Question

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 Mass of filter paper Mass of filter paper with copper (ml) (g) (g)A 49.6 0.908 1.694B 48.3 0.922 1.693C 42.2 0.919 1.588Write the empirical formula of copper chloride based on the experimental data.

Answer 1

Answer: The empirical formula for the given compound is $CuCl_3$

Explanation:

We are given:

Mass of copper chloride in 1 L or 1000 mL of solution = 42.62 grams

Taking Trial A:

Volume of solution = 49.6 mL

Applying unitary method:

In 1000 mL of solution, the mass of copper chloride present is 42.62 grams

So, in 49.6 mL of solution, the mass of copper chloride will be = $\frac{42.62}{1000}\times 49.6=2.114g$

We are given:

Mass of filter paper = 0.908 g

Mass of filter paper + copper = 1.694 g

Mass of copper = [1.694 - 0.908] g = 0.786 g

Mass of chlorine in the sample = [2.114 - 0.786]g = 1.328 g

To formulate the empirical formula, we need to follow some steps:

• Step 1: Converting the given masses into moles.

Moles of Copper =$\frac{\text{Given mass of Copper}}{\text{Molar mass of Copper}}=\frac{0.786g}{63.5g/mole}=0.0124moles$

Moles of Chlorine = $\frac{\text{Given mass of Chlorine}}{\text{Molar mass of Chlorine}}=\frac{1.328g}{35.5g/mole}=0.0374moles$

• Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0124 moles.

For Copper = $\frac{0.0124}{0.0124}=1$

For Chlorine = $\frac{0.0374}{0.0124}=3.02\approx 3$

• Step 3: Taking the mole ratio as their subscripts.

The ratio of Cu : Cl = 1 : 3

Hence, the empirical formula for the given compound is $CuCl_3$