Chapter 7, Problem 7.48QA

Interpretation Introduction

To calculate:

The percent composition of the following compounds

a) Sodium sulfate

b) Dinitrogen tetroxide

c) Strontium nitrate

d) Aluminum sulfide

Answer to Problem 7.48QA

Solution:

Percent composition of each element in a given compound is:

a) Sodium sulfate contains $32.37\% Na, 22.57\% S and 45.05\% O$

b) Dinitrogen tetroxide contains $30.45\% N and 69.55\% O$

c) Strontium nitrate contains $13.24\% N, 30.24\% O and 41.40\% Sr$

d) Aluminum sulfide contains $35.94\% Al and 64.06\% S$

Explanation of Solution

1) Concept:

We have to find the percent composition compound of the given compounds. The chemical formula of each of the given compounds is $N{a}_{2}S{O}_4, N_2{O}_4, Sr{\left(N{O}_3\right)}_2 and A{l}_2{S}_3$ respectively. We first find the molar mass of each compound, and then, we find the percent composition for each element in the chemical formula by the percent composition formula.

$\% composition=\frac{mass of an element}{total mass} \times 100$

The chemical formula tells us the ratio of moles of each elements in the formula. For example: one mole of $N_2{O}_4$ contains two moles of $N$ and four moles of $O$.

A mole is the SI unit of amount chemical substance. When writing units, it is written as “mol”.

2) Formula:

$\% composition=\frac{mass of an element}{total mass} \times 100$

3) Given:

Chemical names for four compounds are given as:

a) Sodium sulfate

b) Dinitrogen tetroxide

c) Strontium nitrate

d) Aluminum sulfide

4) Calculations:

Calculating mass of each of the given compounds as follows:

a) Calculating mass of $Na, S and O$ in one mole of $N{a}_{2}S{O}_4$ and the molar mass of $N{a}_{2}S{O}_4$:

$2 mol Na \times 22.99\frac{g}{mol}=45.98 g Na$

$1 mol S \times 32.065\frac{g}{mol}=32.065 g S$

$4 mol O \times 16.0\frac{g}{mol}=64.0 g O$

Molar mass of $N{a}_{2}S{O}_4$= $45.98+32.065+64.00=142.045 g/mol$

Calculating percent composition of $Na, S and O$ as:

$\% Na=\frac{45.98 g Na}{142.045 g} \times 100=32.37 \% Na$

$\% S=\frac{32.065 g S}{142.045 g} \times 100=22.57 \% S$

$\% O=\frac{64.0 g O}{142.045 g} \times 100=45.06 \% O$

b) Calculating mass of $N and O$ in one mole of $N_2{O}_{4 }$ and the molar mass of $N_2{O}_{4 }$:

$2 mol N \times 14.007\frac{g}{mol}=28.014 g N$

$4 mol O \times 16.0\frac{g}{mol}=64.0 g O$

Molar mass of $N_2{O}_{4 }$ = $28.014+64.00=92.014 g/mol$

Calculating percent composition of $N and O$ as:

$\% N=\frac{28.014 g N}{92.014 g} \times 100=30.45 \% N$

$\% O=\frac{64.0 g O}{92.014 g} \times 100=69.55 \% O$

c) Calculating mass of $Sr, N and O$ in one mole of $Sr{\left(N{O}_3\right)}_2$ and the molar mass of $Sr{\left(N{O}_3\right)}_2$:

$2 mol N \times 14.007\frac{g}{mol}=28.014 g N$

$6 mol O \times 16.0\frac{g}{mol}=96.0 g O$

$1 mol Sr \times 87.62\frac{g}{mol}=87.62 g Sr$

Molar mass of $N_2{O}_{4 }$ = $28.014+96.0+87.62=211.654 g/mol$

Calculating percent composition of $Sr, N and O$ as:

$\% N=\frac{28.014 g N}{211.654 g} \times 100=13.24 \% N$

$\% O=\frac{96.0 g O}{211.654g} \times 100=45.36 \% O$

$\% Sr=\frac{87.62 g N}{211.654 g} \times 100=41.40 \% Sr$

d) Calculating mass of $Al and S$ in one mole of $A{l}_2{S}_3$ and the molar mass of $A{l}_2{S}_3$:

$2 mol Al \times 26.982\frac{g}{mol}=53.964 g Al$

$3 mol S \times 32.065\frac{g}{mol}=96.195 g O$

Molar mass of $A{l}_2{S}_3$= $53.964+96.195=150.159 g/mol$

Calculating percent composition of $Al and S$ as:

$\% Al=\frac{53.964 g Al}{150.159 g} \times 100=35.94 \% Al$

$\% S=\frac{96.195 g S}{150.159 g} \times 100=64.06 \% S$

Conclusion:

We can find the percent composition of each element present in a given compound by using atomic weights and mole ratio.