Question

Percentage composition of elements in magnesium nitrate $Mg(N{O}_3{)}_2$ is:

• A. 16.2 % magnesium
• B. 46.6 % nitrogen
• C. 8.9 % nitrogen
• D. 64.8 % oxygen

Answer 1

Answer: (A), (C), (D)

The correct options are
A 16.2 % magnesium
C 8.9 % nitrogen
D 64.8 % oxygen

Molecular weight of ${Mg\left({NO}_2\right)}_2=24+2\times 14+4\times 16$

$=24+28+64$

$=116g$

Let there be $1$ mole of ${Mg\left({NO}_2\right)}_2$

1 mole of ${Mg\left({NO}_2\right)}_2$ contains

$=1$ mole of $Mg=24g$

$=2$ moles of $N=2\times 14=28g$

$=4$ moles of $O=4\times 16=64g$

$\therefore$ Percentage compoistion :

$\ast Mg=\frac{24}{116}\times 100=20.69$%

$\ast N=\frac{28}{116}\times 100=24.14$%

$\ast$ % of $O=\frac{64}{116}\times 100=55.17$%