10. A student investigated the purity of a sample of magnesium chloride, MgCl₂ - Scottish Highers Chemistry - Question 10 - 2019

To find the mass of magnesium chloride, we start by using the molar ratios from the balanced equation:

From the reaction, we have: $MgCl₂ + 2AgNO₃ \rightarrow 2AgCl + Mg(NO₃)₂$

Using the molar mass of silver chloride (AgCl) and magnesium chloride (MgCl₂):

• Molar mass of AgCl = 143·4 g/mol
• Molar mass of MgCl₂ = 95·3 g/mol

We first determine the number of moles of silver chloride produced:

$n = \frac{mass}{molar mass} = \frac{1.393 g}{143.4 g/mol} \approx 0.0097 \text{ mol}$

From the equation, 1 mole of MgCl₂ produces 2 moles of AgCl. Thus, the number of moles of MgCl₂ is half the moles of AgCl:

$n_{MgCl₂} = \frac{0.0097 \text{ mol}}{2} \approx 0.00485 \text{ mol}$

Now, calculate the mass of MgCl₂:

$\text{mass} = n \times \text{molar mass} = 0.00485 ext{ mol} \times 95.3 g/mol \approx 0.462 g$

To calculate the percentage of magnesium chloride in the original sample, we use the formula:

$\% \text{ of MgCl}_2 = \left( \frac{\text{mass of MgCl}_2}{\text{mass of impure sample}} \right) \times 100$.

Using the mass calculated previously: $\% \text{ of MgCl}_2 = \left( \frac{0.462 g}{2.503 g} \right) \times 100 \approx 18.4\%$