Mole Concept and Stoichiometry(B)
Vapour Density and Molecular Weight.
Num. 1. When heated, potassium permanganate decomposes according to the following equation
:
2KMnO4 → K2MnO4 + MnO2 + O2
Some 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 hydrogen under the same condition of temperature and pressure has a mass of 0.0825 g,
calculate the relative molecular mass of oxygen.
Sol : - Mass of 1 lit. of oxygen = 1.32 g
Mass of 1 lit. of hydrogen = 0.0825 g
Vapour density of oxygen = mass of 1 lit. of oxygen / mass of 1 lit. of hydrogen
= 1.32 /0.0825 = 16 g.
Relative molecular mass of oxygen = 2 × vapour density = 2 × 16 = 32.
Q. 2. Name the term which defines the mass of a given volume of a gas compared to the mass of
an equal volume of hydrogen.
Ans.: - Vapour density.
Num. 2. Find the relative molecular mass of a gas, 0.546 g of which occupies 360 cm3 at 87 ºC and
380 mm Hg pressure. [1 litre of hydrogen at s.t.p. weigh 0.09 g].
Sol : - P1 = 380 mm of Hg, V1 = 360 cm3 = 360 ml, T1 = 87 + 273 = 360 K.
P2 = 760 mm of Hg, V2 = x ml, T2 = 273 K
Using, P1 V1 /T1 = P2 V2 / T2 we get
(380 × 360) / 360 = (760 × x) / 273 => x = (273 × 380) / 760 = 136.5 l
Mass of 136.5 ml = 0.546 g
=> mass of 1000 ml = (0.546/136.5) × 1000 = 4 g
Vapour density = mass of 1 lit. of the gas / mass of 1 lit. of hydrogen
= 4/0.09 = 44.44
Molecular mass = 2 × 44.44 = 88.88 [Ans.]
Percentage Composition.
Num 1. Find the total percentage of oxygen in magnesium nitrate crystal :
Mg(NO3)2.6H2O [H = 1, N = 14, O = 16, Mg = 24].
Sol :- gram molecular mass of Mg(NO3)2.6H2O = 24 + (14 + 48) × 2 + 6 × 18 = 256.
Mass of oxygen in magnesium nitrate = 192 g
% of oxygen = (192 / 256) × 100 = 75 %.
Num 2. What is the mass of nitrogen in 1000 kg of urea [CO(NH 2)2]. [Answer to the nearest kg.]
Sol.: - Molecular weight of urea [CO(NH2)2] = 12 + 16 + (14 + 2) × 2 = 60.
60 kg of urea contains 28 kg of nitrogen
1000 kg of urea will contain (28/60) × 1000 = 467kg of nitrogen .
Num 3. Calculate the percentage of boron [B] in borax Na2B4O7.10H2O [H = 1, B = 11, O = 16, Na =
23].
Sol : - Gram molecular weight of Borax = 46 + 44 + 112 + 180 = 382 g
Weight of Boron in Borax = 44 g.
Percentage of Boron = (44/382) × 100 = 11.5 %.
Num 4. If a crop of wheat removes 20 kg of nitrogen per hectare of soil, what mass of the fertilizer
calcium nitrate Ca(NO3)2would be required to replace the nitrogen in a 10 hectare field? [N = 14; O
= 16; Ca = 40] (Give your answer to the nearest kg).
Sol : - Amount of nitrogen removed from 10 hectare field = 10 × 20kg = 200 kg.
Molar mass of Calcium nitrate Ca(NO3)2 = 40 + (14 + 16 × 3) × 2 = 164
Mass of nitrogen in 164 kg of Ca(NO3)2 is 28 kg
For 28 kg of nitrogen, amount of calcium nitrate required is 164 kg
For 200 kg of nitrogen, amount of calcium nitrate required = (164/28) × 200 kg
= 1171 kg.
Empirical Formula and Molecular Formula.
Num.1. Determine the empirical formula of the compound whose composition by mass is 42 %
nitrogen, 48 % oxygen and 9 % hydrogen. (H = 1: N = 14; O = 16).
Sol.: Element
Percentage
Composition
Atomic
weight
Relative No. of
Atoms
Simplest ratio
Nitrogen
42
14
42/14 = 3.0
3.0/3.0 = 1
Oxygen
48
16
48/16 = 3.0
3.0/3.0 = 1
Hydrogen
9
1
9/1 = 9.0
9.0/3.0 = 3
Thus empirical formula of the compound is NOH3.
Num. 2. A metal M forms a volatile chloride containing 65.5% chlorine. If the density of the
chloride relative to hydrogen is 162.5, find the molecular formula of the chloride.
(M = 56, Cl = 35.5).
Sol.: - Percentage of M in the chloride = 100 – 65.5 = 34.5
Element
Percentage
Atomic
weight
Relative No. of
Atoms
Simplest ratio
M
34.5
56
34.5/56 = 0.61
0.61/0.61 = 1
Cl
65.5
35.5
65.5/35.5 = 1.84
1.84/0.61 = 3
Empirical formula of the chloride = MCl3
Empirical formula mass = 56 + 35.5 × 3 = 162.5 g
Vapour density = 162.5
Molar mass of the chloride = 2 × V.D. = 2 × 162.5 = 325.0 g
Molecular formula = (Empirical formula)n
Or, n = Molar mass /Empirical formula mass = 325.0 g / 165.5 g = 2
Molecular formula = (MCl3)2 = M2Cl6
Num. 3. The percentage composition of sodium phosphate as determined by analysis is 42.1 %
sodium, 18.9 % phosphorus and 39 % oxygen. Find the empirical formula of the compound. [Na =
23, P = 31, O = 16].
Sol : Element
Percentage
composition
Atomic
weight
Relative No. of
atoms
Simplest ratio
Sodium
42.1
23
42.1/23 = 1.83
1.83/0.61 = 3
Phosphorus
18.9
31
18.9/31 = 0.61
0.61/0.61 = 1
Oxygen
39.0
16
39.0/16 = 2.44
2.44/0.61 = 4
Thus empirical formula is Na3PO4.
Num. 4. An experiment showed that in a lead chloride solution, 6.21g of lead combined with 4.26 g
of chlorine. What is the empirical formula of this chloride? [Pb = 207; Cl = 35.5].
Sol : Element
Mass
Atomic
weight
Relative No. of
atoms
Simplest ratio
Lead
6.21 g
207
6.21/207 = 0.03
0.03/0.03 = 1
Chlorine
4.26 g
35.5
4.26/35.5 = 0.12
0.12/0.03 = 4
Hence, empirical formula is PbCl4.
Num.5. Determine the empirical formula of a compound containing 47.9% potassium, 5.5%
beryllium and 46.6% fluorine bymass.[Atomic weight of Be = 9; F = 19; K = 39]. Work to one
decimal place.
Sol : Element
Percentage
composition
Atomic
weight
Relative No. of
atoms
Simplest ratio
Potassium
47.9
39
47.9/39 = 1.2
1.2/0.6 = 2
Beryllium
5.5
9
5.5/9 = 0.6
0.6/0.6 = 1
Fluorine
46.6
19
46.6/19 = 2.4
2.4/0.6 = 4
Hence empirical formula is K2BeF4.
Num.6. A compound has the following percentage composition by mass : 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 its
molecular formula. [C = 12; H = 1; Cl = 35.5]
Sol : –
Element
Percentage
composition
Atomic
weight
Relative No. of
atoms
Simplest ratio
Carbon
14.4
12
14.4/12 = 1.2
1.2/1.2 = 1
Hydrogen
1.2
1
1.2/1 = 1.2
1.2/1.2 = 1
Chlorine
84.5
35.5
84.5/35.5 = 2.38
2.38/1.2 = 2
Hence empirical formula is CHCl2.
n = Molecular mass/Empirical formula mass
= 168/(12 + 1 + 70)
= 168/83
= 2.
Therefore, molecular formula = (CHCl2)2 = C2H2Cl4.
Q.1.What is the empirical formula of octane ?
Ans.: - C8H18.
Problem based on Chemical Equations
Num.1. Given that the relative molecular mass [molecular weight] of copper oxide is 80, what
volume of ammonia [measured at s.t.p.]is required to completely reduce 120 g of copper oxide.
The equation for the reaction is :
3CuO + 2NH3 → 3Cu + 3H2O + N2.
[Vol. occupied by 1 mole of gas at s.t.p. is 22.4 lit.]
Sol : - 3CuO + 2 NH3 → 3Cu + 3H2O + N2
3 mole 2 mole
3×80 g 2×22.4 l
= 240 g = 44.8 l
For 240 g of CuO, ammonia consumed is 44.8 l
For 120 g of CuO, ammonia consumed = (44.8/240)/120 = 22.4 lit.
Num. 2. The equations given below relate to the manufacture of sodium carbonate [Mol. Wt. of
Na2CO3= 106]
(i) NaCl + NH3 + CO2 + H2O → NaHCO3 + NH4Cl
(ii) 2NaHCO4→ Na2CO3 + H2O + CO22.
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 NaHCO2 = 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.
Sol : - (a) 2NaHCO3 → Na2CO3 + H2O + CO2
2×84 = 168 g 106 g
106 g of Na2CO3 is obtained from 168 g of NaHCO3
21.2 g of Na2CO3 is obtained from (168/106)×21.2 = 33.6 g of NaHCO 3.
(b) NaCl + NH3 + CO2 + H2O → NaHCO3 + NH4Cl
1 mole 1 mole
22.4 l 84 g
84 g of NaHCO3 requires 22.4 l of NH4Cl
33.6 g of NaHCO3requires (22.4/84)×33.6 = 8.96 l of CO2.
Num. 3. The reaction of potassium permanganate with acidified iron [II] sulphate is given below :
2KMnO4 + 10FeSO4 + 8H2SO4 → K2SO4 + 2MnSO4 + 5Fe2(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].
Sol : - Molecular weight of KMnO4 = 39 + 55 + 16 × 4 = 158
Molecular weight of K2SO4 = 2 × 39 + 32 + 16 × 4 = 174
Molecular weight of FeSO4 = 56 + 32 + 16 × 4 = 152
2 × 158 g of KMnO4 yields 174 g of K2SO4
15.8 g of KMnO4 will yield [(174)/(2 × 158)] × 15.8 g = 8.7 g of K2SO4.
174 g of K2SO4 yields 152 g of FeSO4
8.7 g of K2SO4 will yield (152/174) × 8.7 = 7.6 g of FeSO4.
Hence, 7.6 g of Iron (II) sulphate is used .
Num. 4. 10 g of a mixture of sodium chloride and anhydrous sodium sulphate is dissolved in
water. An excess of barium chloride solution is added and 6.99 g of barium sulphate is
precipitated according to the equation :
Na2SO4 + BaCl2 → BaSO4 + 2NaCl
Calculate the percentage of sodium sulphate in the original mixture. [O = 16, Na = 23, S = 32, Ba =
137].
Sol : - Na2SO4 + BaCl2 → BaSO4 + 2NaCl
Molecular weight of BaSO4 = 137 + 32 + 16 × 4 = 233
Molecular weight of Na2SO4 = 2 × 23 + 32 + 16 × 4 = 142
233 g of BaSO4 is obtained from 142 g of Na2SO4
6.99 g of BaSO4 will be obtained from (142/233) × 6.99 g = 4.26 g of Na2SO4
Percentage of Na2SO4 in 10 g of mixture = (4.26/10) × 100 = 42.6%.
Num. 5. From the equation :
C + 2H2SO4 → CO2 _ 2H2O + 2SO2.
Calculate :
i.
ii.
The mass of carbon dioxide by 49 g of sulphuric acid [C = 12, rel. mol. Mass of H 2SO4 = 98]
The volume of sulphur dioxide measured at s.t.p., liberated at the same time. [Volume
occupied by 1 mole of a gas at s.t.p is 22.4 dm3].
Sol : – C + 2H2SO4 → CO2 + 2H2O + 2SO2
Molecular mass of CO2 = 12 + 32 = 44.
As, 98×2 g of sulphuric acid produces 44 g of carbon dioxide.
Therefore, 49 g of sulphuric acid will provide (44/196)×49 = 11 g of CO2. [Ans.]
2 moles of sulphur dioxide is produced. Volume = 222.4 dm3 = 44.8 dm 3.
As, 196 g of sulphuric acid provide 44.8 dm 3 of SO2
Therefore, 49 g of sulphuric acid will provide (44/196)×44.8 dm 3 = 11.2 dm3. [Ans.]