Question 1

Calculate the molecular mass of the following:(i) H₂O (ii) CO₂ (iii) CH₄

Accepted Answer

(i) H₂O: Molecular weight of water, H₂O = (2 × Atomic weight of hydrogen) + (1 × Atomic weight of oxygen) = [2(1.008 u) + 1(16.00 u)] = 2.016 u + 16.00 u = 18.016 u  (ii) CO2: Molecular weight of carbon dioxide, CO₂  = (1 × Atomic weight of carbon) + (2 × Atomic weight of oxygen) = [1(12.011 u) + 2(16.00 u)] = 12.011 u + 32.00 u = 44.011 u  (iii) CH₄ (Methane): Molecular weight of methane, CH₄ = (1 × Atomic weight of carbon) + (4 × Atomic weight of hydrogen) = [1(12.011 u) + 4(1.008 u)] = 12.011

Question 2

Calculate the mass percent of different elements present in Sodium Sulphate (Na₂SO₄).

Accepted Answer

Molar mass of Na₂SO₄ = [(2 x 23.0) + (32.00) + 4(16.00)] = 142 g Mass percent of an element = (Mass of that element in compound / Molar mass of that compound) × 100 Mass percent of sodium (Na): (46/142) × 100 = 32.39% Mass percent of sulphur (S): (32/142) × 100 = 22.54% Mass percent of oxygen (O): (64/142) × 100 = 45.07%

Question 3

Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.

Accepted Answer

% of iron by mass = 69.9 % [Given] % of oxygen by mass = 30.1 % [Given] Atomic mass of iron = 55.85 amu Atomic mass of oxygen = 16.00 amu Relative moles of iron = 69.9/55.85 = 1.25 Relative moles of oxygen = 30.01/16 = 1.88 Simplest molar ratio = 1.25/1.25 : 1.88/1.25 ⇒ 1 : 1.5 = 2 : 3∴  The empirical formula of the iron oxide is Fe₂O₃.

Question 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

Accepted Answer

The balanced reaction is: C(s) + O₂(g) → CO₂(g) (i) 1 mole of carbon is burnt in air: Since the combustion is complete, 1 mole of carbon dioxide is produced.  (ii) 1 mole of carbon is burnt in 16 g of dioxygen: Oxygen acts as a limiting reagent (16 g = 0.5 mole). It will react with 0.5 mole of carbon to give 22 g of carbon dioxide.  (iii) 2 moles of carbon are burnt in 16 g of dioxygen: Oxygen again acts as a limiting reagent (0.5 mole). It can combine with only 0.5 mole of carbon to give 22 g o

Question 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.

Accepted Answer

0.375 M aqueous solution means 1000 mL solution contains 0.375 moles of sodium acetate. ∴ No. of moles of sodium acetate in 500 mL = (0.375/1000) × 500 = 0.375/2 = 0.1875 moles Molar mass of sodium acetate = 82.0245 g mol⁻¹ ∴ Mass of sodium acetate required = 0.1875 × 82.0245 g = 15.380 g

Question 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%.

Accepted Answer

Mass percent of 69% means 100g of nitric acid solution contains 69 g of nitric acid by mass. Molar mass of nitric acid (HNO₃) = 1+14+48 = 63 g mol⁻¹ Number of moles in 69 g of HNO₃ = 69/63 moles = 1.095 moles Volume of 100g nitric acid solution = 100/1.41 mL = 70.92 mL = 0.07092 L ∴ Concentration of HNO₃ in moles per litre = 1.095/0.07092 = 15.44 M

Question 7

How much copper can be obtained from 100 g of copper sulphate (CuSO₄)?

Accepted Answer

1 mole of CuSO₄ contains 1 mole of copper. Molar mass of CuSO₄ = (63.5) + (32.00) + 4(16.00) = 63.5 + 32.00 + 64.00 = 159.5 g 159.5 g of CuSO₄ contains 63.5 g of copper. ∴ Copper obtained from 100 g of copper sulphate = (63.5/159.5) × 100 = 39.81 g

Question 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. Given that the molar mass of the oxide is 159.69 g mol⁻¹.

Accepted Answer

% of iron by mass = 69.9 % [Given] % of oxygen by mass = 30.1 % [Given] Atomic mass of iron = 55.85 amu Atomic mass of oxygen = 16.00 amu Relative moles of iron = 69.9/55.85 = 1.25 Relative moles of oxygen = 30.01/16 = 1.88 Simplest molar ratio = 1.25/1.25 : 1.88/1.25 ⇒ 1 : 1.5 = 2 : 3 ∴ The empirical formula of the iron oxide is Fe₂O₃. Mass of Fe₂O₃ = (2×55.85) + (3×16.00) = 159.7 g mol⁻¹ n = Molar mass / Empirical formula mass = 159.7 / 159.6 = 1 (approx) Thus, Molecular formula is same as Emp

Question 9

Calculate the atomic mass (average) of chlorine using the following data:
Isotope % Natural Abundance Molar Mass
35Cl &nbs

Accepted Answer

Fractional Abundance of 35Cl = 0.7577 and Molar mass = 34.9689 Fractional Abundance of 37Cl = 0.2423 and Molar mass = 36.9659 ∴ Average Atomic mass = (0.7577 × 34.9689) amu + (0.2423 × 36.9659) = 26.4959 + 8.9568 = 35.4527

Question 10

In three moles of ethane (C₂H₆), calculate the following: (i) Number of moles of carbon atoms. (ii) Number of moles of hydrogen atoms. (iii) Number of molecules of ethane.

Accepted Answer

(i) 1 mole of C₂H₆ contains 2 moles of Carbon atoms ∴ 3 moles of C₂H₆ will contain 6 moles of Carbon atoms. (ii) 1 mole of C₂H₆ contains 6 moles of Hydrogen atoms ∴ 3 moles of C₂H₆ will contain 18 moles of Hydrogen atoms. (iii) 1 mole of C₂H₆ contains Avogadro's no. (6.02 × 10²³ molecules) ∴ 3 moles of C₂H₆ will contain ethane molecules = 3 × 6.02 × 10²³ = 18.06 × 10²³ molecules.

Some Basic Concepts of Chemistry