Question

Question #9bb2b

Answer 1

`C_4H_10`

Explanation:

%C = 82.6% => 82.6g => (86.2g/12g/mol) => 6.88 mole
%H = 17.4% => 17.4g => (17.4g/1g/mol) => 17.4 mole

C:H => 6.88:17.4 => Reduces to 1:2.5 => Emp Ratio => 2:5
Emp Formula => `C_2H_5`
Emp Formula Wt = 2C + 5H = [2(12) + 5(1)] g/mole = 29 g/mole

2.9 gram sample of hydrocarbon => 1.12 `dm^3` = 1.12 Liters at STP
moles of hydrocarbon sample = (1.12L/22.4L/mol) = 0.05 mole
mole wt = (2.9 g/ 0.05 mole) = 58 g/mole

(Emp. Wt) N = Mole Wt. => (29)N = 58 => N = 2

`(C_2H_5)_N`; N = 2 => Molecular Formula => `C_4H_10`