Answer to Question #123471 in Physics for Bob

Question

Hydrated copper (iii) sulphate with mass of 499g is heated. This results in the mass of the anhydrous copper (iii) sulphate being 319g.

Determine the empirical formula for the hydrated copper (iii) sulphate.

Answer 1

Assume that the formula is $\text{Cu}_2(\text{SO}_4)_3\cdot n\text{H}_2\text{O}$.

Find the number of moles of Cu₂(SO₄)₃:

$$n_1=\frac{m_1}{\Mu_1}=\frac{319}{2\cdot64+3(32+16\cdot4)}=0.767\text{ mol}.$$

The amount of substance of water was

$$n_2=\frac{m_2}{\Mu_2}=\frac{499-319}{2\cdot1+16}=10\text{ mol}.$$

From the formula $\text{Cu}_2(\text{SO}_4)_3\cdot n\text{H}_2\text{O}$ and the previous calculations, we see the ratio:

$$\frac{1}{n}=\frac{n_1}{n_2}=\frac{0.767}{10},\\\space\\ n=\frac{10}{0.767}=13.$$

Hence, the formula is

$$\text{Cu}_2(\text{SO}_4)_3\cdot 13\text{H}_2\text{O}$$