A compound of phosphorus oxide has [tex]$43.6 \%$[/tex] of oxygen. What is its empirical formula?

A. [tex]$P _2 O _5$[/tex]
B. [tex]$P _2 O _3$[/tex]
C. [tex]$P _3 O _2$[/tex]
D. [tex]$PO _3$[/tex]



Answer :

To determine the empirical formula of a compound given that it contains 43.6% oxygen by mass, follow these steps:

1. Calculate the percentage of phosphorus in the compound:
Since the compound contains 43.6% oxygen, the remaining percentage must be phosphorus:
[tex]\[ 100\% - 43.6\% = 56.4\% \][/tex]

2. Set up the ratio of oxygen to phosphorus by mass:
The molar mass of oxygen (O) is 16 g/mol, and the molar mass of phosphorus (P) is 30.97 g/mol.

3. Calculate the mass-to-mole ratio for each element:
- For phosphorus:
[tex]\[ \text{Moles of phosphorus} = \frac{56.4}{30.97} \approx 1.821 \][/tex]
- For oxygen:
[tex]\[ \text{Moles of oxygen} = \frac{43.6}{16} \approx 2.725 \][/tex]

4. Determine the simplest mole ratio between the elements:
We calculate the ratio of moles of oxygen to moles of phosphorus:
[tex]\[ \text{Oxygen to Phosphorus ratio} = \frac{2.725}{1.821} \approx 1.496 \][/tex]

5. Convert the mole ratio to a small whole number ratio:
The ratio 1.496 is approximately 1.5 suggesting that for every phosphorus (P) atom, there are roughly one and a half oxygen (O) atoms. To express this as a simple whole number ratio, we multiply both sides by 2:
[tex]\[ 1.5 \text{ (O)} \approx 3 \text{ (P)} \rightarrow 3 \times 2 \approx 2 \text{ (O)} \times 5 \rightarrow \text {P}_2\text{O}_5 \][/tex]

Hence, the empirical formula of the compound is:
[tex]\[ \boxed{P_2O_5} \][/tex]