Question

In an experiment, a compound was determined to contain 68.94% oxygen and 31.06% of an unknown element by weight. The molecular weight of the compound is 69.7 g/mol. What is this compound?\

Answer 1

Answer is: the compound is B₂O₃.

ω(O) = 68.94% ÷ 100%.

ω(O) = 0.6894; percentage of oxygen in the compound.

ω(X) = 31.06% ÷ 100%.

ω(X) = 0.3106; percentage of unknown element in the compound.

If we take 69.7 grams of the compound:

M(compound) = 69.7 g/mol.

n(compound) = 69.7 g ÷ 69.7 g/mol.

n(compound) = 1 mol.

n(O) = (69.7 g · 0.6894) ÷ 16 g/mol.

n(O) = 3 mol.

M(compound) = n(O) · M(O) + n(X) · M(X).

n(X) = 1 mol ⇒ M(X) = 21.7 g/mol; there is no element with this molecular weight.

n(X) = 2 mol ⇒ M(X) = 10.85 g/mol; this element is boron (B).