Question

Calculate the percentage composition of oxygen, nitrogen and calcium in Calcium nitrate $\left(\mathrm{Ca}{\left({\mathrm{NO}}_3\right)}_2\right)$.

Given : Atomic weight of O = 16u, Ca = 40u, N = 14u

Answer 1

Step 1: Given data

Atomic weight of O = 16u, Ca = 40u, N = 14u

Step 2: Finding the mass for the given data

The molecular mass of $\left(\mathrm{Ca}{\left({\mathrm{NO}}_3\right)}_2\right)$ = (Number of calcium atoms x atomic mass of calcium) + (Number of nitrogen atoms x atomic mass of nitrogen) + (Number of oxygen atoms x atomic mass of oxygen)

The molecular mass of $\left(\mathrm{Ca}{\left({\mathrm{NO}}_3\right)}_2\right)$ = $\left(1\times 40\right)+\left(2\times 14\right)+\left(6\times 16\right)$

The molecular mass of $\left(\mathrm{Ca}{\left({\mathrm{NO}}_3\right)}_2\right)$ = $40+28+96=164$u

Step 3: Finding the mass for each element

Mass of $\mathrm{Ca}$ = (Number of calcium atoms x atomic mass of calcium) $=\left(1\times 40\right)=40u$

Mass of $\mathrm{N}$ = (Number of nitrogen atoms x atomic mass of nitrogen) $=\left(2\times 14\right)=28u$

Mass of $\mathrm{O}$ = (Number of oxygen atoms x atomic mass of oxygen) $=\left(6\times 16\right)=96u$

Step 4: Finding the % composition

% composition is the proportion by mass of each element present in a compound.

% composition of $\mathrm{Ca}=\frac{40}{164}=24.39\%$

% composition of $\mathrm{N}=\frac{28}{164}=17.07\%$

% composition of $\mathrm{O}=\frac{96}{164}=58.54\%$

Thus, the percentage composition of oxygen, nitrogen and calcium in Calcium nitrate $\left(\mathrm{Ca}{\left({\mathrm{NO}}_3\right)}_2\right)$ is $58.54\%,17.07\%\mathrm{and}24.39\%$respectively.