Question

In the formula mass of $\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2},$ find the approximate percentage of oxygen?
(A) 28
(B) 42
(C)58
(D) 68
(E) 84

Answer 1

Hint: Calcium nitrate is an inorganic compound with the formula $\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} .$ This colourless salt absorbs moisture from the air and is commonly found as a tetrahydrate. The formula to be used here would be $\%$ age of oxygen $=($ Weight of $\mathrm{O} \mathrm{X} 100) /$ Total molar mass of $\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}$

Complete step by step solution:
-Let us see the formula of Calcium nitrate i.e., $\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} .$ Now, let us calculate the molecular mass of $\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}$
-We know the molecular masses of individual atoms. Let us note it down now.
Calcium is $40 \mathrm{g} / \mathrm{mol}$. Nitrogen is $14 \mathrm{g} / \mathrm{mol}$. Oxygen is $16 \mathrm{g} / \mathrm{mol}$. Then according to the formula $\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2},$ it is $40+2 \times(14+3(16))=40+2 \times(62)=164 \mathrm{g} / \mathrm{mol}$
-1 mole (164.1 g) of calcium nitrate will contain 6 moles of O atoms.
-The atomic mass of oxygen is $16 \mathrm{g} / \mathrm{mol}$. 6 moles of oxygen atoms corresponds to $6 \times 16=96 \mathrm{g}$
-Discussing Option $\mathrm{C},$ The percent of oxygen in calcium nitrate $=(96 \times 100) / 164$ which gives us
$58 \%$. Clearly, it is $58 \%$.

Hence, the answer is (C).

Note: As in the formula of calcium nitrate, the percentage of oxygen is $58 \% .$ The percentage of calcium in the compound is $(40 \times 100) / 164$ i.e., $, 24 \% .$ Similarly, let us find out the percentage composition of nitrogen in the compound i.e., $18 \% .$ Hence this is the total percentage composition of the elements in calcium nitrate.