A compound contains 78.14% B and 21.86% H by mass and has a molar mass of 27.67 g/mol. What is the molecular formula of the compound

The molecular formula of the compound is B₂H₆

Explanation:

From the question given above, the following data were obtained:

Composition of Boron (B) = 78.14%

Composition of Hydrogen (H) = 21.86%

Molar mass of compound = 27.67 g/mol

Molecular formula =?

Next, we shall determine the empirical formula of the compound. This can be obtained as follow:

B = 78.14%

H = 21.86%

Divide by their molar mass

B = 78.14 / 11 = 7.104

H = 21.86 / 1 = 21.86

Divide by the smallest

B = 7.104 / 7.104 = 1

H = 21.86 / 7.104 = 3

Thus, the empirical formula of the compound is BH₃

Finally, we shall determine the molecular formula of the compound. This can be obtained as illustrated below:

Molecular formula = empirical formula × n

Molecular formula = [BH₃]ₙ

Thus, we shall determine the value of n to obtain the molecular formula. The value of n can be obtained as follow:

[BH₃]ₙ = 27.67

[11 + (3×1)]n = 27.67

[11 + 3]n = 27.67

14n = 27.67

Divide both side by 14

n = 27.67 / 14

n = 2

Molecular formula = [BH₃]ₙ

Molecular formula = [BH₃]₂

Molecular formula = B₂H₆

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