• # question_answer A compound contains $C=90%$ and $H=10%$. Empirical formula of the compound is [NCERT 1976; EAMCET 1978] A) ${{C}_{3}}{{H}_{10}}$ B) $C{{H}_{2}}$ C) ${{C}_{3}}{{H}_{2}}$ D) ${{C}_{3}}{{H}_{4}}$

Elements       No. of Moles         Simple ratio  C = 90% 90/12 = 7.5 7.5/7.5 = 1$\times$3 = 3 H = 10% 10/1 = 10 10/7.5 = 1.33 × 3 = 4
\ Empirical formula = ${{C}_{3}}{{H}_{4}}$