Question

One way in which the useful metal copper is produced is by dissolving the mineral azurite, which contains copper(II) carbonate, in concentrated sulfuric acid. The sulfuric acid reacts with the copper(II) carbonate to produce a blue solution of copper(II) sulfate. Scrap iron is then added to this solution, and pure copper metal precipitates out because of the following chemical reaction:Fe(s)+CuSO4(aq)→Cu(s)+FeSO4(aq)Suppose an industrial quality-control chemist analyzes a sample from a copper processing plant in the following way. He adds powdered iron to a 200.mL copper(II) sulfate sample from the plant until no more copper will precipitate. He then washes, dries, and weighs the precipitate, and finds that it has a mass of 95.mg. Calculate the original concentration of copper(II) sulfate in the sample.

Answer 1

Answer:

0.0075 M

Explanation:

Mass of copper metal recovered:

$m_o = 95 mg = 95\cdot 10^{-3} g$

Molar mass of copper should be used in this problem in order to convert the mass of copper into moles of copper:

$M_o = 63.546 g/mol$

According to the balanced chemical equation, 1 mole of copper(II) sulfate produces 1 mole of copper metal, this means, when we find the number of moles of copper metal, this will be equivalent to the number of moles of copper(II) sulfate. Dividing the number of moles of copper(II) sulfate by the volume of the solution in liters will yield the molarity needed.

Firstly, the number of moles of copper (and copper(II) sulfate):

$n_o = \frac{m_o}{M_o}$

Given a volume of:

$V = 200 mL = 0.200 L$

We obtain a molarity of:

$c_o = \frac{n_o}{V} = \frac{m_o}{M_o V} = \frac{95\cdot 10^{-3} g}{63.546 g/mol\cdot 0.200 L} = 0.0075 M$