Answer to Question #123471 in Physics for Bob

Question #123471
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.
1
Expert's answer
2020-06-23T13:01:45-0400

Assume that the formula is Cu2(SO4)3nH2O\text{Cu}_2(\text{SO}_4)_3\cdot n\text{H}_2\text{O}.

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


n1=m1M1=319264+3(32+164)=0.767 mol.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

n2=m2M2=49931921+16=10 mol.n_2=\frac{m_2}{\Mu_2}=\frac{499-319}{2\cdot1+16}=10\text{ mol}.

From the formula Cu2(SO4)3nH2O\text{Cu}_2(\text{SO}_4)_3\cdot n\text{H}_2\text{O} and the previous calculations, we see the ratio:


1n=n1n2=0.76710, n=100.767=13.\frac{1}{n}=\frac{n_1}{n_2}=\frac{0.767}{10},\\\space\\ n=\frac{10}{0.767}=13.

Hence, the formula is


Cu2(SO4)313H2O\text{Cu}_2(\text{SO}_4)_3\cdot 13\text{H}_2\text{O}


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