Mole Concept and Stoichiometry

State:

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

(b) Avogadro's law

Answer

(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."

(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.

Answer

(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

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.

Answer

(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

According to Avogadro's Law:
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.

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

Answer

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.

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

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

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

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

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 .

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

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

[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

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

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

(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.

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

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

[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.

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

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

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.

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

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

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, oxygen is 1⁄5^th 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

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

Answer

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

[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

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

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

[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.

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

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

[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

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.

Answer

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.

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

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

[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

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

Answer

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.

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)

Answer

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

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?

Justify your answers.

Answer

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.

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.

Answer

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.

(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

Answer

(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.

(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.?

Answer

(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.

Define or explain the terms:

(a) Vapour density

(b) Molar volume

(c) Relative atomic mass

(d) Relative molecular mass

(e) Avogadro's number

(f) Gram atom

(g) Mole

Answer

(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.

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

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

Answer

(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.

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

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

Answer

(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.

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.

Answer

(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.

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

Answer

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.

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

Answer

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.

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

Answer

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

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

Answer

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

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

Answer

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

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

Answer

1 mole of aluminium has mass = 27 g

0.2 mole of aluminium has mass

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

= 5.4 g

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

Answer

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

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

Answer

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

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

Answer

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

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?

Answer

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

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

Answer

1 mole of SO₂ has volume = 22.4 litres

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

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

Answer

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

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

Answer

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

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

Answer

(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

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)

Answer

(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

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

Answer

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

Calculate the number of gram atoms in 4.6 gram of sodium

Answer

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

Calculate the number of moles in 12 g of oxygen gas

Answer

32 g of oxygen = 1 mole

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

= 0.375 mole

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

Answer

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.

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) 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

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

Answer

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

Calculate the mass of nitrogen supplied to soil by 5 kg of urea [CO(NH₂)₂]

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

Answer

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

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

[S = 32; O = 16]

Answer

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

(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?

Answer

(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

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

(C = 12, O = 16)

Answer

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

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

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

Answer

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

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

(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 × 10²² ≈ 1.2 × 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

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

Answer

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

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.

Answer

(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

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

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

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

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

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

Answer

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

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

Answer

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.

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

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

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

Answer

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

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.

Answer

(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.

Give three kinds of information conveyed by the formula H₂O.

Answer

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.

Explain the terms empirical formula and molecular formula.

Answer

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.

Give the empirical formula of:

(a) C₆H₆

(b) C₆H₁₈O₃

(c) C₂H₂

(d) CH₃COOH

Answer

(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

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

Answer

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%

Calculate the percentage of phosphorus in

(a) Calcium hydrogen phosphate Ca(H₂PO₄)₂

(b) Calcium phosphate Ca₃(PO₄)₂

Answer

(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 %

Calculate the percent composition of Potassium chlorate KClO₃.

Answer

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%

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

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

Answer

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

Hence, empirical formula is Pb(NO₃)₂.

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

Answer

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

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

Answer

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

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

Answer

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

Hence, empirical formula is NH₃

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

Answer

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

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

Answer

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

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

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

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.

Answer

(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)

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

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

Answer

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

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

Hence, empirical formula is SiCl₃

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

Answer

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

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.

Answer

(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

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

Answer

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.

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).

Answer

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%

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.

Answer

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.

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

Answer

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

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

Answer

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

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)

Answer

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

Hence, empirical formula is AB₄

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

Answer

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

Hence, empirical formula is NH₂

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

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.

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 + ............... ⟶ ............... + ...............

Answer

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

(b) Grams: 42g + 36g ⟶ 74g + 4g

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

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.

Answer
(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

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.

Answer
(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.

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.

Answer

(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

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

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

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.

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

(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

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

Answer

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

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

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.

Answer

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

(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

(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?

Answer

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