$1.279\; g$ of a substance sample containing $C$. $H$. $O$ and $N$ produce by combustion $1.60\;g\;CO_2$ and $0.77\;g\;H_2O$.
A sample of $1.625\;g $ produces by a separate analysis $195\;cm^3\;N_2$ at $740\;mmHg$ and $27\;^oC$.
Calculate the percentage of elements in this substance.
1st analysis:
$n_{CO_2}$ obtained = $\frac{1.60}{44}$ $=$ $0.0364$
$n_C$ in the sample = $0.0364$
$m_C $ in the sample = $0.0364\cdot 12$ $=$ $0.437\; g $
$n_{H_2O}$ obtained = $\frac{0.77}{18}$ $=$ $0.0428$
$n_H$ in the sample = $0.0956$
$m_H $ in the sample = $0.0956\cdot 1$ $=$ $0.0956\; g $
$\%_C$ $=$ $\frac{0.437\cdot 100}{1.279}$ $=$ $34.2$
$\%_H$ $=$ $\frac{0.0956\cdot 100}{1.279}$ $=$ $7.5 $
2nd analysis:
$n_{N_2}$ obtained =
$\frac{p\cdot V}{R\cdot T}$ =
$\frac{740\cdot 0.195}{760\cdot 0.082\cdot 300.15}$=
$ \; 0.00771 $
$n_N$ in the sample = $0.0154$
$m_N$ in the sample = $0.0154\cdot 14$ $=$ $0.216\; g$
$\%_N$ $=$ $\frac{0.216\cdot 100}{1.625}$ $=$ $13.3$
$\%_O$ $=$ $100-34.2-7.5-13.3=$ $45.0$
