Question

$20 \cdot 0{\text{ mg}}$ of a Magnesium Carbonate sample decomposes on heating to give carbon dioxide, and $8 \cdot 0{\text{ mg }}$ magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample?
[ Atomic weight of ${\text{Mg = 24}}$]

Answer 1

Hint:Percentage purity of a substance in a sample is calculated by dividing the mass of the pure compound by the total mass of the sample given, then multiplied by ${\text{100}}$ to make it a percentage.

Complete step by step answer:
Step (1):
The reaction involved in this question will be: $MgC{O_3} \to MgO + C{O_2}{\text{ }}........{\text{(1)}}$,
where $MgC{O_3}$ is magnesium carbonate,
${\text{MgO}}$ is magnesium oxide,
and, $C{O_2}$ is carbon dioxide.
Step (2):
Total mass of the magnesium carbonate sample given = $20 \cdot 0{\text{ mg}}$
Molar mass of magnesium carbonate, $MgC{O_3}$ is equal to:
= $1 \times {\text{ Mass of Mg atom }} + 1 \times {\text{ Mass of C atom + 3}} \times {\text{ Mass of O atom }}$
= $1 \times 24 + 1 \times 12 + 3 \times 16$

= $84{\text{ g mo}}{{\text{l}}^{ - 1}}$
Mass of magnesium oxide formed on heating is given = $8 \cdot 0{\text{ mg }}$.
Molar mass of magnesium oxide, ${\text{MgO}}$, is equal to:
= $1 \times {\text{ Mass of Mg atom }} + 1 \times {\text{ Mass of O atom }}$
= $1 \times 24 + 1 \times 16$
= $40{\text{ g mo}}{{\text{l}}^{ - 1}}$
Step (3):
From the chemical equation ${\text{(1)}}$, we can see that, ${\text{1 mol}}$ of $MgC{O_3}$ gives one mole of ${\text{MgO}}$.
In terms of molar mass, $84{\text{ g}}$ of $MgC{O_3}$ gives $40{\text{ g}}$ of ${\text{MgO}}$.
Now, in question, it is given that $20 \cdot 0{\text{ mg}}$ of the $MgC{O_3}$ sample decomposes on heating to give $8 \cdot 0{\text{ mg}}$ of ${\text{MgO}}$. So, we have to calculate, how much pure $MgC{O_3}$ is needed to produce $8 \cdot 0{\text{ mg}}$ of ${\text{MgO}}$ on heating.
Since, $40 \times 1000{\text{ mg}}$ of $MgO$ is given by $84 \times 1000{\text{ mg}}$ of pure $MgC{O_3}$,
So, ${\text{1 mg}}$ of ${\text{MgO}}$ is given by $\dfrac{{84 \times 1000{\text{ mg}}}}{{40 \times 1000{\text{ mg}}}}$ of pure $MgC{O_3}$
Or, ${\text{8 mg}}$ of ${\text{MgO}}$ is given by $\dfrac{{84 \times 1000{\text{ mg}}}}{{40 \times 1000{\text{ mg}}}} \times 8{\text{ mg}}$ of pure $MgC{O_3}$
$ \Rightarrow {\text{8 mg}}$ of ${\text{MgO}}$ is given by $16 \cdot 8{\text{ mg}}$of pure $MgC{O_3}$
Step (4):
Now, we have to find the percentage purity of $MgC{O_3}$ using the formula:
Percentage purity = $\dfrac{{{\text{Mass of pure }}MgC{O_3}{\text{ in the sample}}}}{{{\text{Total mass of the sample}}}} \times 100$
= $\dfrac{{16 \cdot 8}}{{20 \cdot 0}} \times 100$
= $80\% $
Hence, option (C) is the correct answer, $80\% $ is the percentage purity of magnesium carbonate in the sample.


Note:Purity, and yield are two different things. Yield is the amount of substance obtained from the reaction, and purity is the actual amount of substance in a sample. Percentage yield is given by:
${\text{Percentage yield = }}\dfrac{{{\text{Actual yield}}}}{{{\text{Theoretical yield}}}} \times 100\% $