Question

What is the empirical formula of a compound 0.2801 gm of which gave on complete combustion 0.9482 gm of carbon dioxide and 0.1939 gm of water?

Answer 1

Let the compound be $CxHy$

$CxHy+(2x+\frac{y}{2})⟶xC{O}_2+\frac{y}{2}{H}_{2}O$

Moles of $C{O}_2=\frac{0.9482}{44}=0.02155$

Moles of $H_{2}O=\frac{0.1939}{18}=0.01077$

$1$ $mole$ $CxHy$ give $x$ $mole$ $C{O}_2$ and $\frac{y}{2}{H}_{2}O$

No. of moles of $CxHy=\frac{0.02155}{x}=\frac{0.01077}{y/2}$

$⟹x=y$

Empirical Formula $CH$