Chemistry - Mole Concept Question with Solution | TestHub

$\mathrm{C}=85.71\%=\frac{85.71}{12}=7.14;\frac{7.14}{7.14}=1$ $\mathrm{H}=14.29\%=\frac{14.29}{1}=14.29,\frac{14.29}{7.14}=2$ $\therefore$ Empirical formula $={\mathrm{CH}}_2$ and, empirical formulaweight $=12+2=14$ Nowsincemolecular formula weight $=2\times$ vapour density $$=2\times 14=28$$ $\therefore n=\frac{28}{14}=2$ $\therefore$ Molecular formula $={\left({\mathrm{CH}}_2\right)}_2={\mathrm{C}}_2{\mathrm{H}}_4$