Chapter 7, Problem 7.47QA

Interpretation Introduction

To calculate:

Percent composition of following compounds:

a) $N{a}_{2}O$

b) $NaOH$

c) $NaHC{O}_3$

d) $N{a}_{2}C{O}_3$

Answer to Problem 7.47QA

Solution:

The Percent composition of following compounds are:

a) $N{a}_{2}O$

$74.19\% Na, 25.81\% O$

b) $NaOH$

$57.48\% Na, 40.00\% O, 2.52\% H$

c) $NaHC{O}_3$

$27.37\% Na, 14.30\% C, 57.13\% O, 1.20\% H$

d) $N{a}_{2}C{O}_3$

$43.38\% Na, 45.29\% O, 11.33 \% C$

Explanation of Solution

1) Concept:

We are asked to calculate the percent composition of the given four compounds. The chemical formula of each of the given compounds gives us the stoichiometry between the compound and its corresponding elements. That is, for example, one mole of $N{a}_{2}O$ contains two moles of $Na$ and one mole of $O$. To find thepercent composition values based on mass, we must first convert the mole values to mass values by multiplying each value by the appropriate molar mass. Summing these element masses and then dividing the mass of each element by the total mass, and expressing the results as percentages gives us the percent composition on each element in the sample.

2) Formula:

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

3) Given:

Molecular formulas for four compounds are:

(a) $N{a}_{2}O$

(b) $NaOH$

(c) $NaHC{O}_3$

(d) $N{a}_{2}C{O}_3$

4) Calculations:

Calculating molar mass of each of the given compounds as follows:

(a) Calculating mass of $Na and O$ in one mole of $N{a}_{2}O$ and the molar mass of $N{a}_{2}O$

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

$1 mol O \times 16.0\frac{g}{mol}=16.0 g O$

Molar mass of $N{a}_{2}O$ = $45.98+16.0=61.98\frac{g}{mol}$

Calculating percent composition of $Na and O$ as:

$\% Na=\frac{45.98 g Na}{61.98 g} \times 100=74.19 \% Na$

$\% O=\frac{16.0 g O}{61.98 g} \times 100=25.81 \% O$

(b) Calculating mass of $N, O and H$ in one mole of $NaOH$ and the molar mass of $NaOH$:

$1 mol Na \times 22.99\frac{g}{mol}=22.99 g Na$

$1 mol O \times 16.0\frac{g}{mol}=16.0 g O$

$1 mol H\times 1.008\frac{g}{mol}=1.008 g H$

Molar mass of $NaOH$= $22.99+16.0+1.008=39.998\frac{g}{mol}$

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

$\% Na=\frac{22.99 g Na}{39.998g} \times 100=57.48 \% Na$

$\% O=\frac{16.0 g O}{39.998 g} \times 100=40.00 \% O$

$\% H=\frac{1.008 g O}{39.998 g} \times 100=2.52 \% H$

(c) Calculating mass of $N, O, C and H$ in one mole of $NaHC{O}_3$ and the molar mass of $NaHC{O}_3$

$1 mol Na \times 22.99\frac{g}{mol}=22.99 g Na$

$3 mol O \times 16.0\frac{g}{mol}=48.0 g O$

$1 mol H\times 1.008\frac{g}{mol}=1.008 g H$

$1 mol C\times 12.01\frac{g}{mol}=12.01 g C$

Molar mass of $NaHC{O}_3$= $22.99+48.0+1.008+12.01=84.008\frac{g}{mol}$

Calculating percent composition of $Na, O, C and H$ as:

$\% Na=\frac{22.99 g N}{84.008 g} \times 100=27.37 \% Na$

$\% O=\frac{48.0 g O}{84.008 g} \times 100=57.13 \% O$

$\% C=\frac{12.01 g C}{84.008 g} \times 100=14.30 \% C$

$\% H=\frac{1.008 g H}{84.008 g} \times 100=1.20 \% H$

(d) Calculating mass of $N, O, and C$ in one mole of $N{a}_{2}C{O}_3$ and the molar mass of $N{a}_{2}C{O}_3$

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

$3 mol O \times 16.0\frac{g}{mol}=48.0 g O$

$1 mol C\times 12.01\frac{g}{mol}=12.01 g C$

Molar mass of $N{a}_{2}C{O}_3$= $45.98+48.0+12.01=105.99\frac{g}{mol}$

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

$\% Na=\frac{45.98 g Na}{105.99 g} \times 100=43.38 \% Na$

$\% O=\frac{48.0 g O}{105.99 g} \times 100=45.29 \% O$

$\% C=\frac{12.01 g C}{105.99 g} \times 100=11.33 \% C$

Conclusion:

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