Question

The percentage composition in a sample of $CaC{O}_3$ is given as: $Ca=40\%;C=12\%;O=48\%$. What will be the weight of $Ca$ in $4\text{g}$ of a sample of $CaC{O}_3$ obtained from another source, if the law of constant proportions is true?

• A. $0.016\text{g}$
• B. $0.16\text{g}$
• C. $16\text{g}$
• D. $1.6\text{g}$

Answer 1

The correct option is D $1.6\text{g}$

The law of definite proportions states that a given chemical compound always contains its component elements in a fixed ratio by mass and does not depend on its source and the method of preparation.
As the molar mass of $CaC{O}_3$ is $100\text{g/mol}$, $Ca$ present is $40\text{g}$ in 1 mole.
So, if $4\text{g}$ of $CaC{O}_3$ is present, then $Ca=\frac{40\times 4}{100}=1.6\text{g}$