Question

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?

Answer 1

Answer:

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

Answer 2

The molecular formula of a compound composed of 3.3% H, 19.3% C, and 77.4% O and a molar mass of approximately 60 g/mol is CH2O3.

HOW TO CALCULATE MOLECULAR FORMULA:

• The molecular formula of an unknown compound can be calculated from the empirical formula. The empirical formula can be calculated as follows:

• C = 19.3% = 19.3g
• H = 3.3% = 3.3g
• O = 77.4% = 77.4g

• First, we calculate the number of moles in each element by dividing by their respective atomic mass.

1. C = 19.3g ÷ 12g/mol = 1.61mol
2. H = 3.3g ÷ 1g/mol = 3.3mol
3. O = 77.4g ÷ 16g/mol = 4.84mol

• Next, we divide by the smallest mole value.

1. C = 1.61mol ÷ 1.61 = 1
2. H = 3.3mol ÷ 1.61 = 2.05
3. O = 4.84mol ÷ 1.61 = 3.01

• Approximately, the empirical ratio of CHO is 1:2:3. Hence, the empirical formula is CH2O3.

• {CH2O3}n = 60g/mol

• {12 + 2 + 48}n = 60

• 62n = 60

• n = 60 ÷ 62

• n = 0.968

• n = 1

• Therefore, the molecular formula of a compound composed of 3.3% H, 19.3% C, and 77.4% O and a molar mass of approximately 60 g/mol is CH2O3.

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