Question

Determine percentage of oxygen in ammonium nitrate.
A.60%
B.80%
C.70%
D.55%

Answer 1

Hint:We can calculate the percentage of oxygen in ammonium nitrate by calculating the required mass of oxygen atoms from the number of moles of oxygen and the molar mass of ammonium nitrate. The mass of oxygen atoms is calculated by multiplying the number of moles of oxygen atom present in the ammonium nitrate and atomic weight of oxygen atom.
Formula used: We can calculate the mass percent using the formula,
${\text{Mass}}\,{\text{percentage = }}\dfrac{{{\text{Grams}}\,{\text{of}}\,{\text{solute}}}}{{{\text{Grams}}\,{\text{of}}\,{\text{solution}}}}{\text{ X 100% }}$

Complete step by step answer:
Let us first calculate the molar mass of ammonium nitrate compound.
We know that,
The molar mass of hydrogen is \[1{\text{ }}g/mol\].
The molar mass of Nitrogen is \[14.01{\text{ }}g/mol\].
The molar mass of oxygen is \[16{\text{ }}g/mol\].
We can see in the formula of ammonium nitrate $\left( {N{H_4}N{O_3}} \right)$ that there is one hydrogen atom, two nitrogen atoms and three oxygen atoms.
We can find the total molar mass of the compound by multiplying the number of atoms with their respective molar mass and then adding them together.
The mass of hydrogen atom=$4\,mol\,H X \dfrac{{1\,g\,H}}{{1\,mol\,H}} = 4\,g\,H$
The mass of nitrogen atom=$2\,mol\,N X \dfrac{{14.01\,g\,N}}{{1\,mol\,N}} = 28.02\,g\,N$
The mass of oxygen atom=$3\,mol\,O X \dfrac{{16.00\,g\,O}}{{1\,mol\,O}} = 48\,g\,O$
By summing up, all the individual masses we get the molar mass.
Molar mass of $N{H_4}N{O_3}$=$4\,g + 28.02\,g + 48.00\,g$
Molar mass of $N{H_4}N{O_3}$=$80\,g/mol$
The molar mass of $N{H_4}N{O_3}$ is $80\,g/mol$.
One mole of ammonium nitrate contains three moles of oxygen atoms.
We know that oxygen has an atomic mass of $16\,g/mol$.
Therefore, six moles of oxygen atoms will correspond to $3\,mol X \dfrac{{16\,g}}{{1\,mol}} = 48\,g$
We have calculated that three moles of oxygen atom contains $48\,g$.
From the calculated mass, let us now calculate the mass percent of oxygen in ammonium nitrate.
Mass percent=$\dfrac{{{\text{Mass}}\,\left( {{\text{in}}\,{\text{grams}}} \right)}}{{{\text{Total}}\,{\text{mass}}\left( {{\text{in}}\,{\text{grams}}} \right)}}{\text{ X 100% }}$
The mass of the oxygen atom is $48\,g$.
The total mass of the compound is $80\,g$.
Let us substitute the values in the equation of mass percent.
Mass percent=$\dfrac{{{\text{Mass}}\,\left( {{\text{in}}\,{\text{grams}}} \right)}}{{{\text{Total}}\,{\text{mass}}\left( {{\text{in}}\,{\text{grams}}} \right)}}{\text{ X 100% }}$
Mass percent=$\dfrac{{{\text{48}}\,{\text{g}}}}{{{\text{80 g}}}}{\text{ X 100% }}$
Mass percent=${\text{60% }}$
The mass percent of oxygen in ammonium nitrate is ${\text{60% }}$.
Therefore, Option (A) is correct.

Note:
We can prepare ammonium nitrate by the acid base reaction of ammonia with nitric acid. It is an essential fertilizer. Ammonium nitrate is used in cold packs, as its dissolution in water is extremely endothermic. Pure ammonium nitrate does not burn, but acts as a strong oxidizer in supporting and accelerating the combustion of organic material. It is less concentrated than urea and thus, transportation of ammonium nitrate is at a slight disadvantage.