A) \[{{C}_{3}}{{H}_{6}}\]
B) \[{{C}_{3}}{{H}_{8}}\]
C) \[C{{H}_{2}}\]
D) \[{{C}_{2}}{{H}_{2}}\]
Correct Answer: A
Solution :
Weight of empirical formula \[C{{H}_{2}}=12+(1\times 2)\] \[=12+2\] \[=14\] Mass of one mole of the compound = its molecular weight = 42 \[n=\frac{Mol.\text{ }wt.}{Empirical\text{ }formula\text{ }wt.}=\frac{42}{14}=3\] \[\therefore \]Mol. formula == (Empirical formula)\[\times n\] \[=(C{{H}_{2}})\times 3={{C}_{3}}{{H}_{6}}\].
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