An organic compound contains 40%40\% carbon, 6.67%6.67\% hydrogen, and the rest is oxygen. If the molar mass of the compound is 180 g/mol180 \text{ g/mol}, its molecular formula is:

Model Answer & Options

Source: Extra Practice

C_6H_{12}O_6

CH_2O

C_2H_4O_2

C_3H_6O_3

Explanation

Percentage of Oxygen = 100−(40+6.67)=53.33%100 - (40 + 6.67) = 53.33\%. Calculate the relative number of atoms: C=40/12=3.33C = 40/12 = 3.33, H=6.67/1=6.67H = 6.67/1 = 6.67, O=53.33/16=3.33O = 53.33/16 = 3.33. The simple whole number ratio C:H:OC:H:O is 1:2:11:2:1. Thus, the empirical formula is CH2OCH_2O. Empirical formula mass = 12+(2×1)+16=3012 + (2 \times 1) + 16 = 30. To find the molecular formula, calculate n=Molar mass/Empirical formula mass=180/30=6n = \text{Molar mass} / \text{Empirical formula mass} = 180 / 30 = 6. Molecular formula = n×(Empirical formula)=6×(CH2O)=C6H12O6n \times (\text{Empirical formula}) = 6 \times (CH_2O) = C_6H_{12}O_6. Option 2 is the empirical formula, and Options 3 and 4 have incorrect molar masses (6060 and 9090 respectively).

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