Question

$\_{}^{'}R_{}^{'}\overset{1200 K}{ \rightarrow }CaO+CO_{2}$
Calculate the mass (in $mg$ ) of carbon in $1.5g$ sample of pure $'R$ '. [Given: At. $Wt$ . of $C=12,O=16,Ca=40u\left]\right.$

Answer 1

$'R'$ is $CaCO_{3}$
$\%$ of $C$ in $CaCO_{3}=\frac{12}{100}\times 100=12\%$
i.e., $100gCaCO_{3}=12gC$
$\therefore 1.5gCaCO_{3}=\frac{12 \times 1 . 5}{100}=0.18g$ of $C$
$=180mg$ of $C$
Calcium carbonate is a chemical compound with the formula CaCO3. It is a common substance found in rocks as the minerals calcite and aragonite.
Percent composition is calculated from a molecular formula by dividing the mass of a single element in one mole of a compound by the mass of one mole of the entire compound. This value is presented as a percentage.