Question

What percentage of oxygen is present in the compound $CaC{O_3} \cdot 3C{a_3}{(P{O_4})_2}$ ?
A.$23.3\%$
B.$45.36\%$
C.$41.94\%$
D.$17.08\%$

Answer 1

In order to solve this question first of all we need to calculate the total mass of the compound then we calculate the mass of oxygen present in the compound in terms of their molar weight and then take out the percentage of total weight of oxygen to the total weight of the compound.

Complete answer:
In the above given question, we are asked to calculate the oxygen percentage in the given compound $CaC{O_3} \cdot 3C{a_3}{(P{O_4})_2}$ .
Now the given compound $CaC{O_3} \cdot 3C{a_3}{(P{O_4})_2}$ has too many elements present in it. There is calcium present in the compound, there is carbon, phosphorus and oxygen,
First, we need to calculate the total molecular weight of the compound which would be equal to the individual molecular weights of calcium, carbon, phosphorus and oxygen multiplied by their respective number of units in the compound which we get to know by seeing their lower coefficients.
Total molecular weight of the compound is equal to
$molecular\,weight = Ca + C + O \times 3 + 3 \times [Ca \times 3 + 2(P + 4 \times O)]$
$\Rightarrow molecular\,weight = 40 + 12 + 3 \times 16 + 3[40 \times 3 + 2(31 + 4 \times 16)]$
$\Rightarrow molecular\,weight = 100 + 3[120 + 190]$
$\Rightarrow molecular\,weight = 1030\dfrac{g}{{mol}}$
We have got the total molecular weight of the compound now we have to calculate the total mass of oxygen in the molecular compound $CaC{O_3} \cdot 3C{a_3}{(P{O_4})_2}$
Only calculating the total molar mass of oxygen
$weight\,of\,O = 3 \times 16 + 3[2(4 \times 16)]$
$\Rightarrow weight\,of\,O = 48 + 3 \times 128$
$\Rightarrow weight\,of\,O = 432\dfrac{g}{{mol}}$
Now let’s calculate the oxygen percentage in the compound by dividing the weight of oxygen with the weight of the whole compound $CaC{O_3} \cdot 3C{a_3}{(P{O_4})_2}$
Therefore, the percentage of oxygen is equal to
$\% \,of\,O = \dfrac{{weighto\,of\,oxygen}}{{weight\,of\,CaC{O_3} \cdot 3C{a_3}{{(P{O_4})}_2}}} \times 100$
$\Rightarrow \% \,of\,O = \dfrac{{432}}{{1030}} \times 100$
$\Rightarrow \% \,of\,O = 41.94\%$
Thus, from the above given solution we can say that the percentage of oxygen in the compound $CaC{O_3} \cdot 3C{a_3}{(P{O_4})_2}$ is $41.94\%$.

Note:
Oxygen is the second most abundant element in our atmosphere. It helps in the formation of major compounds and also used vividly for various purposes. Fun fact to know is that oxygen is what ignites a fire therefore extinguishers have a carbon compound which shields fire from the atmosphere and then the fire extinguishers.