Question
$$0.46\,g$$ of an organic compound was analysed. The increase in mass of $$CaC{l_2}$$ $$U$$ - tube was $$0.54\,g$$ and potash bulb was $$0.88\,g.$$ The percentage composition of the compound is
A.
$$C = 52.17\% ,H = 13.04\% ,$$ $$O = 34.79\% $$
B.
$$C = 50\% ,H = 50\% $$
C.
$$C = 32.19\% ,H = 18.01\% ,$$ $$O = 49.8\% $$
D.
$$C = 72\% ,H = 28\% $$
Answer :
$$C = 52.17\% ,H = 13.04\% ,$$ $$O = 34.79\% $$
Solution :
$$\eqalign{
& {\text{Mass of}}\,\,{H_2}O = 0.54\,g \cr
& {\text{Mass of}}\,\,C{O_2} = 0.88\,g \cr
& \% \,{\text{of}}\,\,C = \frac{{12}}{{44}} \times \frac{{0.88}}{{0.46}} \times 100 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 52.17\% \cr
& \% \,{\text{of}}\,\,H = \frac{2}{{18}} \times \frac{{0.54}}{{0.46}} \times 100 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 13.04\% \cr
& \% \,{\text{of}}\,\,O = 100 - \left( {52.17 + 13.04} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 34.79\% \cr} $$