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  • 120 g of an organic compound that contains only carbon and hydrogen gives 330 g of , and

120 g of an organic compound that contains only carbon and hydrogen gives 330 g of \mathrm{CO_{2}}, and 270 g of water on complete combustion, The percentage of carbon and hydrogen, respectively are

Option: 1

25 and 75


Option: 2

40 and 60


Option: 3

60 and 40


Option: 4

75 and 25


Answers (1)

best_answer

\begin{aligned} &\mathrm{{C}_{x}} \mathrm{{H}_{y}}+\mathrm{\left(x+\frac{y}{4}\right) \mathrm{O}_{2} \longrightarrow \mathrm{xCO}_{2}+ \frac{y}{2}\mathrm{H}_{2} \mathrm{O} }\\ &120 \mathrm{~g}\; \; \; \; \; \; \; \; \; \;\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; 330 \mathrm{~g} \; \; \; \; \; 270 \mathrm{~g} \end{aligned}

So, \mathrm{moles\ of\ CO_2=\frac{330}{44}=\frac{15}{2} \mathrm{~mol}=x}     

\text{and, moles of }\mathrm{H_2O}=\frac{270}{18}=15 \mathrm{~mol}=\mathrm{\frac{y}{2}}

\therefore The hydrocarbon is \mathrm{C_{15}H_{60}}   \mathrm{\left ( \because C_{\frac{15}{2}}H_{30} = C_{15}H_{60} \right )}

Now,

\mathrm{mass\ \%\ of\ C = \frac{180}{240}\times 100 = 75\%}\mathrm{mass\ \%\ of\ H= \frac{60}{240}\times 100 = 25\%}

Hence, the correct option (4)

Posted by

Divya Prakash Singh

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