Question

A sample of bismuth weighing 0.687 g was converted to bismuth chloride by reacting it first with HNO3, and then with HCl, followed by careful evaporation to dryness. The mass of the bismuth chloride obtained was 1.032 g. What is its empirical formula

Answer 1

Answer:

BiCl₃

Explanation:

The molar mass of bismuth is 208.987g/mol. Thus, moles of 0.687g of Bi are:

0.687g Bi × (1mol / 208.987g) = 3.287x10⁻³ moles of Bi

The mass of Bismuth is bismuth chloride is also 0.687g. Thus, mass of chloride is:

Cl mass = 1.032g - 0.687g = 0.345g of Cl.

As molar mass of Cl is 35.45g/mol:

0.345g Cl ₓ (1mol / 35.45g) = 9.732x10⁻³ moles of Cl

Empirical formula is defined as the simplest ratio of atoms in a compound. Dividing each moles in the lower number of moles (3.287x10⁻³ moles)

Ratio of Bi: 3.287x10⁻³ moles / 3.287x10⁻³ moles = 1

Ratio of Cl:9.732x10⁻³ moles of Cl / 3.287x10⁻³ moles = 3

Thus, empirical formula is:

BiCl₃