A sample of substance containing $C$. $H$ and $N$ gives by its combustion $7.922\;g\;CO_2$. $4.325g\;H_2O$ and $0.840\;g\;N_2$.
Calculate the percentage of elements in this substance.
$n_{CO_2}$ obtained = $\frac{7.922}{44}$ $=$ $0.180 $
$n_C$ in the sample = $0.180$
$m_C $ in the sample = $0.180\cdot 12$ $=$ $2.16\; g $
$n_{H_2O}$ obtained = $\frac{4.325}{18}$ $=$ $0.240 $
$n_H$ in the sample = $0.480$
$m_H $ in the sample = $0.480\cdot 1 = 0.48\; g $
$n_{N_2}$ obtained = $\frac{0.840}{28}$ $=$ $0.0300$
$n_N$ in the sample = $0.0600$
$m_N $ in the sample = $0.0600\cdot 14$ $=$ $0.84 \; g $
Mass of the sample = $2.16+0.48+0.84$ $=$ $3.48\; g $
$\%_C$ $=$ $\frac{2.16\cdot 100}{3.48}$ $=$ $62.1$
$\%_H$ $=$ $\frac{0.48\cdot 100}{3.48}$ $=$ $13.8 $
$\%_N$ $=$ $100-62.1-13.8$ $=$ $24.1 $
