A sample of non-stoichiometric cuprous sulfide is found to have the composition $Cu_{1.8} S$ . This sample is actually a mixture of $Cu_{2} S$ and CuS . Calculate the mass fraction of $Cu^{2 +}$ ions relative to the total mass of the sample.
(Atomic masses: Cu = 63.5, S = 32)

Question

A sample of non-stoichiometric cuprous sulfide is found to have the composition $Cu_{1.8} S$ . This sample is actually a mixture of $Cu_{2} S$ and CuS . Calculate the mass fraction of $Cu^{2 +}$ ions relative to the total mass of the sample.

(Atomic masses: Cu = 63.5, S = 32)

TestHub · Accepted Answer

Explain the question:

Find the mass fraction of Cu²⁺ ions in the non-stoichiometric compound Cu₁․₈S.

Concept:

Principle of electrical neutrality: the sum of positive charges from Cu⁺ and Cu²⁺ ions must equal the negative charge from S²⁻ ions.

Solution:

In the formula Cu₁․₈S, let's assume we have 1 mole of S²⁻ ions, which carries a total negative charge of 2⁻. To maintain electrical neutrality, the 1.8 moles of copper ions must provide a total positive charge of 2⁺.

Let the number of moles of Cu²⁺ ions be $x$ .

The number of moles of Cu⁺ ions will be $(1.8 - x)$ .

Setting up the charge balance equation:

$2 (x) + 1 (1.8 - x) = 2$

$x = 2 - 1.8 = 0.2 m o l$

Thus, in one mole of the compound, there are 0.2 moles of Cu²⁺ ions.

Now, calculate the respective masses:

Mass of Cu²⁺ ions = $0.2 m o l \times 63.5 g / m o l = 12.7 g$

Total molar mass of Cu₁․₈S = $(1.8 \times 63.5) + 32 = 114.3 + 32 = 146.3 g / m o l$

The mass fraction of Cu²⁺ ions relative to the total mass of the sample is:

$M a ss f r a c t i o n = \frac{ma sso f C u ^{2 +}}{t o t a l ma sso f s am pl e} = \frac{12.7 g}{146.3 g} \approx 0.0868$

Rounding to three decimal places, we get 0.087.

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