Question

1 g of an impure sample of magnesium carbonate (containing no thermally decomposable impurities) on complete thermal decomposition gave 0.44 g of carbon dioxide gas. The percentage of impurity in the sample is
(a) 0 %
(b) 4.4 %
(c) 16 %
(d) 8.4 %

Answer 1

MgCO3 = MgO + CO2

Molar mass of MgCO3 = 24 + 12 + 16x3 = 84

Molar mass of CO2 = 12 + 16x 2 = 44


44g of CO2 is obtained from 84 g of MgCO3

∴ 0.44 g of CO2 is obtained from = (0.44x84)/44

                                                     = 0.84 g of MgCO3


Mass of impurities = 1 - 0.84 = 0.16

Percentage of impurities = (0.16/1) x 100 

                                        = 16%


Choice C 

Answer 2

Answer:

Explanation:

MgCO3 = MgO + CO2

Molar mass of MgCO3 = 24 + 12 + 16x3 = 84

Molar mass of CO2 = 12 + 16x 2 = 44

44g of CO2 is obtained from 84 g of MgCO3

∴ 0.44 g of CO2 is obtained from = (0.44x84)/44

= 0.84 g of MgCO3

Mass of impurities = 1 - 0.84 = 0.16

Percentage of impurities = (0.16/1) x 100

= 16%