Question

In the formula mass of $Ca{\left( {N{O_3}} \right)_2}$, find the approximate percentage of oxygen?
A.$28$
B.$42$
C.$58$
D.$68$
E.$84$

Answer 1

Hint: We can calculate the percentage of oxygen in calcium nitrate by calculating the required mass of oxygen atoms from the number of moles of oxygen and the molar mass of calcium nitrate. The mass of oxygen atoms is calculated by multiplying the number of moles of oxygen atom present in the calcium 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 the calcium nitrate compound.
We know that,
The molar mass of Calcium is \[40.1{\text{ }}g/mol\].
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 calcium nitrate $Ca{\left( {N{O_3}} \right)_2}$ that there is one calcium atom, two nitrogen atoms and six 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 calcium atom=$1\,mol\,Ca X \dfrac{{40.1\,g\,Ca}}{{1\,mol\,Ca}} = 40.1\,g\,Ca$
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=$6\,mol\,O X \dfrac{{16.00\,g\,O}}{{1\,mol\,O}} = 96\,g\,O$
By summing up, all the individual masses we get the molar mass.
Molar mass of $Ca{\left( {N{O_3}} \right)_2}$=$40.1\,g + 28.02\,g + 96.00\,g$
Molar mass of $Ca{\left( {N{O_3}} \right)_2}$=$164.12\,g/mol$
The molar mass of $Ca{\left( {N{O_3}} \right)_2}$ is $164.12\,g/mol$.
One mole of calcium nitrate contains six 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 $6\,mol X \dfrac{{16\,g}}{{1\,mol}} = 96\,g$
We have calculated that six moles of oxygen atom contains $96\,g$.
From the calculated mass, let us now calculate the mass percent of oxygen in calcium 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 $96\,g$.
The total mass of the compound is $164.12\,g$.
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=$\dfrac{{{\text{96}}\,{\text{g}}}}{{{\text{164}}{\text{.12 g}}}}{\text{ X 100% }}$
Mass percent=${\text{58\% }}$
The mass percent of oxygen in calcium nitrate is ${\text{58% }}$
Therefore, option (C) is correct.

Note:
We can prepare calcium nitrate using an aqueous solution of ammonium nitrate and calcium hydroxide. It appears colourless and absorbs moisture from the atmosphere. Calcium nitrate is an intermediate product formed in the Odda process. It is used in agriculture, wastewater treatment, cold packs, latex coagulant, concrete and also a molten part for transfer of heat and storage. Calcium nitrate is a major component in fertilizers.