (Na₂CO₃).

The molar mass of Na₂CO₃ is 106 g/mol. Therefore, 21.2 g of sodium carbonate is 21.2 / 106 = 0.2 moles.

To produce 0.2 moles of Na₂CO₃, we need twice the number of moles of NaHCO₃, which is 0.4 moles.

The molar mass of NaHCO₃ is 84 g/mol. Therefore, the mass of NaHCO₃ required is 0.4 × 84 = 33.6 g.

(b) Volume of Carbon Dioxide at STP

The first reaction shows that 1 mole of CO₂ is needed to produce 1 mole of NaHCO₃.

From part (a), we require 0.4 moles of NaHCO₃. This means 0.4 moles of CO₂ are also required.

At STP, 1 mole of any gas occupies 22.4 liters. Therefore, the volume of CO₂ needed is 0.4 × 22.4 = 8.96 liters.

Final Answers:

(a) 33.6 g of sodium hydrogen carbonate must be heated.

(b) 8.96 liters of carbon dioxide at STP would be required.

43. A sample of ammonium nitrate when heated yields 8.96 litres of steam (measured at STP).

NH4NO3 → N2O + 2H2O

(Note: The question seems incomplete in the image; likely asks for the mass of ammonium nitrate that produced this steam.)

Ans: Based on the reaction NH₄NO₃ → N₂O + 2H₂O, heating ammonium nitrate produces water vapour (steam) as a product.

At STP, 1 mole of any gas occupies 22.4 litres. The volume of steam produced is 8.96 litres.

Number of moles of steam = Volume at STP / Molar Volume = 8.96 L / 22.4 L/mol = 0.4 moles.

From the balanced equation, 2 moles of H₂O (steam) are produced from 1 mole of NH₄NO₃.

Therefore, moles of NH₄NO₃ that produced 0.4 moles of steam = (1 mole NH₄NO₃ / 2 moles H₂O) × 0.4 moles H₂O = 0.2 moles.

The molar mass of NH₄NO₃ is 80 g/mol.

Thus, the mass of ammonium nitrate decomposed = Moles × Molar mass = 0.2 mol × 80 g/mol = 16 g.

Final Answer: 16 grams

44. Given that the relative molecular mass of copper oxide is 80, what volume of ammonia (measured at STP) is required to completely reduce 120 g of copper oxide?

The equation for the reaction is:

3CuO+2NH3→3Cu+3H2O+N2

Ans: To find the volume of ammonia needed, we first determine the number of moles of copper oxide (CuO) we have. With a relative molecular mass of 80 g/mol, 120 grams of CuO is equal to 120 / 80 = 1.5 moles.

From the balanced equation:

3CuO + 2NH₃ → 3Cu + 3H₂O + N₂

We see that 3 moles of CuO require 2 moles of NH₃.

Using this ratio, 1.5 moles of CuO will require (2/3) × 1.5 = 1 mole of NH₃.

At STP, one mole of any gas occupies 22.4 liters. Therefore, the volume of ammonia required is 22.4 liters.

45. (a) Calculate the number of moles and the number of molecules present in 1.4 g of ethylene gas. What is the volume occupied by the same amount of ethylene?

(b) What is the vapour density of ethylene?

Ans: (a) Ethylene (C₂H₄)

Molar mass = 28 g/mol

Moles = 0.05 mol

Molecules = 3.011 × 10²²

Volume at STP = 1.12 L

(b) Vapour Density

Vapour density = 14

Final Answer:

(a) 0.05 mol, 3.011 × 10²² molecules, 1.12 L

(b) 14

46. (a) Calculate the percentage of sodium in sodium aluminium fluoride (Na3AlF6) correct to the nearest whole number. (F = 19; Na = 23; Al = 27)

(b) 560 ml of carbon monoxide is mixed with 500 ml of oxygen and ignited. The chemical equation for the reaction is as follows:

2CO+O2→2CO2

Calculate the volume of oxygen used and carbon dioxide formed in the above reaction.

Ans: (a)Molar mass of Na₃AlF₆ = 3(23) + 27 + 6(19)

= 69 + 27 + 114 = 210 g/mol

Mass of sodium = 3 × 23 = 69 g

Percentage of sodium = (69 / 210) × 100 ≈ 33%

Answer: 33%

(b)Reaction: 2CO + O₂ → 2CO₂

From stoichiometry: 2 vol CO reacts with 1 vol O₂ to give 2 vol CO₂

Given: CO = 560 mL, O₂ = 500 mL

O₂ required for 560 mL CO = 560 / 2 = 280 mL

(O₂ is in excess)

Volume of O₂ used = 280 mL

Volume of CO₂ formed = volume of CO used = 560 mL (since 2 vol CO → 2 vol CO₂)

Answer: O₂ used = 280 mL, CO₂ formed = 560 mL

47. (a) A gas cylinder of capacity of 20 dm3 is filled with gas X the mass of which is 10 g. When the same cylinder is filled with hydrogen gas at the same temperature and pressure the mass of the hydrogen is 2 g, hence the relative molecular mass of the gas is:

(i) 5

(ii) 10

(iii) 15

(iv) 20

(b) (i) Calcium carbide is used for the artificial ripening of fruits. Actually the fruit ripens because of the heat evolved while calcium carbide reacts with moisture. During this reaction calcium hydroxide and acetylene gas is formed. If 200 cm3 of acetylene is formed from a certain mass of calcium carbide, find the volume of oxygen required and carbon dioxide formed during the complete combustion. The combustion reaction can be represented as below:

2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)

(ii) A gaseous compound of nitrogen and hydrogen contains 12.5% hydrogen by mass. Find the molecular formula of the compound if its relative molecular mass is 37. [N = 14, H = 1]

(c) (i) A gas cylinder contains 24×1021 molecules of nitrogen gas. If Avogadro’s number is 6×1023 and the relative atomic mass of nitrogen is 14, calculate:

(1) Mass of nitrogen gas in the cylinder

(2) Volume of nitrogen at STP in dm³

(ii) Commercial sodium hydroxide weighing 30 g has some sodium chloride in it. The mixture on dissolving in water and subsequent treatment with excess silver nitrate solution formed a precipitate weighing 14.3 g. What is the percentage of sodium chloride in the commercial sample of sodium hydroxide? The equation for the reaction is:

NaCl+AgNO3→AgCl+NaNO3

[Relative molecular mass of NaCl = 58; AgCl = 143]

(iii) A certain gas ‘X’ occupies a volume of 100 cm³ at STP, and weighs 0.5 g. Find its relative molecular mass.

Ans: (a) Moles of H₂ = 1 mol

Moles of X = 10/M

Since moles are equal:

10/M = 1 → M = 10 g/mol

Answer: 10 g/mol

(b)(i)

2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O

For 200 cm³ C₂H₂:

O₂ = (5/2) × 200 = 500 cm³

CO₂ = (4/2) × 200 = 400 cm³

Answer: O₂ = 500 cm³, CO₂ = 400 cm³

(b)(ii)

H = 12.5%, N = 87.5%

Mole ratio N : H = 6.25 : 12.5 = 1 : 2

Empirical formula = NH₂, empirical mass = 16

Molecular mass = 32 → n = 2

Molecular formula = N₂H₄

Answer: N₂H₄

(c)(i)

Molecules of N₂ = 2.4 × 10²²

Moles = 0.04 mol

Mass = 0.04 × 28 = 1.12 g

Volume at STP = 0.04 × 22.4 = 0.896 dm³

Answer: Mass = 1.12 g, Volume = 0.896 dm³

(c)(ii)

AgCl mass = 14.3 g, molar mass = 143 g/mol

Moles of AgCl = 0.1 mol = moles of NaCl

Mass of NaCl = 0.1 × 58 = 5.8 g

% NaCl = (5.8 / 30) × 100 = 19.33%

Answer: 19.33%

(c)(iii)

Mass of 100 cm³ gas = 0.5 g

Mass of 22400 cm³ = (0.5 / 100) × 22400 = 112 g

Answer: 112 g/mol

2010

48. (a) (i) LPG stands for liquefied petroleum gas. Varieties of LPG are marketed including a mixture of propane (60%) and butane (40%). If 10 litre of this mixture is burnt, find the total volume of carbon dioxide gas added to the atmosphere. Combustion reactions can be represented as:

C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)

2C4H10(g)+13O2(g)→8CO2(g)+10H2O(g)

(ii) Calculate the percentage of nitrogen and oxygen in ammonium nitrate. [Relative molecular mass of ammonium nitrate is 80, H = 1, N = 14, O = 16]

(b) 4.5 moles of calcium carbonate are reacted with dilute hydrochloric acid.

(i) Write the equation for the reaction.

(ii) What is the mass of 4.5 moles of calcium carbonate? (Relative molecular mass of calcium carbonate is 100)

(iii) What is the volume of carbon dioxide liberated at STP?

(iv) What mass of calcium chloride is formed? (Relative molecular mass of calcium chloride is 111)

(v) How many moles of HCl are used in this reaction?

Ans: (a)(i)

LPG: 6 L C₃H₈ → 18 L CO₂, 4 L C₄H₁₀ → 16 L CO₂

Total CO₂ = 34 L

(a)(ii)

NH₄NO₃ → N = 35%, O = 60%

(b)(i)

CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

(b)(ii)

Mass of 4.5 moles CaCO₃ = 450 g

(b)(iii)

Volume of CO₂ at STP = 100.8 L

(b)(iv)

Mass of CaCl₂ = 499.5 g

(b)(v)

Moles of HCl required = 9.0 moles

2011

49. (a) (i) Calculate the volume of 320 g of SO₃ at STP. (Atomic mass: S = 32 and O = 16)

(ii) State Gay-Lussac’s Law of combining volumes.

(iii) Calculate the volume of oxygen required for the complete combustion of 8.8 g of propane (C₃H₈). (Atomic mass: C = 12, O = 16, H = 1, Molar Volume = 22.4 dm³ at STP)

(b) (i) An organic compound with vapour density = 94 contains C = 12.67%, H = 2.13%, and Br = 85.11%. Find the molecular formula. [Atomic mass: C = 12, H = 1, Br = 80]

(ii) Calculate the mass of:

(1) 1022 atoms of sulphur

(2) 0.1 mole of carbon dioxide

[Atomic mass: S = 32, C = 12 and O = 16 and Avogadro’s Number = 6×1023]

Ans: (a) (i) Volume of 320 g of SO₃ at STP

Step 1: Molar mass of SO₃

Atomic mass of S = 32 g/mol

Atomic mass of O = 16 g/mol

Molar mass of SO₃ = 32 + (3 × 16) = 32 + 48 = 80 g/mol

Step 2: Number of moles of SO₃

Mass = 320 g

Number of moles = Mass / Molar mass = 320 / 80 = 4 moles

Step 3: Volume at STP

At STP, 1 mole of gas occupies 22.4 dm³

Volume = 4 × 22.4 = 89.6 dm³

Final Answer: The volume of 320 g of SO₃ at STP is 89.6 dm³.

(a) (ii) Gay-Lussac’s Law of Combining Volumes

Gay-Lussac’s Law states that when gases react, they do so in volumes that are in simple whole-number ratios to each other and to the volumes of the gaseous products, provided all measurements are taken at the same temperature and pressure.

(a) (iii) Volume of Oxygen for Combustion of 8.8 g of Propane (C₃H₈)

Step 1: Balanced chemical equation

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

1 mole of propane requires 5 moles of oxygen.

Step 2: Molar mass of propane (C₃H₈)

C = 12 g/mol, H = 1 g/mol

Molar mass = (3 × 12) + (8 × 1) = 36 + 8 = 44 g/mol

Step 3: Moles of propane

Mass = 8.8 g

Moles = 8.8 / 44 = 0.2 moles

Step 4: Moles of oxygen required

From the equation, moles of O₂ = 5 × 0.2 = 1.0 mole

Step 5: Volume of oxygen at STP

Volume = 1.0 × 22.4 = 22.4 dm³

Final Answer: The volume of oxygen required is 22.4 dm³.

(b) (i) Molecular Formula from Vapour Density and Percentage Composition

Step 1: Assume 100 g of the compound

Mass of C = 12.67 g

Mass of H = 2.13 g

Mass of Br = 85.11 g

Step 2: Moles of each element

Moles of C = 12.67 / 12 ≈ 1.0558

Moles of H = 2.13 / 1 = 2.13

Moles of Br = 85.11 / 80 ≈ 1.0639

Step 3: Simplest mole ratio

Divide by the smallest value (1.0558):

C: 1.0558 / 1.0558 = 1

H: 2.13 / 1.0558 ≈ 2.017 ≈ 2

Br: 1.0639 / 1.0558 ≈ 1.007 ≈ 1

Empirical formula = CH₂Br

Step 4: Empirical formula mass

= 12 + (2 × 1) + 80 = 94 g/mol

Step 5: Molecular mass from vapour density

Vapour density = 94

Molecular mass = 2 × 94 = 188 g/mol

Step 6: Molecular formula

n = Molecular mass / Empirical mass = 188 / 94 = 2

Molecular formula = (CH₂Br)₂ = C₂H₄Br₂

Final Answer: The molecular formula is C₂H₄Br₂.

(b) (ii) Mass Calculations

(1) Mass of 10²² atoms of sulphur

Avogadro’s number = 6 × 10²³ atoms/mol

Moles of S = 10²² / (6 × 10²³) = 1 / 60 moles

Atomic mass of S = 32 g/mol

Mass = (1 / 60) × 32 = 32 / 60 ≈ 0.533 g

Final Answer (1): The mass is 0.533 g.

(2) Mass of 0.1 mole of carbon dioxide

Molar mass of CO₂ = 12 + (2 × 16) = 44 g/mol

Mass = 0.1 × 44 = 4.4 g

Final Answer (2): The mass is 4.4 g.

2012

50. (a) Concentrated nitric acid oxidises phosphorus to phosphoric acid according to the following equation:

P+5HNO3(conc.)→H3PO4+H2O+5NO2

If 9.3 g of phosphorus was used in the reaction, calculate:

(i) Number of moles of phosphorus taken

(ii) The mass of phosphoric acid formed

(iii) The volume of nitrogen dioxide produced at STP

(b) (i) 67.2 litres of hydrogen combines with 44.8 litres of nitrogen to form ammonia under specific conditions as:

N2(g)+3H2(g)→2NH3(g)

Calculate the volume of ammonia produced. What is the other substance, if any, that remains in the resultant mixture?

(ii) The mass of 5.6 dm³ of a certain gas at STP is 12.0 g. Calculate the relative molecular mass of the gas.

(iii) Find the total percentage of Magnesium in magnesium nitrate crystals, Mg(NO₃)₂·6H₂O.

[Mg = 24, N = 14, O = 16 and H = 1]

Ans: (a) Given: P + 5HNO₃(conc.) → H₃PO₄ + H₂O + 5NO₂

Mass of P = 9.3 g

Molar mass of P = 31 g/mol

(i) Moles of Phosphorus taken:

Moles = Mass / Molar mass = 9.3 / 31 = 0.3 mol

(ii) Mass of Phosphoric acid formed:

From the equation, 1 mole of P produces 1 mole of H₃PO₄.

So, moles of H₃PO₄ formed = 0.3 mol

Molar mass of H₃PO₄ = 3(1) + 31 + 4(16) = 98 g/mol

Mass = Moles × Molar mass = 0.3 × 98 = 29.4 g

(iii) Volume of Nitrogen dioxide at STP:

From the equation, 1 mole of P produces 5 moles of NO₂.

So, moles of NO₂ produced = 0.3 × 5 = 1.5 mol

Volume at STP = Moles × 22.4 L = 1.5 × 22.4 = 33.6 L

(b) (i) N₂(g) + 3H₂(g) → 2NH₃(g)

Volumes given: H₂ = 67.2 L, N₂ = 44.8 L

According to Gay-Lussac’s law, volume ratios are same as molar ratios.

From the equation, 1 vol N₂ reacts with 3 vol H₂.

To react with 67.2 L of H₂, N₂ required = (67.2 / 3) = 22.4 L.

But we have 44.8 L of N₂. This means N₂ is in excess.

Thus, H₂ is the limiting reagent.

From the equation, 3 vol H₂ produces 2 vol NH₃.

So, Volume of NH₃ produced = (2/3) × 67.2 = 44.8 L

Substance remaining: Excess N₂ = 44.8 L – 22.4 L = 22.4 L of Nitrogen

(ii) Mass of 5.6 dm³ of gas at STP = 12.0 g

5.6 dm³ at STP = 5.6 / 22.4 = 0.25 moles

Mass = 12.0 g

Molar mass (Relative molecular mass) = Mass / Moles = 12.0 / 0.25 = 48 g/mol

(iii) For Mg(NO₃)₂·6H₂O:

Molar mass = 24 + 2[14 + 3(16)] + 6[2(1) + 16]

= 24 + 2(62) + 6(18)

= 24 + 124 + 108 = 256 g/mol

Mass of Magnesium = 24 g

% of Magnesium = (24 / 256) × 100 = 9.375%

2013

51. (a) (i) What volume of oxygen is required to burn 90 dm³ of butane under similar conditions of temperature and pressure?

2C4H10+13O2→8CO2+10H2O

(ii) The vapour density of a gas is 8. What would be the volume occupied by 24.0 g of the gas at STP?

(iii) A vessel contains X number of molecules of hydrogen gas at a certain temperature and pressure. How many molecules of nitrogen gas would be present in the same vessel under the same conditions of temperature and pressure?

(b) O2 is evolved by heating KClO3using MnO2as a catalyst.

2KClO3→MnO22KCl+3O2

(i) Calculate the mass of KClO3 required to produce 6.72 litre of O2 at STP.

[Atomic masses of K=39,Cl=35.5,O=16

(ii) Calculate the number of moles of oxygen present in the above volume and also the number of molecules.

(iii) Calculate the volume occupied by 0.01 mole of CO2 at STP.

Ans: (a)

(i) Volume of oxygen = 585 dm³

(ii) Volume at STP = 33.6 dm³

(iii) Molecules of nitrogen = X

(b)

(i) Mass of KClO₃ = 24.5 g

(ii) Moles of O₂ = 0.3 mol, Molecules of O₂ = 1.8066 × 10²³

(iii) Volume of 0.01 mole CO₂ at STP = 0.224 L (224 cm³)

2014

52. (a) (i) Oxygen oxidises ethyne to carbon dioxide and water as shown by the equation:

2C2H2+5O2→4CO2+2H2O

What volume of ethyne gas at STP is required to produce 8.4 dm3 of carbon dioxide at STP?

[H = 1, C = 12, O = 16]

(ii) A compound made up of two elements X and Y has an empirical formula X2Y. If the atomic weight of X is 10 and that of Y is 5 and the compound has a vapour density (V.D.) 25, find its molecular formula.

(b) A cylinder contains 68 g of Ammonia gas at STP

(i) What is the volume occupied by this gas?

(ii) How many moles of ammonia are present in the cylinder?

(iii) How many molecules of ammonia are present in the cylinder?

Ans: (a) (i) Volume of Ethyne

From the reaction:

2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O

2 volumes of C₂H₂ give 4 volumes of CO₂.

So, 1 volume of C₂H₂ gives 2 volumes of CO₂.

Volume of C₂H₂ needed for 8.4 dm³ CO₂ = 8.4 ÷ 2 = 4.2 dm³

(ii) Molecular Formula

Empirical formula = X₂Y

Empirical mass = 2×10 + 1×5 = 25

Vapour density = 25

Molecular mass = 2 × V.D. = 50

n = Molecular mass ÷ Empirical mass = 50 ÷ 25 = 2

Molecular formula = (X₂Y)₂ = X₄Y₂

(b) Ammonia Gas (NH₃) at STP

Molar mass of NH₃ = 17 g/mol

Moles of NH₃ = 68 ÷ 17 = 4 moles

(i) Volume at STP = 4 × 22.4 = 89.6 dm³

(ii) Number of moles = 4 moles

(iii) Number of molecules = 4 × 6.022 × 10²³ = 2.4088 × 10²⁴ molecules

Mole Concept and Stoichiometry

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