Question
Question: What percentage of oxygen is present in the compound\(\text{ CaC}{{\text{O}}_{\text{3}}}\text{ }\tex...
What percentage of oxygen is present in the compound CaCO3 .3 Ca3(PO4)2 ?
(A) 23.30 0/0
(B) 45.36 0/0
(C) 41.94 0/0
(D) 17.08 0/0
Solution
The percentage composition of the element in the compound will be,
Percentage of atom in compound = Molecular mass of compoundmolar mass of atom ×1000/0
The molecular mass of the molecule is calculated by adding the molar mass of the atoms present in the compound.
Complete step by step solution:
The percentage composition is a very simple representation of the percentage of the atom in the compound. It tells about the mass of each element forming a respective compound. For convenience. We assume that the entire mass percentage of a compound is always equal to the 1000/0 . So that, an atom of the element of the compound may have its share in the formation of the compound out of the 1000/0 .
Following are the steps which can be used to calculate the percentage composition of elements in the compound:
- First, find out the molar masses of all the elements constituting the compound in terms of grams per mole.
- Second step is to find out the molecular mass of the compound. This can be calculated by adding the molar masses of all atoms in the compound.
- Third step is to obtain the ratio of the molar mass of the atom of interest to the entire molecular mass. It generally gives the number which is in between the 0 to 1.
- In the last step, multiply the obtained ratio by the 1000/0 to get the percentage composition of the atom of interest.
We have given the compound CaCO3 .3 Ca3(PO4)2 . We are interested to find out the percentage of oxygen in the compound.
Let’s first write down the molar mass of all elements present in the compound.
Element | Molar mass ( in gram per mole) |
---|---|
Ca | 40 |
C | 16 |
O | 12 |
P | 30 |
Let's calculate the molecular mass of the entire compound CaCO3 .3 Ca3(PO4)2 .
\begin{aligned}
& \text{=10 (40) + 1 (12) + 27 (16) + 6 (31)} \\
& \text{= 400 + 12 + 432 + 186} \\
& \text{= 1030 g mo}{{\text{l}}^{-1}} \\
\end{aligned}
So, the total number of the oxygen atom in the compound is equal to the 27. Therefore, the total mass of the oxygen atoms in the compound will be = 27 × 16 = 432 g mol−1
Let's find out the 0/0 composition of the oxygen in the compound.
composition !! !! of !! !! oxygen !! !! =mass !! !! of !! !! compoundmass !! !! of !! !! oxygen !! !! in !! !! compound=1030 g mol−1 432 g mol−1= 0.419
Now the 0/0 of oxygen is,
0/0 composition !! !! of !! !! oxygen = 0.419 × 1000/0 = 41.94 0/0
Therefore, the percentage composition of the oxygen in the CaCO3 .3 Ca3(PO4)2 compound is equal to41.94 0/0.
So, (C) is the correct option.
Note: Note that, the percentage composition of all the elements in the compound is always equal to the 1000/0 . In a binary compound AxBy , if the 0/0 is the composition of the first element then the composition of the second element is equal to the 1000/0 − ( 0/0 of the first element) .