Question
An organic compound gave $$0.4655\,g$$ of $$C{O_2}$$ on complete combustion. If the mass of the compound taken was $$0.2115\,g,$$ what is the percentage of $$C$$ in it?
A.
$$13.30\% $$
B.
$$26.67\% $$
C.
$$60.03\% $$
D.
$$28.80\% $$
Answer :
$$60.03\% $$
Solution :
$${\text{Mass of }}C{O_2}{\text{ formed}} = 0{\text{ }}{\text{.4655}}\,{\text{g}}$$
$${\text{Mass of organic compound taken}}$$ $$ = 0.2115{\text{ }}g$$
$$\eqalign{
& \% \,{\text{of}}\,C = \frac{{12}}{{44}} \times \frac{{{\text{mass}}\,\,{\text{of}}\,\,C{O_2}}}{{{\text{ mass of compound }}}} \times 100 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{12}}{{44}} \times \frac{{0.4655}}{{0.2115}} \times 100 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 60.03\% \cr} $$