Question

A sample of chalk contains clay as impurity. The clay impurity loses 11% of its weight as moisture on prolong heating. 5 gram sample of chalk on heating shows a loss in weight (due to evolution of CO2 and water) by 1.1 g. Calculate % of chalk (CaCO3) in the sample. [Hint : Chalk (CaCO3) releases CO2 on heating]

Answer 1

33.33%

Answer 2

Let the mass of chalk (CaCO₃) in the sample be $m_{CaCO₃}$ grams and the mass of clay impurity be $m_{clay}$ grams. The total mass of the sample is 5 g. So, $m_{CaCO₃} + m_{clay} = 5$ g.

When the sample is heated, two processes contribute to the loss in weight:

1. Decomposition of CaCO₃: $CaCO₃(s) \rightarrow CaO(s) + CO₂(g)$ The mass loss from CaCO₃ is due to the evolution of CO₂ gas. The molar mass of CaCO₃ is approximately 100 g/mol (Ca=40, C=12, O=16). The molar mass of CO₂ is approximately 44 g/mol (C=12, O=16). According to the stoichiometry, 1 mole of CaCO₃ (100 g) produces 1 mole of CO₂ (44 g). So, the mass loss from $m_{CaCO₃}$ grams of CaCO₃ is $\frac{44}{100} \times m_{CaCO₃} = 0.44 \times m_{CaCO₃}$ grams.

2. Loss of moisture from clay: The clay impurity loses 11% of its weight as moisture. The mass loss from $m_{clay}$ grams of clay is $11\%$ of $m_{clay} = 0.11 \times m_{clay}$ grams.

The total mass loss on heating is given as 1.1 g. Total mass loss = Mass loss from CaCO₃ + Mass loss from clay $1.1 = 0.44 \times m_{CaCO₃} + 0.11 \times m_{clay}$.

We now have a system of two linear equations with two variables:

1. $m_{CaCO₃} + m_{clay} = 5$
2. $0.44 \times m_{CaCO₃} + 0.11 \times m_{clay} = 1.1$

From equation (1), we can express $m_{clay}$ as $m_{clay} = 5 - m_{CaCO₃}$. Substitute this expression for $m_{clay}$ into equation (2): $0.44 \times m_{CaCO₃} + 0.11 \times (5 - m_{CaCO₃}) = 1.1$ $0.44 \times m_{CaCO₃} + 0.11 \times 5 - 0.11 \times m_{CaCO₃} = 1.1$ $0.44 \times m_{CaCO₃} + 0.55 - 0.11 \times m_{CaCO₃} = 1.1$ Combine the terms involving $m_{CaCO₃}$: $(0.44 - 0.11) \times m_{CaCO₃} + 0.55 = 1.1$ $0.33 \times m_{CaCO₃} = 1.1 - 0.55$ $0.33 \times m_{CaCO₃} = 0.55$ $m_{CaCO₃} = \frac{0.55}{0.33} = \frac{55}{33} = \frac{5}{3}$ grams.

The mass of chalk (CaCO₃) in the sample is $\frac{5}{3}$ grams. The percentage of chalk in the sample is calculated as: $\% \text{ of } \text{CaCO}_3 = \left( \frac{\text{Mass of } \text{CaCO}_3}{\text{Total mass of sample}} \right) \times 100$ $\% \text{ of } \text{CaCO}_3 = \left( \frac{5/3 \text{ g}}{5 \text{ g}} \right) \times 100$ $\% \text{ of } \text{CaCO}_3 = \left( \frac{5}{3 \times 5} \right) \times 100$ $\% \text{ of } \text{CaCO}_3 = \left( \frac{1}{3} \right) \times 100$ $\% \text{ of } \text{CaCO}_3 = \frac{100}{3} \%$

The percentage of chalk in the sample is $\frac{100}{3} \%$. This is approximately 33.33%.