Mole Concept and Stoichiometry: ICSE Class 10 Chemistry

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Summary

This chapter teaches us about how scientists count and measure very tiny things like atoms and molecules, especially in gases and chemical reactions. It begins with some rules about gases. Gay-Lussac’s Law says that when gases combine or are produced in a chemical reaction, they do so in simple ratios by volume, as long as the temperature and pressure don’t change. Avogadro’s Law helps us understand this better. It states that if you have equal volumes of any gases at the same temperature and pressure, they will contain the same number of molecules. This idea also helps us find out how many atoms are in a single molecule of common gases.

Because atoms and molecules are so small, we need a special way to count them in large groups. Scientists use a very big number called Avogadro’s number for this, which is 6.022 x 10^23. A group containing this many particles (atoms, molecules, or ions) is called one “mole.” A mole is just a counting unit, like a “dozen” means 12 things or a “gross” means 144 things. The mass of one mole of a substance’s atoms is its gram atomic mass, and the mass of one mole of its molecules is its gram molecular mass. For any gas at standard conditions of temperature and pressure (STP), one mole will always take up 22.4 litres of space.

The chapter also explains how we compare the masses of different atoms and molecules. We use the carbon-12 atom as a standard. The relative atomic mass tells us how heavy an atom is compared to 1/12th the mass of a carbon-12 atom. Similarly, relative molecular mass tells us the mass of a molecule. Vapour density is a property of gases that is related to their molecular mass; the molecular mass of a gas is simply twice its vapour density.

We learn how to figure out the chemical formula of a compound. This starts with finding the percentage composition, which is the percentage by weight of each element in the compound. From this information, we can calculate the empirical formula, which shows the simplest whole-number ratio of atoms of each element in the compound. If we also know the compound’s molecular mass, we can find its molecular formula, which gives the actual number of atoms of each element in one molecule.

Lastly, the chapter discusses stoichiometry. This involves using balanced chemical equations to make calculations. A balanced equation shows the exact proportions of reactants (starting substances) and products (substances formed) in moles. Using these proportions, we can calculate how much of one substance will react with another, or how much product will be made. These calculations can be about the mass of substances or, for gases, their volumes.

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Workbook Solutions (Concise/Selina)

Exercise A

1. (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
(b) 200 cm³ of hydrogen and 150 cm³ of oxygen are mixed and ignited, as per the following reaction,
2H₂ + O₂ → 2H₂O
What volume of oxygen remains unreacted?

Answer:

(a) Solution:

Given reaction:
2CO + O₂ → 2CO₂

According to Gay-Lussac’s Law, the volumes of reacting gases are in a simple ratio.
2 vols : 1 vol : 2 vols

From the equation, 2 volumes of carbon monoxide require 1 volume of oxygen.
Therefore, 100 litres of carbon monoxide will require:
(1/2) × 100 = 50 litres of oxygen.

The volume of oxygen required is 50 litres.

(b) Solution:

Given reaction:
2H₂ + O₂ → 2H₂O

According to Gay-Lussac’s Law:
2 vols : 1 vol : 2 vols

From the equation, 2 volumes of hydrogen react with 1 volume of oxygen.
Therefore, 200 cm³ of hydrogen will react with:
(1/2) × 200 = 100 cm³ of oxygen.

Initial volume of oxygen = 150 cm³
Volume of oxygen used = 100 cm³
Volume of oxygen remaining unreacted = Initial volume – Used volume
= 150 – 100 = 50 cm³

50 cm³ of oxygen remains unreacted.

2. 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.

Answer:

Given:
Initial volume of CH₄ = 24 cc
Initial volume of O₂ = 106 cc
Final gaseous volume (after cooling) = 82 cc
Volume of unreacted O₂ in final mixture = 58 cc

Solution:
The balanced equation for the combustion of marsh gas (methane) is:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
1 vol : 2 vols : 1 vol : (negligible vol)

First, let’s calculate the theoretical volumes based on the reaction:
Volume of O₂ required to burn 24 cc of CH₄:
From the ratio (1 vol CH₄ : 2 vols O₂), 24 cc of CH₄ requires 2 × 24 = 48 cc of O₂.

Volume of O₂ remaining unreacted:
Initial O₂ – Reacted O₂ = 106 cc – 48 cc = 58 cc.
This matches the experimental data.

Volume of CO₂ produced from 24 cc of CH₄:
From the ratio (1 vol CH₄ : 1 vol CO₂), 24 cc of CH₄ produces 1 × 24 = 24 cc of CO₂.

Total volume of gaseous mixture after cooling:
The mixture contains unreacted O₂ and produced CO₂.
Total Volume = Volume of unreacted O₂ + Volume of CO₂
= 58 cc + 24 cc = 82 cc.
This also matches the experimental data.

The experiment supports Gay-Lussac’s Law of Combining Volumes.
Explanation: The calculations show that the volumes of the gaseous reactant methane (24 cc), the gaseous reactant oxygen (48 cc), and the gaseous product carbon dioxide (24 cc) are in the simple whole-number ratio of 24:48:24, which simplifies to 1:2:1. This is consistent with Gay-Lussac’s Law.

3. What volume of oxygen would be required to burn completely 400 mL of acetylene [C₂H₂]? Also, calculate the volume of carbon dioxide formed.
2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O(l)

Answer:

Given:
Volume of acetylene (C₂H₂) = 400 mL

Given reaction:
2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O(l)
2 vols : 5 vols : 4 vols

Solution:
To find the volume of oxygen required:
From the equation, 2 volumes of C₂H₂ require 5 volumes of O₂.
Therefore, 400 mL of C₂H₂ will require:
(5/2) × 400 = 1000 mL of oxygen.

To find the volume of carbon dioxide formed:
From the equation, 2 volumes of C₂H₂ produce 4 volumes of CO₂.
Therefore, 400 mL of C₂H₂ will produce:
(4/2) × 400 = 800 mL of carbon dioxide.

The volume of oxygen required is 1000 mL and the volume of carbon dioxide formed is 800 mL.

4. 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.

Answer:

Balanced Equation:
H₂S(g) + Cl₂(g) → 2HCl(g) + S(s)
1 vol : 1 vol : 2 vols : (negligible vol)

Given:
Volume of H₂S = 112 cm³
Volume of Cl₂ = 120 cm³

Solution:
From the equation, 1 volume of H₂S reacts with 1 volume of Cl₂.
Therefore, 112 cm³ of H₂S will react with 112 cm³ of Cl₂.
Since there is 120 cm³ of Cl₂, H₂S is the limiting reactant.

(i) Volume of gaseous product (HCl) formed:
From the equation, 1 volume of H₂S produces 2 volumes of HCl.
Therefore, 112 cm³ of H₂S will produce:
2 × 112 = 224 cm³ of HCl.

(ii) Composition of the resulting mixture:
The resulting mixture will contain the product HCl and the unreacted reactant Cl₂.
Volume of HCl formed = 224 cm³
Volume of unreacted Cl₂ = Initial Cl₂ – Reacted Cl₂
= 120 cm³ – 112 cm³ = 8 cm³

The volume of gaseous product (HCl) is 224 cm³, and the resulting mixture contains 224 cm³ of HCl and 8 cm³ of unreacted Cl₂.

5. 1250 cc of oxygen was burnt with 300 cc of ethane [C₂H₆]. Calculate the volume of unused oxygen and the volume of carbon dioxide formed:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

Answer:

Given:
Volume of ethane (C₂H₆) = 300 cc
Volume of oxygen (O₂) = 1250 cc

Given reaction:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
2 vols : 7 vols : 4 vols

Solution:
First, determine the volume of oxygen required to burn 300 cc of ethane.
From the equation, 2 volumes of C₂H₆ require 7 volumes of O₂.
Therefore, 300 cc of C₂H₆ will require:
(7/2) × 300 = 1050 cc of O₂.

To find the volume of unused oxygen:
Since only 1050 cc of O₂ is required and 1250 cc is available, oxygen is in excess.
Volume of unused oxygen = Initial O₂ – Used O₂
= 1250 cc – 1050 cc = 200 cc.

To find the volume of carbon dioxide formed:
From the equation, 2 volumes of C₂H₆ produce 4 volumes of CO₂.
Therefore, 300 cc of C₂H₆ will produce:
(4/2) × 300 = 600 cc of CO₂.

The volume of unused oxygen is 200 cc and the volume of carbon dioxide formed is 600 cc.

6. 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

Answer:

Given:
Initial conditions for ethylene (C₂H₄):
V₁ = 11 litres
T₁ = 273°C = 273 + 273 = 546 K
P₁ = 380 mm of Hg

STP conditions:
T₂ = 273 K
P₂ = 760 mm of Hg

Solution:
Step 1: Convert the volume of ethylene to STP.
Using the combined gas equation: (P₁V₁)/T₁ = (P₂V₂)/T₂
V₂ = (P₁ × V₁ × T₂) / (P₂ × T₁)
V₂ = (380 × 11 × 273) / (760 × 546)
V₂ = (1 × 11 × 1) / (2 × 2) = 11/4 = 2.75 litres
So, the volume of ethylene at STP is 2.75 litres.

Step 2: Calculate the volume of oxygen required.
Given reaction:
C₂H₄ + 3O₂ → 2CO₂ + 2H₂O
1 vol : 3 vols

From the equation, 1 volume of C₂H₄ requires 3 volumes of O₂.
Therefore, 2.75 litres of C₂H₄ at STP will require:
3 × 2.75 = 8.25 litres of O₂ at STP.

The volume of oxygen required at STP is 8.25 litres.

7. 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

Answer:

Given:
Volume of methane (CH₄) = 40 mL

Given reaction:
CH₄ + 2Cl₂ → CH₂Cl₂ + 2HCl
1 vol : 2 vols : 1 vol : 2 vols

Solution:
To find the volume of chlorine gas required:
From the equation, 1 volume of CH₄ requires 2 volumes of Cl₂.
Therefore, 40 mL of CH₄ will require:
2 × 40 = 80 mL of Cl₂.

To find the volume of HCl gas formed:
From the equation, 1 volume of CH₄ forms 2 volumes of HCl.
Therefore, 40 mL of CH₄ will form:
2 × 40 = 80 mL of HCl.

80 mL of chlorine gas is required and 80 mL of HCl gas is formed.

8. 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

Answer:

Given:
Volume of air used = 500 cm³
Oxygen content in air = 1/5th or 20%

Solution:
Step 1: Calculate the volume of oxygen in the air.
Volume of O₂ = (1/5) × Volume of air
= (1/5) × 500 = 100 cm³.

Step 2: Calculate the volume of propane burnt.
Given reaction:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
1 vol : 5 vols

From the equation, 5 volumes of O₂ burn 1 volume of C₃H₈.
Therefore, 100 cm³ of O₂ will burn:
(1/5) × 100 = 20 cm³ of propane.

20 cm³ of propane is burnt.

9. 450 cm³ of nitrogen monoxide and 200 cm³ of oxygen are mixed together and ignited. Calculate the composition of resulting mixture.
2NO + O₂ → 2NO₂

Answer:

Given:
Volume of nitrogen monoxide (NO) = 450 cm³
Volume of oxygen (O₂) = 200 cm³

Given reaction:
2NO + O₂ → 2NO₂
2 vols : 1 vol : 2 vols

Solution:
First, determine the limiting reactant.
From the equation, 2 vols of NO react with 1 vol of O₂.
Volume of O₂ required to react with 450 cm³ of NO = (1/2) × 450 = 225 cm³.
Since only 200 cm³ of O₂ is available, oxygen is the limiting reactant.

Now, calculate based on the limiting reactant (200 cm³ O₂):
Volume of NO that reacts with 200 cm³ of O₂ = 2 × 200 = 400 cm³.
Volume of NO₂ formed from 200 cm³ of O₂ = 2 × 200 = 400 cm³.

Composition of the resulting mixture:
Unreacted NO = Initial NO – Reacted NO = 450 – 400 = 50 cm³.
Unreacted O₂ = 200 – 200 = 0 cm³.
NO₂ formed = 400 cm³.

The resulting mixture consists of 50 cm³ of nitrogen monoxide (NO) and 400 cm³ of nitrogen dioxide (NO₂).

10. 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.

Answer:

Given:
Volume of hydrogen (H₂) = 6 litres
Volume of chlorine (Cl₂) = 4 litres

Solution:
Step 1: Reaction between H₂ and Cl₂.
H₂ + Cl₂ → 2HCl
1 vol : 1 vol : 2 vols

From the equation, 1 volume of H₂ reacts with 1 volume of Cl₂.
Since there are 6 litres of H₂ and 4 litres of Cl₂, chlorine is the limiting reactant.
Volume of H₂ that reacts = 4 litres.
Volume of HCl formed = 2 × 4 = 8 litres.
Volume of unreacted H₂ = 6 – 4 = 2 litres.
After the explosion, the gaseous mixture contains 8 litres of HCl and 2 litres of H₂.

Step 2: Water is added.
Hydrogen chloride (HCl) gas is extremely soluble in water and will dissolve completely. Hydrogen (H₂) gas is insoluble.
Therefore, the only gas remaining is the unreacted hydrogen.

Volume of the residual gas = 2 litres.

11. 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?

Answer:

Given reaction:
4NH₃ + 5O₂ → 4NO + 6H₂O
4 vols : 5 vols : 4 vols

Solution:
From the equation, the total volume of gaseous reactants (NH₃ and O₂) is:
4 vols NH₃ + 5 vols O₂ = 9 volumes of reactants.
These 9 volumes of reactants produce 4 volumes of nitrogen monoxide (NO).

Ratio of (Total Reactant Volume) to (Product NO Volume) = 9 : 4

Given that 27 litres of reactants are consumed.
Volume of NO produced = (4/9) × Total volume of reactants consumed
= (4/9) × 27 = 4 × 3 = 12 litres.

12 litres of nitrogen monoxide is produced.

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

Answer:

Given:
Total initial volume (H₂ + Cl₂) = 36 cm³
Volume of unreacted H₂ = 4 cm³

Reaction:
H₂ + Cl₂ → 2HCl

Solution:
Let the initial volume of H₂ be ‘V_H₂’ and the initial volume of Cl₂ be ‘V_Cl₂’.
V_H₂ + V_Cl₂ = 36 cm³ —(i)

After the reaction, the mixture is shaken with water. HCl dissolves, and the unreacted gas remains.
The unreacted gas is 4 cm³ of hydrogen. This means hydrogen was in excess and chlorine was the limiting reactant (it reacted completely).
Volume of H₂ that reacted = Initial H₂ – Unreacted H₂ = V_H₂ – 4.
From the stoichiometry (1 vol H₂ : 1 vol Cl₂), the volume of H₂ that reacted is equal to the initial volume of Cl₂.
So, V_Cl₂ = V_H₂ – 4 —(ii)

Now, substitute equation (ii) into equation (i):
V_H₂ + (V_H₂ – 4) = 36
2V_H₂ – 4 = 36
2V_H₂ = 40
V_H₂ = 20 cm³

Now find the volume of Cl₂ using equation (ii):
V_Cl₂ = 20 – 4 = 16 cm³

The composition of the initial mixture was 20 cm³ of hydrogen and 16 cm³ of chlorine.

13. What volume of air (containing 20% O₂ by volume) will be required to burn completely 10 cm³ each of methane and acetylene.
CH₄ + 2O₂ → CO₂ + 2H₂O
2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O

Answer:

Given:
Volume of methane (CH₄) = 10 cm³
Volume of acetylene (C₂H₂) = 10 cm³
Oxygen content in air = 20%

Solution:
Step 1: Calculate O₂ required for methane.
CH₄ + 2O₂ → CO₂ + 2H₂O
1 vol : 2 vols
Volume of O₂ for CH₄ = 2 × 10 = 20 cm³.

Step 2: Calculate O₂ required for acetylene.
2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O
2 vols : 5 vols
Volume of O₂ for C₂H₂ = (5/2) × 10 = 25 cm³.

Step 3: Calculate total O₂ required.
Total O₂ = (O₂ for CH₄) + (O₂ for C₂H₂) = 20 + 25 = 45 cm³.

Step 4: Calculate the volume of air required.
Volume of Air = (Volume of O₂) / (Percentage of O₂ in air)
= 45 / (20/100)
= 45 / 0.20 = 225 cm³.

225 cm³ of air will be required.

14. LPG has 60% propane and 40% butane: 10 litres of this mixture is burnt. Calculate the volume of carbon dioxide added to atmosphere.
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

Answer:

Given:
Total volume of LPG = 10 litres
Composition: 60% Propane (C₃H₈), 40% Butane (C₄H₁₀)

Solution:
Step 1: Calculate the volume of each gas in the mixture.
Volume of Propane = 60% of 10 litres = 0.60 × 10 = 6 litres.
Volume of Butane = 40% of 10 litres = 0.40 × 10 = 4 litres.

Step 2: Calculate CO₂ produced from propane combustion.
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
1 vol : 3 vols
Volume of CO₂ from propane = 3 × 6 = 18 litres.

Step 3: Calculate CO₂ produced from butane combustion.
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
2 vols : 8 vols (or 1 vol : 4 vols)
Volume of CO₂ from butane = 4 × 4 = 16 litres.

Step 4: Calculate total CO₂ produced.
Total CO₂ added to atmosphere = (CO₂ from propane) + (CO₂ from butane)
= 18 + 16 = 34 litres.

34 litres of carbon dioxide is added to the atmosphere.

15. 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 2.5 times increase in pressure (temp. remaining constant). What volume will H₂ now occupy?
(c) Taking the volume of H₂ calculated in 15(b), what changes must be made in kelvin (absolute) temperature to return the volume to 2500 cm³ pressure remaining constant.

Answer:

(a) Solution:
From the equation: 2H₂O → 2H₂ + O₂, the volume ratio of products is 2 vols H₂ : 1 vol O₂.
Volume of O₂ liberated = (1/2) × Volume of H₂ produced
= (1/2) × 2500 = 1250 cm³.

(b) Solution:
Using Boyle’s Law (P₁V₁ = P₂V₂), at constant temperature.
Let P₁ = P and V₁ = 2500 cm³.
The pressure is increased 2.5 times, so P₂ = 2.5 P.
V₂ = (P₁V₁) / P₂ = (P × 2500) / (2.5 P) = 2500 / 2.5 = 1000 cm³.
The new volume will be 1000 cm³.

(c) Solution:
Using Charles’s Law (V₁/T₁ = V₂/T₂), at constant pressure.
We want to change the volume from V₁ = 1000 cm³ to V₂ = 2500 cm³.
Let the initial Kelvin temperature be T₁. We need to find the final temperature T₂.
T₂ = (V₂ × T₁) / V₁ = (2500 × T₁) / 1000 = 2.5 T₁.
The Kelvin (absolute) temperature must be increased to 2.5 times its original value.

16. 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.

Gas

Volume (in litres)

Number of molecules

Chlorine

Nitrogen

Ammonia

Sulphur dioxide

Answer: According to Avogadro’s Law, at the same temperature and pressure, the volume of a gas is directly proportional to the number of molecules.
We are given that 20 litres of Nitrogen contain ‘x’ molecules.

Chlorine: Volume = 10 litres. Since 10 L is half of 20 L, it will contain half the number of molecules. Number of molecules = x/2.
Ammonia: Volume = 20 litres. Since the volume is the same as Nitrogen, it will contain the same number of molecules. Number of molecules = x.
Sulphur dioxide: Volume = 5 litres. Since 5 L is one-fourth of 20 L, it will contain one-fourth the number of molecules. Number of molecules = x/4.

Gas

Volume (in litres)

Number of molecules

Chlorine

x/2

Nitrogen

Ammonia

Sulphur dioxide

x/4

17. (i) If 150 cc of gas A contains x molecules, how many molecules of gas B will be present in 75 cc of gas B? The gases A and B are under the same conditions of temperature and pressure?
(ii) Name the law on which the above problem is based.

Answer: (i) Solution: According to Avogadro’s Law, under the same conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules.
This means the number of molecules is directly proportional to the volume.

Given that 150 cc of gas A contains x molecules.
This implies that 150 cc of gas B would also contain x molecules.

We need to find the number of molecules in 75 cc of gas B.
Since 75 cc is half the volume of 150 cc, it will contain half the number of molecules.
Number of molecules in 75 cc of gas B = (75 / 150) × x = 1/2 × x = x/2.

There will be x/2 molecules of gas B present.

(ii) The problem is based on Avogadro’s Law.

Exercise B

1. What are the relative molecular masses of the following compounds?

(a) Ammonium chloroplatinate (NH₄)₂PtCl₆
(b) Potassium chlorate, KClO₃
(c) Copper sulphate crystals, CuSO₄·5H₂O
(d) Ammonium sulphate, (NH₄)₂SO₄
(e) Sodium acetate, CH₃COONa
(f) Chloroform, CHCl₃
(g) Ammonium dichromate, (NH₄)₂Cr₂O₇

Answer: (a)
Given:
Formula = (NH₄)₂PtCl₆
Relative atomic masses: N = 14, H = 1, Pt = 195, Cl = 35.5

To find:
Relative molecular mass of (NH₄)₂PtCl₆

Solution:
Relative molecular mass = 2 × (N + 4 × H) + Pt + 6 × Cl
=> Relative molecular mass = 2 × (14 + 4 × 1) + 195 + 6 × 35.5
=> Relative molecular mass = 2 × (18) + 195 + 213
=> Relative molecular mass = 36 + 195 + 213
=> Relative molecular mass = 444 a.m.u.

(b)
Given:
Formula = KClO₃
Relative atomic masses: K = 39, Cl = 35.5, O = 16

To find:
Relative molecular mass of KClO₃

Solution:
Relative molecular mass = K + Cl + 3 × O
=> Relative molecular mass = 39 + 35.5 + 3 × 16
=> Relative molecular mass = 39 + 35.5 + 48
=> Relative molecular mass = 122.5 a.m.u.

To find:
Relative molecular mass of CuSO₄·5H₂O

Solution:
Relative molecular mass = Cu + S + 4 × O + 5 × (2 × H + O)
=> Relative molecular mass = 63.5 + 32 + (4 × 16) + 5 × (2 × 1 + 16)
=> Relative molecular mass = 63.5 + 32 + 64 + 5 × (18)
=> Relative molecular mass = 159.5 + 90
=> Relative molecular mass = 249.5 a.m.u.

(d)
Given:
Formula = (NH₄)₂SO₄
Relative atomic masses: N = 14, H = 1, S = 32, O = 16

To find:
Relative molecular mass of (NH₄)₂SO₄

Solution:
Relative molecular mass = 2 × (N + 4 × H) + S + 4 × O
=> Relative molecular mass = 2 × (14 + 4 × 1) + 32 + 4 × 16
=> Relative molecular mass = 2 × (18) + 32 + 64
=> Relative molecular mass = 36 + 32 + 64
=> Relative molecular mass = 132 a.m.u.

(e)
Given:
Formula = CH₃COONa
Relative atomic masses: C = 12, H = 1, O = 16, Na = 23

To find:
Relative molecular mass of CH₃COONa

Solution:
Relative molecular mass = C + 3 × H + C + 2 × O + Na
=> Relative molecular mass = 12 + 3 × 1 + 12 + 2 × 16 + 23
=> Relative molecular mass = 12 + 3 + 12 + 32 + 23
=> Relative molecular mass = 82 a.m.u.

(f)
Given:
Formula = CHCl₃
Relative atomic masses: C = 12, H = 1, Cl = 35.5

To find:
Relative molecular mass of CHCl₃

Solution:
Relative molecular mass = C + H + 3 × Cl
=> Relative molecular mass = 12 + 1 + 3 × 35.5
=> Relative molecular mass = 12 + 1 + 106.5
=> Relative molecular mass = 119.5 a.m.u.

(g)
Given:
Formula = (NH₄)₂Cr₂O₇
Relative atomic masses: N = 14, H = 1, Cr = 52, O = 16

To find:
Relative molecular mass of (NH₄)₂Cr₂O₇

Solution:
Relative molecular mass = 2 × (N + 4 × H) + 2 × Cr + 7 × O
=> Relative molecular mass = 2 × (14 + 4 × 1) + 2 × 52 + 7 × 16
=> Relative molecular mass = 2 × (18) + 104 + 112
=> Relative molecular mass = 36 + 104 + 112
=> Relative molecular mass = 252 a.m.u.

2. What are the following values?

(a) the number of molecules in 73 g of HCl.
(b) the weight of 0.5 mole of O₂.
(c) the number of molecules in 1.8 g of H₂O.
(d) the number of moles in 10 g of CaCO₃.
(e) the weight of 0.2 mole of H₂ gas.
(f) the number of molecules in 3.2 g of SO₂.

Answer: (a)
Given:
Mass of HCl = 73 g
Molar mass of HCl = 1 + 35.5 = 36.5 g/mol
Avogadro’s number (Nₐ) = 6.022 x 10²³ molecules/mol

To find:
Number of molecules in 73 g of HCl

Solution:
Number of moles = Given mass / Molar mass
=> Number of moles = 73 g / 36.5 g/mol = 2 moles
Number of molecules = Number of moles × Nₐ
=> Number of molecules = 2 × 6.022 x 10²³
=> Number of molecules = 12.044 x 10²³ = 1.2044 x 10²⁴ molecules.

(b)
Given:
Moles of O₂ = 0.5 mole
Molar mass of O₂ = 2 × 16 = 32 g/mol

To find:
Weight of 0.5 mole of O₂

Solution:
Weight = Number of moles × Molar mass
=> Weight = 0.5 mol × 32 g/mol
=> Weight = 16 g.

(c)
Given:
Mass of H₂O = 1.8 g
Molar mass of H₂O = (2 × 1) + 16 = 18 g/mol
Avogadro’s number (Nₐ) = 6.022 x 10²³ molecules/mol

To find:
Number of molecules in 1.8 g of H₂O

Solution:
Number of moles = Given mass / Molar mass
=> Number of moles = 1.8 g / 18 g/mol = 0.1 moles
Number of molecules = Number of moles × Nₐ
=> Number of molecules = 0.1 × 6.022 x 10²³
=> Number of molecules = 6.022 x 10²² molecules.

(d)
Given:
Mass of CaCO₃ = 10 g
Molar mass of CaCO₃ = 40 + 12 + (3 × 16) = 100 g/mol

To find:
Number of moles in 10 g of CaCO₃

Solution:
Number of moles = Given mass / Molar mass
=> Number of moles = 10 g / 100 g/mol
=> Number of moles = 0.1 mole.

(e)
Given:
Moles of H₂ = 0.2 mole
Molar mass of H₂ = 2 × 1 = 2 g/mol

To find:
Weight of 0.2 mole of H₂

Solution:
Weight = Number of moles × Molar mass
=> Weight = 0.2 mol × 2 g/mol
=> Weight = 0.4 g.

(f)
Given:
Mass of SO₂ = 3.2 g
Molar mass of SO₂ = 32 + (2 × 16) = 64 g/mol
Avogadro’s number (Nₐ) = 6.022 x 10²³ molecules/mol

To find:
Number of molecules in 3.2 g of SO₂

Solution:
Number of moles = Given mass / Molar mass
=> Number of moles = 3.2 g / 64 g/mol = 0.05 moles
Number of molecules = Number of moles × Nₐ
=> Number of molecules = 0.05 × 6.022 x 10²³
=> Number of molecules = 0.3011 x 10²³ = 3.011 x 10²² molecules.

3. Which of the following would weigh most: (a) 1 mole of H₂O, (b) 1 mole of CO₂, (c) 1 mole of NH₃, or (d) 1 mole of CO?

Answer: To find:
The substance with the greatest mass among the given options.

Solution:
The mass of 1 mole of a substance is equal to its gram molecular mass. We will calculate the molar mass for each substance.
(a) Molar mass of H₂O = (2 × 1) + 16 = 18 g
(b) Molar mass of CO₂ = 12 + (2 × 16) = 44 g
(c) Molar mass of NH₃ = 14 + (3 × 1) = 17 g
(d) Molar mass of CO = 12 + 16 = 28 g

Comparing the masses: 44 g > 28 g > 18 g > 17 g.
Therefore, 1 mole of CO₂ would weigh the most.

4. Which of the following contains the maximum number of molecules: (a) 4 g of O₂, (b) 4 g of NH₃, (c) 4 g of CO₂, or (d) 4 g of SO₂?

Answer: To find:
The substance with the maximum number of molecules for a given mass (4 g).

Solution:
Number of molecules = (Given mass / Molar mass) × Nₐ.
Since the given mass and Avogadro’s number (Nₐ) are constant for all options, the substance with the lowest molar mass will have the maximum number of molecules.

(a) Molar mass of O₂ = 2 × 16 = 32 g/mol
(b) Molar mass of NH₃ = 14 + (3 × 1) = 17 g/mol
(c) Molar mass of CO₂ = 12 + (2 × 16) = 44 g/mol
(d) Molar mass of SO₂ = 32 + (2 × 16) = 64 g/mol

Comparing the molar masses: 17 g/mol is the lowest.
Therefore, 4 g of NH₃ contains the maximum number of molecules.

5. Calculate the number of the following?
(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.

Answer: (a)
Given:
Number of moles = 0.1 mole
Avogadro’s number (Nₐ) = 6.022 x 10²³ particles/mol

To find:
Number of particles

Solution:
Number of particles = Number of moles × Nₐ
=> Number of particles = 0.1 × 6.022 x 10²³
=> Number of particles = 6.022 x 10²² particles.

(b)
Given:
Moles of H₂SO₄ = 0.1 mole

To find:
Number of hydrogen atoms

Solution:
One molecule of H₂SO₄ contains 2 atoms of hydrogen.
Therefore, 1 mole of H₂SO₄ contains 2 moles of hydrogen atoms.
Moles of hydrogen atoms in 0.1 mole of H₂SO₄ = 0.1 × 2 = 0.2 moles.
Number of hydrogen atoms = Moles of H atoms × Nₐ
=> Number of hydrogen atoms = 0.2 × 6.022 x 10²³
=> Number of hydrogen atoms = 1.2044 x 10²³ atoms.

To find:
Number of molecules in 1 kg of CaCl₂

Solution:
Number of moles = Given mass / Molar mass
=> Number of moles = 1000 g / 111 g/mol ≈ 9.009 moles
Number of molecules = Number of moles × Nₐ
=> Number of molecules = 9.009 × 6.022 x 10²³
=> Number of molecules ≈ 54.25 x 10²³ = 5.425 x 10²⁴ molecules.

6. How many grams of the following substances are present?
(a) Al in 0.2 mole of it.
(b) HCl in 0.1 mole of it.
(c) H₂O in 0.2 mole of it.
(d) CO₂ in 0.1 mole of it.

Answer: (a)
Given:
Moles of Al = 0.2 mole
Atomic mass of Al = 27 g/mol

To find:
Mass of Al

Solution:
Mass = Number of moles × Atomic mass
=> Mass = 0.2 mol × 27 g/mol
=> Mass = 5.4 g.

(b)
Given:
Moles of HCl = 0.1 mole
Molar mass of HCl = 1 + 35.5 = 36.5 g/mol

To find:
Mass of HCl

Solution:
Mass = Number of moles × Molar mass
=> Mass = 0.1 mol × 36.5 g/mol
=> Mass = 3.65 g.

To find:
Mass of H₂O

Solution:
Mass = Number of moles × Molar mass
=> Mass = 0.2 mol × 18 g/mol
=> Mass = 3.6 g.

(d)
Given:
Moles of CO₂ = 0.1 mole
Molar mass of CO₂ = 12 + (2 × 16) = 44 g/mol

To find:
Mass of CO₂

Solution:
Mass = Number of moles × Molar mass
=> Mass = 0.1 mol × 44 g/mol
=> Mass = 4.4 g.

7. (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?

Answer: (a)
Given:
Volume of gas at STP = 5.6 L
Mass of gas = 12 g
Molar volume at STP = 22.4 L/mol

To find:
Molar mass of the gas

Solution:
The mass of 22.4 L of any gas at STP is its molar mass.
Using proportion:
Mass / Volume = Molar mass / Molar volume
=> 12 g / 5.6 L = Molar mass / 22.4 L
=> Molar mass = (12 g × 22.4 L) / 5.6 L
=> Molar mass = 12 g × 4
=> Molar mass = 48 g/mol .
The relative molecular mass is 48.

(b)
Given:
Moles of SO₂ = 2 moles
Molar volume at STP = 22.4 L/mol

To find:
Volume of SO₂ at STP

Solution:
Volume at STP = Number of moles × Molar volume
=> Volume at STP = 2 mol × 22.4 L/mol
=> Volume at STP = 44.8 L.

8. Calculate the number of moles of the following?
(a) CO₂ which contains 8.00 g of O₂.
(b) methane in 0.80 g of methane.

Answer: (a)
Given:
Mass of oxygen in CO₂ = 8.00 g
Molar mass of CO₂ = 44 g/mol

To find:
Number of moles of CO₂

Solution:
One mole of CO₂ contains 2 moles of oxygen atoms, which is 2 × 16 = 32 g of oxygen.
Using proportion:
1 mole of CO₂ contains 32 g of oxygen
x moles of CO₂ contain 8 g of oxygen
=> x = (1 mole × 8 g) / 32 g
=> x = 0.25 moles of CO₂.

(b)
Given:
Mass of methane (CH₄) = 0.80 g
Molar mass of CH₄ = 12 + (4 × 1) = 16 g/mol

To find:
Number of moles of methane

Solution:
Number of moles = Given mass / Molar mass
=> Number of moles = 0.80 g / 16 g/mol
=> Number of moles = 0.05 moles.

9. Calculate the weight/mass of the following?
(a) an atom of oxygen.
(b) an atom of hydrogen.
(c) a molecule of NH₃.
(d) 10²² atoms of carbon.
(e) the molecule of oxygen.
(f) 0.25 gram atom of calcium.

Answer: (a) Mass of one atom = Gram atomic mass / Nₐ
=> Mass = 16 g / (6.022 x 10²³)
=> Mass ≈ 2.657 x 10⁻²³ g.

(b) Mass of one atom = Gram atomic mass / Nₐ
=> Mass = 1 g / (6.022 x 10²³)
=> Mass ≈ 1.66 x 10⁻²⁴ g.

(c) Molar mass of NH₃ = 17 g/mol
Mass of one molecule = Molar mass / Nₐ
=> Mass = 17 g / (6.022 x 10²³)
=> Mass ≈ 2.823 x 10⁻²³ g.

(d) Moles of C = Number of atoms / Nₐ
=> Moles = 10²² / (6.022 x 10²³) ≈ 0.0166 moles
Mass = Moles × Atomic mass
=> Mass = 0.0166 mol × 12 g/mol ≈ 0.199 g ≈ 0.2 g.

(e) Molar mass of O₂ = 32 g/mol
Mass of one molecule = Molar mass / Nₐ
=> Mass = 32 g / (6.022 x 10²³)
=> Mass ≈ 5.314 x 10⁻²³ g.

(f) 1 gram atom = 1 mole of atoms
Moles of Ca = 0.25 moles
Atomic mass of Ca = 40 g/mol
Mass = Moles × Atomic mass
=> Mass = 0.25 mol × 40 g/mol
=> Mass = 10 g.

10. Calculate the mass of 0.1 mole of each of the following (Ca = 40, Na = 23, Mg = 24, S = 32, C = 12, Cl=35 .5, O = 16, H = 1)?
(a) CaCO₃
(b) Na₂SO₄·10H₂O
(c) CaCl₂
(d) Mg

Answer: (a) Molar mass = 40 + 12 + (3 × 16) = 100 g/mol
Mass = 0.1 mol × 100 g/mol = 10 g.

(b) Molar mass = (2×23) + 32 + (4×16) + 10×((2×1)+16) = 46 + 32 + 64 + 10×18 = 142 + 180 = 322 g/mol
Mass = 0.1 mol × 322 g/mol = 32.2 g.

(d) Atomic mass = 24 g/mol
Mass = 0.1 mol × 24 g/mol = 2.4 g.

11. Calculate the number of the following?
(a) oxygen atoms in 0.10 mole of Na₂CO₃·10H₂O.
(b) gram atoms in 4.6 gram of sodium.
(c) moles in 12 g of oxygen gas.

Answer: (a) One formula unit of Na₂CO₃·10H₂O contains 3 + 10 = 13 oxygen atoms.
Moles of O atoms in 0.10 mole of compound = 0.10 × 13 = 1.3 moles.
Number of O atoms = 1.3 × Nₐ = 1.3 × 6.022 x 10²³ = 7.8286 x 10²³ atoms.

(b) Gram atoms = number of moles.
Moles = Given mass / Atomic mass = 4.6 g / 23 g/mol = 0.2 moles.
So, there are 0.2 gram atoms of sodium.

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

Answer: Given:
Mass of S = 3.2 g
Atomic mass of S = 32 g/mol
Atomic mass of Ca = 40 g/mol

To find:
Mass of Ca with the same number of atoms.

Solution:
First, calculate the moles of S atoms.
Moles of S = 3.2 g / 32 g/mol = 0.1 moles.
To have the same number of atoms, we need 0.1 moles of Ca.
Mass of Ca = Moles × Atomic mass
=> Mass of Ca = 0.1 mol × 40 g/mol = 4.0 g.

13. 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.

Answer: (a) Number of atoms = Moles × Nₐ
=> Number of atoms = 52 × 6.022 x 10²³ = 313.144 x 10²³ = 3.131 x 10²⁵ atoms.

(b) The atomic mass of one He atom is 4 amu.
Number of atoms = Total mass in amu / Mass of one atom in amu
=> Number of atoms = 52 amu / 4 amu = 13 atoms.

(c) Moles = Given mass / Atomic mass = 52 g / 4 g/mol = 13 moles.
Number of atoms = Moles × Nₐ
=> Number of atoms = 13 × 6.022 x 10²³ = 78.286 x 10²³ = 7.8286 x 10²⁴ atoms.

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

Answer: Given:
Mass of Na₂CO₃ = 5.3 g
Molar mass of Na₂CO₃ = (2×23) + 12 + (3×16) = 106 g/mol

To find:
Number of Na, C, and O atoms.

Solution:
Moles of Na₂CO₃ = 5.3 g / 106 g/mol = 0.05 moles.
In 1 mole of Na₂CO₃, there are:

2 moles of Na atoms
1 mole of C atoms
3 moles of O atoms
In 0.05 moles of Na₂CO₃, there are:
Moles of Na = 0.05 × 2 = 0.1 moles
Moles of C = 0.05 × 1 = 0.05 moles
Moles of O = 0.05 × 3 = 0.15 moles
Number of atoms = Moles × Nₐ
No. of Na atoms = 0.1 × 6.022 x 10²³ = 6.022 x 10²² atoms
No. of C atoms = 0.05 × 6.022 x 10²³ = 3.011 x 10²² atoms
No. of O atoms = 0.15 × 6.022 x 10²³ = 9.033 x 10²² atoms

15. (a) Calculate the mass of nitrogen supplied to soil by 5 kg of urea [CO(NH₂)₂]. (b) Calculate the volume occupied by 320 g of sulphur dioxide at STP?

Answer: (a)
Given:
Mass of urea = 5 kg = 5000 g
Formula = CO(NH₂)₂

To find:
Mass of nitrogen (N).

Solution:
Molar mass of urea = 12 + 16 + 2 × (14 + 2×1) = 28 + 2 × 16 = 60 g/mol .
Mass of nitrogen in 1 mole of urea = 2 × 14 = 28 g.
Percentage of N in urea = (Mass of N / Molar mass of urea) × 100
=> % N = (28 / 60) × 100 = 46.67%.
Mass of nitrogen supplied = 46.67% of 5 kg
=> Mass of N = 0.4667 × 5 kg = 2.33 kg.

(b)
Given:
Mass of SO₂ = 320 g

To find:
Volume of SO₂ at STP.

Solution:
Molar mass of SO₂ = 32 + (2 × 16) = 64 g/mol .
Moles of SO₂ = 320 g / 64 g/mol = 5 moles.
Volume at STP = Moles × 22.4 L/mol
=> Volume = 5 × 22.4 L = 112 L.

16. (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?

Answer: (a) The statement means that a certain volume of carbon dioxide gas is 22 times heavier than the same volume of hydrogen gas, when both are measured under identical conditions of temperature and pressure. It also allows calculation of the molecular mass: Molecular Mass = 2 × Vapour Density = 2 × 22 = 44.

(b) Chlorine exists as a diatomic molecule (Cl₂).
Molecular mass of Cl₂ = 2 × 35.5 = 71.
Vapour Density = Molecular Mass / 2
=> Vapour Density = 71 / 2 = 35.5.

17. What is the mass of 56 cm³ of carbon monoxide at STP?

Answer: Given:
Volume of CO = 56 cm³ = 0.056 L

To find:
Mass of CO.

Solution:
Moles of CO = Volume at STP / 22.4 L/mol
=> Moles = 0.056 L / 22.4 L/mol = 0.0025 moles.
Molar mass of CO = 12 + 16 = 28 g/mol .
Mass = Moles × Molar mass
=> Mass = 0.0025 × 28 g = 0.07 g.

18. Determine the no. of molecules in a drop of water which weighs 0.09 g?

Answer: Given:
Mass of H₂O = 0.09 g

To find:
Number of molecules.

Solution:
Molar mass of H₂O = 18 g/mol .
Moles = 0.09 g / 18 g/mol = 0.005 moles.
Number of molecules = Moles × Nₐ
=> Number of molecules = 0.005 × 6.022 x 10²³ = 0.03011 x 10²³ = 3.011 x 10²¹ molecules.

19. 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?

Answer: Given:
Mass of S₈ = 5.12 g

Solution:
Molar mass of S₈ = 8 × 32 = 256 g/mol .
(a) Moles of sulphur (S₈):
Moles = 5.12 g / 256 g/mol = 0.02 moles.

(b) Molecules and atoms:
Number of molecules = Moles × Nₐ
=> No. of molecules = 0.02 × 6.022 x 10²³ = 1.2044 x 10²² molecules.
Number of atoms = Number of molecules × 8
=> No. of atoms = 1.2044 x 10²² × 8 = 9.6352 x 10²² atoms.

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

Answer: Given:
Mass of P₄ = 100 g
Atomic mass of P = 31

Solution:
Molar mass of P₄ = 4 × 31 = 124 g/mol .
Moles = 100 g / 124 g/mol ≈ 0.806 moles.

21. Calculate the following?
(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.

Answer: (a) Gram molecular mass is the mass of 22400 cm³ of the gas at STP.
Using proportion:
0.979 g / 308 cm³ = GMM / 22400 cm³
=> GMM = (0.979 g × 22400 cm³) / 308 cm³
=> GMM ≈ 71.2 g.

(b) Assuming standard temperature (0°C or 273 K).
Moles of H₂ = 4 g / 2 g/mol = 2 moles.
At 1 atm (STP), Volume = 2 moles × 22.4 L/mol = 44.8 L.
Using Boyle’s Law (P₁V₁ = P₂V₂):
1 atm × 44.8 L = 4 atm × V₂
=> V₂ = 44.8 L / 4 = 11.2 L (or 11.2 dm³).

(c) Moles of CO₂ = 2.2 L / 22.4 L/mol ≈ 0.0982 moles.
1 mole of CO₂ contains 32 g of oxygen.
Mass of oxygen = Moles of CO₂ × 32 g/mol
=> Mass of O = 0.0982 × 32 g ≈ 3.14 g.

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

Answer:
Moles of C = Mass / Atomic mass = 10⁻¹² g / 12 g/mol .
Number of atoms = Moles × Nₐ
=> No. of atoms = (10⁻¹² / 12) × 6.022 x 10²³
=> No. of atoms ≈ 0.5018 x 10¹¹ = 5.018 x 10¹⁰ atoms.

23. 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?

Answer: Using the formula PM = dRT.
P = 1140 mm Hg = 1140/760 atm = 1.5 atm
d = 3 g/L
R = 0.0821 L·atm/mol·K
T = 273°C = 546 K
M = dRT / P
=> M = (3 × 0.0821 × 546) / 1.5
=> M ≈ 89.6 g/mol .

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

Answer: Molar mass of sugar = (12×12) + (22×1) + (11×16) = 144 + 22 + 176 = 342 g/mol .
Molar mass in kg = 0.342 kg/mol .
Cost per mole = Molar mass (kg) × Cost per kg
=> Cost per mole = 0.342 × ₹40 = ₹13.68.

25. Calculate the number of molecules in one kg of NaOH?

Answer: Mass = 1 kg = 1000 g.
Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol .
Moles = 1000 g / 40 g/mol = 25 moles.
Number of molecules = 25 × Nₐ = 25 × 6.022 x 10²³ = 150.55 x 10²³ = 1.5055 x 10²⁵ molecules.

26. Calculate the number of atoms present in (a) 10 g of Chlorine and (b) 10 g of Nitrogen?

Answer: (a) Assuming Chlorine gas (Cl₂). Molar mass = 71 g/mol .
Moles of Cl₂ = 10 g / 71 g/mol ≈ 0.1408 moles.
No. of molecules = 0.1408 × Nₐ ≈ 0.848 x 10²³.
No. of atoms = 2 × 0.848 x 10²³ ≈ 1.696 x 10²³ atoms.

(b) Assuming Nitrogen gas (N₂). Molar mass = 28 g/mol .
Moles of N₂ = 10 g / 28 g/mol ≈ 0.357 moles.
No. of molecules = 0.357 × Nₐ ≈ 2.15 x 10²³.
No. of atoms = 2 × 2.15 x 10²³ = 4.3 x 10²³ atoms.

27. Are the following statements correct? If not, correct them.
(a) Equal volumes of any gas, under similar conditions, contain an equal number of atoms.
(b) 22 g of CO₂ occupies 22.4 litres at STP.
(c) The unit of atomic weight is grams.

Answer: (a) Correct Statement: Equal volumes of all gases, under similar conditions of temperature and pressure, contain an equal number of molecules. (Avogadro’s Law)

(b) Molar mass of CO₂ = 44 g. One mole (44 g) of CO₂ occupies 22.4 L at STP.
22 g of CO₂ is 0.5 moles.
Correct Statement: 22 g of CO₂ occupies 11.2 litres at STP.

(c) The unit of atomic weight (or relative atomic mass) is the atomic mass unit (amu or u). Grams (g) is the unit for gram atomic mass (the mass of one mole of atoms).
Correct Statement: The unit of atomic weight is the atomic mass unit (amu).

Exercise C

1. Give the empirical formula of the following:

(a) C₆H₆

Answer: Given:

Molecular Formula = C₆H₆

To find:

Empirical Formula = ?

Solution:

The ratio of atoms in the molecule is C : H = 6 : 6.
To find the simplest whole-number ratio, we divide the subscripts by their greatest common divisor, which is 6.
=> C : H = 6/6 : 6/6
=> C : H = 1 : 1
The empirical formula is CH.

(b) C₆H₁₂O₆

Answer: Given:

Molecular Formula = C₆H₁₂O₆

To find:

Empirical Formula = ?

Solution:

The ratio of atoms in the molecule is C : H : O = 6 : 12 : 6.
The greatest common divisor of the subscripts (6, 12, 6) is 6.
Dividing the subscripts by 6:
=> C : H : O = 6/6 : 12/6 : 6/6
=> C : H : O = 1 : 2 : 1
The empirical formula is CH₂O.

(c) C₂H₂

Answer: Given:

Molecular Formula = C₂H₂

To find:

Empirical Formula = ?

Solution:

The ratio of atoms in the molecule is C : H = 2 : 2.
The greatest common divisor of the subscripts (2, 2) is 2.
Dividing the subscripts by 2:
=> C : H = 2/2 : 2/2
=> C : H = 1 : 1
The empirical formula is CH.

(d) CH₃COOH

Answer: Given:

Molecular Formula = CH₃COOH

To find:

Empirical Formula = ?

Solution:

First, find the total number of each type of atom in the molecule.
Carbon (C) = 1 + 1 = 2
Hydrogen (H) = 3 + 1 = 4
Oxygen (O) = 1 + 1 = 2
The molecular formula can be written as C₂H₄O₂.
The ratio of atoms is C : H : O = 2 : 4 : 2.
The greatest common divisor of the subscripts (2, 4, 2) is 2.
Dividing the subscripts by 2:
=> C : H : O = 2/2 : 4/2 : 2/2
=> C : H : O = 1 : 2 : 1
The empirical formula is CH₂O.

2. Find the percentage of water of crystallisation in CuSO₄·5H₂O. (At. mass Cu = 64, H = 1, O = 16, S = 32)

Answer: Given:

Compound = CuSO₄·5H₂O
Atomic masses: Cu = 64, S = 32, O = 16, H = 1

To find:

Percentage of water of crystallisation = ?

Solution:

First, calculate the relative molecular mass (molar mass) of CuSO₄·5H₂O.
Molar Mass = (Mass of Cu) + (Mass of S) + 4×(Mass of O) + 5×(Mass of H₂O)
=> Molar Mass = 64 + 32 + (4 × 16) + 5 × ((2 × 1) + 16)
=> Molar Mass = 64 + 32 + 64 + 5 × (18)
=> Molar Mass = 160 + 90
=> Molar Mass = 250 a.m.u.

Now, calculate the mass of water of crystallisation (5H₂O).
Mass of 5H₂O = 5 × 18 = 90 a.m.u.

Percentage of water = (Mass of water / Total molar mass) × 100
=> Percentage of water = (90 / 250) × 100
=> Percentage of water = 0.36 × 100
=> Percentage of water = 36%

3. Calculate the percentage of phosphorus in:

(a) Calcium hydrogen phosphate Ca(H₂PO₄)₂

Answer: Given:

Compound = Ca(H₂PO₄)₂
Atomic masses: Ca = 40, H = 1, P = 31, O = 16

To find:

Percentage of phosphorus (% P) = ?

Solution:

First, calculate the molar mass of Ca(H₂PO₄)₂.
Molar Mass = (Mass of Ca) + 2 × [(Mass of H) + (Mass of P) + 4×(Mass of O)]
=> Molar Mass = 40 + 2 × [2×1 + 31 + 4×16]
=> Molar Mass = 40 + 2 × [2 + 31 + 64]
=> Molar Mass = 40 + 2 × [97]
=> Molar Mass = 40 + 194
=> Molar Mass = 234 a.m.u.

Now, calculate the total mass of phosphorus in the compound.
Mass of P = 2 × 31 = 62 a.m.u.

Percentage of phosphorus = (Mass of P / Total molar mass) × 100
=> % P = (62 / 234) × 100
=> % P = 0.2649 × 100
=> % P ≈ 26.5%

(b) Calcium phosphate Ca₃(PO₄)₂

Answer: Given:

Compound = Ca₃(PO₄)₂
Atomic masses: Ca = 40, P = 31, O = 16

To find:

Percentage of phosphorus (% P) = ?

Solution:

First, calculate the molar mass of Ca₃(PO₄)₂.
Molar Mass = 3×(Mass of Ca) + 2 × [(Mass of P) + 4×(Mass of O)]
=> Molar Mass = (3 × 40) + 2 × [31 + (4 × 16)]
=> Molar Mass = 120 + 2 × [31 + 64]
=> Molar Mass = 120 + 2 × [95]
=> Molar Mass = 120 + 190
=> Molar Mass = 310 a.m.u.

Now, calculate the total mass of phosphorus in the compound.
Mass of P = 2 × 31 = 62 a.m.u.

Percentage of phosphorus = (Mass of P / Total molar mass) × 100
=> % P = (62 / 310) × 100
=> % P = 0.2 × 100
=> % P = 20%

4. Calculate the percentage composition of: Potassium chlorate, KClO₃.

Answer: Given:

Compound = KClO₃
Atomic masses: K = 39, Cl = 35.5, O = 16

To find:

Percentage composition of K, Cl, and O.

Solution:

First, calculate the molar mass of KClO₃.
Molar Mass = (Mass of K) + (Mass of Cl) + 3×(Mass of O)
=> Molar Mass = 39 + 35.5 + (3 × 16)
=> Molar Mass = 39 + 35.5 + 48
=> Molar Mass = 122.5 a.m.u.

Now, calculate the percentage of each element.
Percentage of Potassium (K) = (Mass of K / Molar Mass) × 100
=> % K = (39 / 122.5) × 100 ≈ 31.84%

Percentage of Chlorine (Cl) = (Mass of Cl / Molar Mass) × 100
=> % Cl = (35.5 / 122.5) × 100 ≈ 28.98%

Percentage of Oxygen (O) = (Mass of O / Molar Mass) × 100
=> % O = (48 / 122.5) × 100 ≈ 39.18%

5. Find the empirical formula of the compounds with the following percentage composition: Pb = 62.5%; N = 8.5%; O = 29.0%.

Answer: Given:

Percentage composition: Pb = 62.5%, N = 8.5%, O = 29.0%
Atomic masses: Pb = 207, N = 14, O = 16

To find:

Empirical Formula = ?

Solution:
We will use a table to determine the simplest ratio of atoms.

Element	Percentage	Atomic Mass	Atomic Ratio (Moles)	Simplest Ratio
Pb	62.5	207	62.5 / 207 = 0.302	0.302 / 0.302 = 1
N	8.5	14	8.5 / 14 = 0.607	0.607 / 0.302 ≈ 2
O	29.0	16	29.0 / 16 = 1.8125	1.8125 / 0.302 ≈ 6

The simplest whole-number ratio of atoms Pb : N : O is 1 : 2 : 6.
The empirical formula is PbN₂O₆, which can be written as Pb(NO₃)₂.

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

Answer: Given:

Total mass of iron ore = 10 kg
Percentage of pure ferric oxide (Fe₂O₃) in ore = 80%
Atomic masses: Fe = 56, O = 16

To find:

Mass of iron (Fe) in the ore = ?

Solution:

Step 1: Calculate the mass of pure ferric oxide in the ore.
Mass of Fe₂O₃ = 80% of 10 kg
=> Mass of Fe₂O₃ = (80 / 100) × 10 kg = 8 kg.

Step 2: Calculate the molar mass of ferric oxide (Fe₂O₃).
Molar Mass of Fe₂O₃ = (2 × 56) + (3 × 16)
=> Molar Mass of Fe₂O₃ = 112 + 48 = 160 g/mol .

Step 3: Calculate the mass of iron in one mole of Fe₂O₃.
Mass of Fe in Fe₂O₃ = 2 × 56 = 112 g.

Step 4: Calculate the percentage of iron in Fe₂O₃.
% Fe = (Mass of Fe / Molar Mass of Fe₂O₃) × 100
=> % Fe = (112 / 160) × 100 = 70%.

Step 5: Calculate the mass of iron in 8 kg of pure Fe₂O₃.
Mass of Fe = 70% of 8 kg
=> Mass of Fe = (70 / 100) × 8 kg = 5.6 kg.

7. If the empirical formula of two compounds is CH and their vapour densities are 13 and 39 respectively, find their molecular formula.

Answer: Given:

Empirical Formula (EF) = CH
Vapour Density of Compound 1 (V.D.₁) = 13
Vapour Density of Compound 2 (V.D.₂) = 39

To find:

Molecular formulas of both compounds.

Solution:

First, calculate the Empirical Formula Mass (EFM) of CH.
EFM = (1 × 12) + (1 × 1) = 13 a.m.u.

For Compound 1:
Molecular Mass (MM₁) = 2 × V.D.₁
=> MM₁ = 2 × 13 = 26 a.m.u.
Now, find the integer ‘n’.
n₁ = MM₁ / EFM = 26 / 13 = 2
Molecular Formula₁ = (EF)n₁ = (CH)₂ = C₂H₂.

For Compound 2:
Molecular Mass (MM₂) = 2 × V.D.₂
=> MM₂ = 2 × 39 = 78 a.m.u.
Now, find the integer ‘n’.
n₂ = MM₂ / EFM = 78 / 13 = 6
Molecular Formula₂ = (EF)n₂ = (CH)₆ = C₆H₆.

8. Find the empirical formula of a compound containing 17.64% hydrogen and 82.35% nitrogen.

Answer: Given:

Percentage composition: H = 17.64%, N = 82.35%
Atomic masses: H = 1, N = 14

To find:

Empirical Formula = ?

Solution:
We will use a table to determine the simplest ratio of atoms.

Element	Percentage	Atomic Mass	Atomic Ratio (Moles)	Simplest Ratio
H	17.64	1	17.64 / 1 = 17.64	17.64 / 5.88 ≈ 3
N	82.35	14	82.35 / 14 = 5.88	5.88 / 5.88 = 1

The simplest whole-number ratio of atoms N : H is 1 : 3.
The empirical formula is NH₃.

9. 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.

Answer: Given:

Percentage composition: C = 54.54%, H = 9.09%, O = 36.36%
Vapour Density (V.D.) = 44
Atomic masses: C = 12, H = 1, O = 16

To find:

(a) Empirical Formula
(b) Molecular Formula

Solution:

(a) Empirical Formula:

Element	Percentage	Atomic Mass	Atomic Ratio (Moles)	Simplest Ratio
C	54.54	12	54.54 / 12 = 4.545	4.545 / 2.2725 ≈ 2
H	9.09	1	9.09 / 1 = 9.09	9.09 / 2.2725 ≈ 4
O	36.36	16	36.36 / 16 = 2.2725	2.2725 / 2.2725 = 1

The simplest whole-number ratio C : H : O is 2 : 4 : 1.
The empirical formula is C₂H₄O.

(b) Molecular Formula:
First, calculate the Empirical Formula Mass (EFM).
EFM of C₂H₄O = (2 × 12) + (4 × 1) + (1 × 16) = 24 + 4 + 16 = 44 a.m.u.

Next, calculate the Molecular Mass (MM).
MM = 2 × V.D. = 2 × 44 = 88 a.m.u.

Now, find the integer ‘n’.
n = MM / EFM = 88 / 44 = 2.

Molecular Formula = (Empirical Formula)n = (C₂H₄O)₂ = C₄H₈O₂.

10. 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.

Answer: Given:

V.D. = 45
Percentage composition: H = 2.22%, O = 71.19%
Atomic masses: H = 1, O = 16, C = 12

To find:

(a) Empirical Formula
(b) Molecular Formula

Solution:

First, calculate the percentage of Carbon (C).
% C = 100 – (% H + % O)
=> % C = 100 – (2.22 + 71.19) = 100 – 73.41 = 26.59%.

(a) Empirical Formula:

Element	Percentage	Atomic Mass	Atomic Ratio (Moles)	Simplest Ratio
C	26.59	12	26.59 / 12 = 2.216	2.216 / 2.216 ≈ 1
H	2.22	1	2.22 / 1 = 2.22	2.22 / 2.216 ≈ 1
O	71.19	16	71.19 / 16 = 4.449	4.449 / 2.216 ≈ 2

The simplest whole-number ratio C : H : O is 1 : 1 : 2.
The empirical formula is CHO₂.

(b) Molecular Formula:
First, calculate the Empirical Formula Mass (EFM).
EFM of CHO₂ = 12 + 1 + (2 × 16) = 13 + 32 = 45 a.m.u.

Next, calculate the Molecular Mass (MM).
MM = 2 × V.D. = 2 × 45 = 90 a.m.u.

Now, find the integer ‘n’.
n = MM / EFM = 90 / 45 = 2.

Molecular Formula = (Empirical Formula)n = (CHO₂)₂ = C₂H₂O₄.

11. 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.

Answer: Given:

Molar Mass (MM) = 98.96
Percentage composition: H = 4.07%, Cl = 71.65%
Atomic masses: H = 1, Cl = 35.5, C = 12

To find:

(a) Empirical Formula
(b) Molecular Formula

Solution:

First, calculate the percentage of Carbon (C).
% C = 100 – (% H + % Cl)
=> % C = 100 – (4.07 + 71.65) = 100 – 75.72 = 24.28%.

(a) Empirical Formula:

Element	Percentage	Atomic Mass	Atomic Ratio (Moles)	Simplest Ratio
C	24.28	12	24.28 / 12 = 2.023	2.023 / 2.018 ≈ 1
H	4.07	1	4.07 / 1 = 4.07	4.07 / 2.018 ≈ 2
Cl	71.65	35.5	71.65 / 35.5 = 2.018	2.018 / 2.018 = 1

The simplest whole-number ratio C : H : Cl is 1 : 2 : 1.
The empirical formula is CH₂Cl.

(b) Molecular Formula:
First, calculate the Empirical Formula Mass (EFM).
EFM of CH₂Cl = 12 + (2 × 1) + 35.5 = 12 + 2 + 35.5 = 49.5 a.m.u.

Now, find the integer ‘n’.
n = MM / EFM = 98.96 / 49.5 ≈ 2.

Molecular Formula = (Empirical Formula)n = (CH₂Cl)₂ = C₂H₄Cl₂.

12. 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.

Answer: Given:

Mass of carbon = 4.8 g
Mass of hydrogen = 1 g
Vapour Density (V.D.) = 29
Atomic masses: C = 12, H = 1

To find:

(a) Gram atom (moles) of each
(b) Empirical formula
(c) Molecular formula

Solution:

(a) Gram atom of each:
Gram atom of Carbon = Mass / Atomic Mass = 4.8 / 12 = 0.4 moles.
Gram atom of Hydrogen = Mass / Atomic Mass = 1 / 1 = 1 mole.

(b) Empirical formula:
The ratio of gram atoms is C : H = 0.4 : 1.
To get the simplest whole-number ratio, divide by the smallest number (0.4).
C = 0.4 / 0.4 = 1
H = 1 / 0.4 = 2.5
Multiply by 2 to get whole numbers: C = 2, H = 5.
The empirical formula is C₂H₅.

(c) Molecular formula:
First, calculate the Empirical Formula Mass (EFM).
EFM of C₂H₅ = (2 × 12) + (5 × 1) = 24 + 5 = 29 a.m.u.

Next, calculate the Molecular Mass (MM).
MM = 2 × V.D. = 2 × 29 = 58 a.m.u.

Now, find the integer ‘n’.
n = MM / EFM = 58 / 29 = 2.

Molecular Formula = (Empirical Formula)n = (C₂H₅)₂ = C₄H₁₀.

13. 0.2 g atom of silicon combine with 21.3 g of chlorine. Find the empirical formula of the compound formed.

Answer: Given:

Gram atoms (moles) of silicon (Si) = 0.2
Mass of chlorine (Cl) = 21.3 g
Atomic masses: Si = 28, Cl = 35.5

To find:

Empirical Formula = ?

Solution:

First, calculate the gram atoms (moles) of chlorine.
Moles of Cl = Mass / Atomic Mass = 21.3 / 35.5 = 0.6 moles.

Now, find the ratio of moles of Si to Cl.
Ratio Si : Cl = 0.2 : 0.6
To get the simplest whole-number ratio, divide by the smallest number (0.2).
Si = 0.2 / 0.2 = 1
Cl = 0.6 / 0.2 = 3
The empirical formula is SiCl₃.

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

Answer: Given:

Percentage of Carbon (% C) = 82.76%
V.D. = 29
Atomic masses: C = 12, H = 1

To find:

Molecular Formula = ?

Solution:

Step 1: Find the Empirical Formula.
Since it is a hydrocarbon, it contains only C and H.
Percentage of Hydrogen (% H) = 100 – 82.76 = 17.24%.

Element	Percentage	Atomic Mass	Atomic Ratio (Moles)	Simplest Ratio
C	82.76	12	82.76 / 12 = 6.897	6.897 / 6.897 = 1
H	17.24	1	17.24 / 1 = 17.24	17.24 / 6.897 ≈ 2.5

The ratio C : H is 1 : 2.5. To make it a whole number, multiply by 2.
C : H = 2 : 5.
The empirical formula is C₂H₅.

Step 2: Find the Molecular Formula.
Empirical Formula Mass (EFM) = (2 × 12) + (5 × 1) = 29 a.m.u.
Molecular Mass (MM) = 2 × V.D. = 2 × 29 = 58 a.m.u.
n = MM / EFM = 58 / 29 = 2.
Molecular Formula = (C₂H₅)₂ = C₄H₁₀.

15. 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:

Answer: Given:

Mass of Mg = 18 g
Mass of N = 7 g
Atomic masses: Mg = 24, N = 14

(a) How many gram-atoms of magnesium are equal to 18g?
Solution:
Gram-atoms of Mg = Mass / Atomic Mass
=> Gram-atoms of Mg = 18 / 24 = 0.75 moles.

(b) How many gram-atoms of nitrogen are equal to 7g of nitrogen?
Solution:
Gram-atoms of N = Mass / Atomic Mass
=> Gram-atoms of N = 7 / 14 = 0.5 moles.

(c) Calculate simple ratio of gram-atoms of magnesium to gram-atoms of nitrogen and hence the simplest formula of the compound formed.
Solution:
The ratio of gram-atoms is Mg : N = 0.75 : 0.5.
To get the simplest ratio, divide by the smaller value (0.5).
Mg = 0.75 / 0.5 = 1.5
N = 0.5 / 0.5 = 1
The ratio is 1.5 : 1. To get whole numbers, multiply by 2.
The simplest whole-number ratio is 3 : 2.
The simplest formula (empirical formula) is Mg₃N₂.

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

Answer: Given:

Percentage of water in BaCl₂·xH₂O = 14.8%
Atomic masses: Ba = 137, Cl = 35.5, H = 1, O = 16

To find:

The value of ‘x’.

Solution:

Percentage of water = 14.8%
Percentage of anhydrous BaCl₂ = 100 – 14.8 = 85.2%.

Let’s consider 100 g of the crystal.
Mass of H₂O = 14.8 g
Mass of BaCl₂ = 85.2 g

Molar mass of H₂O = 18 g/mol .
Molar mass of BaCl₂ = 137 + (2 × 35.5) = 137 + 71 = 208 g/mol .

Now, calculate the moles of each component.
Moles of H₂O = 14.8 / 18 ≈ 0.822
Moles of BaCl₂ = 85.2 / 208 ≈ 0.410

Find the simple ratio of moles of H₂O to moles of BaCl₂.
Ratio = Moles of H₂O / Moles of BaCl₂ = 0.822 / 0.410 ≈ 2.
The ratio is 2:1, which means for every 1 molecule of BaCl₂, there are 2 molecules of H₂O.
So, x = 2.
The formula is BaCl₂·2H₂O.

17. 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).

Answer: Given:

Formula of urea = CON₂H₄
Atomic masses: C = 12, O = 16, N = 14, H = 1

To find:

Percentage of nitrogen (% N) in urea.

Solution:

First, calculate the molar mass of urea.
Molar Mass of CON₂H₄ = 12 + 16 + (2 × 14) + (4 × 1)
=> Molar Mass = 12 + 16 + 28 + 4 = 60 g/mol .

Next, find the mass of nitrogen in one mole of urea.
Mass of N = 2 × 14 = 28 g.

Now, calculate the percentage of nitrogen.
% N = (Mass of N / Molar Mass of urea) × 100
=> % N = (28 / 60) × 100
=> % N = 46.67%.

18. 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.

Answer: Given:

Number of C atoms in molecule = 12
% H = 6.48%, % O = 51.42%
Atomic masses: C = 12, H = 1, O = 16

To find:

Molecular formula of the compound.

Solution:

Step 1: Find the percentage of Carbon.
% C = 100 – (% H + % O)
=> % C = 100 – (6.48 + 51.42) = 100 – 57.9 = 42.1%.

Step 2: Find the Empirical Formula.

Element	Percentage	Atomic Mass	Atomic Ratio (Moles)	Simplest Ratio
C	42.1	12	42.1 / 12 = 3.508	3.508 / 3.214 ≈ 1
H	6.48	1	6.48 / 1 = 6.48	6.48 / 3.214 ≈ 2
O	51.42	16	51.42 / 16 = 3.214	3.214 / 3.214 = 1

The empirical formula is CH₂O.

Step 3: Find the Molecular Formula.
The molecular formula is (CH₂O)n.
The number of carbon atoms in the molecular formula is n × 1.
We are given that the number of carbon atoms is 12.
So, n × 1 = 12, which means n = 12.
Molecular Formula = (CH₂O)₁₂ = C₁₂H₂₄O₁₂.

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

Answer: Given:

Empirical Formula (EF) = AB₂
Vapour Density (V.D.) = Empirical Formula Mass (EFM)

To find:

Molecular Formula.

Solution:
Let the Empirical Formula Mass be EFM.
Molecular Mass (MM) = 2 × V.D.
Since V.D. = EFM, then MM = 2 × EFM.
The integer ‘n’ is calculated as n = MM / EFM.
=> n = (2 × EFM) / EFM = 2.
Molecular Formula = (EF)n = (AB₂)₂ = A₂B₄.

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

Answer: Given:

Empirical Formula (EF) = AB
Vapour Density (V.D.) = 3 × Empirical Formula Mass (EFM)

To find:

Molecular Formula.

Solution:
Let the Empirical Formula Mass be EFM.
Molecular Mass (MM) = 2 × V.D.
Since V.D. = 3 × EFM, then MM = 2 × (3 × EFM) = 6 × EFM.
The integer ‘n’ is calculated as n = MM / EFM.
=> n = (6 × EFM) / EFM = 6.
Molecular Formula = (EF)n = (AB)₆ = A₆B₆.

(c) 10.47 g of a compound contained 6.25 g of metal A and rest non metal B. Calculate the empirical formula of the compound [At. wt of A = 207, B = 35.5].

Answer: Given:

Mass of compound = 10.47 g
Mass of metal A = 6.25 g
Atomic weights: A = 207, B = 35.5

To find:

Empirical Formula.

Solution:
First, find the mass of non-metal B.
Mass of B = Mass of compound – Mass of A
=> Mass of B = 10.47 – 6.25 = 4.22 g.

Next, find the moles (gram atoms) of A and B.
Moles of A = 6.25 / 207 ≈ 0.0302
Moles of B = 4.22 / 35.5 ≈ 0.119

Find the simplest ratio of moles.
Ratio A : B = 0.0302 : 0.119
Divide by the smaller value (0.0302).
A = 0.0302 / 0.0302 = 1
B = 0.119 / 0.0302 ≈ 3.94 ≈ 4
The empirical formula is AB₄.

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

Answer: Given:

Percentage of Nitrogen (% N) = 87.5%
Atomic masses: N = 14, H = 1

To find:

Empirical Formula.

Solution:
The compound is a hydride of nitrogen, so it contains N and H.
Percentage of Hydrogen (% H) = 100 – 87.5 = 12.5%.

Element	Percentage	Atomic Mass	Atomic Ratio (Moles)	Simplest Ratio
N	87.5	14	87.5 / 14 = 6.25	6.25 / 6.25 = 1
H	12.5	1	12.5 / 1 = 12.5	12.5 / 6.25 = 2

The simplest whole-number ratio N : H is 1 : 2.
The empirical formula is NH₂.

21. 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.

Answer: Given:

Percentage composition: O=61.32%, S=11.15%, H=4.88%, Zn=22.65%
Molecular Mass (MM) = 287 a.m.u.
All H is present as H₂O.
Atomic masses: Zn = 65, S = 32, O = 16, H = 1

To find:

Molecular Formula.

Solution:

Step 1: Find the Empirical Formula.
We will find the molar ratio of Zn, S, and H₂O.

Element/Compound	Percentage	Atomic/Molar Mass	Moles Ratio	Simplest Ratio
Zn	22.65	65	22.65 / 65 = 0.348	0.348 / 0.348 = 1
S	11.15	32	11.15 / 32 = 0.348	0.348 / 0.348 = 1
H	4.88	1	4.88 / 1 = 4.88	–
H₂O	–	18	Moles of H₂O = 4.88/2 = 2.44	2.44 / 0.348 ≈ 7

The moles of H₂O are half the moles of H, so Moles H₂O = 4.88/2 = 2.44.
Now we need the moles of oxygen that are NOT part of the water.
Moles of O in H₂O = Moles of H₂O = 2.44.
Mass of O in H₂O = 2.44 × 16 = 39.04 g (in a 100g sample).
The total mass of O in 100g sample is 61.32 g.
Mass of O not in H₂O = 61.32 – 39.04 = 22.28 g.
Moles of O not in H₂O = 22.28 / 16 = 1.39.
Now, the simplest ratio for the remaining oxygen: 1.39 / 0.348 ≈ 4.

So the simplest ratio of the components is Zn : S : O : H₂O = 1 : 1 : 4 : 7.
The empirical formula is ZnSO₄·7H₂O.

Step 2: Find the Molecular Formula.
Calculate the Empirical Formula Mass (EFM).
EFM = 65 + 32 + (4 × 16) + 7 × (18)
=> EFM = 65 + 32 + 64 + 126 = 287 a.m.u.

Since the given Molecular Mass (287 a.m.u.) is equal to the Empirical Formula Mass, the molecular formula is the same as the empirical formula.
The molecular formula is ZnSO₄·7H₂O.

Exercise D

1. 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.

Answer:

Given:
Mass of marble (CaCO₃) = 15 g

To find:
(a) Mass of anhydrous calcium nitrate (Ca(NO₃)₂)
(b) Volume of carbon dioxide (CO₂) at STP

Solution:
First, we write the equation with the molecular masses and volumes.
(Atomic masses: Ca=40, C=12, O=16, N=14)

CaCO₃ + 2HNO₃ → Ca(NO₃)₂ + H₂O + CO₂
(40+12+48) g → (40 + 2(14+48)) g → 22.4 L at STP
100 g → 164 g → 22.4 L at STP

(a) Mass of anhydrous calcium nitrate formed:
From the equation, 100 g of CaCO₃ produces 164 g of Ca(NO₃)₂.

Therefore, 15 g of CaCO₃ will produce:
Mass of Ca(NO₃)₂ = (164 / 100) × 15 g
=> Mass of Ca(NO₃)₂ = 1.64 × 15 g
=> Mass of Ca(NO₃)₂ = 24.6 g.

(b) Volume of carbon dioxide evolved at STP:
From the equation, 100 g of CaCO₃ produces 22.4 L of CO₂ at STP.

Therefore, 15 g of CaCO₃ will produce:
Volume of CO₂ = (22.4 / 100) × 15 L
=> Volume of CO₂ = 0.224 × 15 L
=> Volume of CO₂ = 3.36 L.

2. 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.

Answer:

Given:
Mass of ammonium sulphate ((NH₄)₂SO₄) = 66 g

To find:
(a) Mass of ammonia (NH₃)
(b) Volume of ammonia (NH₃) at STP
(c) Mass of sulphuric acid (H₂SO₄)

Solution:
The balanced chemical equation for the reaction is:
2NH₃ + H₂SO₄ → (NH₄)₂SO₄

Now, we write the equation with the molecular masses and volumes.
(Atomic masses: N=14, H=1, S=32, O=16)

2NH₃ + H₂SO₄ → (NH₄)₂SO₄
2 × (14+3) g → (2+32+64) g → (2(14+4)+32+64) g
2 × 17 g → 98 g → 132 g
34 g → 98 g → 132 g
Volume of NH₃: 2 × 22.4 L = 44.8 L at STP

(a) Mass of ammonia required:
From the equation, 132 g of (NH₄)₂SO₄ is produced from 34 g of NH₃.

Therefore, 66 g of (NH₄)₂SO₄ will be produced from:
Mass of NH₃ = (34 / 132) × 66 g
=> Mass of NH₃ = 17 g.

(b) Volume of the gas (ammonia) used at STP:
From the equation, 132 g of (NH₄)₂SO₄ is produced from 44.8 L of NH₃ at STP.

Therefore, 66 g of (NH₄)₂SO₄ will be produced from:
Volume of NH₃ = (44.8 / 132) × 66 L
=> Volume of NH₃ = 22.4 L.

(c) Mass of acid required:
From the equation, 132 g of (NH₄)₂SO₄ is produced from 98 g of H₂SO₄.

Therefore, 66 g of (NH₄)₂SO₄ will be produced from:
Mass of H₂SO₄ = (98 / 132) × 66 g
=> Mass of H₂SO₄ = 49 g.

3. 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.

Answer:

Given:
Mass of red lead (Pb₃O₄) = 6.85 g

To find:
(a) Mass of lead chloride (PbCl₂)
(b) Mass of chlorine (Cl₂)
(c) Volume of chlorine (Cl₂) at STP

Solution:
We write the equation with the molecular masses and volumes.
(Atomic masses: Pb=207, O=16, Cl=35 .5)

Pb₃O₄ + 8HCl → 3PbCl₂ + 4H₂O + Cl₂
(3×207 + 4×16) g → 3×(207 + 2×35.5) g → (2×35.5) g
(621 + 64) g → 3 × (207 + 71) g → 71 g
685 g → 3 × 278 g → 71 g
685 g → 834 g → 71 g
Volume of Cl₂: 22.4 L at STP

(a) Mass of lead chloride formed:
From the equation, 685 g of Pb₃O₄ produces 834 g of PbCl₂.

Therefore, 6.85 g of Pb₃O₄ will produce:
Mass of PbCl₂ = (834 / 685) × 6.85 g
=> Mass of PbCl₂ = 8.34 g.

(b) Mass of chlorine:
From the equation, 685 g of Pb₃O₄ produces 71 g of Cl₂.

Therefore, 6.85 g of Pb₃O₄ will produce:
Mass of Cl₂ = (71 / 685) × 6.85 g
=> Mass of Cl₂ = 0.71 g.

(c) Volume of chlorine evolved at STP:
From the equation, 685 g of Pb₃O₄ produces 22.4 L of Cl₂ at STP.

Therefore, 6.85 g of Pb₃O₄ will produce:
Volume of Cl₂ = (22.4 / 685) × 6.85 L
=> Volume of Cl₂ = 0.224 L.

4. 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

Answer:

Given:
Mass of nitric acid (HNO₃) = 126 kg

To find:
Mass of KNO₃ required.
Comparison of mass of NaNO₃ with KNO₃.

Solution:
(Atomic masses: K=39, N=14, O=16, Na=23, H=1)

Part 1: Using KNO₃
KNO₃ + H₂SO₄ → KHSO₄ + HNO₃
(39+14+48) g → (1+14+48) g
101 g → 63 g

To produce 63 kg of HNO₃, 101 kg of KNO₃ is required.
Therefore, to produce 126 kg of HNO₃:
Mass of KNO₃ = (101 / 63) × 126 kg
=> Mass of KNO₃ = 101 × 2 kg
=> Mass of KNO₃ = 202 kg.

Part 2: Using NaNO₃
NaNO₃ + H₂SO₄ → NaHSO₄ + HNO₃
(23+14+48) g → (1+14+48) g
85 g → 63 g

To produce 63 kg of HNO₃, 85 kg of NaNO₃ is required.
Therefore, to produce 126 kg of HNO₃:
Mass of NaNO₃ = (85 / 63) × 126 kg
=> Mass of NaNO₃ = 85 × 2 kg
=> Mass of NaNO₃ = 170 kg.

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.

5. 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.

Answer:

Given:
Volume of CO₂ collected = 2 L
Temperature (T₁) = 27°C = 273 + 27 = 300 K
Pressure (P₁) = Normal pressure = 760 mm Hg

To find:
(a) Mass of salt (CaCl₂)
(b) Mass of acid (HCl)

Solution:
First, we convert the volume of CO₂ to STP conditions (T₂ = 273 K, P₂ = 760 mm Hg) using the Combined Gas Law.
P₁V₁ / T₁ = P₂V₂ / T₂
(760 × 2) / 300 = (760 × V₂) / 273
V₂ = (760 × 2 × 273) / (300 × 760)
V₂ = (2 × 273) / 300
V₂ = 1.82 L at STP

Now we use the stoichiometric equation.
(Atomic masses: Ca=40, Cl=35 .5, H=1, C=12, O=16)

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
2 × (1+35.5) g → (40 + 2×35.5) g → 22.4 L at STP
2 × 36.5 g → (40 + 71) g → 22.4 L at STP
73 g → 111 g → 22.4 L at STP

(a) Mass of salt (CaCl₂) required:
From the equation, 22.4 L of CO₂ at STP is produced along with 111 g of CaCl₂.

Therefore, for 1.82 L of CO₂ at STP:
Mass of CaCl₂ = (111 / 22.4) × 1.82 g
=> Mass of CaCl₂ ≈ 9.01 g.

(b) Mass of the acid (HCl) required:
From the equation, 22.4 L of CO₂ at STP is produced from 73 g of HCl.

Therefore, for 1.82 L of CO₂ at STP:
Mass of HCl = (73 / 22.4) × 1.82 g
=> Mass of HCl ≈ 5.93 g.

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

Answer:

Given:
Moles of water = 1 mole
Mass of water = 18 g

To find:
Mass of oxygen (O₂)
Volume of oxygen (O₂) at STP

Solution:
The balanced equation for the electrolysis of water is:
2H₂O → 2H₂ + O₂

From the equation, 2 moles of water produce 1 mole of oxygen.
Therefore, 1 mole of water will produce:
Moles of O₂ = (1 / 2) × 1 mole
=> Moles of O₂ = 0.5 moles.

Mass of oxygen:
Molar mass of O₂ = 2 × 16 = 32 g/mol
Mass of O₂ = Moles × Molar mass
=> Mass of O₂ = 0.5 × 32 g
=> Mass of O₂ = 16 g.

Volume of oxygen at STP:
1 mole of any gas at STP occupies 22.4 L.
Volume of O₂ = Moles × 22.4 L
=> Volume of O₂ = 0.5 × 22.4 L
=> Volume of O₂ = 11.2 L.

7. 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.

Answer:

Given:
Mass of sodium peroxide (Na₂O₂) = 1.56 g

To find:
(a) Mass of NaOH
(b) Volume of O₂ at STP
(c) Mass of O₂

Solution:
We use the stoichiometric equation with molecular masses.
(Atomic masses: Na=23, O=16, H=1)

2Na₂O₂ + 2H₂O → 4NaOH + O₂
2 × (2×23 + 2×16) g → 4 × (23+16+1) g → (2×16) g
2 × (46 + 32) g → 4 × 40 g → 32 g
2 × 78 g → 160 g → 32 g
156 g → 160 g → 32 g
Volume of O₂: 22.4 L at STP

(a) Mass of sodium hydroxide formed:
From the equation, 156 g of Na₂O₂ produces 160 g of NaOH.

Therefore, 1.56 g of Na₂O₂ will produce:
Mass of NaOH = (160 / 156) × 1.56 g
=> Mass of NaOH = 1.6 g.

(b) Volume of oxygen liberated at STP:
From the equation, 156 g of Na₂O₂ produces 22.4 L of O₂ at STP.

Therefore, 1.56 g of Na₂O₂ will produce:
Volume of O₂ = (22.4 / 156) × 1.56 L
=> Volume of O₂ = 0.224 L or 224 cm³.

(c) Mass of oxygen liberated:
From the equation, 156 g of Na₂O₂ produces 32 g of O₂.

Therefore, 1.56 g of Na₂O₂ will produce:
Mass of O₂ = (32 / 156) × 1.56 g
=> Mass of O₂ = 0.32 g.

8. (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?

Answer:

Given:
Mass of ammonium chloride (NH₄Cl) = 21.4 g

To find:
(a) Mass of ammonia (NH₃)
(b) Volume of ammonia (NH₃) at STP

Solution:
We use the stoichiometric equation with molecular masses.
(Atomic masses: N=14, H=1, Cl=35 .5)

2NH₄Cl + Ca(OH)₂ → CaCl₂ + 2H₂O + 2NH₃
2 × (14+4+35.5) g → 2 × (14+3) g
2 × 53.5 g → 2 × 17 g
107 g → 34 g
Volume of NH₃: 2 × 22.4 L = 44.8 L at STP

(a) Mass of ammonia:
From the equation, 107 g of NH₄Cl produces 34 g of NH₃.

Therefore, 21.4 g of NH₄Cl will produce:
Mass of NH₃ = (34 / 107) × 21.4 g
=> Mass of NH₃ = 34 × 0.2 g
=> Mass of NH₃ = 6.8 g.

(b) Volume of ammonia at STP:
From the equation, 107 g of NH₄Cl produces 44.8 L of NH₃ at STP.

Therefore, 21.4 g of NH₄Cl will produce:
Volume of NH₃ = (44.8 / 107) × 21.4 L
=> Volume of NH₃ = 44.8 × 0.2 L
=> Volume of NH₃ = 8.96 L.

9. Aluminium carbide reacts with water according to the following equation:
Al₄C₃ + 12H₂O → 3CH₄ + 4Al(OH)₃
(a) What mass of aluminium hydroxide is formed from 12 g of aluminium carbide?
(b) What volume of methane is obtained from 12 g of aluminium carbide?

Answer:

Given:
Mass of aluminium carbide (Al₄C₃) = 12 g

To find:
(a) Mass of aluminium hydroxide (Al(OH)₃)
(b) Volume of methane (CH₄) at STP

Solution:
We use the stoichiometric equation with molecular masses.
(Atomic masses: Al=27, C=12, O=16, H=1)

Al₄C₃ + 12H₂O → 3CH₄ + 4Al(OH)₃
(4×27 + 3×12) g → 3 × (12+4) g → 4 × (27 + 3(16+1)) g
(108 + 36) g → 3 × 16 g → 4 × (27 + 51) g
144 g → 48 g → 4 × 78 g
144 g → 48 g → 312 g
Volume of CH₄: 3 × 22.4 L = 67.2 L at STP

(a) Mass of aluminium hydroxide:
From the equation, 144 g of Al₄C₃ produces 312 g of Al(OH)₃.

Therefore, 12 g of Al₄C₃ will produce:
Mass of Al(OH)₃ = (312 / 144) × 12 g
=> Mass of Al(OH)₃ = 26 g.

(b) Volume of methane:
From the equation, 144 g of Al₄C₃ produces 67.2 L of CH₄ at STP.

Therefore, 12 g of Al₄C₃ will produce:
Volume of CH₄ = (67.2 / 144) × 12 L
=> Volume of CH₄ = 5.6 L.

10. 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.

Answer:

Given:
Moles of MnO₂ = 0.02 moles

To find:
(a) mass of MnO₂, (b) moles of MnCl₂, (c) mass of MnCl₂, (d) moles of Cl₂, (e) mass of Cl₂, (f) volume of Cl₂, (g) moles of HCl, (h) mass of HCl

Solution:
The stoichiometric ratios from the balanced equation are:
1 mole MnO₂ : 4 moles HCl : 1 mole MnCl₂ : 1 mole Cl₂

(Atomic masses: Mn=55, O=16, Cl=35 .5, H=1)

(a) Mass of MnO₂ used:
Molar mass of MnO₂ = 55 + 2×16 = 87 g/mol
Mass = Moles × Molar mass = 0.02 × 87 g = 1.74 g.

(b) Moles of salt (MnCl₂) formed:
Ratio MnO₂ : MnCl₂ is 1:1.
Moles of MnCl₂ = 0.02 moles.

(c) Mass of salt (MnCl₂) formed:
Molar mass of MnCl₂ = 55 + 2×35.5 = 126 g/mol
Mass = Moles × Molar mass = 0.02 × 126 g = 2.52 g.

(d) Moles of chlorine gas formed:
Ratio MnO₂ : Cl₂ is 1:1.
Moles of Cl₂ = 0.02 moles.

(e) Mass of chlorine gas formed:
Molar mass of Cl₂ = 2×35.5 = 71 g/mol
Mass = Moles × Molar mass = 0.02 × 71 g = 1.42 g.

(f) Volume of chlorine gas formed at STP:
Volume = Moles × 22.4 L = 0.02 × 22.4 L = 0.448 L (or 0.448 dm³).

(g) Moles of acid (HCl) required:
Ratio MnO₂ : HCl is 1:4.
Moles of HCl = 4 × Moles of MnO₂ = 4 × 0.02 = 0.08 moles.

(h) Mass of acid (HCl) required:
Molar mass of HCl = 1 + 35.5 = 36.5 g/mol
Mass = Moles × Molar mass = 0.08 × 36.5 g = 2.92 g.

11. 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.

Answer:

Given:
Mass of H₂ = 1000 g
Mass of N₂ = 2000 g

To find:
(a) The excess reactant and its remaining mass.
(b) Mass of ammonia (NH₃) formed.

Solution:
This is a limiting reactant problem. First, we convert the masses of reactants to moles.
(Atomic masses: N=14, H=1)
Molar mass of N₂ = 28 g/mol
Molar mass of H₂ = 2 g/mol

Moles of N₂ available = 2000 g / 28 g/mol ≈ 71.43 moles
Moles of H₂ available = 1000 g / 2 g/mol = 500 moles

From the balanced equation, the stoichiometric ratio is: 1 mole N₂ reacts with 3 moles H₂.

Let’s find the limiting reactant.

Method 1: Using all N₂
To react with 71.43 moles of N₂, we need:
Moles of H₂ required = 71.43 × 3 = 214.29 moles.
We have 500 moles of H₂, which is more than enough. Therefore, N₂ is the limiting reactant and H₂ is the excess reactant.

Method 2: Using all H₂
To react with 500 moles of H₂, we need:
Moles of N₂ required = 500 / 3 = 166.67 moles.
We only have 71.43 moles of N₂. Therefore, N₂ is the limiting reactant.

(a) Excess reactant and its remaining mass:
The excess reactant is H₂.
Moles of H₂ reacted = 3 × moles of N₂ = 3 × 71.43 = 214.29 moles.
Mass of H₂ reacted = 214.29 moles × 2 g/mol ≈ 428.6 g.

Mass of H₂ remaining = Initial mass – Reacted mass
=> Mass of H₂ remaining = 1000 g – 428.6 g = 571.4 g.
So, Yes, hydrogen will remain unreacted. Its mass is 571.4 g.

(b) Mass of ammonia (NH₃) formed:
The amount of product formed depends on the limiting reactant (N₂).
From the equation, 1 mole of N₂ produces 2 moles of NH₃.
Moles of NH₃ formed = 2 × moles of N₂ = 2 × 71.43 ≈ 142.86 moles.

Molar mass of NH₃ = 14 + 3 = 17 g/mol .
Mass of NH₃ formed = Moles × Molar mass
=> Mass of NH₃ formed = 142.86 × 17 g ≈ 2428.6 g.
The mass of ammonia formed is approximately 2428.6 g (or 2.43 kg).

Miscellaneous Exercise

MCQs

1. Which of the following weighs the least ?

(a) 2 g atom of N
(b) 3 x 10²⁵ atoms of carbon
(c) 1 mole of sulphur
(d) 7g silver

Answer: (d) 7g silver

2. Four grams of caustic soda contains:

(a) 6-02 x 10²³ atoms of it,
(b) 4 g atom of sodium,
(c) 6-02 x 10²² molecules
(d) 4 moles of NaOH.

Answer: (c) 6-02 x 10²² molecules

3. The number of molecules in 4-25 g of ammonia is:

(a) 1-0 x 10²³
(b) 1.5 x 10²³,
(c) 2-0 x 10²³
(d) 3.5 x 10²³

Answer: (b) 1.5 x 10²³

4. 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 and the mass of the hydrogen is 2 g, the relative molecular mass of the gas will be :

(a) 5
(b) 10
(c) 15
(d) 20

Answer: (b) 10

5. Twice the vapour density gives:

(a) actual vapour density
(b) relative vapour density
(c) molecular mass
(d) molar volume

Answer: (c) molecular mass

6. The empirical formula of the compound is CH₂O, the possible molecular formula can be :

(a) C₃H₆O₃
(b) C₂H₄O
(c) C₄H₈O₂
(d) C₄H₆O₂

Answer: (a) C₃H₆O₃

7. If relative molecular mass of butane (C₄H₁₀) is 58, then its vapour density will be :

(a) 5
(b) 29
(c) 32
(d) 16

Answer: (b) 29

8. If the empirical mass of the formula PQ₂ is 10 and the relative molecular mass is 30, then the molecular formula will be :

(a) PQ₂
(b) P₃Q₂
(c) P₆Q₃
(d) P₃Q₆

Answer: (d) P₃Q₆

9. The ratio between the number of molecules in 2 g of hydrogen and 32 g of oxygen is : [Atomic mass: H = 1, O = 16]

(a) 1:2
(b) 1:0-01
(c) 1:1
(d) 0-01:1

Answer: (c) 1:1

10. One mole of sulphur dioxide represents which of the following? P 22-4 litres at STP Q 6-02 x 10²³ atoms R 6-02 x 10²³ molecules

(a) Only P
(b) Only Q
(c) Both P and Q
(d) Both P and R

Answer: (d) Both P and R

11. Assertion (A): According to gas equation, P₁V₁/T₁ = P₂V₂/T₂
Reason (R): On combining Boyle’s law and Charle’s law, ‘volume of a given mass of a dry gas varies inversely as the pressure and directly as the absolute temperature.’