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 42.02 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)
b	48.3	0.922	1.893
c	42.2	0.919	1.888

part a
enter the empirical formula of copper chloride based on the experimental data. express your answer as a chemical formula.
view available hint(s)

incorrect; try again
your answer implies that there are two moles of chloride per mole of copper. the given data will allow you to calculate the actual mole - ratio of copper to chloride in this experiment.
you may want to review hint 5. determine the average number of moles of chloride.

Explanation:

Step1: Calculate mass of copper in each trial

For trial A: Mass of copper = 1.894 - 0.908 = 0.986 g.
For trial B: Mass of copper = 1.893 - 0.922 = 0.971 g.
For trial C: Mass of copper = 1.588 - 0.919 = 0.669 g.

Step2: Calculate moles of copper in each trial

The molar - mass of copper (Cu) is approximately 63.55 g/mol.
For trial A: Moles of Cu = $\frac{0.986}{63.55}\approx0.0155$ mol.
For trial B: Moles of Cu = $\frac{0.971}{63.55}\approx0.0153$ mol.
For trial C: Moles of Cu = $\frac{0.669}{63.55}\approx0.0105$ mol.

Step3: Calculate average moles of copper

Average moles of Cu = $\frac{0.0155 + 0.0153+0.0105}{3}=\frac{0.0413}{3}\approx0.0138$ mol.

Step4: Calculate moles of chloride

We know that the copper - halide solution contains 42.02 g of copper chloride per liter. First, we need to find the mass of copper chloride in each volume.
For trial A: Mass of copper chloride = $\frac{42.02}{1000}\times49.6 = 2.0842$ g.
For trial B: Mass of copper chloride = $\frac{42.02}{1000}\times48.3 = 2.0296$ g.
For trial C: Mass of copper chloride = $\frac{42.02}{1000}\times42.2 = 1.7732$ g.
The mass of chloride in each sample can be found by subtracting the mass of copper from the mass of copper chloride.
For trial A: Mass of Cl = 2.0842 - 0.986 = 1.0982 g.
For trial B: Mass of Cl = 2.0296 - 0.971 = 1.0586 g.
For trial C: Mass of Cl = 1.7732 - 0.669 = 1.1042 g.
The molar - mass of chlorine (Cl) is approximately 35.45 g/mol.
For trial A: Moles of Cl = $\frac{1.0982}{35.45}\approx0.031$ mol.
For trial B: Moles of Cl = $\frac{1.0586}{35.45}\approx0.0299$ mol.
For trial C: Moles of Cl = $\frac{1.1042}{35.45}\approx0.0311$ mol.
Average moles of Cl = $\frac{0.031 + 0.0299+0.0311}{3}=\frac{0.092}{3}\approx0.0307$ mol.

Step5: Find the mole - ratio of Cu to Cl

Mole - ratio of Cu to Cl = $\frac{0.0138}{0.0307}\approx\frac{1}{2.22}\approx\frac{1}{2}$ (rounded to the nearest whole - number ratio).
The empirical formula of copper chloride is CuCl.

Answer:

CuCl