Question

A 20.69 g sample of impure magnesium car- bonate was heated to complete decomposition according to the equation MgCO3(s) → MgO(s) + CO2(g) . After the reaction was complete, the solid residue (consisting of MgO and the original impurities) had a mass of 17.48 g. Assum- ing that only the magnesium carbonate had decomposed, what was the percent of magne- sium carbonate in the original sample? Answer in units of %.

Answer 1

Answer:

$\boxed{29.7 \, \%}$

Explanation:

We will need an equation with molar masses, so let’s gather all the information in one place.  

M_r:     84.31                    44.01

          MgCO₃ ⟶ MgO + CO₂

The mass lost was that of CO₂.

1. Mass of CO₂

Mass of CO₂ = 20.69 g - 17.48 g = 3.21 g CO₂

2. Moles of CO₂

$\text{Moles of CO}_{2} = \text{ 3.21 g CO}_{2} \times \dfrac{\text{1 mol CO}_{2}}{\text{ 44.01g CO}_{2}} = \text{0.072 94 mol CO}_{2}$

3. Moles of MgCO₃

The molar ratio is 1 mol MgCO₃:1 mol CO₂

$\text{Moles of MgCO}_{3} =\text{0.072 94 mol CO}_{2} \times \dfrac{\text{1 mol MgCO}_{3}}{\text{1 mol CO}_{2}} = \text{0.072 94 mol MgCO}_{3}$

4. Mass of MgCO₃

$\text{Mass of MgCO}_{3} = \text{0.072 94 mol MgCO}_{3} \times \dfrac{\text{84.31 g MgCO}_{3}}{\text{1 mol MgCO}_{3}} =\textbf{6.149 g MgCO}_{3}$

5. % MgCO₃

$\text{Percent by mass} = \dfrac{\text{mass of component}}{\text{total mass}}\times 100 \, \% = \dfrac{\text{6.149 g}}{\text{20.69 g}} \times 100 \, \%\\ = \mathbf{29.7 \, \%}\\\text{The percent of magnesium carbonate in the sample was \boxed{\mathbf{29.7 \, \%}}}$