Chapter 10.4, Problem 54PP

Interpretation Introduction

Interpretation:

The percent composition of phosphoric acid needs to be determined.

Concept introduction:

The molar mass of any substance (which is $1$ mole of substance) contain different elements present in different quantity. So percent composition in Stochiometry tells us the mass /volume/molar composition in any compound.

$\text{\%composition = }\frac{\text{mass of component}}{\text{total mass }\left(\text{molar mass}\right)\text{of the compound}}\times 100$

Answer to Problem 54PP

$\begin{array}{l}\text{Hydrogen }\Rightarrow \text{3}\cdot \text{08\% }\\ \text{Phosphorous}\Rightarrow \text{31}\cdot \text{6\% }\\ \text{Oxygen}\Rightarrow \text{65}\cdot \text{32\%}\end{array}$

Explanation of Solution

The given compound is phosphoric acid ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$

To find the $\%$ composition of hydrogen, phosphorous and oxygen atom first calculate the number of moles.

Molar mass of Hydrogen $=1\text{ g/mol}$

Molar mass of phosphorus $=31\text{ g/mol}$

Molar mass of oxygen $=16\text{ g/mol}$

Total molar mass of ${\text{H}}_3{\text{PO}}_4\text{ =3}\times 1+31+\text{64=98 g/mol}$

Mass of hydrogen ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}=3\times 1=3\text{ g}$

Mass of phosphorous ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}=1\times 31=31\text{ g}$

Mass of oxygen ${\text{H}}_3{\text{PO}}_{\text{4}}=4\times 16=16\text{ g}$

$\begin{array}{l}\therefore \text{Mass of atoms in a compound}\Rightarrow \text{number of atom in that compound}\times \text{molar mass of that atom}\\ \left[\%\Rightarrow \text{percent}\right]\end{array}$

Thus,

$\text{\%composition = }\frac{\text{mass of component}}{\text{total mass }\left(\text{molar mass}\right)\text{of the compound}}\times 100$

Putting the values,

$\%$ Composition of hydrogen $=\frac{3}{98}\times 100=3\cdot 06\%$

$\%$ Composition of phosphorus $=\frac{31}{98}\times 100=31\cdot 63\%$

$\%$ Composition of oxygen $=\frac{64}{98}\times 100=65\cdot 30\%$