Question

$\_{}^{{}^{\prime }}{R}_{}^{{}^{\prime }}\stackrel{1200K}{\to }CaO+C{O}_2$
Calculate the mass (in $mg$ ) of carbon in $1.5g$ sample of pure ${}^{\prime }R$ '. [Given: At. $Wt$ . of $C=12,O=16,Ca=40u]$

Answer 1

Answer: 180

${}^{\prime }{R}^{\prime }$ is $CaC{O}_3$
$\%$ of $C$ in $CaC{O}_3=\frac{12}{100}\times 100=12\%$
i.e., $100gCaC{O}_3=12gC$
$\therefore 1.5gCaC{O}_3=\frac{12\times 1.5}{100}=0.18g$ 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.