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?

Options

A. CH
B. `CH_(2)`
C. `CH_(3)`
D. `C_(2)H_(2)`

Answer

B

Explanation

Step 1: calculate the percentage of C and H
`c%=12/44xx("Weight of "CO_(2)xx100)/("Weight of organic compund")``85.7%`
`H%=2/18xx("Weight of water"xx100)/("weight of organic compound")`
` 2/18xx(5.143xx100)/(4)=14.3%`
step @: Calculate the mole ratio of C to H
` C:H=(85.7)/(12):(14.3)/(1)= 7.15:14.3`
step divide by the lowest value obatined in step2
`7.15/7.15:14.3/7.15=1:2`
therefore Emprical formula=`CH_(2)`