# Stoichiometry — Percentage Composition — Empirical & Molecular Formula — Chemical Equations

## Percentage Composition

#### Question 1

Calculate the percentage by weight of :

(a) C in carbon dioxide

(b) Na in sodium carbonate

(c) Al in aluminium nitride.

[C = 12, O = 16, H = 1, Na = 23, Al = 27, N = 14]

(a) Molecular weight of carbon dioxide CO2 = 12 + (2 x 16) = 12 + 32 = 44

44 g of CO2 contains 12 g of carbon

∴ 100 g of CO2 contains $\dfrac{12}{44}$ x 100 = 27.27% of carbon

Hence, percentage by weight of C in CO2 is 27.3%

(b) Molecular weight of sodium carbonate Na2CO3 = (2 x 23) + 12 + (3 x 16) = 46 + 12 + 48 = 106

106 g of sodium carbonate Na2CO3 contains 46 g of sodium

∴ 100 g of Na2CO3 contains $\dfrac{46}{106}$ x 100 = 43.39% ≈ 43.4% of sodium.

Hence, percentage by weight of sodium in sodium carbonate is 43.4%

(c) Molecular weight of Aluminium Nitride = AlN = 27 + 14 = 41

41 g of Aluminium Nitride contains 27 g of Aluminium

∴ 100 g of Aluminium Nitride contains $\dfrac{27}{41}$ x 100 = 65.85% of Aluminium

Hence, percentage by weight of Aluminium in Aluminium Nitride is 65.85%

#### Question 2

Calculate the percentage of iron in K3Fe(CN)6. [K = 39, Fe = 56, C = 12, N = 14]

Molecular weight of K3Fe(CN)6 = 3(39) + 56 + 6(12 + 14) = 117 + 56 + 156 = 329 g

329 g of K3Fe(CN)6 contains 56 g of iron

∴ 100 g of K3Fe(CN)6 contains $\dfrac{56}{329}$ x 100 = 17.02% of iron.

Hence, percentage by weight of iron in K3Fe(CN)6 is 17.02%

#### Question 3

Calculate which of the following — calcium nitrate or ammonium sulphate has a higher % of nitrogen. [Ca = 40, , O = 16, S = 32, N = 14]

Molecular weight of calcium nitrate Ca[NO3]2 = 40 + 2[(14 + 3(16))] = 40 + 2[14 + 48] = 164 g

164 g of calcium nitrate contains 28 g of nitrogen

∴ 100 g of calcium nitrate contains $\dfrac{28}{164}$ x 100 = 17.07% of nitrogen

Percentage by weight of nitrogen in calcium nitrate is 17.07%

Molecular weight of ammonium sulphate [NH4]2SO4 = 2[14 + 4(1)] + 32 + 4(16) = 36 + 32 + 64 = 132

132 g of ammonium sulphate contains 28 g of nitrogen

∴ 100 g of ammonium sulphate contains $\dfrac{28}{132}$ x 100 = 21.21% of nitrogen

Percentage by weight of nitrogen in ammonium sulphate is 21.21 %

Hence, ammonium sulphate has higher concentration of nitrogen than calcium nitrate.

#### Question 4

Calculate the percentage of pure aluminium in 10 kg. of aluminium oxide [Al2O3] of 90% purity. [Al = 27, O = 16]

90% of 10 kg = 9000 g

Molecular weight of aluminium oxide [Al2O3] = 2(27) + 3(16) = 54 + 48 = 102 g

102 g of aluminium oxide contains 54 g of aluminium

∴ 9000 g of aluminium oxide contains $\dfrac{54}{102}$ x 9000 = 4764 g = 4.764 kg of aluminium

∴ Percentage of pure Al in 10 kg. of aluminium oxide [Al2O3] = $\dfrac{4.764}{10}$ x 100 = 47.64%

Hence, percentage of pure aluminium in 10 kg. of aluminium oxide [Al2O3] of 90% purity is 47.64%

#### Question 5

State which of the following are better fertilizers —

(i) Potassium phosphate [K3PO4] or potassium nitrate [KNO3]

(ii) Urea [NH2CONH2] or ammonium phosphate [(NH4)3PO4]

[K = 39, P = 31, O = 16, N = 14, H = 1]

(i) Molecular weight of K3PO4 = 3(39) + 31 + 4(16) = 117 + 31 + 64 = 212 g

Percentage of K in K3PO4 = $\dfrac{117}{212}$ x 100 = 55.18%

Molecular weight of [KNO3] = 39 + 14 + 3(16) = 39 + 14 + 48 = 101 g

Percentage of K in KNO3 = $\dfrac{39}{101}$ x 100 = 38.61%

Hence, potassium phosphate (% of K = 55.18 %) is better fertilizer than potassium nitrate (% of K = 38.61 %).

(ii) Molecular weight of NH2CONH2 = 14 + 2(1) + 12 + 16 + 14 + 2(1) = 14 + 2 + 12 + 16 + 14 + 2 = 60 g

Percentage of N in NH2CONH2 = $\dfrac{28}{60}$ x 100 = 46.67%

Molecular weight of (NH4)3PO4 = 3[14 + 4(1)] + 31 + 4(16) = 54 + 31 + 64 = 149 g

Percentage of N in (NH4)3PO4 = $\dfrac{42}{149}$ x 100 = 28.19%

Hence, Urea is a better fertilizer as percentage of N in urea is 46.67% and in (NH4)3PO4 is 28.19 %

#### Question 6

Calculate the percentage of carbon in a 55% pure sample of carbon carbonate. [Ca = 40, C = 12, O = 16]

Molecular weight of CaCO3 = 40 + 12 + 3(16) = 40 + 12 + 48 = 100 g

100 g of CaCO3 contains 12 g of carbon.

Given, purity is 55% therefore,

55% of 12 g = $\dfrac{55}{100}$ x 12 = 6.6%

Hence, percentage of carbon is 6.6%.

#### Question 7

Calculate the percentage of water of crystallization in hydrated copper sulphate [CuSO4.5H2O].

[Cu = 63.5, S = 32, O = 16, H = 1]

Molecular weight of hydrated copper sulphate CuSO4.5H2O = 63.5 + 32 + 4(16) + 5[2(1) + 16] = 63.5 + 32 + 64 + 5 = 63.5 + 32 + 64 + 90 = 249.5 g

249.5 g of CuSO4.5H2O contains 90 g of water of crystallization.

∴ Percentage of water of crystallization = $\dfrac{90}{249.5}$ x 100 = 36%

Hence, percentage of water of crystallization is 36%

#### Question 8

Hydrated calcium sulphate [CaSO4.xH2O] contains 21% of water of crystallization. Calculate the number of molecules of water of crystallization i.e. 'X' in the hydrated compound.

[Ca = 40, S = 32, O = 16, H = 1]

Given,

CaSO4.xH2O contains 21% of water of crystallization, so % of CaSO4 = 100 - 21 = 79%

Molecular weight of anhydrous calcium sulphate CaSO4 = 40 + 32 + 4(16) = 40 + 32 + 64 = 136 g

i.e. 79% of CaSO4.xH2O = 136 g

∴ CaSO4.xH2O = $\dfrac{136}{79}$ x 100 = 172.15 g

21% of CaSO4.xH2O = $\dfrac{21}{100}$ x 172.15 = 36.15 g

Molecular weight of water H2O = 2(1) + 16 = 18

Number of molecules (x) = $\dfrac{36.15}{18}$ = 2 molecules

CaSO4.xH2O = CaSO4.2H2O

## Empirical & Molecular Formula

#### Question 1

A compound gave the following data : C = 57.82%, O = 38.58% and the rest hydrogen. It's vapour density is 83. Find it's empirical and molecular formula. [C = 12, O = 16, H = 1]

Element% compositionAt. wt.Relative no. of atomsSimplest ratio
Carbon57.8212$\dfrac{57.82}{12}$ = 4.81$\dfrac{4.81}{2.41}$ = 2
Oxygen38.5816$\dfrac{38.58}{16}$ = 2.41$\dfrac{2.41}{2.41}$ = 1
Hydrogen3.601$\dfrac{3.60}{1}$ = 3.6$\dfrac{3.6}{2.41}$ = $\dfrac{3}{2}$

C : O : H = 2 : 1 : $\dfrac{3}{2}$ = 4 : 2 : 3

Simplest ratio of whole numbers = 4 : 2 : 3

Hence, empirical formula is C4O2H3 or C4H3O2

Empirical formula weight = (4 x 12) + (3 x 1) + (2 x 16) = 48 + 3 + 32 = 83

Vapour density (V.D.) = 83

Molecular weight = 2 x V.D. = 2 x 83 = 166

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{166}{83} = 2$

Molecular formula = n[E.F.] = 2[C4H3O2] = C8H6O4

Molecular formula = C8H6O4

#### Question 2

Four g of a metallic chloride contains 1.89 g of the metal 'X'. Calculate the empirical formula of the metallic chloride. [At. wt. of 'X' = 64, Cl = 35.5]

4 g of metallic chloride contains 1.89 g of metal X

∴ 100 g of metallic chloride contains $\dfrac{1.89}{4}$ x 100 = 47.25 g of metal X

∴ % of Cl in the metallic chloride = 100 - 47.25 = 52.75%.

Element% compositionAt. wt.Relative no. of atomsSimplest ratio
X47.2564$\dfrac{47.25}{64}$ = 0.738$\dfrac{0.738}{0.738}$ = 1
Cl52.7535.5$\dfrac{52.75}{35.5}$ = 1.48$\dfrac{1.48}{0.738}$ = 2

X : Cl = 1 : 2

Hence, empirical formula is XCl2

#### Question 3

Calculate the molecular formula of a compound whose empirical formula is CH2O and vapour density is 30.

Empirical formula is CH2O

Empirical formula weight = 12 + 2(1) + 16 = 12 + 2 + 16 = 30

Vapour density (V.D.) = 30

Molecular weight = 2 x V.D. = 2 x 30 = 60

$\text{n} = \dfrac{\text{Molecular weight}}{\text{empirical formula weight}} \\[0.5em] = \dfrac{60}{30} = 2$

Molecular formula = n[E.F.] = 2[CH2O] = C2H4O2

Hence, Molecular formula is C2H4O2

#### Question 4

A compound has the following percentage composition. Al = 0.2675 g.; P = 0.3505 g.; O = 0.682 g. If the molecular weight of the compound is 122 and it's original weight which on analysis gave the above results 1.30 g. Calculate the molecular formula of the compound. [Al = 27, P = 31, O = 16]

Original weight = 1.30 g

Element% compositionAt. wt.Relative no. of atomsSimplest ratio
Al$\dfrac{0.2675}{1.30}$ x 100 = 20.5727$\dfrac{20.57}{27}$ = 0.7618$\dfrac{0.7618}{0.7618}$ = 1
P$\dfrac{0.3505}{1.30}$ x 100 = 26.9631$\dfrac{26.96}{31}$ = 0.8696$\dfrac{0.8696}{0.7618}$ = 1.14 = 1
O$\dfrac{0.682}{1.30}$ x 100 = 52.4616$\dfrac{52.46}{16}$ = 3.278$\dfrac{3.278}{0.7618}$ = 4.30 = 4

Al : P : O = 1 : 1 : 4

Hence, Simplest ratio of whole numbers = 1 : 1 : 4

Hence, empirical formula is AlPO4

Empirical formula weight = 27 + 31+ 4(16) = 27 + 31 + 64 = 122

Molecular weight = 122

$\text{n} = \dfrac{\text{Molecular weight}}{\text{empirical formula weight}} \\[0.5em] = \dfrac{122}{122} = 1$

∴ Molecular formula = n[E.F.] = 1[AlPO4] = AlPO4

Hence, Molecular formula = AlPO4

#### Question 5

Two organic compounds 'X' and 'Y' containing carbon and hydrogen only have vapour densities 13 and 39 respectively. State the molecular formula of 'X' and 'Y' [C = 12, H = 1]

Given,

Vapour densityX = 13

Vapour densityY = 39

Molecular weightX = 2 x V.D. = 2 x 13 = 26

Molecular weightY = 2 x V.D. = 2 x 39 = 78

Let formula for X be CaHb

where a and b are simple whole numbers.

∴ Molecular mass of X = 12a + b = 26

Hence, a = 2 and b = 2 (only values which hold true for the above equation)

So, Molecular formula of X = C2H2

Let formula for Y be CcHd

where c and d are simple whole numbers.

∴ Molecular mass of Y = 12c + d = 78

Hence, c = 6 and d = 6 (only values which hold true for the above equation)

So, Molecular formula of X = C6H6

#### Question 6

A compound has the following % composition. Zn = 22.65%; S = 11.15%; O = 61.32% and H = 4.88%. It's relative molecular mass is 287 g. Calculate it's molecular formula assuming that all the hydrogen in the compound is present in combination with oxygen as water of crystallization. [Zn = 65, S = 32, O = 16, H = 1]

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.

Hence, molecular formula is ZnSO4.7H2O.

#### Question 7

A hydrocarbon contains 82.8% of carbon. Find it's molecular formula if it's vapour density is 29 [H = 1, C = 12]

Element% compositionAt. wt.Relative no. of atomsSimplest ratio
Carbon82.812$\dfrac{82.8}{12}$ = 6.9$\dfrac{6.9}{6.9}$ = 1
Hydrogen100 - 82.8 = 17.21$\dfrac{17.2}{1}$ = 17.2$\dfrac{17.2}{6.9}$ = 2.5

C : H = 1 : 2.5 = 1 : $\dfrac{5}{2}$ = 2 : 5

Simplest ratio of whole numbers = 2 : 5

Hence, empirical formula is C2H5

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

Vapour density (V.D.) = 29

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

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

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

Hence, Molecular formula = C4H10

#### Question 8

An organic compound on analysis gave H = 6.48% and O = 51.42%. Determine it's empirical formula if the compound contains 12 atoms of carbon. [C = 12, H = 1, O = 16]

Element% compositionAt. wt.Relative no. of atomsSimplest ratio
Oxygen51.4216$\dfrac{51.42}{16}$ = 3.213$\dfrac{3.213}{3.213}$ = 1
Hydrogen6.481$\dfrac{6.48}{1}$ = 6.48$\dfrac{6.48}{3.213}$ = 2.016 = 2
Carbon100 - (51.42 + 6.48) = 42.112$\dfrac{42.1}{12}$ = 3.508$\dfrac{3.508}{3.213}$ = 1.091 = 1

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

As number of carbon atoms = 12

Therefore, O : H : C = 12 : 24 : 12

Hence, empirical formula is C12H24O12

#### Question 9

A hydrated salt contains Cu = 25.50%, S = 12.90%, O = 25.60% and the remaining % is water of crystallization. Calculate the empirical formula of the salt. [Cu = 64, S = 32, O = 16, H = 1]

Element% compositionAt. wt.Relative no. of atomsSimplest ratio
Cu25.5064$\dfrac{25.50}{64}$ = 0.398$\dfrac{0.398}{0.398}$ = 1
S12.9032$\dfrac{12.90}{32}$ = 0.403$\dfrac{0.403}{0.398}$ = 1.01 = 1
O25.6016$\dfrac{25.60}{16}$ = 1.6$\dfrac{1.6}{0.398}$ = 4.020 = 4
H2O100 - (25.50 + 12.90 + 25.60) = 3618$\dfrac{36}{18}$ = 2$\dfrac{2}{0.398}$ = 5.025 = 5

Hence, Simplest ratio of whole numbers = Cu : S : O : H2O = 1 : 1 : 4 : 5

Hence, empirical formula is CuSO4.5H2O

#### Question 10

A gaseous hydrocarbon weights 0.70 g. and contains 0.60 g. of carbon. Find the molecular formula of the compound if it's molecular weight is 70. [C = 12, H = 1]

H = 0.70 - 0.60 = 0.1 g

Element% compositionAt. wt.Relative no. of atomsSimplest ratio
Carbon$\dfrac{0.60}{0.70}$ x 100 = 85.7112$\dfrac{85.71}{12}$ = 7.142$\dfrac{ 7.142}{7.142}$ = 1
Hydrogen$\dfrac{0.1}{0.70}$ x 100 = 14.281$\dfrac{14.28}{1}$ = 14.28$\dfrac{14.28}{7.142}$ = 2

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

Hence, empirical formula is CH2

Empirical formula weight = 12 + 2(1) = 14

Molecular weight = 70

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{70}{14} = 5$

Molecular formula = n[E.F.] = 5[CH2] = C5H10

Hence, Molecular formula = C5H10

#### Question 11

A salt has the following % composition — Al = 10.50%, K = 15.1%, S = 24.8% and the remaining oxygen. Calculate the empirical formula of the salt. [Al = 27, K = 39, S = 32, O = 16]

Element% compositionAt. wt.Relative no. of atomsSimplest ratio
Al10.5027$\dfrac{10.50}{27}$ = 0.388$\dfrac{0.388}{0.387}$ = 1
K15.139$\dfrac{15.1}{39}$ = 0.387$\dfrac{0.387}{0.387}$ = 1
S24.832$\dfrac{24.8}{32}$ = 0.775$\dfrac{0.775}{0.387}$ = 2
O100 - (10.50 + 15.1 + 24.8) = 49.616$\dfrac{49.6}{16}$ = 3.1$\dfrac{3.1}{0.387}$ = 8

Simplest ratio of whole numbers = Al : K : S : O = 1 : 1 : 2 : 8

Hence, empirical formula is AlKS2O8 or AlK(SO4)2

## Chemical Equations

#### Question 1

What mass of silver chloride will be obtained by adding an excess of hydrochloric acid to a solution of 0.34 g of silver nitrate. [Cl = 35.5, Ag = 108, N = 14, O = 16, H = 1 ]

Molecular Weight of Silver Nitrate (AgNO3) = 108 + 14 + 48 = 170 g

Molecular Weight of Silver Chloride (AgCl) = 108 + 35.5 = 143.5 g

Chemical Equation:

$\underset{170 g}{\text{AgNO}_3} + \text{HCl} \longrightarrow \underset{143.5 g}{\text{AgCl}} + \text{HNO}_{3}$

170 g of AgNO3 gives 143.5 g of AgCl

∴ 0.34 g of AgNO3 will give $\dfrac{143.5}{170}$ x 0.34 = 0.287 g of AgCl

Hence, 0.287 g of silver chloride will be obtained

#### Question 2

What volume of oxygen at s.t.p. will be obtained by the action of heat on 20 g KClO3 [K = 39, Cl = 35.5, O = 16]

$\begin{matrix} 2\text{KClO}_3 & \xrightarrow{\space \Delta \space} & 2\text{KCl} & + & 3\text{O}_2 \\ 2[39 + 35.5 & & & & 3 \times 22.4 \\ + 48] = 245 & & & & = 67.2 \\ \end{matrix}$

245 g of KClO3 gives 67.2 lit of O2

∴ 20 g of KClO3 will give $\dfrac{67.2}{245}$ x 20 = 5.486 lit of O2

Hence, 5.486 lit of oxygen will be obtained

#### Question 3

From the equation : 3Cu + 8HNO3 ⟶ 3Cu(NO3)2+ 4H2O + 2NO. Calculate

(i) the mass of copper needed to react with 63 g of nitric acid

(ii) the volume of nitric oxide collected at the same time. [Cu = 64, H = 1, O = 16, N = 14]

$\begin{matrix} 3\text{Cu} & + & 8\text{HNO}_3 & \longrightarrow & 3\text{Cu[NO}_3]_2 & + & 4\text{H}_2\text{O} & + & 2\text{NO} \\ 3(64) & & 8[1 + 14 + 48] \\ = 192 \text{ g} & & = 8 \times 63 = 504 \text{ g} \\ \end{matrix}$

504 g of nitric acid reacts with 192 g of copper

∴ 63 g of nitric acid will react with $\dfrac{192}{504}$ x 63 = 24 g of copper

Hence, 24 g of copper is needed to react with 63 g of nitric acid

(ii)

$3\text{Cu} + \underset{504 \text{ g}}{8\text{HNO}_3} \longrightarrow 3\text{Cu[NO}_3]_2 + 4\text{H}_2\text{O} + \underset{2(22.4) \text {lit.}}{2\text{NO}}$

504 g of nitric acid liberates (2 x 22.4) lit of Nitric Oxide

∴ 63 g of nitric acid will liberate $\dfrac{2 \times 22.4}{504}$ x 63 = 5.6 lits of Nitric Oxide

Hence, 5.6 lits. of nitric oxide is collected.

#### Question 4

Zinc blende [ZnS] is roasted in air. Calculate :

(a) The number of moles of sulphur dioxide liberated by 776 g of ZnS and

(b) The weight of ZnS required to produce 22.4 lits. of SO2 at s.t.p. [S = 32, Zn = 65, O = 16]

$\underset{65 + 32 = 97 \text{ g}}{\text{ZnS}} + \text{O}_2 \longrightarrow \underset{1 \text{ mole}}{\text{SO}_2}$

97 g of ZnS gives 1 mole of SO2

∴ 776 g of ZnS will give $\dfrac{1}{97}$ x 776 = 8 moles of SO2

Hence, 8 moles of sulphur dioxide is liberated by 776 g of ZnS

(ii)

$\underset{65 + 32 = 97 \text{ g}}{\text{ZnS}} + \text{O}_2 \longrightarrow \underset{1 \text{ mole}}{\text{SO}_2}$

97 g of ZnS liberates 1 mole of SO2

As 1 mole occupies 22.4 lit volume at s.t.p.

22.4 lit of SO2 is produced by 97 g of ZnS.

#### Question 5

Ammonia reacts with sulphuric acid to give the fertilizer ammonium sulphate. Calculate the volume of ammonia [at s.t.p.] used to form 59 g of ammonium sulphate.
[N = 14, H = 1, S = 32, O = 16]

$\begin{matrix} 2\text{NH}_3 & + & \text{H}_2\text{SO}_4 & \longrightarrow & [\text{NH}_4]_2\text{SO}_4 \\ 2(22.4 \text{ lit.}) & & & & 2[14 + (4 \times 1)] \\ & & & & + 32 + (4 \times 16) \\ & & & & = 132 \text{g} \end{matrix}$

132 g of ammonium sulphate is produced by 2 x 22.4 lit of Ammonia

∴ 59 g of ammonium sulphate is produced by $\dfrac{2 \times 22.4}{132}$ x 59 = 20.02 lit. of Ammonia

Hence, 20.02 lit. of ammonia [at s.t.p.] is used to form 59 g of ammonium sulphate

#### Question 6

Heat on lead nitrate gives yellow lead [II] oxide, nitrogen dioxide and oxygen. Calculate the total volume of NO2 and O2 produced on heating 8.5 of lead nitrate. [Pb = 207, N = 14, O = 16]

$\begin{matrix} 2\text{Pb[NO}_3]_2 & \xrightarrow{\space \Delta \space} & 2\text{PbO} & + & 4\text{NO}_2 & + & \text{O}_2 \\ 2[207 + 2(14 + 48)] & & & & (4 \times 22.4) & & 22.4 \text{ lit} \\ = 2[207 + 124] & & & & = 89.6 \text{ lit} & & \\ = 662 \text{ g} & & & & & & \end{matrix}$

662 g of lead nitrate gives 89.6 lit. of NO2

∴ 8.5 g of lead nitrate will give $\dfrac{89.6}{662}$ x 8.5 = 1.15 lit. of NO2

Hence, NO2 produced is 1.15 lit

662 g of lead nitrate gives 22.4 lit. of O2

∴ 8.5 g of lead nitrate will give $\dfrac{22.4}{662}$ x 8.5 = 0.287 lit of O2

Hence, O2 produced is 0.287 lit

∴ Total volume produced = 1.15 + 0.287 = 1.437 lits.

Hence, total volume of NO2 and O2 produced on heating 8.5 of lead nitrate is 1.437 lits.

#### Question 7

2KClO3 ⟶ 2KCl + 3O2 ; C + O2 ⟶ CO2. Calculate the amount of KClO3 which on thermal decomposition gives 'X' vol. of O2, which is the volume required for combustion of 24 g. of carbon. [K = 39, Cl = 35.5, O = 16, C = 12].

$\begin{matrix} \text{C} & + & \text {O}_2 & \xrightarrow{\Delta} & \text{CO}_2 \\ 12 \text{ g} & & & & 22.4 \text{ lit} \end{matrix}$

12 g of C gives 22.4 lit of CO2

∴ 24 g of C will give $\dfrac{22.4}{12}$ x 24 = 44.8 lit. of CO2

Hence, 24 g of C will give 44.8 of CO2

$\begin{matrix} 2\text{KClO}_3 & \xrightarrow{\space \Delta \space} & & 2\text{KCl} & + & 3\text{O}_2 \\ 2[39 + 35.5 & & & & & (3 \times 22.4) \\ + 48] = 245 \text{ g} & & & & & = 67.2 \text{ lit} \\ \end{matrix}$

67.2 lit of O2 is produced by 245 g of KClO3

∴ 44.8 lit of O2 is produced by $\dfrac{245}{67.2}$ x 44.8 = 163.33 g

Hence, amount of KClO3 is 163.33 g.

#### Question 8

Calculate the weight of ammonia gas.

(a) Required for reacting with sulphuric acid to give 78 g. of fertilizer ammonium sulphate.

(b) Obtained when 32.6 g. of ammonium chloride reacts with calcium hydroxide during the laboratory preparation of ammonia.

[2NH4Cl + Ca(OH)2 ⟶ CaCl2 + 2H2O + 2NH3]

[N = 14, H = 1, O = 16, S = 32, Cl = 35.5].

(a) Molecular Weight of Ammonia (NH3) = 14 + 3(1) = 17 g

Molecular Weight of Ammonium Sulphate ((NH4)2SO4) = 2[14 + 4(1)] + 32 + 4(16) = 132 g

$\underset{2(17) = 34 \text{ g}}{2\text{NH}_3} + \text{H}_2\text{SO}_4 \longrightarrow \underset{132 \text{ g}}{(\text{NH}_4)_2\text{SO}_4}$

132 g of ammonium sulphate is given by 34 g of ammonia

∴ 78 g of ammonium sulphate is given by $\dfrac{34}{132}$ x 78 = 20.09 g of ammonia

Hence, 20.09 g of ammonia gas is required.

(b) Molecular Weight of Ammonium Chloride (NH4Cl) = 14 + 4(1) + 35.5 = 53.5 g

$\underset{2(53.5) = 107 \text{ g}}{2\text{NH}_4\text{Cl}} + \text{Ca}(\text{OH})_2 \longrightarrow \text{CaCl}_2 + 2\text{H}_2\text{O} + \underset{2(17) = 34 \text{ g}}{2\text{NH}_3}$

107 g of ammonium chloride gives 34 g of ammonia

∴ 32.6 g of ammonium chloride will give $\dfrac{34}{107}$ x 32.6 = 10.36 g of ammonia

Hence, 10.36 g of ammonia gas is obtained.

#### Question 9

Sodium carbonate reacts with dil. H2SO4 to give the respective salt, water and carbon dioxide. Calculate the mass of pure salt formed when 300 g. of Na2CO3 of 80% purity reacts with dil. H2SO4. [Na = 23, C = 12, O = 16, H = 1, S = 32].

Given,

300 g. of Na2CO3 is of 80% purity

∴ Mass of pure Na2CO3 present in it = $\dfrac{80}{100}$ x 300 = 240 g

We have,

$\begin{matrix} \text{Na}_2\text{CO}_3 & + & \text{H}_2\text{SO}_4 & \longrightarrow & \text{Na}_2\text{SO}_4 & + & \text{H}_2\text{O} & + & \text{CO}_2 \\ (2 \times 23) + 12 & & & & (2 \times 23) + 32 \\ + (3 \times 16 ) & & & & + (4 \times 16) \\ = 106 \text{ g} & & & & = 142 \text{ g} \end{matrix}$

106 g of Na2CO3 gives 142 g of Na2SO4

∴ 240 g of Na2CO3 gives $\dfrac{142}{106}$ x 240 = 321.51 g

Hence, mass of pure salt formed is 321.51 g

#### Question 10

Sulphur burns in oxygen to give sulphur dioxide. If 16 g. of sulphur burns in 'x' cc. of oxygen, calculate the amount of potassium nitrate which must be heated to produce 'x' cc. of oxygen. [S = 32, K = 39, N = 14, O = 16].

$\begin{matrix} 2\text{KNO}_3 & \longrightarrow & 2\text{KNO}_2 & + & \text{O}_2 \\ 2[39 + 14 & & & & 2 \times 16 \\ + (16 \times 3)] & & & & = 32 \text{ g} \\ = 202 \text{ g} & & & & \end{matrix}$

$\begin{matrix} \text{S} & + & \text{O}_2 & \longrightarrow & \text{SO}_2 \\ 32 \text {g} \\ \end{matrix}$

32 g of sulphur require 202 g of potassium nitrate to be heated

∴ 16 g of sulphur will require $\dfrac{202}{32}$ x 16 = 101 g of potassium nitrate.

Hence, amount of potassium nitrate which must be heated is 101 g

#### Question 11

Sample of impure magnesium is reacted with dilute sulphuric acid to give the respective salt and hydrogen. If 1 g. of the impure sample gave 298.6 cc. of hydrogen at s.t.p. Calculate the % purity of the sample. [Mg = 24, H = 1].

$\begin{matrix} \text{Mg} & + & \text{H}_2\text{SO}_4 & \longrightarrow & \text{MgSO}_4 & + & \text{H}_2 \\ 24 \text{g} \\ \end{matrix}$

22400 cc of H2 is produced from 24 g of pure magnesium

∴ 298.6 cc. of H2 is produced from $\dfrac{24}{22400}$ x 298.6 = 0.3199 g of pure magnesium

∴ Percentage Purity = 0.3199 x 100 = 31.99 %

Hence, the % purity of the sample is 31.99 %

## Gay Lussac's Law

#### Question 1(2012)

67.2 litres of H2 combines with 44.8 litres of N2 to form NH3 :

N2(g) + 3H2(g) ⟶ 2NH3(g).

Calculate the vol. of NH3 produced. What is the substance, if any, that remains in the resultant mixture ?

[By Lussac's law]

$\begin{matrix} \text{N}_2 & + & 3\text{H}_2 & \longrightarrow & 2 \text{NH}_3 \\ 1 \text{ vol.} & : & 3 \text{ vol.} & \longrightarrow & 2\text{ vol.} \end{matrix}$

To calculate the volume of ammonia gas formed.

$\begin{matrix} \text{H}_2 & : & \text{NH}_3 \\ 3 & : & 2 \\ 67.2 & : & x \end{matrix}$

$\therefore \dfrac{2}{3} \times 67.2 = 44.8 \text{ lit}$

Hence, volume of NH3 formed is 44.8 lit.

We know,

$\begin{matrix}\text{H}_2 & : & \text{N}_2 \\ 3 & : & 1 \\ 67.2 & : & x \end{matrix}$

$\therefore \dfrac{1}{3} \times 67.2 = 22.4 \text{ lit}$

∴ nitrogen left = 44.8 - 22.4 = 22.4 lit.

Hence, 22.4 lit of nitrogen remains in the resultant mixture.

#### Question 1(2013)

What volume of oxygen is required to burn completely 90 dm3 of butane under similar conditions of temperature and pressure?

2C4H10 + 13O2 ⟶ 8CO2 + 10H2O

[By Lussac's law]

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

To calculate the volume of oxygen

$\begin{matrix}\text{C}_4\text{H}_{10} & : & \text{O}_2 \\ 2 & : & 13 \\ 90 & : & x \end{matrix}$

$\dfrac{13}{2} \times 90 = 585 \text{ dm}^3$

Hence, volume of oxygen is required is 585 dm3

#### Question 1(2014)

What volume of ethyne gas at s.t.p. is required to produce 8.4 dm3 of carbon dioxide at s.t.p.?

2C2H2 + 5O2 ⟶ 4CO2 + 2H2O

[H = 1, C = 12, O = 16]

[By Lussac's law]

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

To calculate the volume of ethyne gas

$\begin{matrix}\text{CO}_2 & : & \text{C}_2\text{H}_2 \\ 4 & : & 2 \\ 8.4 & : & x \end{matrix}$

$\dfrac{2}{4} \times 8.4 = 4.2 \text{ dm}^3$

Hence, volume of ethyne gas required = 4.2 dm3.

#### Question 1(2015)

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.

[By Lussac's law]

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

To calculate the amount of hydrogen used:

$\begin{matrix}\text{Cl}_2 & : & \text{H}_2 \\ 1 \text{ vol.} & : & 1 \text{ vol.} \\ 4 \text{ lit.} & : & \text{x} \end{matrix}$

$x = \dfrac{1}{1} \times 4 = 4 \text{ lit.}$

Remaining hydrogen = 6 - 4 = 2 lit.

To calculate the amount of HCl :

$\begin{matrix}\text{Cl}_2 & : & \text{HCl} \\ 1 \text{ vol.} & : & 2 \text{ vol.} \\ 4 \text{ lit.} & : & \text{x} \end{matrix}$

$x = \dfrac{2}{1} \times 4 = 8 \text{ lit.}$

Hence, after reaction 8 lit of HCl is formed which dissolves in water and 2 lit of hydrogen is left.

2 lits. of hydrogen is left.

#### Question 1(2016)

The equation :

4NH3 + 5O2 ⟶ 4NO + 6H2O,

represents the catalytic oxidation of ammonia. If 100 cm3 of ammonia is used calculate the volume of oxygen required to oxidize the ammonia completely.

[By Lussac's law]

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

To calculate the volume of oxygen required :

$\begin{matrix}\text{NH}_3 & : & \text{O}_2 \\ 4 \text{ vol.} & : & 5 \text{ vol.} \\ 100 \text{ cm.}^3 & : & \text{x} \end{matrix}$

$\therefore x = \dfrac{5}{4} \times 100 = 125 \text{ cm.}^3$

Hence, volume of oxygen required = 125 cm3

#### Question 1(2017)

Propane burns in air according to the following equation :

C3H8 + 5O2 ⟶ 3CO2 + 4H2O.

What volume of propane is consumed on using 1000 cm3 of air, considering only 20% of air contains oxygen.

Given, 20% of air contains oxygen

∴ 20% of 1000 cm3 = $\dfrac{20}{100}$ x 1000 = 200 cm3

[By Lussac's law]

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

To calculate the volume of propane consumed :

$\begin{matrix}\text{O}_2 & : & \text{C}_3\text{H}_8 \\ 5 \text{ vol.} & : & 1 \text{ vol.} \\ 200 \text{ cm.}^3 & : & \text{x} \end{matrix}$

$\therefore x = \dfrac{1}{5} \times 200 = 40 \text{ cm.}^3$

Hence, volume of propane is consumed is 40 cm3

#### Question 1(2018)

Ethane burns in O2 to form CO2 and H2O

2C2H6 + 7O2 ⟶ 4CO2 + 6H2O

If 1250 cc. of oxygen is burnt with 300 cc of ethane.

Calculate :

(i) volume of CO2 formed.

(ii) volume of unused O2.

[By Lussac's law]

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

(i) To calculate the volume of CO2 formed :

$\begin{matrix}\text{C}_2\text{H}_6 & : & \text{CO}_2 \\ 2 \text{ vol.} & : & 4 \text{ vol.} \\ 300 \text{ cc} & : & \text{x} \end{matrix}$

$x = \dfrac{4}{2} \times 300 = 600 \text{ cc}$

Hence, volume of carbon dioxide formed = 600 cc

(ii) (i) To calculate the volume of unused O2 :

$\begin{matrix}\text{C}_2\text{H}_6 & : & \text{O}_2 \\ 2 \text{ vol.} & : & 7 \text{ vol.} \\ 300 \text{ cc} & : & \text{x} \end{matrix}$

$x = \dfrac{7}{2} \times 300 = 1050 \text{ cc}$

Unused oxygen = 1250 - 1050 = 200 cc.

Hence, volume of unused oxygen = 200 cc

#### Question 1(2020)

Calculate the amount of each reactant required to produce 750 ml of carbon dioxide as per the equation.

2CO + O2 ⟶ 2CO2

State the law associated in the question.

[By Lussac's law]

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

To calculate the amount of CO required :

$\begin{matrix}\text{CO}_2 & : & \text{CO} & \\ 2 \text{ vol.} & : & 2 \text{ vol.} \\ 750 \text{ ml} & : & \text{x} \end{matrix}$

$\therefore x = \dfrac{2}{2} \times 750 = 750 \text{ ml}$

To calculate the amount of O2 required :

$\begin{matrix}\text{CO}_2 & : & \text{O}_2 & \\ \text{ 2 vol.} & : & 1 \text{ vol.} \\ 750 \text{ cm.}^3 & : & \text{x} \end{matrix}$

$\therefore x = \dfrac{1}{2} \times 750 = 375 \text{ ml}$

Hence, CO required is 750 ml and O2 required = 375 ml

Gay Lussac's law is associated in the question.

## Mole concept — Avogadro's No.

#### Question 1(2005)

The volumes of gases A, B, C and D are in the ratio, 1 : 2 : 2 : 4 under the same conditions of temp. and press.

(i) Which sample contains the maximum no. of molecules. If the temp, and pressure of gas A are kept constant, what will happen to the volume of A when the no. of molecules is doubled.

(ii) If the volume of 'A' is 5.6 dm3 at s.t.p., calculate the no. of molecules in the actual vol. of 'D' at s.t.p. [Avog no. is 6 × 1023).

Using your answer, state the mass of 'D' if the gas is N2O

[N = 14, O = 16]

Volume is directly proportional to the number of molecules, hence gas D will have maximum no. of molecules as its volume is maximum.

If number of molecules of gas A is doubled, the volume will also be doubled.

(ii) 1 mole contains 6 x 1023 number of molecules and occupies 22.4 lit. vol.

Given, volume of 'A' is 5.6 dm3 at s.t.p.

∴ vol. of D will be 4 × 5.6 = 22.4 lit.
No. of molecules in 22.4 lit. of D = 6 x 1023 (Avogadro no.)

As D is 1 mole hence, mass of 1 mole of D (N2O) = 2(14) + 16 = 28 + 16 = 44 g

Hence, mass of N2O = 44 g

#### Question 1(2006)

Calculate the no. of moles and the no. of molecules present in 1.4 g of ethylene gas (C2H4). What is the vol. occupied by the same amount of C2H4. State the vapour density of C2H4.

(Avog. No. = 6 × 1023 ; C = 12, H = 1]

Gram molecular mass of C2H4
= 2(12) + 4(1)
= 24 + 4 = 28 g

As,

28 g of C2H4 = 1 mole
∴ 1.4 g of C2H4 = $\dfrac{1}{28}$ x 1.4 = 0.05 moles

1 mole = 6 × 1023 molecules
∴ 0.05 moles = 6 × 1023 x 0.05 = 3 x 1022 molecules

Hence, no. of moles is 0.05 and no. of molecules is 3 x 1022.

Vol. occupied by 1 mole = 22.4 lit
∴ Vol. occupied by 0.05 moles = 22.4 x 0.05 = 1.12 lit.

$\text{Vapour density} = \dfrac{\text{Molecular weight}}{2} = \dfrac{28}{2} = 14$

Hence, vol. occupied is 1.12 lit and vapour density is 14

#### Question 1(2008)

The equation for the burning of octane is :

2C8H18 + 25O2 ⟶ 16CO2 + 18H2O

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

(ii) What volume, at s.t.p., is occupied by the number of moles determined in (i) ?

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

(i) 2 moles of octane produce 16 moles of carbon dioxide

∴ 1 mole of octane will produce $\dfrac{16}{2}$ = 8 moles of carbon dioxide.

(ii) 1 moles occupies 22.4 lit. of vol.

∴ 8 mole will occupy $\dfrac{22.4}{1}$ x 8 = 179.2 lit. of vol.

(iii) To calculate the mass of CO2 produced :

$\begin{matrix} \text{C}_8 \text{H}_{18} & : & \text{CO}_2 & \\ \text{ 2 moles} & : & 16 \text{ moles} \\ \end{matrix}$

Mass of 1 mole of CO2 = 44 g

∴ Mass of 16 moles of CO2 = 44 x 16 = 704 g

Hence, 8 moles of carbon dioxide are produced, vol occupied is 179.2 lit. and mass of carbon dioxide produced is 704 g

#### Question 1(2009)

Define the term Mole.

A gas cylinder contains 24 × 1024 molecules of nitrogen gas. If Avogadro's number is 6 × 1023 and the relative atomic mass of nitrogen is 14, calculate :

(i) Mass of nitrogen gas in the cylinder.

(ii) Volume of nitrogen at STP in dm3

Mole — A mole is the amount of substance which contains the same number of units as the number of atoms in 12,000 g of carbon - 12 [6C12]

(i) Gram molecular mass of N2 = 2(14) = 28 g

6 × 1023 molecules of N2 weighs 28 g

∴ 24 × 1024 molecules will weigh $\dfrac{28}{6 \times 10^{23}}$ x 24 × 1024 = 1120 g

(ii) 6 × 1023 molecules of N2 occupies 22.4 dm3

∴ 24 × 1024 molecules of N2 will occupy $\dfrac{22.4}{6 \times 10^{23}}$ x 24 × 1024 = 896 dm3

Hence, Mass of nitrogen gas = 1120 g and vol. occupied is 896 dm3

#### Question 2(2009)

Gas 'X' occupies a volume of 100 cm3 at S.T.P and weighs 0.5 g. Find it's relative molecular mass.

100 cm3 weighs 0.5 g .

22400 cm3 will weigh $\dfrac{0.5}{100}$ x 22400 = 112 g

Hence, relative molecular mass = 112 g.

#### Question 1(2010)

Dilute HCl is reacted with 4.5 moles of calcium carbonate.

Calculate :

(i) The mass of 4.5 moles of CaCO3.

(ii) The volume of CO2 liberated at stp.

(iii) The mass of CaCl2 formed ?

(iv) The number of moles of the acid HCl used in the reaction [relative molecular mass of CaCO3 is 100 and of CaCl2 is 111.]

CaCO3 + 2HCl ⟶ CaCl2 + H2O + CO2

(i) Given,

1 mole of CaCO3 = molecular mass of CaCO3 = 100 g

∴ 4.5 moles of CaCO3 weighs $\dfrac{100}{1}$ x 4.5 = 450 g

(ii) 1 mole of CaCO3 produces 1 mole of CO2 and 1 mole occupies 22.4 l of volume.

∴ 4.5 moles of CaCO3 will produce 4.5 moles of CO2 and 4.5 moles will occupy 22.4 x 4.5 = 100.8 l

(iii) 1 mole CaCO3 produces 111 g. CaCl2

∴ 4.5 moles of CaCO3 will produce = 111 x 4.5 = 499.5 g.

(iv)

$\begin{matrix}\text{CaCO}_3 & : & \text{HCl} & \\ \text{ 1 mole} & : & 2 \text{ moles} \\ \text{ 4.5 mole} & : & x \text{ moles} \\ \end{matrix}$

∴ number of moles of HCl used = 2 x 4.5 = 9 moles.

Hence, mass of CaCO3 = 450 g, vol. of CO2 liberated = 100.8 lits., mass of Cl2 formed = 499.5 g, moles of HCl used = 9 moles.

#### Question 1(2011)

Calculate the mass of :

(i) 1022 atoms of sulphur.

(ii) 0.1 mole of carbon dioxide.

[S = 32, C = 12 and O = 16 and Avogadro's Number 6 × 1023]

(i) gram molecular mass of S = 32

6 × 1023 atoms weigh = 32 g

1022 atoms will weigh = $\dfrac{32}{6 \times 10^{23}}$ x 1022 = 0.533 g

(ii) 1 mole of carbon dioxide weighs = C + 2(O) = 12 + 2(16) = 12 + 32 = 44 g

∴ 0.1 mole of carbon dioxide weighs = 44 x 0.1 = 4.4 g.

Hence, mass of 1022 atoms of sulphur = 0.533 g. and mass of 0.1 mole of carbon dioxide = 4.4 g.

#### Question 2(2011)

Calculate the volume of 320 g of SO2 at stp.

[S = 32 and O = 16]

gram molecular mass of SO2 = 32 + 2(16) = 32 + 32 = 64 g

64 g of SO2 occupy 22.4 lit of vol.

320 g of SO2 will occupy = $\dfrac{22.4}{64}$ x 320 = 112 lit.

Hence, volume of 320 g of SO2 = 112 lit.

#### Question 1(2012)

The mass of 5.6 dm3 of a certain gas at STP is 12.0 g. Calculate the relative molecular mass of the gas.

5.6 dm3 weighs 12 g

∴ 22.4 dm3 weighs $\dfrac{12}{5.6}$ x 22.4 = 48 g

Hence, relative molecular mass of the gas = 48 g.

#### Question 1(2013)

The vapour density of a gas is 8. What would be the volume occupied by 24.0 g of the gas at STP ?

Given, V.D. = 8

Gram molecular mass = V.D. x 2 = 8 x 2 = 16 g.

16 g occupies 22.4 lit.

∴ 24 g. will occupy = $\dfrac{22.4}{16}$ x 24 = 33.6 lit.

Hence, volume occupied by gas = 33.6 lit.

#### Question 2(2013)

Calculate the volume occupied by 0.01 mole of CO2 at STP.

1 mole occupies 22.4 lit.

0.01 mole will occupy = 22.4 x 0.01 = 0.224 lit.

Hence, vol. occupied is 0.224 lit.

#### Question 1(2014)

A cylinder contains 68 g of ammonia gas at s.t.p.

(i) What is the volume occupied by this gas?

(ii) How many moles and how many molecules of ammonia are present in the cylinder?

[N = 14, H = 1]

By Avogadro's law: Under the same conditions of temperature and pressure equal volumes of all gases contain the same number of molecules.

(i) Gram molecular mass of ammonia = N + 3(H) = 14 + 3 = 17 g.

17 g. occupies 22.4 lit. of vol.

∴ 68 g will occupy = $\dfrac{22.4}{17}$ x 68 = 89.6 lit.

Hence, volume occupied by this gas = 89.6 lit.

(ii) 17 g = 1 mole

∴ 68 g = $\dfrac{1}{17}$ x 68 = 4 moles.

1 mole = 6 × 1023

∴ 4 moles = 4 x 6.023 × 1023 molecules.

Hence, Moles = 4 and molecules = 4 x 6.023 × 1023 molecules

#### Question 1(2015)

From A, B, C, D, which weighs the least —

A : 2 g. atoms of Nitrogen

B : 1 mole of Silver

C : 22.4 lits. of oxygen gas at 1 atmos. press. and 273 K

D : 6.02 x 1023 atoms of carbon.

[Ag = 108, N = 14, O = 16, C = 12]

D : 6.02 x 1023 atoms of carbon.

Reason
Gram molecular mass of carbon is least.

#### Question 2(2015)

Calculate the mass of Calcium that will contain the same number of atoms as are present in 3.2 gm of sulphur. [S = 32, Ca = 40]

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

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

Hence, mass = 4g.

#### Question 1(2016)

Select the correct answer from A, B, C and D : The ratio between the number of molecules in 2 g of hydrogen and 32 g of oxygen is:

(A) 1 : 2

(B) 1 : 0.01

(C) 1 : 1

(D) 0.01 : 1

[H = 1, O = 16]

(C) 1 : 1
Reason
1 mole of any substance contains same (Avogardo No.) of molecules. Hence, ratio is same.

#### Question 2(2016)

A gas of mass 32 gms has a volume of 20 litres at S.T.P. Calculate the gram mol. weight of the gas.

20 lit. of gas weighs = 32 g.

1 mole = 22.4 lit. of gas will weigh = $\dfrac{32}{20}$ x 22.4 = 35.84 g.

Hence, gram mol. weight of the gas = 35.84 g.

#### Question 3(2016)

A gas cylinder contains 12 x 1024 molecules of oxygen gas.

Calculate : the mass of O2 present in the cylinder.

(ii) the volume of O2 at S.T.P. present in the cylinder.

[O = 16] Avog. no. is 6 × 1023

Gram molecular mass of oxygen = 32 g = 1 mole and contains 6 x 1023 molecules i.e., 6 x 1023 molecules weigh 32 g

∴ 12 x 1024 molecules will weigh = $\dfrac{32}{ 6\times 10^{23}}$ x 12 x 1024 = 640 g

(ii) 6 x 1023 molecules occupy 22.4 lit. of vol.

∴ 12 x 1024 molecules will occupy = $\dfrac{22.4}{ 6\times 10^{23}}$ x 12 x 1024 = 448 lit.

Hence, mass of O2 present in the cylinder = 640 g and volume of O2 at S.T.P. present in the cylinder = 448 lit.

#### Question 1(2017)

Calculate the number of gram atoms in 4.6 grams of sodium [Na = 23]

23 g of Na = 1 g. atom

∴ 4.6 g. of Na = $\dfrac{1}{23}$ x 4.6 = 0.2 g

Hence, 0.2 g. atoms

#### Question 2(2017)

The mass of 11.2 litres of a certain gas at s.t.p. is 24 g. Find the gram molecular mass of the gas.

11.2 lit. weighs 24 g

∴ 22.4 lit. will weigh = $\dfrac{24}{11.2}$ x 22.4 = 48 g

Hence, gram molecular mass of the gas = 48 g.

#### Question 1(2019)

Calculate :

(i) The number of moles in 12 g. of oxygen gas. [O = 16]

(ii) The weight of 1022 atoms of carbon.

[C = 12, Avogadro's No. = 6 x 1023]

(i) Gram molecular mass of oxygen = 32 g = 1 mole

32 g = 1 mole

∴ 12 g = $\dfrac{1}{32}$ x 12 = 0.375 moles.

(ii) 6 x 1023 atoms of carbon weigh = 12 g

∴ 1022 atoms of carbon will weigh = $\dfrac{12}{ 6\times 10^{23}}$ x 1022 = 0.2 g.

Hence, number of moles = 0.375 moles and weight of 1022 atoms of carbon = 0.2 g.

#### Question 1(2020)

Calculate the volume occupied by 80 g. of carbon dioxide at s.t.p.

Gram molecular mass of CO2 = 12 + 32 = 44 g and occupies 22.4 lit. vol.

∴ 80 g. will occupy = $\dfrac{22.4}{44}$ x 80 = 40.73 lit.

Hence, volume occupied = 40.73 lits.

#### Question 2(2020)

Calculate the number of molecules in 4.4 gm. of CO2

[C = 12, O = 16]

Gram molecular mass of CO2 = 12 + 32 = 44 g and contains 6.023 x 1023 molecules.

∴ 4.4 gm. of CO2 will contain = $\dfrac{6.023 \times 10^{23}}{44}$ x 4.4 = 6.023 x 1022 molecules.

Hence, number of molecules = 6.023 x 1022 molecules.

#### Question 3(2020)

Fill in the blank from the choices given in bracket : The number of moles in 11 g. of nitrogen gas is ............... [0.39, 0.49, 0.29] [N = 14]

The number of moles in 11 g. of nitrogen gas is 0.39

Working

Gram molecular mass of nitrogen gas (N2) = 2(14) = 28

28 g of nitrogen gas = 1 mole

∴ 11 g of nitrogen gas = $\dfrac{1}{28}$ x 11 = 0.39

#### Question 4(2020)

(i) State the volume occupied by 40 gm. of methane at stp if it's vapour density is 8.

(ii) Calculate the number of moles present in 160 g of NaOH.

[Na = 23, H = 1, O = 16]

(i) Gram molecular mass of methane = V.D. x 2 = 8 x 2 = 16 g

16 g of methane occupies 22.4 lit.

∴ 40 g of methane will occupy = $\dfrac{22.4}{16}$ x 40 = 56 lits.

(ii) Gram molecular mass of NaOH = 23 + 16 + 1 = 40 g

40 g of NaOH = 1 mole

∴ 160 g of NaOH = $\dfrac{160}{40}$ = 4 moles.

## Mole Concept — Avogadro's Law

#### Question 1(2002)

Samples of O2, N2, CO and CO2 under the same conditions of temperature and pressure contain the same number of molecules represented by X. The molecules of oxygen occupy V litres and have a mass of 8 g. Under the same conditions of temperature and pressure, what is the volume occupied by :

(i) X molecules of N2.

(ii) 3X molecules of CO.

(iii) What is the mass of CO2 in grams.

(iv) In answering the above questions, whose law has been used.

[C = 12, N = 14, O = 16]

By Avogadro's law: Under the same conditions of temperature and pressure equal volumes of all gases contain the same number of molecules.

∴ If gases under the same conditions have same number of molecules then they must have the same volume.

(i) So, X molecules of N2 occupy V litres.

(ii) 3X molecules of CO will occupies 3V litres.

(iii) Molar mass of CO2 = C + 2(O) = 12 + 2(16) = 44 g

1 mole of O2 weighs 32 g and occupy 22.4 lit. volume

∴ 8 g of O2 will occupy $\dfrac{22.4}{32}$ x 8 = 5.6 lit. vol. = Molar volume

If 8 g of O2 occupies 5.6 lit. vol.

And 1 mole of CO2 occupy 22.4 lit and weighs = 44 g at s.t.p.

∴ 5.6 lit. vol. of CO2 will weigh = $\dfrac{44}{22.4}$ x 5.6 = 11 g

Hence, mass of CO2 = 11 g

(iv) Avogadro's Law is used above.

#### Question 1(2005)

Define the term 'atomic weight'.

Atomic weight is the number of times one atom of an element is heavier than 1⁄12 the mass of an atom of carbon (C12)

#### Question 1(2008)

The gas law which relates the volume of a gas to the number of molecules of the gas is :

2. Gay-Lussac's Law
3. Boyle's Law
4. Charles' Law

#### Question 1(2009)

Correct the following — Equal masses of all gases under identical conditions contain the same number of molecules.

Equal volumes of all gases under identical conditions contain the same number of molecules.

#### Question 1(2013)

A vessel contains X number of molecules of H2 gas at a certain temperature & pressure. Under the same conditions of temperature & pressure, how many molecules of N2 gas would be present in the same vessel.

According to Avogadro's law, equal volume of all gases under similar conditions of temperature and pressure contain equal number of molecules.

Hence, number of molecules of N2 = Number of molecules of H2 = X

#### Question 1(2017)

A gas cylinder can hold 1 kg. of H2 at room temp. & press. :

(i) Find the number of moles of hydrogen present.

(ii) What weight of CO2 can the cylinder hold under similar conditions to temp. & press.

(iii) If the number of molecules of hydrogen in the cylinder is X, calculate the number of CO2 molecules in the cylinder under the same conditions of temp. & press.

(iv) State the law that helped you to arrive at the above result.

[H = 1, C = 12, O = 16]

(i) 2 g of hydrogen gas = 1 mole

∴ 1000 g of hydrogen gas = $\dfrac{1000}{2}$ = 500 moles.

(ii) Molar mass of CO2 = C + 2(O) = 12 + 2(16) = 44 g

1 mole of carbon dioxide = 44 g

∴ 500 moles of carbon dioxide = 44 x 500 = 22,000 g = 22 kg.

Hence, Weight of carbon dioxide in cylinder = 22 kg

(iii) According to Avogadro's law — Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.

∴ molecules in the cylinder of carbon dioxide = X.

(iv) Avogadro's law — Equal volume of all gases under similar conditions of temperature and pressure contain equal number of molecules.

#### Question 1(2018)

If 150 cc of a 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. Name the law on which the problem is based.

According to Avogadro’s law : equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.

Given,

150 cc of a gas A contains X molecules

Therefore, 150 cc of B will also contain X molecule.

Hence, 75 cc of B will contain X/2 molecules.

The problem is based on Avogadro's law.

## Vapour Density & Molecular Weight

#### Question 1(1996)

Find the relative molecular mass of a gas, 0.546 g of which occupies 360 cm3 at 87°C and 380 mm Hg pressure. (1 litre of hydrogen at s.t.p. weighs 0.09 g)

Given,

Initial ConditionsFinal Conditions (s.t.p.)
P1 = 380 mm of HgP2 = 760 mm of Hg
V1 = 360 cm3V2 = x
T1 = 87 + 273 K = 360 KT2 = 273 K

Using the gas equation,

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

Substituting the values we get,

$\dfrac{380 \times 360}{360} = \dfrac{760 \times x}{273} \\[0.5em] x = \dfrac{38 \times 273}{76 } \\[0.5em] x = \dfrac{10,374}{76 } \\[0.5em] x = 136.5 \text{ cm}^3 = 0.1365 \text{ lit.}$

At s.t.p. 0.1365 lit weighs 0.546 g

Therefore, 1 lit of gas weighs = $\dfrac{0.546}{0.1365}$ x 1 = 4 g

Vapour density of gas =

$\dfrac{\text{Wt. of certain volume of gas }}{\text {Wt. of same volume of H}_2} \\[0.5em] = \dfrac{4}{0.09} \\[0.5em] = 44.444$

Molecular weight = 2 x Vapour density
= 2 x 44.444 = 88.888 = 88.89 g

Hence, relative molecular mass of gas 88.89 g

#### Question 1(2001)

State the term defined by: The mass of a given volume of gas compared to the mass of an equal volume of H2.

Vapour Density.

#### Question 1(2004)

2KMnO4 ⟶ K2MnO4 + MnO2 + O2

K2MnO4 + MnO2 is the solid residue.

Potassium permanganate was heated in a test tube. After collecting one litre of oxygen at room temperature, it was found that the test tube had undergone a loss in mass of 1.32 g. If one litre of H2 under the same conditions of temperature and pressure has a mass of 0.0825 g. Calculate the relative molecular mass of oxygen.

2KMnO4 ⟶ K2MnO4 + MnO2 + O2

Loss in mass = 1.32 g = 1 lit of oxygen

Vapour density of gas =

$\dfrac{\text{Wt. of certain volume of gas }}{\text {Wt. of same volume of H}_2} \\[0.5em] = \dfrac{1.32}{0.0825} \\[0.5em] = 16 \text{ g}$

Molecular weight = 2 x Vapour density
= 2 x 16 = 32 g

Hence, relative molecular mass of oxygen is 32 g

#### Question 1(2009)

A gas cylinder of capacity of 20 dm3 is filled with gas X the mass of which is 10 g. When the same cylinder is filled with hydrogen gas at the same temperature and pressure the mass of the hydrogen is 2 g, hence the relative molecular mass of the gas is :

1. 5
2. 10
3. 15
4. 20

10

Working

Vapour density of gas =

$\dfrac{\text{Wt. of certain volume of gas }}{\text {Wt. of same volume of H}_2} \\[0.5em] = \dfrac{10}{2} \\[0.5em] = 5 \text{ g}$

Molecular weight = 2 x Vapour density
= 2 x 5 = 10 g

Hence, relative molecular mass of gas is 10 g

#### Question 1(2012)

The vapour density of carbon dioxide [C = 12, O = 16] is

1. 32
2. 16
3. 44
4. 22

22

Working

Molecular weight of carbon dioxide = 12 + 2(16) = 12 + 32 = 44 g

Vapour density of gas =

$\dfrac{\text{Molecular weight}}{2} \\[0.5em] = \dfrac{44}{2} \\[0.5em] = 22 \text{ g}$

Hence, vapour density of carbon dioxide is 22 g

#### Question 1(2014)

Give one word or phrase for : The ratio of the mass of a certain volume of gas to the mass of an equal volume of hydrogen under the same conditions of temperature and pressure.

Vapour density.

## Percentage Composition

#### Question 1(1996)

Find the total percentage of oxygen in magnesium nitrate crystals : Mg(NO3)2. 6H2O

[O = 16, N = 14, H = 1, Mg = 24]

Molecular weight of magnesium nitrate Mg(NO3)2. 6H2O

= 24 + 2[14 + 3(16)] + 6[2(1) + 16]
= 24 + 2[14 + 48] + 6(18)
= 24 + 2(62) + 6(18)
= 24 + 124 + 108
= 256 g

Mass of oxygen in Mg(NO3)2. 6H2O
= 6(16) + 6(16)
= 96 + 96 = 192

256 g of magnesium nitrate crystals contains 192 g of oxygen

∴ 100 g of magnesium nitrate crystals contains = $\dfrac{192}{256}$ x 100 = 75%

Hence, total percentage of Oxygen in magnesium nitrate crystals is 75%

#### Question 1(1997)

What is the mass of nitrogen in 1000 kg of urea [CO(NH2)2].

[C = 12] (Answer to the nearest kg.)

Molecular weight of urea [CO(NH2)2]
= 12 + 16 + 2[14 + 2(1)]
= 12 + 16 + 2 = 28 + 32
= 60 g

60 g of urea contains 28 g of nitrogen

∴ 1000 kg of urea will contain = $\dfrac{28}{60}$ x 1000 = 466.66 g ≈ 467 g

Hence, mass of nitrogen is 467 g

#### Question 1(1998)

Calculate the % of boron [B] in borax Na2B4O7.10H2O.

[H = 1, B = 11, O = 16, Na = 23]

Molecular weight of borax Na2B4O7.10H2O
= 2(23) + 4(11) + 7(16) + 10[2(1)+ 16]
= 46 + 44 + 112 + 180 = 382 g

382 g of borax contains 44 g of boron

∴ 100 g of borax will contain = $\dfrac{44}{382}$ x 100 = 11.5%

Hence, Percentage of boron is 11.5%

#### Question 1(1999)

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

[N = 14 ; O = 16 ; Ca = 40]

Molecular weight of Ca(NO3)2
= 40 + 2[14 + 3(16)]
= 40 + 2[14 + 48]
= 40 + 2(62)
= 40 + 124 = 164 g

Total nitrogen that has to be replaced in 10 hectares of land = 20 x 10 = 200 kg

28 g of N is present in 164 kg of calcium nitrate

∴ 200 kg of N is present in = $\dfrac{164}{28}$ x 200 = 1171 kg

Hence, 1171 kg of fertilizer will be required

#### Question 1(2001)

Calculate the percentage of phosphorus in the fertilizer superphosphate Ca(H2PO4)2. (correct to 1 dp)

[H = 1 ; O = 16 ; P = 31 ; Ca = 40]

Molecular weight of Ca(H2PO4)2
= 40 + 2[2(1) + 31 + 4(16)]
= 40 + 2[2 + 31 + 64]
= 40 + 2
= 40 + 194
= 234 g

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

∴ 100 g of Ca(H2PO4)2 will contain = $\dfrac{62}{234}$ x 100 = 26.49% = 26.5%

Hence, 26.5% phosphorous is present in superphosphate Ca(H2PO4)2

#### Question 1(2002)

Calculate the percentage of platinum in ammonium chloroplatinate (NH4)2 PtCl6

[N = 14, H = 1, Cl = 35.5, Pt = 195]

Molecular weight of ammonium chloroplatinate (NH4)2PtCl6
= 2[14 + 4(1)] + 195 + 6(35.5)
= 2 + 195 + 213
= 444 g

444 g of ammonium chloroplatinate contains 195 g of Pt

∴ 100 g of ammonium chloroplatinate will contain = $\dfrac{195}{444}$ x 100 = 43.91% = 44%

Hence, 44% percentage of platinum is present in ammonium chloroplatinate

#### Question 1(2005)

Calculate the percentage of nitrogen in aluminium nitride.

[Al = 27, N = 14]

Molecular weight of aluminium nitride AlN = 27 + 14
= 41 g

41 g of aluminium nitride contains 14 g of N

∴ 100 g of aluminium nitride will contain = $\dfrac{14}{41}$ x 100 = 34.15%

Hence, percentage of nitrogen in aluminium nitride is 34.15%

#### Question 1(2006)

Calculate the percentage of sodium in sodium aluminium fluoride (Na3AlF6) correct to the nearest whole number.

[F = 19 ; Na = 23 ; Al = 27]

Molecular weight of sodium aluminium fluoride (Na3AlF6)
= 3(23) + 27 + 6(19)
= 69 + 27 + 114
= 210 g

210 g of sodium aluminium fluoride contains 69 g of Na

∴ 100 g of sodium aluminium fluoride will contain = $\dfrac{69}{210}$ x 100 = 32.85% = 33%

Hence, percentage of sodium in sodium aluminium fluoride (Na3AlF6) is 33%

#### Question 1(2007)

Determine the percentage of oxygen in ammonium nitrate

[O = 16]

Molecular weight of ammonium nitrate NH4NO3 = 14 + 4(1) + 14 + 3(16)
= 32 + 48 = 80 g

80 g of ammonium nitrate contains 48 g of oxygen

∴ 100 g of ammonium nitrate will contain = $\dfrac{48}{80}$ x 100 = 60%

Hence, percentage of oxygen in ammonium nitrate is 60%

#### Question 1(2010)

If the relative molecular mass of ammonium nitrate is 80, calculate the percentage of nitrogen and oxygen in ammonium nitrate.

[N = 14, H = 1, O = 16]

Given,

molecular mass of ammonium nitrate = 80

mass of nitrogen in 1 mole of ammonium nitrate (NH4NO3) is 2(14) = 28 g

percentage of nitrogen in ammonium nitrate = $\dfrac{28}{80}$ x 100 = 35%

mass of oxygen in 1 mole of ammonium nitrate (NH4NO3) is 3(16) = 48 g

percentage of oxygen in ammonium nitrate = $\dfrac{48}{80}$ x 100 = 60%

Hence, percentage of nitrogen is 35% and percentage of oxygen is 60%

#### Question 1(2012)

Find the total percentage of magnesium in magnesium nitrate crystals, Mg(NO3)2.6H2O

[Mg = 24; N = 14; O = 16 and H = 1]

Molecular weight of Mg(NO3)2.6H2O = 24 + 2[14 + 3(16)] + 6[2(1) + 16]
= 24 + 2[14 + 48] + 6
= 24 + 2(62) + 108
= 24 + 124 + 108 = 256 g

256 g of magnesium nitrate crystals contains 24 g of magnesium

∴ 100 g of magnesium nitrate crystals will contain = $\dfrac{24}{256}$ x 100 = 9.375% = 9.38%

Hence, percentage of magnesium in magnesium nitrate crystals is 9.38%

#### Question 1(2017)

Calculate the percentage of water of crystallization in CuSO4.5H2O

[H = 1, O = 16, S = 32, Cu = 64]

Molecular weight of CuSO4.5H2O = 64 + 32 + 4(16) + 5[2(1) + 16]
= 64 + 32 + 64 + 5(18) = 160 + 90 = 250 g

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

∴ 100 g of CuSO4.5H2O will contain = $\dfrac{90}{250}$ x 100 = 36%

Hence, percentage of water of crystallization is 36%

#### Question 1(2020)

Calculate the percentage of (i) Fluorine (ii) Sodium (iii) Aluminium in sodium aluminium fluoride [Na3AlF6] to the nearest whole number

[Na = 23, Al = 27, F = 19]

(i) Molecular weight of sodium aluminium fluoride (Na3AlF6)
= 3(23) + 27 + 6(19)
= 69 + 27 + 114
= 210 g

210 g of sodium aluminium fluoride contains 114 g of fluorine

∴ 100 g of sodium aluminium fluoride will contain = $\dfrac{114}{210}$ x 100 = 54.28% = 54%

(ii) 210 g of sodium aluminium fluoride contains 69 g of sodium

∴ 100 g of sodium aluminium fluoride will contain = $\dfrac{69}{210}$ x 100 = 32.85% = 33%

(iii) 210 g of sodium aluminium fluoride contains 27 g of aluminium

∴ 100 g of sodium aluminium fluoride will contain = $\dfrac{27}{210}$ x 100 = 12.85% = 13%

Hence, percentage of F = 54%, Na = 33%, Al = 13%

## Empirical & Molecular Formula

#### Question 1(2008)

What is the empirical formula of octane (C8H18)

Molecular Formula of octane = C8H18 = (C4H9)2

Molecular formula = n[Empirical formula]

n = 2

Hence, Empirical formula = C4H9

#### Question 2(2008)

A compound contains — Carbon 14.4%, hydrogen 1.2% and chlorine 84.5%. Determine the empirical formula of this compound. Work correct to 1 decimal place. The relative molecular mass of this compound is 168, so what is it's molecular formula?

[C = 12; H = 1; Cl = 35.5]

Element% compositionAt. wt.Relative no. of atomsSimplest ratio
Carbon14.412$\dfrac{14.4}{12}$ = 1.2$\dfrac{1.2}{1.2}$ = 1
Hydrogen1.21$\dfrac{1.2}{1}$ = 1.2$\dfrac{1.2}{1.2}$ = 1
chlorine84.535.5$\dfrac{84.5}{35.5}$ = 2.38$\dfrac{2.38}{1.2}$ = 1.98 = 2

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

Hence, empirical formula is CHCl2

Empirical formula weight = 12 + 1 + 2(35.5) = 84 g

Relative molecular mass = 168

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{168}{84} = 2$

Molecular formula = n[E.F.] = 2[CHCl2] = C2H2Cl4

Molecular formula = C2H2Cl4

#### Question 1(2009)

A gaseous compound of nitrogen and hydrogen contains 12.5% hydrogen by mass. Find the molecular formula of the compound if it's relative molecular mass is 37.

[N =14, H = 1].

Element% compositionAt. wt.Relative no. of atomsSimplest ratio
Nitrogen100 - 12.5 = 87.514$\dfrac{87.5}{14}$ = 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

Empirical formula weight = 14 + 2(1) = 16

Relative molecular mass = 37

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{37}{16} = 2.31 = 2$

∴ Molecular formula = n[E.F.] = 2[NH2] = N2H4

#### Question 1(2011)

An organic compound has vapour density 94. It contains C = 12.67%, H = 2.13%, and Br = 85.11%. Find the molecular formula of the organic compound.

[C = 12, H = 1, Br = 80]

Element% compositionAt. wt.Relative no. of atomsSimplest ratio
C12.6712$\dfrac{12.67}{12}$ = 1.05$\dfrac{1.05}{1.05}$ = 1
H2.131$\dfrac{2.13}{1}$ = 2.13$\dfrac{2.13}{1.05}$ = 2.02 = 2
Br85.1180$\dfrac{85.11}{80}$ = 1.06$\dfrac{1.06}{1.05}$ = 1

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

Hence, empirical formula is CH2Br

Empirical formula weight = 12 + 2(1) + 80 = 94

Vapour density (V.D.) = 94

Molecular weight = 2 x V.D. = 2 x 94

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{2 \times 94}{94} = 2$

∴ Molecular formula = n[E.F.] = 2[CH2Br] = C2H4Br2

#### Question 1(2014)

A compound having empirical formula X2Y is made of two elements X and Y. Find it's molecular formula if the atomic weight of X is 10 and that of Y is 5 and the compound has a vapour density 25.

Empirical formula is X2Y

Empirical formula weight = 2(10) + 5 = 25

Vapour density (V.D.) = 25

Molecular weight = 2 x V.D. = 2 x 25

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{2 \times 25}{25} = 2$

∴ Molecular formula = n[E.F.] = 2[X2Y] = X4Y2

#### Question 1(2015)

If the empirical formula of a compound is CH and it has a vapour density of 13, find the molecular formula of the compound.

[C = 12, H = 1]

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

#### Question 1(2016)

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

[C = 12, H = 1]

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$\dfrac{17.24}{6.89}$ = 2.5

Simplest ratio of whole numbers = C : H = 1 : 2.5 = 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 1(2017)

A compound of X and Y has the empirical formula XY2. It's vapour density is equal to it's empirical formula weight. Determine it's molecular formula.

Empirical formula = XY2

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[XY2] = X2Y4

#### Question 1(2018)

The percentage composition of a gas is : Nitrogen 82.35%, Hydrogen 17.64%. Find the empirical formula of the gas.

[N = 14, H = 1]

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 1(2019)

Molecular formula of a compound is C6H18O3. Find it's empirical formula.

Molecular formula of a compound = C6H18O3 = 3[C2H6O]

As, molecular formula = n[E.F.] and n = 3

Hence, empirical formula = C2H6O

#### Question 2(2019)

Give the appropriate term defined by the statement given: The formula that represents the simplest ratio of the various elements present in one molecule of the compound.

Empirical formula

#### Question 3(2019)

Find the empirical and molecular formula of an organic compound from the data given: C = 75.92%, H = 6.32% and N = 17.76%. The vapour density of the compound is 39.5

[C = 12, H = 1, N = 14]

Element% compositionAt. wt.Relative no. of atomsSimplest ratio
Carbon75.9212$\dfrac{75.92}{12}$ = 6.32$\dfrac{ 6.32 }{ 1.26}$ = 5
Hydrogen6.321$\dfrac{6.32}{1}$ = 6.32$\dfrac{6.32 }{ 1.26}$ = 5
Nitrogen17.7614$\dfrac{ 17.76}{14}$ = 1.26$\dfrac{ 1.26}{ 1.26}$ = 1

Simplest ratio of whole numbers = C : H : N = 5 : 5 : 1

Hence, empirical formula is C5H5N

Empirical formula weight = 5(12) + 5(1) + 14 = 79

V.D. = 39.5

Molecular weight = 2 x V.D. = 2 x 39.5 = 79

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{79}{79} = 1$

∴ Molecular formula = n[E.F.] = 1[C5H5N] = C5H5N

## Chemical Equations

#### Question 1(2000)

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

$\begin{matrix} \text{Na}_2\text{CO}_3.10\text{H}_2\text{O} & \xrightarrow{\Delta} & \text{Na}_2\text{CO}_3 & + & 10\text{H}_2\text{O} \\ 2(23) + 12 + 3 (16) & & 2(23) + 12 & & 10[2(1) \\ + 10[2(1) + 16] & & + 3(16) & & + 16] \\ = 46 + 12 + 48 & & = 46 + 12 & & = 180 \text{ g} \\ + 10(18) & & + 48 \\ = 286 \text{ g} & & = 106 \text{ g} \\ \end{matrix}$

286 g of washing soda had 106 g of anhydrous sodium carbonate

∴ 57.2 g will have = $\dfrac{106}{286}$ x 57.2 = 21.2 g anhydrous sodium carbonate.

Hence, 21.2 g anhydrous sodium carbonate is left.

#### Question 2(2000)

Na2SO4 + Pb(NO3)2 ⟶ PbSO4 + 2NaNO3. When excess lead nitrate solution was added to a solution of sodium sulphate, 15.15 g of lead sulphate were precipitated. What mass of sodium sulphate was present in the original solution.

[H = 1 ; C = 12 ; O = 16 ; Na = 23; S = 32 ; Pb = 207]

$\begin{matrix} \text{Na}_2\text{SO}_4 & + & \text{Pb}(\text{NO}_3)_2 & \longrightarrow &\text{PbSO}_4 & + & 2\text{NaNO}_3 \\ 2(23) + 32 & & & & 207 + 32 \\ + 4(16) & & & & + 4(16) \\ = 46 + 32 & & & & = 207 + 32 \\ + 64 & & & & + 64 \\ = 142 \text{ g} & & & & = 303 \text{ g} \\ \end{matrix}$

303 g of lead sulphate was obtained from 142 g of Na2SO4

∴ 15.15 g of lead sulphate will be precipitated from $\dfrac{142}{303}$ x 15.15 = 7.1 g.

Hence, 7.1 g of sodium sulphate was present in the original solution.

#### Question 1(2001)

From the equation : (NH4)2 Cr2O7 ⟶ Cr2O3 + 4H2O + N2 Calculate :

(i) the vol. of nitrogen at STP, evolved when 63 g. of ammonium dichromate is heated.

(ii) the mass of Cr2O3 formed at the same time.

[N = 14, H = 1, Cr = 52, O = 16]

$\begin{matrix} (\text{NH}_4)_2\text{Cr}_2\text{O}_7 & \xrightarrow{\Delta} & \text{Cr}_2\text{O}_3 & + & 4\text{H}_2\text{O} & + \text{ N}_2 \\ 2[14 + 4(1)] & & 2(52) & & & 1 \text{ mol.} \\ + 2(52) + 7(16) & & + 3(16) & & & \text{vol.} \\ = 36 + 104 + 112 & & = 104 + 48 & & & = 22.4 \text{ lit} \\ = 252 \text{ g} & & = 152 \text{ g} \\ \end{matrix}$

252 g of ammonium dichromate gives 22.4 lit of nitrogen

∴ 63 g of ammonium dichromate will give $\dfrac{22.4}{252}$ x 63 = 5.6 lit of nitrogen.

Hence, 5.6 lit of nitrogen is evolved.

(ii) 252 g of ammonium dichromate gives 152 g Cr2O3

∴ 63 g of ammonium dichromate will give $\dfrac{152}{252}$ x 63 = 38 g

Hence, 38 g of Cr2O3 is formed.

#### Question 1(2003)

10 g. of a mixture of NaCl and anhydrous Na2SO4 is dissolved in water. An excess of BaCl2 soln. is added and 6.99 g. of BaSO4 is precipitated according to the equation :

Na2SO4 + BaCl2 → BaSO4↓ + 2NaCl.

Calculate the percentage of Na2SO4 in the original mixture.

[O = 16 ; Na = 23 ; S = 32; Ba = 137]

$\begin{matrix} \text{Na}_2\text{SO}_4 & + & \text{BaCl}_2 & \longrightarrow & \text{BaSO}_4 \downarrow & + & 2\text{NaCl} \\ 2(23) + 32 & & & & 137 + 32 \\ + 4(16) & & & & + 4(16) \\ = 46 + 32 & & & & = 137 + 32 \\ + 64 & & & & + 64 \\ = 142 \text{ g} & & & & = 233 \text{ g} \\ \end{matrix}$

233 g of BaSO4 is obtained from 142 g of Na2SO4

6.99 g of BaSO4 will be obtained from $\dfrac{142}{233}$ x 6.99 = 4.26 g of Na2SO4.

∴ In 10 g mixture $\dfrac{4.26}{10}$ x 100 = 42.6% Na2SO4 is present.

Hence, 42.6% Na2SO4 is present in the original mixture

#### Question 1(2004)

The reaction of potassium permanganate with acidified iron (II) sulphate is given below :

2KMnO4 + 10FeSO4 + 8H2SO4 ⟶ K2SO4 + 2MnSO4 + 5 Fe2(SO4)3 + 8H2O.

If 15.8 g. of potassium permanganate was used in the reaction, calculate the mass of iron (II) sulphate used in the above reaction.

[K = 39, Mn = 55, Fe = 56, S = 32, O = 16]

$\begin{matrix} 2\text{KMnO}_4 & + 10 \text{FeSO}_4 & + & 8\text{H}_2\text{SO}_4 \\ 2[39 + 55 & 10[56+ 32 \\ + 4(16)] & + 4(16)] \\ = 2[39 + & 10[56 + \\ 55 + 64] & 32 + 64] \\ = 316 \text{ g} & = 1520\text{ g} \\ \end{matrix}$

$\longrightarrow \text{K}_2\text{SO}_4 + 2\text{MnSO}_4 + 5 \text{Fe}_2(\text{SO}_4)_3 + 8\text{H}_2\text{O}$

316 g of potassium permanganate was used with 1520 g of iron (II) sulphate

∴ 15.8 g. of potassium permanganate will be used with $\dfrac{1520}{316}$ x 15.8 = 76 g of iron (II) sulphate

Hence, 76 g of iron (II) sulphate was used.

#### Question 1(2005)

The equations given below relate to the manufacture of sodium carbonate [Mol. wt. of Na2CO3 = 106]

(i) NaCl + NH3 + CO2 + H2O ⟶ NaHCO3 + NH4Cl

(ii) 2NaHCO3 ⟶ Na2CO3 + H2O + CO2

Questions (a) and (b) are based on the production of 21.2 g. of sodium carbonate.

(a) What mass of sodium hydrogen carbonate must be heated to give 21.2 g. of sodium carbonate [Molecular weight of NaHCO3 = 84].

(b) To produce the mass of sodium hydrogen carbonate calculated in (a), what volume of carbon dioxide, measured at s.t.p., would be required.

$\begin{matrix} 2\text{NaHCO}_3 & \longrightarrow & \text{Na}_2\text{CO}_3 & + & \text{H}_2\text{O} + \text{CO}_2 \\ 2 \text{mol.} & & 1 \text{mol.} \\ 2 \times 84 & & 106\text{ g.} \\ = 168\text{ g.} & & \end{matrix}$

106 g of Na2CO3 is obtained from 168 g of NaHCO3

∴ 21.2 g. of Na2CO3 is obtained from $\dfrac{168}{106}$ x 21.2 = 33.6 g of NaHCO3

(ii)

$\begin{matrix} \text{NaCl} & + & \text{NH}_3 & + & \text{CO}_2 & + & \text{H}_2\text{O} \\ & & & & 1\text{mol.} & \\ & & & & 22.4 \text{lit.} \\ \longrightarrow & \text{NaHCO}_3 & + & \text{NH}_4\text{Cl} \\ & 1 \text{mol.} \\ & 84\text{ g.} \end{matrix}$

84 g of NaHCO3 is obtained from 22.4 lit of CO2

∴ 33.6 g. of NaHCO3 is obtained from $\dfrac{22.4}{84}$ x 33.6 = 8.96 lit.

Hence, 8.96 lit of CO2 is required.

#### Question 1(2006)

The relative molecular mass [mol. wt.] of copper oxide is 80. What vol. of NH3 (measured at s.t.p.) is required to completely reduce 120 g of CuO.

[3CuO + 2NH3 ⟶ 3Cu + 3H2O + N2].

$\begin{matrix} 3\text{CuO} & + & 2\text{NH}_3 \\ 3\text{ mol.} & & 2 \text{ mol.} \\ = 3 \times 80 \text{ g.} & & = 2 \times 22.4 \text{ lit.} \\ = 240 \text{ g.} & & = 44.8\text{ lit.} \\ \end{matrix}$

$\longrightarrow 3\text{Cu} + 3\text{H}_2\text{O} + \text{N}_2$

240 g of CuO is reduced by 44.8 lit of NH3

∴ 120 g of CuO is reduced by $\dfrac{44.8}{240}$ x 120 = 22.4 lit.

Hence, 22.4 lit of NH3 is required

#### Question 1(2007)

A sample of ammonium nitrate when heated yields 8.96 litres of steam (measured at stp.)

NH4NO3 ⟶ N2O + 2H2O

(i) What volume of di nitrogen oxide is produced at the same time as 8.96 litres of steam.

(ii) What mass of ammonium nitrate should be heated to produce 8.96 litres of steam [Relative molecular mass of NH4NO3 is 80]

$\begin{matrix} \text{NH}_4\text{NO}_3 &\xrightarrow{\Delta} & \text{N}_2\text{O} & + & 2\text{H}_2\text{O} \\ 1\text{ vol.} = 80 \text{g} & & 1 \text{ vol.} & & 2\text{ vol.} \\ \end{matrix}$

(i) Given,

1 vol. of di nitrogen produced at the same time as 8.96 litres of steam (2 vol)

Hence, Vol of di nitrogen = $\dfrac{8.96}{2}$ = 4.48 lit.

Hence, volume of di nitrogen oxide produced = 4.48 lit.

(ii) 2 vol = (2 x 22.4) lit steam is produced from 80 g NH4NO3

∴ 8.96 lit of steam will be produced by $\dfrac{80}{2 \times 22.4}$ x 8.96 = 16 g

Hence, 16 g of ammonium nitrate is required to be heated.

#### Question 1(2008)

From the equation : C + 2H2SO4 ⟶ CO2 + 2H2O + 2SO2

Calculate :

(i) The mass of carbon oxidized by 49 g of sulphuric acid [C = 12 ; relative molecular mass of H2SO4 = 98].

(ii) The volume of SO2 measured at s.t.p., liberated at the same time.

$\underset{12 \text{g}}{\text{C}} + \underset{2(98) = 196 \text{ g}}{2\text{H}_2\text{SO}_4} \longrightarrow \text{CO}_2 + 2\text{H}_2\text{O} + 2\text{SO}_2$

(i) 196 g of sulphuric acid oxidizes 12 g carbon

∴ 49 g of sulphuric acid will oxidize = $\dfrac{12}{196}$ x 49 = 3 g

Hence, 3 g of carbon is oxidized.

(ii) 12 g carbon liberates 2 vol = (2 x 22.4) lit of SO2

∴ 3 g of carbon will liberate $\dfrac{2 \times 22.4}{12}$ x 3 = 11.2 lit of SO2.

Hence, 11.2 lit of SO2 is liberated.

#### Question 1(2009)

Commercial NaOH weighing 30 g. has some NaCl in it. The mixture on dissolving in water and treatment with excess AgNO3 soln. formed a precipitate weighing 14.3 g. What is the percentage of NaCl in the commercial sample of NaOH.

NaCl + AgNO3 ⟶ AgCl + NaNO3

[Relative molecular mass of NaCl = 58 ; AgCl = 143]

$\underset{58 \text{g}}{\text{NaCl}} + \text{AgNO}_3 \longrightarrow \underset{143 \text{ g}}{\text{AgCl}} + \text{NaNO}_3$

(i) 143 g AgCl is formed by 58 g NaCl

∴ 14.3 g of AgCl will be formed by $\dfrac{58}{143}$ x 14.3 = 5.8 g

Percentage of NaCl = $\dfrac{5.8}{30}$ x 100 = 19.33%

Hence, percentage of NaOH is 19.33%

#### Question 1(2011)

Calculate the volume of oxygen required for the complete combustion of 8.8 g of propane [C3H8].

[C = 12, O = 16, H = 1, Molar Volume = 22.4 dm3 at s.t.p.]

$\begin{matrix} \text{C}_3\text{H}_8 & + &5\text{O}_2 & \longrightarrow & 3\text{CO}_2 & + & 4\text{H}_2\text{O} \\ 3(12) + 8(1) & & 5 \text{ vol} \\ = 44 \text{ g}& & 5 (22.4) \text{ lit} \\ \end{matrix}$

(i) 44 g propane requires 5 x 22.4 lit of oxygen

∴ 8.8 g of propane will require $\dfrac{5 \times 22.4}{44}$ x 8.8 = 22.4 lit.

Hence, 22.4 lit of Oxygen is required.

#### Question 1(2012)

P + 5HNO3 [conc.] ⟶ H3PO4 + H2O + 5NO2.

If 9.3 g of phosphorous was used in the reaction, calculate :

(i) Number of moles of phosphorous taken.

(ii) The mass of phosphoric acid formed.

(iii) The volume of NO2 produced at s.t.p.

[H = 1, N = 14, P = 31, O = 16]

$\begin{matrix} \text{P} & + & 5\text{HNO}_3 & \longrightarrow & \text{H}_3\text{PO}_4 & + & \text{H}_2\text{O} & + & 5\text{NO}_2 \\ 31 \text{ g} & & [conc.] & & 3(1) + 31 \\ & & & & + 4(16) \\ & & & & = 98 \text{ g} \\ \end{matrix}$

(i) 31 g of P = 1 mole

∴ 9.3 g of P = $\dfrac{1}{31}$ x 9.3 = 0.3 moles.

Hence, 0.3 moles of phosphorous was taken for the reaction.

(ii) 31 g of P forms 98 g of phosphoric acid

∴ 9.3 g will form $\dfrac{ 98}{31}$ x 9.3 = 29.4 g.

Hence, 29.4 g. of phosphoric acid is formed

(iii) 31 g of P produces 5 vol = 5 x 22.4 lit.

∴ 9.3 g will produce = $\dfrac{5 \times 22.4}{31}$ x 9.3 = 33.6 lit.

#### Question 1(2013)

2KClO3 $\xrightarrow{\text{MnO}_2}$ 2KCl + 3O2

(i) Calculate the mass of KClO3 required to produce 6.72 lit of O2 at s.t.p.

[K = 39, Cl = 35.5, O = 16]

(ii) Calculate the number of moles of O2 in the above volume and also the number of molecules.

$\begin{matrix} 2\text{KClO}_3 & \xrightarrow{MnO_2} & 2\text{KCl} & + & 3\text{O}_2 \\ 2[39 + 35.5 & & 2[39 & & 3(22.4) \\ + 3(16)] & & + 35.5] & & = 67.2 \text{ lit.} \\ = 245 \text{ g} & & = 149 \text{g} \\ \end{matrix}$

67.2 lit. of O2 is produced by 245 g of KClO3

∴ 6.72 lit of O2 will be obtained from $\dfrac{245}{67.2}$ x 6.72 = 24.5 g.

(ii) 22.4 lit = 1 mole
∴ 6.72 lit = $\dfrac{1}{22.4}$ x 6.72 = 0.3 moles

1 mole = 6.023 x 1023 molecules
∴ 0.3 moles = 0.3 x 6.023 x 1023 molecules.

#### Question 1(2015)

From the equation :

(NH₄)₂Cr₂O₇ ⟶ N₂(g) + 4H₂O(g) + Cr₂O₃

Calculate:

(i) the quantity in moles of (NH₄)₂Cr₂O₇ if 63 gm of (NH₄)₂Cr₂O₇ is heated.

(ii) the quantity in moles of N₂ formed.

(iii) the volume in litres or dm3 of N₂ evolved at s.t.p.

(iv) the mass in grams of Cr₂O₃ formed at the same time.

[H = 1, Cr = 52, N = 14]

$\begin{matrix} (\text{NH}_4)_2\text{Cr}_2\text{O}_7 & \xrightarrow{\Delta} & \text{N}_2 & + & 4\text{H}_2\text{O} & + & \text{Cr}_2\text{O}_3 \\ 2[14 + 4(1)] & & & & & & 2(52) \\ + 2(52) + 7(16) & & & & & & + 3(16) \\ = 36 + 104 & & & & & & = 104 + 48 \\ + 112 = 252 \text{ g} & & & & & & = 152 \text{ g} \\ \end{matrix}$

(i) 252 g of (NH4)2Cr2O7 = 1 mole

∴ 63 g of (NH4)2Cr2O7 = $\dfrac{1}{252}$ x 63 = 0.25 moles

Hence, no. of moles = 0.25 moles

(ii)

$\begin{matrix} (\text{NH}_4)_2\text{Cr}_2\text{O}_7 & : & \text{N}_2 \\ 1 \text{ mol.} & : & 1 \text{ mol.} \\ 0.25 \text{ mol.} & : & x \\ \end{matrix}$

Hence, 0.25 moles of (NH4)2Cr2O7 will produce 0.25 moles of nitrogen.

(iii) 1 mole of N₂ occupies 22.4 lit.

∴ 0.25 moles of N₂ will occupy = 22.4 x 0.25 = 5.6 lit.

Hence, volume of N₂ evolved at s.t.p = 5.6 lit.

(iv) 1 mole of Cr2O7 = 152 g.

∴ 0.25 moles of Cr2O7 = 152 x 0.25 = 38 g.

Hence, mass in gms of Cr2O7 formed = 38 g.

#### Question 1(2016)

How much calcium oxide is formed when 82 g. of calcium nitrate is heated. Also find the volume of nitrogen dioxide evolved :

2Ca(NO3)2 ⟶ 2CaO + 4NO2 + O2

[Ca = 40, N = 14, O = 16]

$\begin{matrix} 2\text{Ca}(\text{NO}_3)_2 & \longrightarrow & 2\text{CaO} & + & 4\text{NO}_2 & + & \text{O}_2 \\ 2[(40) + & & 2(40 & & 4[22.4] \\ 2(14 +3(16))] & & + 16) & & \text{ lit.} \\ = 328 \text{ g} & & 112 \text{g} \\ \end{matrix}$

(i) 328 g of Ca(NO3)2 produces 112 g of calcium oxide.

∴ 82 g of Ca(NO3)2 will produce = $\dfrac{112}{328}$ x 82 = 27.99 g = 28 g

Hence, 28 g of calcium oxide is produced.

(ii) 328 g of Ca(NO3)2 produces 4(22.4) lit of nitrogen dioxide.

∴ 82 g of Ca(NO3)2 will produce = $\dfrac{4 \times 22.4}{328}$ x 82 = 22.4 lit.

Hence, vol of nitrogen dioxide evolved = 22.4 lit.

#### Question 1(2018)

Aluminium carbide reacts with water according to the following equation :

Al4C3 + 12H2O ⟶ 4Al(OH)3 + 3CH4

(i) State what mass of aluminium hydroxide is formed from 12 g. of aluminium carbide.

(ii) State the volume of methane at s.t.p. obtained from 12 g of aluminium carbide.

[relative molecular weight of Al4Cl3] = 144 ; Al(OH)3 = 78

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

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 at s.t.p. = 5.6 lit.