Question

In diammonium phosphate ${\left( {N{H_4}} \right)_2}HP{O_4}$, the percentage of ${P_2}{O_5}$ is:
A.35.87
B.46.44
C.51.99
D.53.78

Answer 1

We can calculate the percentage of ${P_2}{O_5}$ using the molar mass of ${P_2}{O_5}$ and twice the molar mass of ${\left( {N{H_4}} \right)_2}HP{O_4}$, whole multiplied by 100.
Formula used: We can calculate the percent using the formula,
Mass percentage$= \dfrac{{{\text{Molar mass of }}{{\text{P}}_2}{{\text{O}}_5}}}{{2 \times {\text{Molar mass }}{{\left( {N{H_4}} \right)}_2}HP{O_4}}} \times 100\%$

Complete step by step answer: Two moles of diammonium hydrogen phosphate gives one mole of phosphorus pentoxide.
We can write the conversion of diammonium phosphate ${\left( {N{H_4}} \right)_2}HP{O_4}$ to phosphorus pentoxide as,
$2{\left( {N{H_4}} \right)_2}HP{O_4} \to {P_2}{O_5}$
Let us now calculate the molar mass of ${\left( {N{H_4}} \right)_2}HP{O_4}$ is,
The molar mass of nitrogen is $14g/mol$.
The molar mass of hydrogen is $1g/mol$.
The molar mass of phosphorus is $31g/mol$.
The molar mass of oxygen is $16g/mol$.
So, the molar mass of ${\left( {N{H_4}} \right)_2}HP{O_4}$ is calculated as,
Molar mass=$2 \times \left( {14 + \left( {1 \times 4} \right)} \right) + \left( {1 + 31 + \left( {4 \times 16} \right)} \right)g/mol$
$\Rightarrow$Molar mass=$2 \times \left( {14 + 4} \right) + \left( {1 + 31 + 64} \right)g/mol$
$\Rightarrow$Molar mass=$\left( {36 + 1 + 31 + 64} \right)g/mol$
$\Rightarrow$Molar mass=$132g/mol$
The molar mass of ${\left( {N{H_4}} \right)_2}HP{O_4}$ is $132g/mol$.
So, the molar mass of ${P_2}{O_5}$ is calculated as,
Molar mass=$\left( {\left( {2 \times 31} \right) + \left( {5 \times 16} \right)} \right)g/mol$
$\Rightarrow$Molar mass=$\left( {62 + 80} \right)g/mol$
$\Rightarrow$Molar mass=$142g/mol$
The molar mass of ${P_2}{O_5}$ is $142g/mol$.
From the calculated molar mass of ${P_2}{O_5}$ and ${\left( {N{H_4}} \right)_2}HP{O_4}$, we can calculate the percentage of ${P_2}{O_5}$.
In the starting part of the explanation, we saw that two moles of diammonium hydrogen phosphate gives one mole of phosphorus pentoxide. So we have to multiply the molar mass of diammonium hydrogen phosphate by two.
We get the molar mass of diammonium hydrogen phosphate as $264g/mol$.
So, from these data let us now calculate the percentage of ${P_2}{O_5}$.
Substituting the values of molar mass of ${P_2}{O_5}$ and ${\left( {N{H_4}} \right)_2}HP{O_4}$, we get the percentage of ${P_2}{O_5}$ as,
Mass percentage$= \dfrac{{{\text{Molar mass of }}{{\text{P}}_2}{{\text{O}}_5}}}{{2 \times {\text{Molar mass }}{{\left( {N{H_4}} \right)}_2}HP{O_4}}} \times 100\%$
Substituting the values in above equation we get,
$\Rightarrow$Mass percentage$= \dfrac{{142g/mol}}{{264g/mol}} \times 100\%$
On simplifying,
$\Rightarrow$Mass percentage$= 53.78\%$
So, the percentage of ${P_2}{O_5}$ is $53.78\%$.
Therefore, the option (D) is correct.

Note: We can also calculate the mass of a substance using mass percentage. An example is given below.
Example: The mass of sodium chloride present in $35g$ of ${\text{3}}{\text{.5\% }}$ solution has to be calculated.
Mass percentage of the solution =$3.5$
Mass of solution =$35.0g$
The mass of sodium chloride is given as,
${\text{Mass percentage}} = \dfrac{{{\text{Grams of sodium chloride}}}}{{{\text{Grams of solution}}}} \times 100\%$
Substituting the values we get,
$3.5\% = \dfrac{{{\text{Grams of sodium chloride}}}}{{3.50g}} \times 100\%$
Grams of sodium chloride =$1.23g$
The mass of sodium chloride present in $35g$ of {\text{3}}{\text{.5% }} solution is $1.23g$.