An organic compound contains 78% (by wt.) carbon and remaining percentage of hydrogen. The right option for the empirical formula of this compound is : [Atomic wt. of C is 12, H is 1]
CH
CH_{2}
CH_{3}
CH_{4}
Element C: Mass percentage 78%, No. of mole = 78/12 = 6.5, Mole ratio = 6.5/6.5 = 1. Element H: Mass percentage 22%, No. of mole = 22/1 = 22, Mole ratio = 22/6.5 = 3.38 ≈ 3. Based on above calculation, possible empirical formula is CH_{3}.
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