A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen. If the molar mass of the compound is 180. Find the molecular formula.
[H = 1, C = 12, O = 16]

A.

C\(_3\)H\(_6\)O\(_3\)

B.

C\(_6\)H\(_6\)O\(_3\)

C.

C\(_6\)H\(_{12}\)O\(_6\)

D.

CH\(_2\)O

Correct answer is C

C → 40/12 ≈ 3

H → 6.7/1 ≈ 6

O → 53.3/16 ≈ 3

dividing through with the lowest value, 3

C\(_1\)H\(_2\)O\(_1\)

[C\(_1\)H\(_2\)O\(_1\)]n = 180

[12 + 2 + 16]m = 180

30n = 180 

n = 6

:C\(_6\)H\(_12\)O\(_6\)