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Chapter 4: Q85E (page 224)

A 0.025-g sample of a compound composed of boron and hydrogen, with a molecular mass of ~28 amu, burns spontaneously when exposed to air, producing 0.063 g of B2O3. What are the empirical and molecular formulas of the compound?

Short Answer

Expert verified

Empirical formula = B1H3

Molecular formula = B2H6

Step by step solution

01

Information given

We can write the equation

Bx Hy + O2鈫払2O3 + H2O

0.025 g 0.063 g

0.063 g of B2O3 x 1 mol of B2O3 x 2 mol B x 10.81g of B

69.62g of B2O3 1 mol B2O3 1 mol B

= 0.01956 g of B

Bx Hy = 0.025 g

So g of H = 0.025 g 鈥� 0.01956 g = 0.00544 g of H

02

Required formula

% Grams \( \to \) mols \( \to \) mol ratio \( \to \) empirical formula \( \to \) molecular formula

0.01956 g of B x (1 mol B/10.81 g mol B) = (0.0081 mol B/0.0081) = 1 mol B

0.00544 g of H x (1 mol H/1g mol H) = (0.00540 mol H/0.0081) = 3 mol H

Empirical formula = B1H3

Molecular mass/ Empirical mass = 28/ 13.83 = 2

Molecular formula = B2H6

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