Question

1 gram of an impure sample of magnesium carbonate (containing no thermally decomposable impurities) on complete thermal decomposition gave 0.44 gram of carbon dioxide gas.The percentage of the impurity of sample is:-
a)0%
b)4.4%
c)16%
d)8.4%

Answer 1

Answer:- 16%

Solution:- On decomposition, metal carbonates gives metal oxide and carbon dioxide. The decomposition reaction for magnesium carbonate is written as:

$MgCO_3\rightarrow MgO+CO_2$

From this equation, there is 1:1 mol ratio between magnesium carbonate and carbon dioxide.

From given mass of carbon dioxide we calculate it's moles and then using mol ratio the moles of magnesium carbonate are calculated that could further be converted to grams. Molar mass of magnesium carbonate is 84.31 gram per mol. The calculations for grams of carbon dioxide to grams of magnesium carbonate are shown as:

$0.44gCO_2(\frac{1molCO_2}{44gCO_2})(\frac{1molMgCO_3}{1molCO_2})(\frac{84.31gMgCO_3}{1molMgCO_3})$

= $0.843gMgCO_3$

Mass of the impure sample given as 1 gram. If we subtract the mass of pure magnesium carbonate from it then we would get the mass of the impurity in the sample:

mass of impurity = 1 - 0.843 = 0.157 g

Now, we could calculate the mass percentage of the impurity of the sample as:

Impurity% = $(\frac{0.157}{1})100$

= 15.7%

It could be rounded to 16%. So, the impurity percentage of the sample is 16%.