four grams of hydrocarbon (C(x)H(y)) on ...

four grams of hydrocarbon `(C_(x)H_(y))` on complete combustion gave 12grams of `CO_(2)`. What is the empirical formula of the hydrocarbon ? `(C=12 , H=1)`

A

`CH_(3)`

B

`C_(4)H_(9)`

C

`CH`

D

`C_(3)H_(8)`

Verified by Experts

The correct Answer is:

D

`%C=(12)/(4)xx(12)/(44)xx100=81.8%`
`%C" is "C_(3)H_(8)=(3xx12)/(44)xx100=81.8%`

Similar Questions

Explore conceptually related problems

C_(40) H_(56) is the empirical formula of

4 g of a hydrocarbon on complete combustion gave 12.571 g of Co, and 5.143 g of water. What is the empirical formula of the hydrocarbon?

C_(40)H_(56) O_(2) is the empirical formula of

100grams of a hydrocarbon has 93.71 g of carbon. What is its empirical formula of the hydrocarbon?

4.4 gms of a hydrocarbon on complete combustion produced 13.2 gms of CO_(2) and 7.2 gms of H_(2)O . What is the hydrocarbon?

0.2 gram -mole of an unsaturated hydrocarbon on complete combustion produces 26.4 g of CO_(2) . The molecular weight of the hydro-carbon is

0.1 moles of Hydrocaron on complete combustion produced 17.6 gms of CO_(2) . How many Carbon atoms are present in each molecule of the hydrocarbon.