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An organic compound containes `C=74.0%, H=8.65%` and `N=17.3%`. Its empirical formul aisA. `C_(5)H_(8)N`B. `C_(10)H_(12)N`C. `C_(5)H_(7)N`D. `C_(10)H_(14)N` |
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Answer» Correct Answer - C `"Element No. of Mol es Simple Ration"` `C=74" "74//12=6.1" "6.1//1.2=5.08 ` or `5` `H=8.65" "8.65//1=8.65" "8.6//1.5=7.16` or `7` `N=1.73" "1.3//14=1.2" "1.2//1.2=1` or `1` Therefore Empirical formula `=C_(5)H_(7)N.` |
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