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]

Answer

(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]

Answer

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]

Answer

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]

Answer

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]

Answer

(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]

Answer

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]

Answer

Molecular weight of hydrated copper sulphate CuSO₄.5H₂O = 63.5 + 32 + 4(16) + 5[2(1) + 16] = 63.5 + 32 + 64 + 5[18] = 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]

Answer

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]

Answer

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]

Answer

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%.

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.

Answer

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]

Answer

Original weight = 1.30 g

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]

Answer

Given,

Vapour density_X = 13

Vapour density_Y = 39

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

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

Let formula for X be C_aH_b

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 C_cH_d

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 Y = 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]

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.

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]

Answer

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]

Answer

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]

Answer

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]

Answer

H = 0.70 - 0.60 = 0.1 g

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]

Answer

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 ]

Answer

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]

Answer

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

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]

Answer

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

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]

Answer

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

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]

Answer

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

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]

Answer

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

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].

Answer

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

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].

Answer

(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].

Answer

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].

Answer

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

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

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].

Answer

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

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 ?

Answer

[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

Answer

[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]

Answer

[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.

Answer

[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.

Answer

[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.

Answer

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.

Answer

[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.

Answer

[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]

Answer

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]

Answer

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 ?

Answer

(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

Answer

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.

Answer

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.]

Answer

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]

Answer

(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]

Answer

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.

Answer

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 ?

Answer

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.

Answer

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)

State Avogadro's Law.

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]

Answer

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]

Answer

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]

Answer

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

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

Answer

(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.

Answer

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

Answer

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]

Answer

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.

Answer

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]

Answer

(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.

Answer

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]

Answer

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]

Answer

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]

Answer

(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]

Answer

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'.

Answer

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 :

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

Answer

Avogadro's Law

Question 1(2009)

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

Answer

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.

Answer

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]

Answer

(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.

Answer

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)

Answer

Given,

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.

Answer

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.

Answer

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

Answer

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

Answer

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.

Answer

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]

Answer

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.)

Answer

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

Converting given weight of urea in grams,
1000 kg of urea = 1000 x 10³ g of urea

60 g of urea contains 28 g of nitrogen

∴ 1000 x 10³ g of urea will contain = $\dfrac{28}{60}$ x 1000 x 10³ = 466.66 x 10³ g ≈ 467 x 10³ g = 467 kg of nitrogen.

Hence, mass of nitrogen is 467 kg

Question 1(1998)

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

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

Answer

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]

Answer

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]

Answer

Molecular weight of Ca(H₂PO₄)₂
= 40 + 2[2(1) + 31 + 4(16)]
= 40 + 2[2 + 31 + 64]
= 40 + 2[97]
= 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

(Give your answer correct to the nearest whole number.

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

Answer

Molecular weight of ammonium chloroplatinate (NH₄)₂PtCl₆
= 2[14 + 4(1)] + 195 + 6(35.5)
= 2[18] + 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]

Answer

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]

Answer

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]

Answer

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]

Answer

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]

Answer

Molecular weight of Mg(NO₃)₂.6H₂O = 24 + 2[14 + 3(16)] + 6[2(1) + 16]
= 24 + 2[14 + 48] + 6[18]
= 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]

Answer

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]

Answer

(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)

Answer

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]

Answer

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].

Answer

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]

Answer

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.

Answer

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]

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

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]

Answer

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.

Answer

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]

Answer

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.

Answer

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.

Answer

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]

Answer

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.

Answer

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

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]

Answer

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

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]

Answer

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

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]

Answer

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