Question

When the mixture of $MgC{{O}_{3}}$and $CaC{{O}_{3}}$was heated for a long time, the weight decreased by 50%. Calculate the percentage composition of the mixture.

Answer 1

Percent composition of any compound can be defined as the ratio of the amount of every individual element to the total amount of individual elements present in the compound multiplied by 100. This helps in chemical analysis of given compounds.

Complete answer:
Hence to find out the percentage composition first consider the total mass will be 100 g then suppose
Weight of $MgC{{O}_{3}}$= x gm
This implies Weight of $CaC{{O}_{3}}$= 100-x gm as total weight is 100 g
Now the reaction of magnesium and calcium carbonate upon heating can be shown as:
$MgC{{O}_{3}}\to MgO+C{{O}_{2}}$
$CaC{{O}_{3}}\to CaO+C{{O}_{2}}$
Number of moles of magnesium carbonate will be given as $\dfrac{x}{84}$
Number of moles of calcium carbonate is given as $\dfrac{100-x}{100}$
Now the weight of $C{{O}_{2}}$evolved in $MgC{{O}_{3}}$ will be calculated by multiplying the number of moles into its molar mass which will be given as follows:
$\dfrac{x}{84}\times 44g$
Similarly the weight of $C{{O}_{2}}$evolved in $CaC{{O}_{3}}$ will be given as follows:
$\dfrac{100-x}{100}\times 44g$
According to question weight loss given is 50%, therefore
$\dfrac{x}{84}\times 44+\dfrac{100-x}{100}\times 44g=50$, here 50 is 50% of 100 g
$\dfrac{x}{84}+\dfrac{100-x}{100}=\dfrac{50}{44}$
Now by solving the above equation the value of x will be 71.59%
This implies that % of $MgC{{O}_{3}}$= $\dfrac{x}{100}\times 100=71.59$
% of $CaC{{O}_{3}}$= 100-x = 28.41%

Note:
Mole is generally represented by the symbol mol. It is generally described as the unit of measurement for the amount of substance in SI where SI stands for International System of units. It is defined on the basis of Avogadro’s number.