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Question: A sample of ammonium phosphate \({\left( {N{H_4}} \right)_3}P{O_4}\)contains 3.18 moles of hydrogen ...

A sample of ammonium phosphate (NH4)3PO4{\left( {N{H_4}} \right)_3}P{O_4}contains 3.18 moles of hydrogen atoms. What is the number of moles of oxygen atoms in the sample?

Explanation

Solution

Hint – Here we will proceed by using molecular mass of ammonium sulphate to find out the number of moles of oxygen atoms in the sample.
Molecular Formula: During the chemical bond formation, every atom forms a bond with other atoms in certain molar ratios to form a compound.

Complete answer:

We know that 1 mole of ammonium phosphate (molecules) has 12 moles of hydrogen (atoms).
Here we need to find out how much ammonium phosphate has 3.18 moles of hydrogen.
By using Unitary method, we will find out –
We find that 0.265 moles of ammonium phosphate has 3.18 moles of hydrogen. We also know that 1 mole of ammonium phosphate 4 moles of oxygen. Again, using the Unitary method, we find that 0.265 moles of ammonium phosphate of 0.265 moles of ammonium phosphate has 1.06 moles of oxygen.
Let’s count individual atoms in (NH4)3PO4{\left( {N{H_4}} \right)_3}P{O_4}
N=3 H=12 P=1 O=4  N = 3 \\\ H = 12 \\\ P = 1 \\\ O = 4 \\\
N- Nitrogen, H- Hydrogen, O- Oxygen, P- Phosphorus.
From this, we come to know that 12 moles of Hydrogen correspond to 3 moles of Oxygen.
Let xx moles of Oxygen correspond to 3.18 moles of Hydrogen.
Therefore x4=3.1812\dfrac{x}{4} = \dfrac{{3.18}}{{12}}
x=3.18×412 =3.183 =1.06moles  \therefore x = \dfrac{{3.18 \times 4}}{{12}} \\\ = \dfrac{{3.18}}{3} \\\ = 1.06moles \\\
Therefore, 1.06 moles of oxygen is present in the sample.

Note – In this particular question, one must know that (NH4)3PO4{\left( {N{H_4}} \right)_3}P{O_4} This compound is called Ammonium sulphate. It contains a total 12 hydrogen atoms with 4 oxygen atoms. According to the formula no. of oxygen atoms is 3 times lower than the number of hydrogen atoms. Then by using the molar mass for ammonium sulphate one can easily solve this question.