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

`\boxed{29.7 \, \%}`

Step-by-step 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) * \frac{\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) * \frac{\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) * \frac{\text{84.31 g MgCO}_(3)}{\text{1 mol MgCO}_(3)} =\textbf{6.149 g MgCO}_(3)`

5. % MgCO₃

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