Solving for a reaction in solution

Question

Suppose an industrial quality chemist analyzes a sample from a copper processing plant in the following way. He added powdered iron to a 150 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 62 mg. What is the original concentration of copper ii sulfate in the sample

Answer 1

First, let's look at the chemical reaction that is taking place:

Fe(s) + CuSO₄(aq) ==> Cu(s) + FeSO₄(aq) .. balanced redox reaction

Note, the precipitate that he is weighing is the Cu(s) that has formed. So, if we know the moles of Cu(s), we can find the moles of CuSO₄ originally present as they are in a 1 to 1 mole ratio. We can then also determine the original concentration of CuSO₄.

Find moles of Cu(s) formed:

62 mg Cu x 1 g / 1000 mg = 0.062 g Cu

0.062 g Cu x 1 mol Cu / 63.55 g = 9.76x10^-4 moles Cu

This is also the moles of CuSO₄ in the original sample.

To find the concentration of CuSO₄, we simply divide the moles by the liters of solution:

moles CuSO₄ = 9.76x10^-4 moles

Liters of solution = 150 mls x 1 L / 1000 ml = 0.150 L

Concentration of CuSO₄ in original sample = 9.76x10^-4 moles / 0.150 L = 6.5x10^-3 mol/L = 6.5x10^-3 M

Answer 2

Hi, this is a question of mole-based calculation.

Step 1. Write and balance the chemical equation.

Fe + CuSO₄ → Cu +FeSO₄

Step 2 Determine the moles of CuSO₄ in the solution.

Based on the above balanced equation, 1 mole CuSO₄ can produce 1 mole Cu; that is to say, they are equal in amount (mole). The mole of Cu is calculated as follows.

62 mg × (1 g/1000 mg) × (1 mol/ 63.55 g) = 9.8 × 10^-4 mol

So the amount of CuSO₄ is 9.8 × 10^-4 mol

Step 3 Determine the original concentration (or molarity) of CuSO₄ in the solution based on the molarity formula.

molarity (M) = moles of solute (CuSO₄)/volume of solution in liters

mole of CuSO₄ = 9.8 × 10^-4 mol

volume of solution (L) = 150 mL × 1 L/1000 mL = 0.15 L

molarity (M) = 9.8 × 10^-4 mol / 0.15 L = 6.5 × 10^-3 mol/L = 6.5 × 10^-3 M

Hope this helps. Let me know if you have any questions.