In a student experiment, the empirical formula of a copper halide was found by adding aluminum metal to an aqueous solution of the halide, displacing copper metal. The copper metal was filtered, washed with distilled water, dried, and weighed; three separate determinations were performed. The copper halide solution contained 42.62 g of copper chloride per liter. The student recorded the following experimental data. Trial Volume of copper chloride solution (mL) Mass of filter paper (g) Mass of filter paper with copper (g) A 49.6 0.908 1.694 B 48.3 0.922 1.693 C 42.2 0.919 1.588 Part A Write the empirical formula of copper chloride based on the experimental data.

Empirical formula of a compound is the formula that shows simplest whole number ratio of the atoms contained in a particular compound.
It may be found when the percentage composition by mass or the mass of each atom in the compound is known.

In this case we first determine the mass of each atom in the compound from data given.

Mass of copper in Trial A = 1.694 - 0.908 = 0.786

Mass of copper in Trial B = 1.693 - 0.922 = 0.771

Mass of copper in Trial C = 1.588 - 0.919 = 0.669

Average mass of copper = (0.786 + 0.771 + 0.669) ÷ 3

= 0.742 g of Copper

Volume in Trial A = 49.6 ml

Volume in Trial B = 48.3 ml

Volume in Trial C = 42.2 ml

Average volume of copper chloride = (49.6 + 48.3 + 42.2) ÷ 3

= 46.7 ml or

= 0.046 L

But we are given 42.62 g of copper chloride in a liter, therefore;

Using proportions

1 L = 42.62 g Copper chloride

0.046 L will contain;

= (42.62 g CuClₓ × 0.0467 L) ÷ 1

= 1.99 g Copper Chloride

Mass of chlorine = Mass of copper chloride – mass of copper

= 1.99 g – 0.742 g

= 1.24 g chlorine

Moles = mass / atomic mass

Moles of copper = 0.742 g / 63.55 g/mol

= 0.0117 moles copper

Moles of Chlorine = 1.24 g/35.45 g/mol

= 0.0350 moles Chlorine

Copper: Chlorine

0.0117 moles : 0.0350 moles

= 1 : 2.98

= 1 : 3 (whole number ratio)

Therefore, the empirical formula of copper chloride is CuCl₃