A bromoalkane contains 35% carbon and 6.57% hydrogen by mass. Calculate the empirical formula of this bromoalkane.

Options

(a) CH₂Br
(b) C₂H₂Br₂
(c) C₄H₄Br
(d) C₄H₉Br

Correct Answer:

C₄H₉Br

Explanation:

1) Assume 100 g of the compound is available:
C ⇒ 35 g
H ⇒ 6.57 g
Br ⇒ 58.43 g (from 100 minus 41.57)
2) Determine moles:
C ⇒ 35 g / 12 gmol = 2.917
H ⇒ 6.57 g / 1 g/mol = 6.57
Br ⇒ 58.43 g / 80 g/mol = 0.730375
3) Divide by smallest to seek lowest whole-number ratio:
C ⇒ 2.917 / 0.730375 = 4
H ⇒ 6.57 / 0.730375 = 9
Br ⇒ 0.730375 / 0.730375 = 1
C₄H₉Br