Question

# Phosphoric acid is used in carbonated beverages containing 3.086% (w/w) hydrogen, 31.61% (w/w) phosphorous and remaining oxygen. If the atomic mass of hydrogen, phosphorus and oxygen are 1.01U, 31.0U and 16U respectively. And if the molar mass of phosphoric acid is 98.03 g/mol. What is the molecular formula of phosphoric acid?

Hint: There are three types of formula in the chemistry i.e. empirical, molecular and structural formula. The empirical formula gives us the simplest ratio of the atoms present in the compound whereas the molecular formula tells us about the total no. of each atom in a molecule and structural formula shows the bonding of an atom.

Formula used:

- In the given question, it is given that the mass percent of hydrogen and phosphorus is 3.086 % and 31.61 % respectively.
- So, we can calculate the mass % of oxygen by subtracting the addition of mass percent of hydrogen and phosphorus from 100 i.e.
$=100\text{ - ( 3}\text{.086 -31}\text{.61)}\ \text{= 65}\text{.304 }\!\!%\!\!\text{ }$
- It is given that the atomic mass of H, P and O are 1.01u, 31.0u and 16u respectively.
-So, we can calculate the empirical formula of phosphoric acid i.e.
 Element Mass Percent Atomic mass Relative no. of atoms Simplest atomic ratio Simplest whole no. ratio H 3.086 1.01 $\frac{3.086}{1.01}\text{ = 3}\text{.055}$ $\frac{3.055}{1.02}\text{ = 3}$ 3 P 31.61 31.0 $\frac{31.61}{31}\text{ = 1}\text{.02}$ $\frac{1.02}{1.02}\text{ = 1}$ 1 O 65.304 16 $\frac{65.304}{16}\text{ = 4}\text{.08}$ $\frac{4.08}{1.02}\text{ = 4}$ 4

- So, the empirical formula will be ${{\text{H}}_{3}}\text{P}{{\text{O}}_{4}}$.
- Now, the empirical mass of the compound will be:
$3\ \cdot \text{ 1}\text{.01 + }31\ \cdot \text{ 1}\ \text{+ 4}\ \cdot \text{ 1}6\text{ = 98}\text{.03u}$
- It is given that the molar mass of the compound is 98.03u so the molecular mass will be calculated by dividing molar mass by empirical mass i.e. $\frac{98.03}{98.03}\ \text{= 1}$.
- Now, by multiplying the 1 by the subscript of the empirical formula we will get the molecular formula and it is ${{\text{H}}_{3}}\text{P}{{\text{O}}_{4}}$.
So, the correct answer is “${{\text{H}}_{3}}\text{P}{{\text{O}}_{4}}$”.

Note: The empirical formula and molecular formula have a great role because they can be used to determine the general formula of a compound & also the type of molecule and balancing the equations respectively.