# Mole Concept and Stoichiometry

## Exercise 5A

#### Question 1

State:

(a) Gay-Lussac's Law of combining volumes.

(a) Gay-Lussac's Law of combining volumes — When gases react, they do so in volumes which bear a simple ratio to one another, and to the volume of the gaseous product, provided that all the volumes are measured at the same temperature and pressure.

(b) Avogadro's law states that "equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules."

#### Question 2

(a) What do you mean by stoichiometry?

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

(c) Differentiate between N2 and 2N.

(a) Stoichiometry measures quantitative relationships, and is used to determine the amount of products/reactants that are produced/needed in a given reaction. Describing the quantitative relationships among substances as they participate in chemical reactions is known as reaction stoichiometry.

(b) Atomicity is the number of atoms in a molecule of an element.

Atomicity of hydrogen is 2, phosphorus is 4 and sulphur is 8.

(c) Difference between N2 and 2N

N22N
It means 1 molecule of nitrogen.It means two atoms of nitrogen.
It can exist independently.It cannot exist independently.

#### Question 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.

(a) Avogadro's Law states that "equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules."

Considering equal volumes of hydrogen and helium,
volume of hydrogen gas = volume of helium gas

n molecules of hydrogen = n molecules of helium gas

i.e., nH2 = nHe

1 molecule of hydrogen has 2 atoms of hydrogen and 1 molecule of helium has 1 atom of helium

∴ 2H = He

∴ atoms in hydrogen are double the atoms of helium.

(b) Since, the volume of a gas changes remarkably with change in temperature and pressure, it becomes necessary to choose standard values of temperature and pressure to which gas volumes can be referred.

(c) According to Boyle's law, the volume of a given mass of dry gas is inversely proportional to its pressure at a constant temperature. When we inflate a balloon, the volume of air keeps increasing and at the same time the pressure of air also increases due to which balloon inflates. As pressure and volume of air increase simultaneously, hence this seems to violate Boyle's law.

## Numerical Problems

#### Question 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

From equation:

$\begin{matrix} 2\text{CO} & + & \text{O}_2 & \longrightarrow & 2 \text{CO}_2 \\ 2 \text{ vol.} & : & 1 \text{ vol.} & \longrightarrow & 2\text{ vol.} \end{matrix}$

[By Gay Lussac's law]

2 V of CO requires = 1V of O2

∴ 100 litres of CO requires = $\dfrac{1}{2}$ x 100 = 50 litres.

Hence, required volume of oxygen is 50 litres.

#### Question 4(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?

$\begin{matrix} 2\text{H}_2 & + & \text{O}_2 & \longrightarrow & 2\text{H}_2\text{O} \\ 2 \text{ vol.} & : & 1 \text{ vol.} & \longrightarrow & 2\text{ vol.} \end{matrix}$

2 Vol. of hydrogen reacts with 1 Vol. of oxygen

∴ 200 cm3 of hydrogen reacts with = $\dfrac{1}{2}$ x 200 = 100 cm3 of oxygen.

Hence, unreacted oxygen is 150 - 100 = 50cm3

#### Question 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.

This experiment supports Gay-Lussac's law of combining volumes.

Since the unchanged oxygen is 58 cc so, used oxygen 106 - 58 = 48cc

According to Gay-Lussac's law, the volumes of gases reacting should be in a simple ratio.

$\begin{matrix} \text{CH}_4 & + & 2\text{O}_2 & \longrightarrow & 2\text{CO}_2 + \text{H}_2\text{O} \\ 1 \text{ vol.} & : & 2 \text{ vol.} \\ 24 \text{ cc} & : & 48 \text{ cc} \end{matrix}$

Hence, methane and oxygen are in the ratio 1:2 .

#### Question 6

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)

$\begin{matrix} 2\text{C}_2\text{H}_2 & + & 5\text{O}_2 & \longrightarrow & 4\text{CO}_2 & + & 2\text{H}_2\text{O} \\ 2 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 4 \text{ vol.} & \\ \end{matrix}$

[By Gay Lussac's law]

2 Vol. of C2H2 requires 5 Vol. of oxygen

∴ 400 ml C2H2 will require $\dfrac{5}{2}$ x 400

= 1000 ml of Oxygen

Hence, required volume of oxygen = 1000 ml

Similarly,

2 Vol. of C2H2 produces 4 Vol. of Carbon dioxide

∴ 400 ml of C2H2 produces $\dfrac{4}{2}$ x 400

= 800 ml of Carbon dioxide

Hence, carbon dioxide produced = 800 ml

#### Question 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

$\begin{matrix} \text{H}_2\text{S} & + & \text{Cl}_2 & \longrightarrow & 2\text{HCl} & + & \text{S} \\ 1\text{ vol.} & : & 1 \text{ vol.} & \longrightarrow & 2 \text{ vol.} & : & 1\text{ vol.} \\ \end{matrix}$

(i) At STP,

1 mole gas occupies = 22.4 L.

1 mole H2S gas produces = 2 moles HCl gas,

∴ 22.4 L H2S gas produces

= 22.4 × 2

= 44.8 L HCl gas.

Hence, 112 cm3 H2S gas will produce

= 112 × 2

= 224 cm3 HCl gas.

Hence, 224 cm3 HCl gas is produced.

(ii) 1 mole H2S gas consumes = 1 mole Cl2 gas.

Hence, 22.4 L H2S gas consumes = 22.4 L Cl2 gas at STP.

∴ 112 cm3 H2S gas consumes = 112 cm3 Cl2 gas.

120 cm3 - 112 cm3 = 8 cm3 Cl2 gas remains unreacted.

Hence, the composition of the resulting mixture is 224 cm3 HCl gas + 8 cm3 Cl2 gas.

#### Question 8

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

$\begin{matrix} 2\text{C}_2\text{H}_6 & + & 7\text{O}_2 & \longrightarrow & 4\text{CO}_2 & + & 6\text{H}_2\text{O} \\ 2 \text{ vol.} & : & 7 \text{ vol.} & \longrightarrow & 4 \text{ vol.} & \\ \end{matrix}$

[By Gay Lussac's law]

2 Vol. of C2H6 requires 7 Vol. of oxygen

∴ 300 cc C2H6 will require $\dfrac{7}{2}$ x 300

= 1050 cc of Oxygen

Hence, unused oxygen = 1250 - 1050 = 200 cc

Similarly,

2 Vol. of C2H6 produces 4 Vol. of carbon dioxide

∴ 300 cc C2H6 produces $\dfrac{4}{2}$ x 300

= 600 cc of Carbon dioxide

Hence, carbon dioxide produced = 600 cc.

#### Question 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

$\begin{matrix} \text{C}_2\text{H}_4 & + & 3\text{O}_2 & \longrightarrow & 2\text{CO}_2 + 2\text{H}_2\text{O} \\ 1 \text{ vol.} & : & 3 \text{ vol.} \\ 11 \text{ lit} & : & 33 \text{ lit} \end{matrix}$

STPGiven Values
P1 = 760 mm of HgP2 = 380 mm of Hg
V1 = x litV2 = 33 lit
T1 = 273 KT2 = 273 + 273 K = 546 K

Using the gas equation,

$\dfrac{P_{1}V_{1}}{T_{1}} = \dfrac{P_{2}V_{2}}{T_{2}}$

Substituting the values we get,

$\dfrac{760 \times x}{273} = \dfrac{380 \times 33}{546} \\[0.5em] x = \dfrac{380 \times 33 \times 273}{546 \times 760 } \\[0.5em] x = \dfrac{3,423,420}{414,960} \\ \\[0.5em] x = 8.25 \text{ lit}$

Hence, volume of oxygen required = 8.25 lit.

#### Question 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

$\begin{matrix} \text{CH}_4 & + & 2\text{Cl}_2 & \longrightarrow & \text{CH}_2\text{Cl}_2 & + & 2\text{HCl} \\ 1 \text{ vol.} & : & 2 \text{ vol.} & \longrightarrow & 1 \text{ vol.} & : & 2 \text{ vol.}\\ \end{matrix}$

volume of HCl gas formed = ?

[By Gay Lussac's law]

1 Vol of methane produces = 2 Vol. HCl

∴ 40 ml of methane produces = 80 ml HCl

volume of chlorine gas required = ?

For 1 Vol of methane = 2V of Cl2 required

∴ for 40 ml of methane = 40 x 2 = 80 ml of Cl2 is required.

Hence, volume of HCl gas formed = 80 ml and chlorine gas required = 80 ml

#### Question 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

Given, oxygen is 1⁄5th of air = $\dfrac{1}{5}$ of 500 = 100 cm3

$\begin{matrix} \text{C}_3\text{H}_8 & + & 5\text{O}_2 & \longrightarrow & 3\text{CO}_2 & + & 4\text{H}_2\text{O} \\ 1 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 3 \text{ vol.} & \\ \end{matrix}$

[By Gay Lussac's law]

5 Vol. of O2 requires 1 Vol. of propane

∴ 100 cm3 of O2 will require = $\dfrac{1}{5}$ x 100 = 20 cm3

Hence, propane burnt = 20 cm3 or 20 cc

#### Question 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

$\begin{matrix} 2\text{NO} & + & \text{O}_2 & \longrightarrow & 2\text{NO}_2 \\ 2 \text{ vol.} & : & 1 \text{ vol.} & \longrightarrow & 2 \text{ vol.} \\ \end{matrix}$

[By Gay Lussac's law]

1 Vol. of O2 reacts with = 2V of NO

200 cm3 oxygen will react with
= 200 × 2
= 400 cm3 of NO

∴ remaining NO is 450 - 400 = 50 cm3

NO2 = ?

1 Vol. of O2 produces 2 Vol. of NO2

∴ 200 cm3 of oxygen produces = $\dfrac{2}{1}$ x 200 = 400cm3

Hence, NO2 produced = 400 cm3 and unused oxygen is 50 cm3, so total mixture = 400 + 50 = 450 cm3

#### Question 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.

$\begin{matrix} \text{H}_2 & + & \text{Cl}_2 & \longrightarrow & 2\text{HCl} \\ 1 \text{ vol.} & : & 1 \text{ vol.} & \longrightarrow & 2 \text{ vol.} \\ \end{matrix}$

[By Gay Lussac's law]

1 Vol. of chlorine reacts with = 1 Vol. of hydrogen

∴ 4 litres of chlorine will react with only 4 litres of hydrogen,

hence, 6 - 4 = 2 litres of hydrogen will remain unreacted.

Since, vol. of HCl gas formed is twice that of chlorine used,

∴ vol.of HCl formed will be 4 x 2 = 8 litres However HCl dissolves in water.

Hence, 2 litres of hydrogen is the residual gas, as HCl formed dissolves in water.

#### Question 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?

$\begin{matrix} 4\text{NH}_3 & + & 5\text{O}_2 & \longrightarrow & 4\text{NO} & + & 6\text{H}_2\text{O} \\ 4 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 4 \text{ vol.} \\ \end{matrix}$

[By Gay Lussac's law]

9 litres of reactants produces = 4 litres of NO

So, 27 litres of reactants will produces

$=\dfrac{4}{9} \times 27 \\[0.5em] = 12 \text{ litres}$

Hence, volume of nitrogen monoxide produced = 12 litres

#### Question 15

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 mixture.

According to Gay lussac's law,

$\begin{matrix} \text{H}_2 & + & \text{Cl}_2 & \longrightarrow & 2\text{HCl} \\ 1 \text{ vol.} & : & 1 \text{ vol.} & \longrightarrow & 2 \text{ vol.} \\ \end{matrix}$

As, 4 cm3 of hydrogen was left behind, hence, 36 - 4 = 32 cm3 of mixture of hydrogen and chlorine exploded.

As, 1 Vol. of hydrogen requires 1 Vol. of oxygen
∴ 16 cm3 hydrogen requires 16 cm3 of oxygen

Mixture is 20 cm3 (i.e., 16 + 4) of hydrogen and 16 cm3 of chlorine.

#### Question 16

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

$\begin{matrix} \text{CH}_4 & + & 2\text{O}_2 & \longrightarrow & \text{CO}_2 & + & 2\text{H}_2\text{O} \\ 1 \text{ vol.} & : & 2 \text{ vol.} & \longrightarrow & 1\text{ vol.} \\ 2\text{C}_2\text{H}_2 & + & 5\text{O}_2 & \longrightarrow & 4\text{CO}_2 & + & 2\text{H}_2\text{O} \\ 2 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 4\text{ vol.} \\ \end{matrix}$

[By Gay Lussac's law]

1 Vol. CH4 requires 2 Vol. of O2

∴ 10 cm3 CH4 will require 2 x 10

= 20 cm3 of O2

Given, air contains 20% O2 by volume.

Let $x$ volume of air contain 20 cm3 of O2

$\Rightarrow \dfrac{20}{100} \times x = 20 \\[1em] \Rightarrow x = \dfrac{100}{20} \times 20 \\[1em] \Rightarrow x = 100 \text{ cm}^3$

∴ 20 cm3 O2 is present in 100 cm3 of air.

Similarly, 2 Vol C2H2 requires 5 Vol. of oxygen

∴ 10 cm3 C2H2 will require $\dfrac{5}{2}$ x 10

= 25 cm3 of oxygen

Given, air contains 20% O2 by volume

Let $x$ volume of air contain 25 cm3 of O2

$\Rightarrow \dfrac{20}{100} \times x = 25 \\[1em] \Rightarrow x = \dfrac{100}{20} \times 25 \\[1em] \Rightarrow x = 125 \text{ cm}^3$

∴ 25 cm3 O2 is present in 125 cm3 of air.

Hence, total volume of air required is 100 + 125 = 225 cm3

#### Question 17

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

C3H8 + 5O2 ⟶ 3CO2 + 4H2O

2C4H10 + 13O2 ⟶ 8CO2 + 10H2O

Given, 10 litres of this mixture contains 60% propane and 40% butane. Hence, propane is 6 litres and butane is 4 litres

$\begin{matrix} \text{C}_3\text{H}_8 & + & 5\text{O}_2 & \longrightarrow & 3\text{CO}_2 & + & 4\text{H}_2\text{O} \\ 1 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 3\text{ vol.} \\ 2\text{C}_4\text{H}_{10} & + & 13\text{O}_2 & \longrightarrow & 8\text{CO}_2 & + & 10\text{H}_2\text{O} \\ 2 \text{ vol.} & : & 13 \text{ vol.} & \longrightarrow & 8\text{ vol.} \\ \end{matrix}$

1 Vol. C3H8 produces carbon dioxide = 3 Vol

So, 6 litres C3H8 will produce carbon dioxide = 3 x 6 = 18 litres

2 Vol. C4H10 produces carbon dioxide = 8 Vol

So, 4 litres C4H10 will produce carbon dioxide = $\dfrac{8}{2}$ x 4 = 16 litres

Hence, 34 (i.e., 18 + 16) litres of CO2 is produced.

#### Question 18

200 cm3 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)

According to Gay lussac's law,

$\begin{matrix} 2\text{C}_2\text{H}_2 & + & 5\text{O}_2 & \longrightarrow & 4\text{CO}_2 & + & 2\text{H}_2\text{O} \\ 2 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 4\text{ vol.} \\ \end{matrix}$

4 Vol of CO2 is collected with 2 Vol. of C2H2

So, 200 cm3 CO2 will be collected with

$=\dfrac{2}{4} \times 200 \\[0.5em] = 100 \text{ cc}$

Similarly, 4 Vol of CO2 is produced by 5 Vol of O2

So, 200 cm3 CO2 will be produced by = $\dfrac{5}{4}$ x 200 = 250 cm3

Hence, acetylene = 100 cm3 and oxygen = 250 cm3

#### Question 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 litres 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?

According to Avogadro's law, equal volumes of all gases under similar conditions of temperature and pressure contain same number of molecules. So, under the same conditions of temperature and pressure, if volume of gas is decreased the number of molecules will also decrease.

Hence,

(a) 5 litres of hydrogen contain the greatest number of molecules as it has the highest volume.

(b) 1 litre of SO2 contains the least number of molecules since it has the smallest volume.

#### Question 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.

GasVolume (in litres)Number of molecules
Chlorine10
Nitrogen20x
Ammonia20
Sulphur dioxide5

GasVolume
(in litres)
Number of
molecules
Chlorine10x/2
Nitrogen20x
Ammonia20x
Sulphur dioxide5x/4

Reason — According to Avogadro's law, equal volumes of all gases under similar conditions of temperature and pressure contain same number of molecules. If 20 lit of nitrogen contains x molecules then 20 lit of ammonia will also contain x molecules. As volume of chlorine is half that of nitrogen so it will contain half the number of molecules of nitrogen i.e., x/2. Similarly, sulphur dioxide will contain x/4 molecules.

#### Question 21

(i) If 150 cc of gas A contains X molecules, how many molecules of gas B will be present in 75 cc of 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

(a) Given, 150 cc of gas A contains X molecules. According to Avogadro's law, 150 cc of gas B will also contain X molecules.

So, 75 cc of gas B will contain $\dfrac{x}{2}$ molecules.

(b) The problem is based on Avogadro's law.

## Exercise 5B

#### Question 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.?

(a) The relative atomic masses of any element is the weighted average of the relative atomic masses of it's natural isotopes. Chlorine consists of a mixture of two isotopes of masses 35 and 37 in the ratio 3 : 1.

The average relative atomic mass of Cl = $\dfrac{(35 \times 3) + (37 \times 1)}{4}$ = 35.5

(b) 6.022 × 1023

(c) The molar volume of a gas is 22.4 dm3 (litre) or 22400 cm3 (ml) at S.T.P.

#### Question 2

Define or explain the terms:

(a) Vapour density

(b) Molar volume

(c) Relative atomic mass

(d) Relative molecular mass

(f) Gram atom

(g) Mole

(a) Vapour density is defined as the ratio between the masses of equal volumes of gas (or vapour) and hydrogen under the same conditions of temperature and pressure.

(b) The molar volume of a gas is the volume occupied by one gram-molecular mass or by one mole of the gas at S.T.P. It is equal to 22.4 dm3.

(c) The Relative atomic mass of an element is the number of times one atom of the element is heavier than $\dfrac{1}{12}$ times of the mass of an atom of carbon-12.

(d) The Relative molecular mass of an element or a compound is the number that represents how many times one molecule of the substance is heavier than $\dfrac{1}{12}$ of the mass of an atom of carbon-12.

(e) Avogadro's number is defined as the number of atoms present in 12 g (gram atomic mass) of C-12 isotope, i.e. 6.022 x1023 atoms.

(f) The quantity of the element which weighs equal to it's gram atomic mass is called one gram atom of that element

(g) A Mole is the amount of pure substance containing the same number of chemical units as there are atoms in exactly 12 grams of carbon-12.

#### Question 3 (a)

(a) What are the main applications of Avogadro's Law?

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

(a) The applications of Avogadro's Law are:

1. It explains Gay-Lussac's law.
2. It predicts atomicity of gases.
3. It determines the molecular formula of gases.
4. It determines the relation between molecular mass and vapour density.
5. It gives the relationship between gram molecular mass and gram molar volume.

(b) According to Avogadro's law under the same conditions of temperature and pressure, equal volumes of different gases have the same number of molecules.

Since substances react in simple ratio by number of molecules, volumes of the gaseous reactants and products will also bear a simple ratio to one another. This is what Gay Lussac's Law says.

$\begin{matrix} \text{H}_2 \space + \text{Cl}_2 & \longrightarrow & 2\text{HCl} \\ 1 \text{ vol.} \phantom{+} 1 \text{ vol.} & \longrightarrow & 2 \text{ vol.} & \small{\text{(by Gay-Lussac's Law)}} \\ \underset{\text{molecules}}{\text{n}} \phantom{+} \underset{\text{molecules}}{\text{n}} & \longrightarrow & \underset{\text{molecules}}{\text{2n}} & \small{\text{(by Avogadro's Law)}} \\ \end{matrix}$

#### Question 4

Calculate the relative molecular masses of :

(a) Ammonium chloroplatinate [(NH4)2PtCl6]

(b) Potassium chlorate [KClO3]

(c) CuSO4.5H2O

(d) (NH4)2SO4

(e) CH3COONa

(f) CHCl3

(g) (NH4)2Cr2O7

(a) (NH4)2PtCl6

= (2N) + (8H) + (Pt) + (6Cl)

= (2 x 14) + (8 x 1) + 195 + (6 x 35.5)

= 28 + 8 + 195 + 213

= 444 a.m.u.

(b) KClO3

= (K) + (Cl) + (3O)

= 39 + 35.5 + (3 x 16)

= 39 + 35.5 + 48

= 122.5 a.m.u.

(c) CuSO4.5H2O

= (Cu) + (S) + (4O) + 5(2H + O)

= 63.5 + 32 + (4 x 16) + 5[(2 x 1) + 16]

= 63.5 + 32 + 64 + (5 x 18)

= 63.5 + 32 + 64 + 90

= 249.5 a.m.u.

(d) (NH4)2SO4

= (2N) + (8H) + (S) + (4O)

= (2 x 14) + (8 x 1) + 32 + (4 x 16)

= 28 + 8 + 32 + 64

= 132 a.m.u.

(e) CH3COONa

= (C) + (3H) + (C) + (2O) + (Na)

= 12 + (3 x 1) + 12 + (2 x 16) + 23

= 12 + 3 + 12 + 32 + 23

= 82 a.m.u.

(f) CHCl3

= (C) + (H) + (3Cl)

= 12 + 1 + (3 x 35.5)

= 12 + 1 + 106.5

= 119.5 a.m.u.

(g) (NH4)2Cr2O7

= (2N) + (8H) + (2Cr) + (7O)

= (2 x 14) + (8 x 1) + (2 x 51.9) + (7 x 16)

= 28 + 8 + 103.8 + 112

= 251.8 ≈ 252 a.m.u.

#### Question 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 H2 gas,

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

(a) Number of molecules in 73 g of HCl —

Molecular wt. of any substance contain 6.022 × 1023 molecules.

Mass of 1 mole of HCl is 1 + 35.5 = 36.5 g

36.5 g of HCl contains 6.022 × 1023 molecules

∴ 73 g of HCl contains $\dfrac{6.022 \times 10^{23} \times 73 }{36.5}$

= 1.2 × 1024 molecules

(b) Weight of 0.5 mole of O2

1 mole of O2 weighs = 2O = 2 x 16 = 32 g

∴ 0.5 moles will weigh = $\dfrac{32}{2}$ = 16 g

(c) Number of molecules in 1.8 g of H2O —

Molecular wt. of any substance contains 6.022 × 1023 molecules.

Mass of 1 mole of H2O is (2 x 1) + 16 = 2 + 16 = 18 g

18 g of H2O contains 6.022 × 1023 molecules

∴ 1.8 g of H2O contains $\dfrac{6.022 \times 10^{23} \times 1.8 }{18}$

= 6.02 × 1022 molecules

(d) Number of moles in 10 g of CaCO3

Mass of 1 mole of CaCO3 = 40 + 12 + 3(16) = 52 + 48 = 100 g

100 g of CaCO3 = 1 mole

∴ 10 g of CaCO3 = $\dfrac{1 \times 10}{100}$

= 0.1 mole

(e) Weight of 0.2 mole H2 gas —

1 mole of H2 weighs = 2 g

∴ 0.2 moles will weigh = $\dfrac{2 \times 0.2}{1}$ = 0.4 g

(f) No. of molecules in 3.2 g of SO2

Molecular wt. of any substance contain 6 × 1023 molecules.

Mass of 1 mole of SO2 is 32 + 2(16) = 32 + 32 = 64 g

64 g of SO2 contains 6 × 1023 molecules

∴ 3.2 g of SO2 contains $\dfrac{6 \times 10^{23} \times 3.2 }{64}$

= 3 x 1022 molecules.

#### Question 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

1 mole of CO2

Reason

Weight of H2O = 2 + 16 = 18 g

Weight of CO2 = 12 + (2 x 16) = 12 + 32 = 44 g

Weight of NH3 = 14 + (3 x 1) = 14 + 3 = 17 g

Weight of CO = 12 + 16 = 28 g

As weight of CO2 is maximum, hence 1 mole of CO2 will weigh the most.

#### Question 7

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

(d) 4 g of SO2

4 g of NH3

Reason

(a) No. of molecules in 4 g of O2

Molecular wt. of any substance contain 6.022 × 1023 molecules.

Mass of 1 mole of O2 is 2(16) = 32 g

32 g of O2 contains 6.022 × 1023 molecules

∴ 4 g of O2 contains $\dfrac{6.022 \times 10^{23} \times 4}{32}$

= 7.5 x 1022 molecules.

Similarly,

(b) 4 g of NH3 [14 + 3 = 17g ] contains $\dfrac{6.022 \times 10^{23} \times 4}{17}$

(c) 4 g of CO [12 + 16 = 28g ] contains $\dfrac{6.022 \times 10^{23} \times 4}{28}$

(d) 4 g of SO2 [32 + 32 = 64g ] contains $\dfrac{6.022 \times 10^{23} \times 4}{64}$

∴ 4g of NH3 having minimum molecular mass contains maximum molecules.

Note : The fraction with lowest denominator gives the highest value. Hence, by observation we can say that 4 g of NH3 has maximum number of molecules.

#### Question 8(a)

Calculate the number of particles in 0.1 mole of any substance.

No. of particles in 1 mole = 6.022 × 1023

∴ No. of particles in 0.1 mole = $\dfrac{6.022 \times 10^{23} \times 0.1}{1}$

= 6.022 × 1022

#### Question 8(b)

Calculate the number of hydrogen atoms in 0.1 mole of H2SO4.

1 mole of H2SO4 contains (2 × 6.022 × 1023) hydrogen atoms

∴ 0.1 mole of H2SO4 contains = $\dfrac{6.022 \times 10^{23} \times 2 \times 0.1}{1}$

= 1.2 × 1023 atoms of hydrogen

#### Question 8(c)

Calculate the number of molecules in one kg of calcium chloride.

Mass of 1 mole of CaCl2 = Ca + 2Cl = 40 + (2 x 35.5) = 40 + 71 = 111 g

111 g of CaCl2 contains 6.022 × 1023 molecules

∴ 1000 g of CaCl2 contains $\dfrac{6.022 \times 10^{23} \times 1000}{111}$

= 5.42 × 1024 molecules

#### Question 9(a)

How many grams of Al are present in 0.2 mole of it?

1 mole of aluminium has mass = 27 g

0.2 mole of aluminium has mass

= $\dfrac{27}{1}$ x 0.2

= 5.4 g

#### Question 9(b)

How many grams of HCl are present in 0.1 mole of it?

1 mole of HCl has mass = 1 + 35.5 = 36.5 g

0.1 mole of HCl has mass

= $\dfrac{36.5}{1}$ x 0.1

= 3.65 g

#### Question 9(c)

How many grams of H2O are present in 0.2 mole of it?

1 mole of H2O has mass = 2(1) + 16 = 2 + 16 = 18 g

0.2 mole of H2O has mass

= $\dfrac{18}{1}$ x 0.2

= 3.6 g

#### Question 9(d)

How many grams of CO2 is present in 0.1 mole of it?

1 mole of CO2 has mass = 12 + 2(16) = 12 + 32 = 44 g

0.1 mole of CO2 has mass

= $\dfrac{44}{1}$ x 0.1

= 4.4 g

#### Question 10(a)

The mass of 5.6 litres of a certain gas at S.T.P. is 12 g. What is the relative molecular mass or molar mass of the gas?

5.6 litres of gas at S.T.P. has mass = 12 g

∴ 22.4 litre (molar volume) has mass

= $\dfrac{12}{5.6}$ x 22.4

= 48 g

#### Question 10(b)

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

1 mole of SO2 has volume = 22.4 litres

∴ 2 moles will have = 22.4 × 2 = 44.8 litre

#### Question 11(a)

Calculate the number of moles of CO2 which contain 8.00 g of O2

Oxygen in 1 mole of CO2 = 2O = (2 x 16) = 32 g

or we can say, 32 g of oxygen is present in 1 mole of CO2

∴ 8 gm of O2 is present in $\dfrac{1}{32}$ x 8

= 0.25 moles

#### Question 11(b)

Calculate the number of moles of methane in 0.80 g of methane.

Molar mass of methane (CH4) = C + 4H = 12 + 4 = 16 g

16 g of methane = 1 mole

∴ 0.80 g of methane = $\dfrac{1}{16}$ x 0.80

= 0.05 moles

#### Question 12

Calculate the weight/mass of :

(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

(a) Number of oxygen atoms in 16 g of atomic oxygen = 6.022 × 1023 atoms

∴ mass of 1 atom of oxygen

= $\dfrac{16}{6.022 \times 10^{23}}$

= 2.657 × 10-23 g

(b) Number of hydrogen atoms in 1 g of atomic hydrogen = 6.022 × 1023 atoms

∴ Mass of 1 atom of hydrogen

= $\dfrac{1}{6.022 \times 10^{23}}$

= 1.666 × 10-24 g

(c) Gram molecular mass of NH3 = 14 + 3 = 17 g

Number of NH3 molecules in 17 g of NH3 = 6.022 × 1023 molecules

Mass of 6.022 × 1023 molecules of NH3 = 17g

∴ Mass of 1 molecule of NH3 = $\dfrac{17}{6.022 \times 10^{23}}$

= 2.823 × 10-23 g

(d) Mass of 6.022 × 1023 atoms of atomic carbon = 12 g

∴ Mass of 1022 atoms of carbon = $\dfrac{12}{6.022 \times 10^{23}} \times 10^{22}$

= 0.2 g

(e) Gram molecular mass of oxygen (O2) = 2 x 16 = 32 g

Mass of 6.022 × 1023 molecules of O2 = 32 g

∴ Mass of 1 molecule of O2 = $\dfrac{32}{6.022 \times 10^{23}}$

= 5.314 × 10-23 g

(f) Atomic weight of calcium = 40 g

Gram atom = $\dfrac{\text{Mass of element}}{\text{Atomic mass}}$

Therefore, 0.25 = $\dfrac{\text{Mass of calcium}}{40}$

Mass of calcium = 40 x 0.25 = 10 g

#### Question 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)

(a) Mass of 1 mole of CaCO3

= Ca + C + 3O = 40 + 12 + (3 x 16) = 52 + 48 = 100 g

∴ Mass of 0.1 mole of CaCO3 = 0.1 x 100 = 10 g

(b) Mass of 1 mole of Na2SO4.10H2O

= 2Na + S + 4O + 10(2H + O) = (2 x 23) + 32 + (4 x 16) + 10(2 + 16) = 46 + 32 + 64 + 180 = 322 g

∴ Mass of 0.1 mole of Na2SO4.10H2O = 0.1 x 322 = 32.2 g

(c) Mass of 1 mole of CaCl2

= Ca + 2Cl = 40 + (2 x 35.5) = 40 + 71 = 111 g

∴ Mass of 0.1 mole of CaCl2 = 0.1 x 111 = 11.1 g

(d) Mass of 1 mole of Mg = 24 g

∴ Mass of 0.1 mole of Mg = 24 x 0.1 = 2.4 g

#### Question 14(a)

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

1 molecule of Na2CO3.10H2O contains 13 atoms of oxygen

∴ 6.022 × 1023 molecules (ie., 1 mole) has 13 × 6.022 × 1023 atoms

∴ 0.1 mole will have atoms = 0.1 × 13 × 6.022 × 1023

= 7.8 × 1023 atoms

#### Question 14(b)

Calculate the number of gram atoms in 4.6 gram of sodium

Atomic mass of Na = 23

23 g of sodium = 1 gram atom of sodium

∴ 4.6 gram of sodium = $\dfrac{\text{{4.6}}}{\text{{23}}}$

= 0.2 gram atom of sodium

#### Question 14(c)

Calculate the number of moles in 12 g of oxygen gas

32 g of oxygen = 1 mole

∴ 12 g of oxygen = $\dfrac{12}{32}$ = $\dfrac{3}{8}$

= 0.375 mole

#### Question 15

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

1 mole of Sulphur weighs 32 g and contains 6.02 x 1023 atoms

∴ 3.2 g of Sulphur will contain = $\dfrac{6.02 \times 10^{23}}{32} \times 3.2$

= 6.02 x 1022 atoms.

6.02 x 1023 atoms of Ca weighs = 40 g

∴ 6.02 x 1022 atoms of Ca will weigh = $\dfrac{40}{ 6.02 \times 10^{23}}$ x 6.02 x 1022 = 4 g.

#### Question 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

(a) No. of atoms = Moles x 6.022 x 1023

= 52 × 6.022 x 1023 = 3.131 × 1025 atoms

(b) 4 amu = 1 atom of He

∴ 52 amu = $\dfrac{52}{4}$ = 13 atoms of He

(c) Mass of 1 mole of He is 4 g

4 g of He contains 6.022 × 1023 atoms

∴ 52 g of He contains $\dfrac{6.022 \times 10^{23}}{4} \times 52$

= 7.828 × 1024 atoms

#### Question 17

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

Molecular mass of Na2CO3 = 2Na + C + 3O = (2 x 23) + 12 + (3 x 16) = 46 + 12 + 48 = 106 g

(i) 106 g of Na2CO3 has = 2 × 6.022 × 1023 atoms of Na

∴ 5.3 g of Na2CO3 will have = $\dfrac{2 \times 6.022 \times 10^{23} \times 5.3 }{106}$ = 6.022 × 1022 atoms of Na

(ii) 106 g of Na2CO3 has = 6.022 × 1023 atoms of carbon

∴ 5.3 g of Na2CO3 will have = $\dfrac{6.022 \times 10^{23} \times 5.3 }{106}$ = 3.01 × 1022 atoms of carbon

(iii) 106 g of Na2CO3 has 3 x 6.022 × 1023 atoms of oxygen

∴ 5.3 g of Na2CO3 will have = $\dfrac{3 \times 6.022 \times 10^{23} \times 5.3 }{106}$ = 9.03 × 1022 atoms of oxygen

#### Question 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 ]

Molar mass of urea [CO(NH2)2] = 12 + 16 + 2(14 + (2 x 1))

= 28 + 2(16)

= 28 + 32

= 60 g

Molar mass of nitrogen = 2 x 14 = 28 g

60 g urea has mass of nitrogen = 28 g

∴ 5000 g urea will have mass

= $\dfrac{28 \times 5000 }{60}$

= 2333 g = 2.33 kg

#### Question 18(b)

Calculate the volume occupied by 320 g of sulphur dioxide at S.T.P.

[S = 32; O = 16]

Molar mass of sulphur dioxide (SO2) = S + 2O = 32 + (2 x 16) = 32 + 32 = 64 g

64 g of sulphur dioxide has volume = 22.4 litre

∴ 320 g of sulphur dioxide will have volume = $\dfrac{22.4\times 320}{64}$

= 112 litres

#### Question 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 it's vapour density?

(a) Vapour density of carbon dioxide is 22 implies that 1 molecule of carbon dioxide is 22 times heavier than 1 molecule of hydrogen.

(b) Vapour density = $\dfrac{\text{Molecular mass}}{2}$

Molecular mass of chlorine Cl2 = 2Cl = 2 x 35.5 = 71 g

Substituting in formula;

Vapour density = $\dfrac{71}{2}$ = 35.5

Hence, vapour density of Chlorine atom is 35.5

#### Question 20

What is the mass of 56 cm3 of carbon monoxide at S.T.P.?

(C = 12, O = 16)

22400 cm3 of CO has mass = 12 + 16 = 28 g

∴ 56 cm3 will have mass = $\dfrac{28}{22400}$ x 56 = 0.07 g

#### Question 21

Determine the number of molecules in a drop of water which weighs 0.09 g.

Molecular wt. of any substance contain 6.022 × 1023 molecules.

Mass of 1 mole of water is 2H + O = 2 + 16 = 18 g

18 g of H2O contains 6.022 × 1023 molecules

∴ 0.09 g of H2O contains $\dfrac{6.022 \times 10^{23} \times 0.09 }{18}$

= 3.01 × 1021 molecules

#### Question 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?

(a) Mass of 1 mole of S8 = 8S = 8 x 32 = 256 g

∴ Moles in 5.12 g of sulphur = $\dfrac{5.12}{256}$ = 0.02 moles

(b) 1 mole = 6.022 × 1023 molecules

∴ 0.02 moles will have = 0.02 × 6.022 × 1023
= 1.2044 × 10221.2 × 1022 molecules

No. of atoms in 1 molecule of S8 = 8

∴ No. of atoms in 1.2044 × 1022 molecules = 1.2044 x 1022 × 8

= 9.635 × 1022 molecules

#### Question 23

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

Mass of 1 mole of P4 = 4P = 4 x 31 = 124 g

124 g of phosphorus (P4) = 1 mole

∴ 100 g of phosphorus (P4) = $\dfrac{1}{124}$ x 100 = 0.806 moles

#### Question 24

Calculate:

(a) The gram molecular mass of chlorine if 308 cm3 of it at S.T.P. weighs 0.979 g

(b) The volume of 4 g of H2 at 4 atmospheres.

(c) The mass of oxygen in 2.2 litres of CO2 at S.T.P.

(a) The mass of 22.4 L of a gas at S.T.P. is equal to it's gram molecular mass.

308 cm3 of chlorine weighs 0.979 g

∴ 22,400 cm3 of chlorine will weigh

= $\dfrac{0.979}{308}$ × 22400 = 71.2 g

(b) Molar mass of H2 = 2H = 2 x 1 = 2 g

2g H2 at 1 atm has volume = 22.4 dm3

∴ 4 g H2 at 1 atm will have volume 2 x 22.4 = 44.8 dm3

Now, For 4 g H2

P1 = 1 atm, V1 = 44.8 dm3

P2 = 4 atm, V2 = ?

Using formula P1V1 = P2V2

$\text{V}_2 = \dfrac{\text{P}_1\text{V}_1}{\text{P}_2} \\[1em] \text{V}_2 = \dfrac{1 \times 44.8}{4} \\[1em] = \bold{11.2} \space \bold{dm^3}$

(c) Molar mass of oxygen in carbon dioxide = 2O = 2 x 16 = 32 g

Mass of oxygen in 22.4 litres of CO2 = 32 g

∴ Mass of oxygen in 2.2 litres of CO2

= $\dfrac{32}{22.4}$ x 2.2 = 3.14 g

#### Question 25

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.

No. of atoms in 12 g C = 6.022 × 1023

∴ no. of carbon atoms in 10-12 g

$\dfrac{6.022 \times 10^{23}}{12}$ x 10-12

= 5.019 × 1010 atoms

#### Question 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?

Given:

P = 1140 mm Hg

Density = D = 3 g per L

T = 273 °C = 273 + 273 = 546 K

gram molecular mass = ?

At S.T.P., the volume of one mole of any gas is 22.4 L

Volume of unknown gas at S.T.P. = ?

By Charle’s law.

V1 = 1 L

T1 = 546 K

T2 = 273 K

V2 = ?

$\dfrac{\text{V}_1}{\text{T}_1}$ = $\dfrac{\text{V}_2}{\text{T}_2}$

Hence, V2 = $\dfrac{1}{546}$ x 273 = 0.5 L

Volume at standard pressure = ?

Apply Boyle’s law.

P1 = 1140 mm Hg

V1 = 0.5 L

P2 = 760 mm Hg

V2 = ?

P1 × V1 = P2 × V2

V2 = $\dfrac{1140 \times 0.5}{760}$ = 0.75 L

Now,

22.4 L = 1 mole of any gas at S.T.P.,

then 0.75 L = $\dfrac{0.75}{22.4}$

= 0.0335 moles

The original mass is 3 g

Molecular mass = $\dfrac{\text{Mass of compound}}{\text{Moles of compound}}$

= $\dfrac{3}{0.0335 }$ = 89.55 ≈ 89.6 g per mole

Hence, the gram molecular mass of the unknown gas is 89.6g

#### Question 27

Cost of Sugar (C12H22O11) is ₹40 per kg; calculate it's cost per mole.

Molar mass of C12H22O11 = 12C + 22H + 11O = (12 x 12) + (22 x 1) + (11 x 16) = 144 + 22 + 176 = 342 g

1000 g of sugar costs = ₹40

∴ 342 g of sugar will cost = $\dfrac{40}{1000}$ x 342 = ₹13.68 per mole

#### Question 28

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) 7 g of silver

7 g of silver

Reason

(a) Weight of 1 g atom of N = 14 g

∴ weight of 2 g atom of N = 28 g

(b) 6.022 × 1023 atoms of C weigh = 12 g

∴ 3 × 1025 atoms will weigh = $\dfrac{12}{6.022 \times 10^{23}}$ × 3 × 1025 = 597.7 g

(c) 1 mole of sulphur weighs = 32 g

(d) 7 g of silver

The weight computed in all other options is greater than the weight in option (d). Hence, 7 grams of silver weighs the least.

#### Question 29

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

6.02 × 1022 molecule

Reason

Molar mass of NaOH = Na + O + H = 23 + 16 + 1 = 40 g

40 g of NaOH contains 6.022 × 1023 molecules

∴ 4 g of NaOH contains

= $\dfrac{6.022 \times 10^{23}}{40}$ x 4

= 6.02 × 1022 molecules

#### Question 30

The number of molecules in 4.25 g of ammonia is:

(a) 1.0 × 1023

(b) 1.5 × 1023

(c) 2.0 × 1023

(d) 3.5 × 1023

1.5 × 1023

Reason

Molar mass of ammonia = N + 3H = 14 + 3 = 17 g

The number of molecules in 17 g of ammonia = 6.022 × 1023

∴ No. of molecules in 4.25 g of ammonia

= $\dfrac{6.022 \times 10^{23}}{17}$ x 4.25

= 1.5 × 1023

#### Question 31

Correct the statements, if required

(a) One mole of chlorine contains 6.023 × 1023 atoms 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 $\dfrac{1}{12}$ 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.

(a) One mole of chlorine contains 6.022 × 1023 atoms of chlorine.

(b) Under similar conditions of temperature and pressure, four 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 atom of an element is heavier than $\dfrac{1}{12}$ 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 molecules.

## Exercise 5C

#### Question 1

Give three kinds of information conveyed by the formula H2O.

Information conveyed by formula [H2O] —

1. One molecule of water (H2O) is made of two atoms of Hydrogen and one atom of Oxygen.
2. As atomic weight of hydrogen is 1 and that of oxygen is 16. Therefore, ratio by weight of hydrogen and oxygen is $\dfrac{2\text{H}}{\text{O}}$ = $\dfrac{2}{16}$ = $\dfrac{1}{8}$
3. Molecular weight of H2O is 2H + O = 2 + 16 = 18g.

#### Question 2

Explain the terms empirical formula and molecular formula.

The empirical formula of a compound is the simplest formula, which gives the simplest ratio in whole numbers of atoms of different elements present in one molecule of the compound.

The molecular formula of a compound denotes the actual number of atoms of different elements present in one molecule of a compound.

#### Question 3

Give the empirical formula of:

(a) C6H6

(b) C6H18O3

(c) C2H2

(d) CH3COOH

(a) Molecular formula is C6H6

∴ Ratio of C and H is 6 : 6

Simple ratio is 1 : 1

Hence, empirical formula = CH

(b) Molecular formula is C6H18O3

∴ Ratio of C, H and O is 6 : 18 : 3

Simple ratio is 2 : 6 : 1

Hence, empirical formula = C2H6O

(c) Molecular formula is C2H2

∴ Ratio of C and H is 2 : 2

Simple ratio is 1 : 1

Hence, empirical formula = CH

(d) Molecular formula is CH3COOH i.e. C2H4O2

∴ Ratio of C, H and O is 2 : 4 : 2

Simple ratio is 1 : 2 : 1

Hence, empirical formula = CH2O

#### Question 4

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

Relative molecular mass of CuSO4.5H2O

= 64 + 32 + (4×16) + [5(2+16)]

= 96 + 64 + 90 = 250

250 g of CuSO4.5H2O contains 90 g of water of crystallisation

∴ 100 g of CuSO4.5H2O contains

= $\dfrac{90}{250}$ x 100 = 36%

#### Question 5

Calculate the percentage of phosphorus in

(a) Calcium hydrogen phosphate Ca(H2PO4)2

(b) Calcium phosphate Ca3(PO4)2

(a) Molecular mass of Ca(H2PO4)2

= Ca + 2[2H + P + 4O]

= 40 + 2[2(1) + 31 + 4(16)]

= 40 + 2[2 + 31 + 64]

= 40 + 194

= 234

234 g of Ca(H2PO4)2 contains 62 g of P

∴ 100 g of Ca(H2PO4)2 contains

= $\dfrac{62}{234}$ x 100 = 26.5%

(b) Molecular mass of Ca3(PO4)2

= 3Ca + 2[P + 4O]

= (3 x 40) + 2[31 + 4(16)]

= 120 + 2[31 + 64]

= 120 + 190

= 310

310 g of Ca3(PO4)2 contains 62 g of P

∴ 100 g of Ca3(PO4)2 contains

= $\dfrac{62}{310}$ x 100 = 20 %

#### Question 6

Calculate the percent composition of Potassium chlorate KClO3.

Molecular mass of KClO3

= K + Cl + 3O

= 39 + 35.5 + (3 x 16)

= 39 + 35.5 + 48

= 122.5 g

% of K = ?

Since, 122.5 g of KClO3 contains 39 g of K

∴ 100 g of KClO3 contains

= $\dfrac{39}{122.5}$ x 100

= 31.83%

Similarly, 122.5 g of KClO3 contains 35.5 g of Cl

∴ 100 g of KClO3 contains

= $\dfrac{35.5}{122.5}$ x 100

= 28.98%

And, 122.5 g of KClO3 contains 48 g of O

∴ 100 g of KClO3 contains

= $\dfrac{48}{122.5}$ x 100

= 39.18%

#### Question 7

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

Pb = 62.5%, N = 8.5%, O = 29.0%

Element% compositionAt. massRelative no. of atomsSimplest ratio
Pb62.5207$\dfrac{62.5 }{207}$ = 0.301$\dfrac{0.301}{0.301 }$ = 1
N8.514$\dfrac{8.5}{14}$ = 0.607$\dfrac{0.607}{0.301 }$ = 2
O2916$\dfrac{29}{16}$ = 1.81$\dfrac{1.81}{0.301 }$ = 6

Hence, Simplest ratio of whole numbers = Pb : N : O = 1 : 2 : 6

Hence, empirical formula is Pb(NO3)2.

#### Question 8

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

Atomic wt. of Fe = 56 and O = 16

Molecular mass of Fe2O3 = 2Fe + 3O

=(2 x 56) + (3 x 16)

= 112 + 48

= 160 g

Iron present in 80% of Fe2O3 = $\dfrac{112}{160}$ x 80

= 56 g

∴ Mass of iron in 100 g of iron ore = 56 g

Hence, mass of iron present in 10 kg (i.e., 10,000 g) of iron ore = $\dfrac{56}{100}$ x 10000

= 5600 g = 5.6 kg

#### Question 9

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

Empirical formula is CH

Empirical formula weight = 12 + 1 = 13

Vapour density (V.D.) = 13

Molecular weight = 2 x V.D. = 2 x 13

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{2 \times 13}{13} = 2$

∴ Molecular formula = n[E.F.] = 2[CH] = C2H2

Similarly,

Empirical formula is CH

Empirical formula weight = 12 + 1 = 13

Vapour density (V.D.) = 39

Molecular weight = 2 x V.D. = 2 x 39

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{2 \times 39}{13} = 6$

∴ Molecular formula = n[E.F.] = 6[CH] = C6H6

#### Question 10

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

Element% compositionAt. wt.Relative no. of atomsSimplest ratio
Nitrogen82.3514$\dfrac{82.35 }{14}$ = 5.88$\dfrac{5.88}{5.88}$ = 1
Hydrogen17.641$\dfrac{17.64}{1}$ = 17.64$\dfrac{17.64}{5.88}$ = 3

Simplest ratio of whole numbers = N : H = 1 : 3

Hence, empirical formula is NH3

#### Question 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) it's empirical formula, and

(b) it's molecular formula

Element% compositionAt. wt.Relative no. of atomsSimplest ratio
Carbon54.5412$\dfrac{54.54}{12}$ = 4.545$\dfrac{ 4.545 }{2.275 }$ = 1.99 = 2
Hydrogen9.091$\dfrac{9.09 }{1}$ = 9.09$\dfrac{9.09 }{ 2.275 }$ = 3.99 = 4
Oxygen36.3616$\dfrac{36.36}{16}$ = 2.275$\dfrac{2.275 }{ 2.275 }$ = 1

Simplest ratio of whole numbers = C : H : O = 2 : 4 : 1

Hence, empirical formula is C2H4O

Empirical formula weight = 2(12) + 4(1) + 16 = 24 + 4 + 16 = 44

V.D. = 44

Molecular weight = 2 x V.D. = 2 x 44 = 88

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{88}{44} = 2$

So, molecular formula = (C2H4O)2 = C4H8O2

#### Question 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) it's empirical formula, and

(b) it's molecular formula

Element% compositionAt. wt.Relative no. of atomsSimplest ratio
Hydrogen2.221$\dfrac{2.22}{1}$ = 2.22$\dfrac{2.22 }{2.21}$ = 1
Oxygen71.1916$\dfrac{71.19}{16}$ = 4.44$\dfrac{4.44}{2.21 }$ = 2
Carbon26.5912$\dfrac{26.59}{12}$ = 2.21$\dfrac{ 2.21 }{2.21}$ = 1

Simplest ratio of whole numbers = H : O : C = 1 : 2 : 1

Hence, empirical formula is CHO2

Empirical formula weight = 12 + 1 + (2 x 16) = 13 + 32 = 45

V.D. = 45

Molecular weight = 2 x V.D. = 2 x 45 = 90

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{90}{45} = 2$

So, molecular formula = 2(CHO2) = C2H2O4

#### Question 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

Element% compositionAt. wt.Relative no. of atomsSimplest ratio
Hydrogen4.071$\dfrac{4.07 }{1}$ = 4.07$\dfrac{4.07 }{2.01}$ = 2
chlorine71.6535.5$\dfrac{ 71.65}{35.5}$ = 2.01$\dfrac{2.01}{2.01 }$ = 1
Carbon24.2812$\dfrac{24.28 }{12}$ = 2.02$\dfrac{ 2.02 }{2.01}$ = 1

Simplest ratio of whole numbers = H : Cl : C = 2 : 1 : 1

Hence, empirical formula is CH2Cl

Empirical formula weight = 12 + (2 x 1) + 35.5 = 49.5

molar mass = 98.96

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{98.96}{49.5} = 1.99 = 2$

So, molecular formula = 2(CH2Cl) = C2H4Cl2

#### Question 14

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

(a) the g atom of each

(b) find the empirical formula

(c) find molecular formula, if it's vapour density is 29.

(a) Given, hydrocarbon contains 4.8 g of carbon per gram of hydrogen

Gram atom = $\dfrac{\text{Mass of element}}{\text{Atomic mass}}$

∴ g atom of carbon = $\dfrac{4.8}{12}$ = 0.4 and

g atom of hydrogen = $\dfrac{1}{1}$ = 1

(b)

ElementMassAt. wt.Gram atomsSimplest ratio
Hydrogen11$\dfrac{1 }{1}$ = 1$\dfrac{1 }{0.4 }$ = $\dfrac{5 }{2}$
Carbon4.812$\dfrac{4.8 }{12}$ = 0.4$\dfrac{ 0.4 }{0.4 }$ = 1

Simplest ratio of whole numbers = H : C = $\dfrac{5 }{2}$ : 1 = 5 : 2

Hence, empirical formula is C2H5

(c) Empirical formula weight = (2 x 12) + (5 x 1) = 24 + 5 = 29

V.D. = 29

Molecular weight = 2 x V.D. = 2 x 29 = 58

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{58}{29} = 2$

So, molecular formula = 2(C2H5) = C4H10

#### Question 15

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

Gram atom = $\dfrac{\text{Mass of element}}{\text{Atomic mass}}$

g atom of silicon = 0.2 = $\dfrac{\text{Mass of silicon}}{28}$

∴ Mass of silicon = 5.6 g and

Mass of chlorine = 21.3 g

ElementMassAt. wt.gram atomsSimplest ratio
Silicon5.628$\dfrac{5.6 }{28}$ = 0.2$\dfrac{0.2 }{0.2}$ = 1
Chlorine21.335.5$\dfrac{ 21.3}{35.5}$ = 0.6$\dfrac{ 0.6 }{0.2}$ = 3

Simplest ratio of whole numbers = Si : Cl = 1 : 3

Hence, empirical formula is SiCl3

#### Question 16

A gaseous hydrocarbon contains 82.76% of carbon. Given that it's vapour density is 29, find it's molecular formula.

Element% compositionAt. wt.Relative no. of atomsSimplest ratio
Carbon82.7612$\dfrac{82.76}{12}$ = 6.89$\dfrac{6.89}{6.89}$ = 1
Hydrogen17.241$\dfrac{17.24}{1}$ = 17.24$\dfrac{17.24}{6.89}$ = $\dfrac{5 }{2}$

Simplest ratio of whole numbers = C : H = 1 : $\dfrac{5 }{2}$ = 2 : 5

Hence, empirical formula is C2H5

Empirical formula weight = 2(12) + 5(1) = 29

V.D. = 29

Molecular weight = 2 x V.D. = 2 x 29

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{2 \times 29}{29} =2$

∴ Molecular formula = n[E.F.] = 2[C2H5] = C4H10

#### Question 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.

(a) Gram atom = $\dfrac{\text{Mass of element}}{\text{Atomic mass}}$

∴ g atom of magnesium = $\dfrac{18}{24}$ = $\dfrac{3}{4}$

Hence, $\dfrac{3}{4}$ gram atoms of magnesium are equal to 18g of magnesium.

(b) g atom of nitrogen = $\dfrac{7}{14}$ = $\dfrac{1}{2}$

Hence, $\dfrac{1}{2}$ gram atoms of nitrogen are equal to 7g of nitrogen.

(c) simple ratio of gram-atoms of magnesium to gram-atoms of nitrogen

= $\dfrac{\dfrac{3}{4}}{\dfrac{1}{2}}$ = $\dfrac{3}{2}$ = magnesium : nitrogen

So, the formula is Mg3N2

#### Question 18

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

Barium chloride = BaCl2.xH2O

Molecular weight of BaCl2.xH2O = Ba + 2Cl + x(2H + O)

= 137 + (2 x 35.5) + x(2+16)

= 137 + (2 x 35.5) + x(2+16)

= 137 + 71 + 18x

= (208 + 18 x)

(208 + 18 x) contains 14.8% of water of crystallisation in BaCl2.x H2O

∴ 14.8% of (208 + 18 x) = 18x

$\Big[\dfrac{14.8}{100}\Big]$ x [208 + 18 x] = 18x

[0.148 x 208 ] + [0.148 x 18x] = 18x

30.784 = 18x - [0.148 x 18x]

30.784 = 18x - 2.664x

30.784 = 15.336x

x = $\dfrac{30.784}{15.336}$ = 2

Hence, Barium chloride crystals contain 2 molecules of water of crystallisation per molecule.

#### Question 19

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

Molar mass of urea (CON2H4) = 12 + 16 + 28 + 4 = 60 g

Molar mass of nitrogen (N2) = 2 x 14 = 28 g

60 g urea has mass of nitrogen = 28 g

∴ 100 g urea will have mass

= $\dfrac{28 \times 100 }{60}$

= 46.67%

#### Question 20

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

Element% compositionAt. wt.Relative no. of atomsSimplest ratio
Oxygen51.4216$\dfrac{51.42 }{16}$ = 3.21$\dfrac{3.21}{3.21}$ = 1
Hydrogen6.481$\dfrac{6.48}{1}$ = 6.48$\dfrac{6.48}{3.21}$ = 2
Carbon42.112$\dfrac{42.1}{12}$ = 3.50$\dfrac{3.50}{3.21}$ = 1

Simplest ratio of whole numbers = O : H : C = 1 : 2 : 1

Hence, empirical formula is CH2O

Since the compound has 12 atoms of carbon, so the formula is C12H24O12.

#### Question 21(a)

A compound with empirical formula AB2, has the vapour density equal to it's empirical formula weight. Find it's molecular formula.

Empirical formula = AB2

Empirical formula weight = V.D.

Molecular weight = 2 x V.D.

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] \text{n} = \dfrac{\text{2 x V.D.}}{\text{V.D.}} \\[0.5em] \text{n} = 2$

∴ Molecular formula = n[E.F.] = 2[AB2] = A2B4

#### Question 21(b)

A compound with empirical formula AB has vapour density three times it's empirical formula weight. Find the molecular formula.

Given,

Empirical formula = AB

V.D. = 3 x Empirical formula weight

Hence, Empirical formula weight = $\dfrac{\text{V.D.}}{3}$

and we know, Molecular weight = 2 x V.D.

Substituting in the formula for n we get,

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] \text{n} = \dfrac{\text{ 2 x V.D.}}{\dfrac{\text{V.D.}}{3}} \\[0.5em] \text{n} = \dfrac{\text{ 3 x 2 x V.D.}}{\text{V.D.}} \\[0.5em] \text{n} = 6$

∴ Molecular formula = n[E.F.] = 6[AB] = A6B6

#### Question 21(c)

10.47 g of a compound contains 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)

Element% compositionAt. wt.Relative no. of atomsSimplest ratio
Metal A6.25207$\dfrac{6.25 }{207}$ = 0.03$\dfrac{0.03}{0.03}$ = 1
Non-metal B4.2235.5$\dfrac{4.22}{35.5}$ = 0.11$\dfrac{0.11}{0.03}$ = 3.96 = 4

Simplest ratio of whole numbers = A : B = 1 : 4

Hence, empirical formula is AB4

#### Question 22

A hydride of nitrogen contains 87.5% percent by mass of nitrogen. Determine the empirical formula of this compound.

Element% compositionAt. wt.Relative no. of atomsSimplest ratio
Nitrogen87.514$\dfrac{87.5 }{14}$ = 6.25$\dfrac{6.25}{6.25 }$ = 1
Hydrogen12.51$\dfrac{12.5}{1}$ = 12.5$\dfrac{12.5}{6.25 }$ = 2

Simplest ratio of whole numbers = N : H = 1 : 2

Hence, empirical formula is NH2

#### Question 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.

Element% compositionAt. wt.Relative no. of atomsSimplest ratio
Zn22.6565$\dfrac{22.65}{65}$ = 0.3484$\dfrac{0.3484}{0.3484}$ = 1
S11.1532$\dfrac{11.15}{32}$ = 0.3484$\dfrac{0.3484}{0.3484}$ = 1
O61.3216$\dfrac{61.32 }{16}$ = 3.832$\dfrac{3.832}{0.3484}$ = 10.99 = 11
H4.881$\dfrac{4.88}{1}$ = 4.88$\dfrac{4.88 }{0.3484}$ = 14

Simplest ratio of whole numbers = Zn : S : O : H = 1 : 1 : 11 : 14

Hence, empirical formula is ZnSO11H14

Molecular weight = 287

Empirical formula weight = 65 + 32 + 11(16) + 14(1) = 65 + 32 + 176 + 14 = 287

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{287}{287} = 1$

Molecular formula = n[E.F.] = 1[ZnSO11H14] = ZnSO11H14

Since all the hydrogen in the compound is in combination with oxygen as water of crystallization .

Therefore, 14 atoms of hydrogen and 7 atoms of oxygen = 7H2O and hence, 4 atoms of oxygen remain.

Molecular formula is ZnSO4.7H2O.

## Exercise 5D

#### Question 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: 42g + ............... ⟶ ............... + ...............

(c) Molecules: 6.02 x 1023 + ............... ⟶ ............... + ...............

(a) Moles: 1 mole + 2 mole1 mole + 2 mole

(b) Grams: 42g + 36g 74g + 4g

(c) Molecules: 6.02 x 1023 + 12.04 × 10236.02 x 1023 + 12.04 × 1023

#### Question 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 S.T.P.

(a)

$\begin{matrix} \text{CaCO}_3 & + \space 2\text{HNO}_3 \longrightarrow & \text{Ca(NO}_3)_2 & + \space \text{H}_2\text{O} \space + \text{CO}_2 \\ 40 + 12 + 3(16) & & 40 + 2(14) + 6(16) \\ = 40 + 12 + 48 & & 40 + 28 + 96 \\ = 100 \text{ g} & & 164 \text{ g} \end{matrix}$

100 g of CaCO3 produces = 164 g of Ca(NO3)2

∴ 15 g CaCO3 will produce = $\dfrac{164}{100}$ x 15

= 24.6 g

Hence, mass of anhydrous calcium nitrate formed = 24.6 g

(b) 100 g of CaCO3 produces = 22.4 litres of carbon dioxide

∴ 15 g of CaCO3 will produce = $\dfrac{22.4}{100}$ x 15

= 3.36 litres of CO2

#### Question 3

66g of 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 S.T.P.

(c) The mass of acid required.

(a) $\begin{matrix} 2\text{NH}_3 & + &\text{H}_2\text{SO}_4 & \longrightarrow & \text{(NH}_4)_2\text{SO}_4 \\ 2[14 + 3(1)] & & 2(1) + 32 + 4(16) & & 2[14 + 4(1)] + 32 + 4(16) \\ = (2 \times 17) & & = 2 + 32 + 64 & & = 36 + 32 + 64 \\ = 34 \text{ g} & & = 98 \text{ g} & & = 132 \text{ g} \\ 2\text{ mole} \end{matrix}$

132 g ammonium sulphate is produced by 34 g of NH3

∴66 g ammonium sulphate is produced by $\dfrac{34}{132}$ x 66 = 17 g of NH3

Hence, 17g of NH3 is required.

(b) 132 g ammonium sulphate uses 2 x 22.4 L of gas

∴ 66 g of ammonium sulphate will use $\dfrac{2 \times 22.4}{132}$ x 66 = 22.4 litres

(c) For 132 g ammonium sulphate 98 g of acid is required

∴ For 66 g ammonium sulphate $\dfrac{98}{132}$ x 66 = 49 g

Hence, 49g of acid is required.

#### Question 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 the chlorine and

(c) the volume of chlorine evolved at S.T.P.

(a)

$\begin{matrix} \text{Pb}_3\text{O}_4 & + &8\text{HCl} & \longrightarrow \\ 3(207) + 4(16) & & 8[1 + 35.5] & \\ = 621 + 64 & & = 8(36.5) & \\ = 685 \text{ g} & & = 292 \text{ g} & \\ & & & \\ 3\text{PbCl}_2 & + & 4\text{H}_2\text{O} & + & \text{Cl}_2 \\ 3[207 + 2(35.5)] &&&& 2(35.5) \\ = 3[207 + 71] &&&& = 71\text{g} \\ = 834 \text{ g} \end{matrix}$

685 g of Pb3O4 gives = 834 g of PbCl2

∴ 6.85 g of Pb3O4 will give

= $\dfrac{834}{685}$ x 6.85 = 8.34 g

(b) 685g of Pb3O4 gives = 71g of Cl2

∴ 6.85 g of Pb3O4 will give

= $\dfrac{71}{685}$ x 6.85 = 0.71 g of Cl2

(c) 685 g of Pb3O4 produces 22.4 L of Cl2

∴ 6.85 g of Pb3O4 will produce

$\dfrac{22.4}{685}$ x 6.85 = 0.224 L of Cl2

#### Question 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

$\begin{matrix} \text{KNO}_3 & + &\text{H}_2\text{SO}_4 & \longrightarrow & \text{KHSO}_4 & + \text{HNO}_3 \\ 39 + 14 + 3(16) & & & & & 1 + 14 + 3(16) \\ = 39 + 14 + 48 & & & & & = 1 + 14 + 48 \\ = 101 \text{ g} & & & & & = 63 \text{ g} \end{matrix}$

63 g of HNO3 is formed by 101 g of KNO3

∴ 126000 g of HNO3 is formed by $\dfrac{101}{63}$ x 126000
= 202000 g = 202 kg

Similarly,

$\begin{matrix} \text{NaNO}_3 & + &\text{H}_2\text{SO}_4 & \longrightarrow & \text{NaHSO}_4 & + \text{HNO}_3 \\ 23 + 14 + 3(16) & & & & & 1 + 14 + 3(16) \\ = 23 + 14 + 48 & & & & & = 1 + 14 + 48 \\ = 85 \text{ g} & & & & & = 63 \text{ g} \end{matrix}$

63 g of HNO3 is formed by 85 g of NaNO3

∴ 126000 g of HNO3 is formed by $\dfrac{85}{63}$ x 126000
= 170000 g = 170 kg

So, a smaller mass of NaNO3 is required.

#### Question 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

(a) Given,

$\begin{matrix} \text{CaCO}_3 & + &2\text{HCl} & \longrightarrow & \text{CaCl}_2 & + \text{H}_2\text{O} + &\text{CO}_2 \\ 40 + 12 + 3(16) & & 2[1 + 35.5] & && & 1\text{ mole} \\ = 40 + 12 + 48 && = 73 \text{ g} \\ = 100 \text{ g} \\ \end{matrix}$

First convert the volume of carbon dioxide to STP:

V1 = 2 L

T1 = 27 + 273 K = 300 K

T2 = 273 K

V2 = ?

Using formula:

$\dfrac{\text{V}_1}{\text{T}_1}$ = $\dfrac{\text{V}_2}{\text{T}_2}$

Substituting in the formula,

$\dfrac{2}{300}$ = $\dfrac{\text{V}_2}{273}$

V2 = $\dfrac{2}{300}$ x 273 = 1.82 L

As, 22.4 L of carbon dioxide is obtained using 100 g CaCO3

∴ 1.82 L of carbon dioxide is obtained from $\dfrac{100}{22.4}$ x 1.82

= 8.125 g of CaCO3

(b) Similarly, 22.4 L of carbon dioxide is obtained using 73 g of acid

∴ 1.82 L of carbon dioxide is obtained from $\dfrac{73}{22.4}$ x 1.82

= 5.93 g of acid

#### Question 7

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

$\begin{matrix} 2\text{H}_2\text{O} & \longrightarrow 2\text{H}_2 \space + & \text{O}_2 \\ 2(2 + 16) & & 2(16) \\ 36 \text{g} & & 32 \text{g} \\ \end{matrix}$

36 g of water produces 32 g of O2

∴ 18 g of water will produced

= $\dfrac{32}{36}$ x 18 = 16 g of O2

$\begin{matrix} 2\text{H}_2\text{O} & \longrightarrow 2\text{H}_2 \space + & \text{O}_2 \\ 2\text{ mole} & & 1\text{ mole} \\ \end{matrix}$

2 moles of water produces 1 mole of oxygen

∴ 1 mole of water will produce $\dfrac{1}{2} \times 1$ = 0.5 moles of O2

1 mole of O2 occupies 22.4 L volume

∴ 0.5 moles will occupy = 22.4 × 0.5

= 11.2 L

#### Question 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 S.T.P.

(c) mass of oxygen liberated.

$\begin{matrix} 2\text{Na}_2\text{O}_2 & + \space 2\text{H}_2\text{O} \longrightarrow & 4\text{NaOH}& & + & \text{O}_2 \\ 2[2(23) + 2(16)] & & 4(23 + 16 + 1) & & & 1 \text{mole} \\ 156 \text{g} & & 160\text{g}& & &32 \text{g} \\ \end{matrix}$

(a) 156 g of sodium peroxide produces 160 g of sodium hydroxide

∴ 1.56 g of sodium peroxide will produce $\dfrac{160}{156}$ x 1.56

= 1.6 g of sodium hydroxide

(b) 156 g of sodium peroxide produces 22.4 L of oxygen

∴ 1.56 g of sodium peroxide will produce $\dfrac{22.4}{156}$ x 1.56

= 0.224 L

Converting L to cm3

As 1 L = 1000 cm3

So, 0.224 L = 224 cm3

(c) 156 g of sodium peroxide produces 32 g of oxygen

∴ 1.56 g of sodium peroxide will produce $\dfrac{32}{156}$ x 1.56 = 0.32 g

#### Question 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 S.T.P?

$\begin{matrix} 2\text{NH}_4\text{Cl} & + \text{ Ca(OH)}_2 \longrightarrow \text{CaCl}_2 \space + \space 2\text{H}_2\text{O} \space + & 2\text{NH}_3 \\ 2[14 + 4(1) + 35.5] & & 2[14 + 3(1)] \\ 107 \text{g} & &34\text{g} \\ & &2\text{ mole} \\ \end{matrix}$

(a) 107 g NH4Cl gives 34 g of NH3

∴ 21.4 g NH4Cl will give $\dfrac{34}{107}$ x 21.4

= 6.8 g of NH3

(b) Volume of ammonia produced by 107 g NH4Cl = 2 x 22.4 L

∴ Volume of ammonia produced by 21.4 g NH4Cl = $\dfrac{2 \times 22.4}{107}$ x 21.4

= 8.96 L

#### Question 10

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 12g of aluminium carbide?

(b) What volume of methane is obtained from 12g of aluminium carbide?

$\begin{matrix} \text{Al}_4\text{C}_3 & + & 12\text{H}_2\text{O} & \longrightarrow & 4\text{Al(OH)}_3 & + & 3\text{CH}_4 \\ 4(27) + 3(12) & & & & 4(78) & & 3(22.4) \\ = 144 \text{ g} & & & & = 312 \text{ g} & & = 67.2 \text{ lit.} \\ \end{matrix}$

144 g of aluminium carbide forms 312 g of aluminium hydroxide.

∴ 12 g of aluminium carbide will form $\dfrac{312}{144}$ x 12 = 26 g of aluminium hydroxide

Hence, 26 g of aluminium hydroxide is formed.

(ii) 144 g of aluminium carbide forms 67.2 lit of methane.

∴ 12 g of aluminium carbide will form $\dfrac{67.2}{144}$ x 12 = 5.6 lit.

Hence, vol. of methane obtained = 5.6 L

#### Question 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 S.T.P.,

(g) moles of acid required,

(h) Mass of acid required.

$\begin{matrix} \text{MnO}_2 & + & 4\text{HCl}& \longrightarrow & \text{MnCl}_2 & + & 2\text{H}_2\text{O} & + & \text{Cl}_2 \\ 1 \text{ mole} && 4 \text{ mole} && 1\text{ mole}&&&& 1\text{ mole} \\ 55 +2(16) && 4[1 + 35.5] & & 55 + 2(35.5) & & &&2(35.5) \\ = 87 \text{ g} & & = 146 \text{ g} & & = 126 \text{ g} & & & & = 71\text{ g} \\ \end{matrix}$

(a) 1 mole of MnO2 weighs 87 g

∴ 0.02 mole will weigh $\dfrac{87}{1}$ x 0.02 = 1.74 g

(b) 1 mole MnO2 gives 1 mole of MnCl2

∴ 0.02 mole MnO2 will give 0.02 mole of MnCl2

(c) As, 1 mole MnCl2 weighs 126 g

∴ 0.02 mole MnCl2 will weigh $\dfrac{126}{1}$ x 0.02 = 2.52 g

(d) 1 mole MnO2 gives 1 mole of Cl2

∴ 0.02 mole MnO2will form 0.02 moles of Cl2

(e) 1 mole of Cl2 weighs 71 g

∴ 0.02 mole will weigh $\dfrac{71}{1}$ x 0.02 = 1.42 g

(f) 1 mole of chlorine gas has volume 22.4 dm3

∴ 0.02 mole will have volume $\dfrac{22.4}{1}$ x 0.02 = 0.448 dm3

(g) 1 mole MnO2 requires 4 moles of HCl

∴ 0.02 mole MnO2 will require $\dfrac{4}{1}$ x 0.02 = 0.08 mole

(e) Mass of 1 mole of HCl = 36.5 g

∴ Mass of 0.08 mole = 0.08 × 36.5 = 2.92 g

#### Question 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 it's mass?

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

$\begin{matrix} \text{N}_2 & + & 3\text{H}_2& \longrightarrow & 2\text{NH}_3 \\ 2(14) && 6(1) & & 2[14 + 3(1)] \\ 28 \text{g} & & 6 \text{g} & & 34 \text{ g} \\ \end{matrix}$

(a) 28 g of nitrogen requires 6 g of hydrogen

∴ 2000 g of nitrogen requires $\dfrac{6}{28}$ x 2000

= 428.57 g of hydrogen.

So mass of hydrogen left unreacted = 1000 - 428.57 = 571.42 g

571.42 g of hydrogen is left unreacted.

(b) 28 g of nitrogen forms 34 g NH3

∴ 2000 g of nitrogen forms $\dfrac{34}{28}$ x 2000

= 2428.57 g of NH3