Exercises
1. Calculate the molar mass of the following: (i) H₂O (ii) CO₂ (iii) CH₄
Answer
(i) H₂O: 2(1.008) + 16.00 = 18.016 g mol⁻¹
(ii) CO₂: 12.01 + 2(16.00) = 44.01 g mol⁻¹
(iii) CH₄: 12.01 + 4(1.008) = 16.042 g mol⁻¹
2. Calculate the mass per cent of different elements present in sodium sulphate (Na₂SO₄).
Answer
Molar mass of Na₂SO₄ = 2(23) + 32 + 4(16) = 142 g mol⁻¹
% Na = (46/142) × 100 = 32.39%; % S = (32/142) × 100 = 22.54%; % O = (64/142) × 100 = 45.07%
3. Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.
Answer
Moles of Fe = 69.9/55.85 = 1.252; Moles of O = 30.1/16.00 = 1.881
Ratio Fe : O = 1 : 1.503 ≈ 2 : 3
⇒ Empirical formula: Fe₂O₃
4. Calculate the amount of carbon dioxide that could be produced when (i) 1 mole of carbon is burnt in air. (ii) 1 mole of carbon is burnt in 16 g of dioxygen. (iii) 2 moles of carbon are burnt in 16 g of dioxygen.
Answer
Reaction: C + O₂
⇒ CO₂ (1 mol C needs 32 g O₂)
(i) Excess air → 1 mol (44 g) CO₂
(ii) 16 g O₂ = 0.5 mol; O₂ is limiting
⇒ 0.5 mol (22 g) CO₂
(iii) Only 0.5 mol O₂ available
⇒ 0.5 mol (22 g) CO₂
5. Calculate the mass of sodium acetate (CH₃COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol⁻¹.
Answer
Moles needed = 0.375 × 0.500 = 0.1875 mol; Mass = 0.1875 × 82.0245 = 15.38 g
6. Calculate the concentration of nitric acid in moles per litre in a sample which has a density 1.41 g mL⁻¹ and the mass per cent of nitric acid in it being 69%.
Answer
Mass of 1 L solution = 1410 g; Mass HNO₃ = 69% × 1410 = 973.9 g
Moles HNO₃ = 973.9/63 = 15.46; Molarity = 15.46 M
7. How much copper can be obtained from 100 g of copper sulphate (CuSO₄)?
Answer
Molar mass CuSO₄ = 159.5 g/mol; Moles = 100/159.5 = 0.627 mol
Mass Cu = 0.627 × 63.5 = 39.8 g
8. Determine the molecular formula of an oxide of iron, in which the mass per cent of iron and oxygen are 69.9 and 30.1, respectively.
Answer
Empirical formula is Fe₂O₃ (from Q1.3); empirical formula mass = 159.7 g/mol = molar mass, so n = 1. Molecular formula: Fe₂O₃
9. Calculate the atomic mass (average) of chlorine: ³⁵Cl 75.77% (34.9689), ³⁷Cl 24.23% (36.9659).
Answer
Average = (75.77/100)(34.9689) + (24.23/100)(36.9659) = 26.496 + 8.953 = 35.45 u
10. In three moles of ethane (C₂H₆), calculate: (i) moles of carbon atoms (ii) moles of hydrogen atoms (iii) number of molecules.
Answer
(i) 3 × 2 = 6 mol carbon atoms; (ii) 3 × 6 = 18 mol hydrogen atoms; (iii) 3 × 6.022 × 10²³ = 18.066 × 10²³ molecules
11. What is the concentration of sugar (C₁₂H₂₂O₁₁) in mol L⁻¹ if 20 g are dissolved to make 2 L solution?
Answer
Molar mass = 342 g/mol; Moles = 20/342 = 0.0585 mol; Molarity = 0.0585/2 = 0.02926 mol L⁻¹
12. If the density of methanol is 0.793 kg L⁻¹, what volume is needed for making 2.5 L of 0.25 M solution?
Answer
Moles needed = 0.25 × 2.5 = 0.625 mol; Mass = 0.625 × 32 = 20 g
Volume = 20/793 L = 0.02523 L = 25.23 mL
13. Mass of air at sea level is 1034 g cm⁻². Calculate the pressure in pascal.
Answer
1034 g/cm² = 10340 kg/m²; Pressure = 10340 × 9.8 = 1.013 × 10⁵ Pa
14. What is the SI unit of mass? How is it defined?
Answer
The SI unit of mass is the kilogram (kg). It is defined as the mass corresponding to the international prototype kilogram. Since 2019, it is defined via the fixed numerical value of Planck's constant.
15. Match the prefixes with their multiples: (i) micro (ii) deca (iii) mega (iv) giga (v) femto
Answer
(i) micro
⇒ 10⁻⁶; (ii) deca
⇒ 10; (iii) mega
⇒ 10⁶; (iv) giga
⇒ 10⁹; (v) femto
⇒ 10⁻¹⁵
16. What do you mean by significant figures?
Answer
Significant figures are all the certain digits plus one uncertain digit in a measured or calculated value. They indicate the precision of a measurement. Zeros between non-zero digits are significant; leading zeros are not; trailing zeros after the decimal point are significant.
17. Level of contamination of CHCl₃ in drinking water was 15 ppm (by mass). (i) Express in per cent by mass. (ii) Determine the molality of chloroform in water.
Answer
(i) 15 ppm = 15/10⁶; % = (15/10⁶) × 100 = 1.5 × 10⁻³ %
(ii) In 10⁶ g solution: 15 g CHCl₃ (M = 119.5) and ~10⁶ g water
Moles CHCl₃ = 15/119.5 = 0.1255; Molality = 0.1255/10³ = 1.25 × 10⁻⁴ m
18. Express in scientific notation: (i) 0.0048 (ii) 234,000 (iii) 8008 (iv) 500.0 (v) 6.0012
Answer
(i) 4.8 × 10⁻³; (ii) 2.34 × 10⁵; (iii) 8.008 × 10³; (iv) 5.000 × 10²; (v) 6.0012 × 10⁰
19. How many significant figures are present in: (i) 0.0025 (ii) 208 (iii) 5005 (iv) 126,000 (v) 500.0 (vi) 2.0034
Answer
(i) 2; (ii) 3; (iii) 4; (iv) 3; (v) 4; (vi) 5
20. Round up to three significant figures: (i) 34.216 (ii) 10.4107 (iii) 0.04597 (iv) 2808
Answer
(i) 34.2; (ii) 10.4; (iii) 0.0460; (iv) 2810
21. Dinitrogen and dioxygen form different compounds (given table). (a) Which law does this data obey? Give its statement. (b) Fill: (i) 1 km = ____mm = ____pm (ii) 1 mg = ____kg = ____ng (iii) 1 mL = ____L = ____dm³
Answer
(a) For fixed 14 g N, O masses are 16 g and 32 g (ratio 1:2). This obeys the Law of Multiple Proportions: when two elements form more than one compound, the masses of one element combining with a fixed mass of the other are in simple whole-number ratios.
(b) (i) 1 km = 10⁶ mm = 10¹⁵ pm; (ii) 1 mg = 10⁻⁶ kg = 10⁶ ng; (iii) 1 mL = 10⁻³ L = 10⁻³ dm³
22. Speed of light = 3.0 × 10⁸ m s⁻¹. Calculate distance covered in 2.00 ns.
Answer
t = 2.00 × 10⁻⁹ s; d = 3.0 × 10⁸ × 2.00 × 10⁻⁹ = 0.60 m
23. In A + B₂ → AB₂, identify the limiting reagent in: (i) 300 atoms A + 200 molecules B₂ (ii) 2 mol A + 3 mol B₂ (iii) 100 atoms A + 100 molecules B₂ (iv) 5 mol A + 2.5 mol B₂ (v) 2.5 mol A + 5 mol B₂
Answer
Stoichiometry is 1:1. (i) B₂ is limiting; (ii) A is limiting; (iii) no limiting reagent; (iv) B₂ is limiting; (v) A is limiting
24. N₂(g) + 3H₂(g) → 2NH₃(g). 2.00 × 10³ g N₂ reacts with 1.00 × 10³ g H₂. (i) Mass of NH₃ produced. (ii) Any unreacted reactant? (iii) Which and what mass?
Answer
Moles N₂ = 71.43; Moles H₂ = 500; Required H₂ = 214.3 mol < 500 mol available
⇒ N₂ is limiting.
(i) NH₃ = 2 × 71.43 mol = 142.86 mol; Mass = 142.86 × 17 = 2428.6 g
(ii) Yes, H₂ remains. (iii) H₂ remaining = (500 − 214.3) × 2 = 571.4 g
25. How are 0.50 mol Na₂CO₃ and 0.50 M Na₂CO₃ different?
Answer
0.50 mol Na₂CO₃ is a fixed amount (53 g) irrespective of volume. 0.50 M Na₂CO₃ is a concentration: 0.50 mol dissolved per litre of solution. Concentration depends on volume, while amount does not.
26. 10 volumes H₂ reacts with 5 volumes O₂. How many volumes of water vapour?
Answer
2H₂ + O₂
⇒ 2H₂O; ratio 2:1:2. 10 volumes H₂ needs 5 volumes O₂ (exact ratio)
⇒ 10 volumes water vapour.
27. Convert to basic units: (i) 28.7 pm (ii) 15.15 pm (iii) 25365 mg
Answer
(i) 2.87 × 10⁻¹¹ m; (ii) 1.515 × 10⁻¹¹ m; (iii) 2.5365 × 10⁻² kg
28. Which has the largest number of atoms? (i) 1 g Au (ii) 1 g Na (iii) 1 g Li (iv) 1 g Cl₂
Answer
Atoms per gram: Au = 0.00508NA; Na = 0.0435NA; Li = 0.143NA; Cl₂ = 0.0282NA. Answer: (iii) 1 g Li
29. Mole fraction of ethanol = 0.040. Calculate molarity.
Answer
χethanol = 0.040; so nethanol = 0.040, nwater = 0.960. Mass water = 17.28 g = 0.01728 L. Molarity = 0.040/0.01728 = 2.31 M
30. What will be the mass of one ¹²C atom in g?
Answer
Mass = 12/(6.022 × 10²³) = 1.993 × 10⁻²³ g
31. How many significant figures in answers of: (i) (0.02856 × 298.15 × 0.112)/0.5785 (ii) 5 × 5.364 (iii) 0.0125 + 0.7864 + 0.0215
Answer
(i) Minimum sig figs = 3 (from 0.112)
⇒ 3 sig figs; (ii) 5 is exact
⇒ 4 sig figs; (iii) Addition limited by 4 decimal places
⇒ 4 sig figs
32. Calculate molar mass of naturally occurring argon: ³⁶Ar (35.96755, 0.337%), ³⁸Ar (37.96272, 0.063%), ⁴⁰Ar (39.9624, 99.600%).
Answer
Average = (0.337 × 35.96755 + 0.063 × 37.96272 + 99.600 × 39.9624)/100 = 39.948 g mol⁻¹
33. Calculate number of atoms in: (i) 52 mol Ar (ii) 52 u of He (iii) 52 g of He.
Answer
(i) 52 × 6.022 × 10²³ = 3.131 × 10²⁵ atoms; (ii) 52/4 = 13 atoms; (iii) 13 mol × 6.022 × 10²³ = 7.83 × 10²⁴ atoms
34. A welding gas (C and H only) burns to give 3.38 g CO₂ and 0.690 g H₂O. 10.0 L (STP) weighs 11.6 g. Calculate (i) empirical formula (ii) molar mass (iii) molecular formula.
Answer
C = 3.38 × 12/44 = 0.920 g; H = 0.690 × 2/18 = 0.0767 g; C:H = 0.920/12 : 0.0767/1 = 0.0767:0.0767 = 1:1
⇒ (i) CH
(ii) Molar mass = (11.6/10.0) × 22.4 = 26 g mol⁻¹
(iii) n = 26/13 = 2 → C₂H₂ (acetylene)
35. CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l). What mass of CaCO₃ reacts with 25 mL of 0.75 M HCl?
Answer
Moles HCl = 0.025 × 0.75 = 0.01875 mol; Moles CaCO₃ = 0.01875/2 = 0.009375 mol; Mass = 0.009375 × 100 = 0.9375 g
36. 4HCl(aq) + MnO₂(s) → 2H₂O(l) + MnCl₂(aq) + Cl₂(g). How many grams of HCl react with 5.0 g of MnO₂?
Answer
Moles MnO₂ = 5.0/87 = 0.0575 mol; Moles HCl = 4 × 0.0575 = 0.2299 mol; Mass HCl = 0.2299 × 36.5 = 8.39 g
Important Extra Questions
What is the difference between atomic mass and molar mass?
Answer
Atomic mass is the mass of one atom in atomic mass units (u), where 1 u = 1/12 the mass of ¹²C. Molar mass is the mass of 6.022 × 10²³ atoms expressed in g/mol. Numerically the two values are equal but the units differ.
State the Law of Conservation of Mass.
Answer
The total mass of reactants equals the total mass of products in a chemical reaction; mass is neither created nor destroyed. This was established by Antoine Lavoisier.
What is a mole? State Avogadro's constant.
Answer
A mole is the SI unit of amount of substance. One mole contains 6.022 × 10²³ entities (NA, Avogadro's constant). One mole of any substance has a mass in grams equal to its relative atomic or molecular mass.
Distinguish between empirical and molecular formula.
Answer
The empirical formula gives the simplest whole-number ratio of atoms in a compound (e.g., CH). The molecular formula gives the actual number of atoms per molecule (e.g., C₆H₆). The molecular formula = n × empirical formula, where n = molar mass / empirical formula mass.
What is molarity? How does it differ from molality?
Answer
Molarity (M) = moles of solute per litre of solution; it changes with temperature. Molality (m) = moles of solute per kg of solvent; it is temperature-independent. For dilute aqueous solutions their values are approximately equal.