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 250.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 89.mg. Calculate the original concentration of copper(II) sulfate in the sample. Be sure your answer has the correct number of significant digits.

Answer 1

Answer:

Molar concentration of $CuSO_4$ solution = 0.0056 moles / liter

Explanation:

Looking at the chemical reaction we realize that the precipitate, formed by adding the iron powder, is cooper.

Then for finding the number of moles of precipitated copper:

$number.of.moles= \frac{mass (g)}{molecular.mass (g/mole)}$

$number.of.moles.of.solid.copper = \frac{0.089}{63.5} =1.4*10^{3}$

From the chemical reaction we deduce that $1.4*10^{3}$ moles of Cu equals to $1.4*10^{3}$ moles of $CuSO_4$ in the initial solution.

So molar concentration is defined as:

$molar.concentration = \frac{number.of.moles}{solution.volume (liters)}$

$molar.concentration.of.CuSO_4.solution= \frac{1.4*10^{3}}{0.250}= 5.6*10^{3} moles/liter = 0.0056 moles/liter$

Answer 2

Answer:

$M=5.6x10^{-3}M$

Explanation:

Hello,

In this case, we first must consider the given already-balanced chemical reaction to realize that 89 mg of copper were recovered, moreover we can relate such mass with the employed moles of copper (II) sulfate via this reaction's stoichiometry as follows:

$n_{CuSO_4}=89mg Cu*\frac{1gCu}{1000mgCu} *\frac{1molCu}{63.546gCu} *\frac{1molCuSO_4}{1molCu} \\n_{CuSO_4}=1.4x10^{-3}molCuSO_4$

Now, if we state the molarity (mol/L) as the required concentration, we apply its mathematical definition as shown below:

$M=\frac{n_{CuSO_4}}{V_{sln}} =\frac{1.4x10^{-3}molCuSO_4}{0.250L} \\M=5.6x10^{-3}M$

Best regards.