$1.500\; g$ of a substance containing $C$. $H$ and $O$ are burned completely .
Calculate the percentages of the elements. knowing that one obtains $1.738\;g\;CO_2$ and $0.711g\;H_2O$.
$n_{CO_2}$ obtained = $\frac{1.738}{44}$ $=$ $0.0395$
$n_C$ in $1.500\; g $ = $0.0395$
$m_C $ in $1.500\; g $ = $0.0395\cdot 12$ $=$ $0.474\; g $
$n_{H_2O}$ obtained = $\frac{0.711}{18}$ $=$ $0.0395$
$n_H$ in $1.500\; g $ = $0.0395\cdot 2$ $=$ $0.0790 $
$m_H $ in $1.500\; g $ = $0.0790\cdot 1$ $=$ $0.0790\; g $
$\%_C$ $=$ $\frac{0.474\cdot 100}{1.500}$ $=$ $31.6$
$\%_H$ $=$ $\frac{0.0790\cdot 100}{1.500}$ $=$ $5.3 $
$\%_O$ $=$ $100-31.6-5.3$ $=$ $63.1 $
