A compound composed of 3.3% H, 19.3% C, and 77.4% O3.3% H, 19.3% C, and 77.4% O has a molar mass of approximately 60 g/mol.. What is the molecular formula of the compound?

Molecular formula of the compound = H₂CO₃

Explanation:

First, the empirical formula of the compound is determined

Percentage by mass of each element is given as shown below:

H = 3.3% ; C = 19.9%; O = 77.4%

Mole ratio of the elements= percentage mass/ molar mass

H = 3.3/ 1 = 3.3

C = 19.3/12 = 1.6

O = 77.4/16 = 4.8

whole number ratio is obtained by dividing through with the smallest ratio

H = 3.3/1.6; C = 1.6/1.6; O = 4.8/1.6

H : C : O = 2 : 1 : 3

Empirical formula = H₂CO₃

Molecular formula/mass = n(empirical formula/mass)

60 = n(2*1 + 12*1 + 16*3)

60 = n(62)

n = 60/62 = 0.96

n is approximately = 1

Therefore, molecular formula of the compound = (H₂CO₃) * 1

Molecular formula of the compound = H₂CO₃

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