Question

A form of phosphorus called red phosphorus is used in match heads. When 0.062 g of red phosphorus burns in air, it forms 0.142 g of phosphorus oxide. Determine the empirical formula of phosphorus oxide.(show work; use labels)

Answer 1

Answer: The empirical formula is $P_2O_5$

Explanation:

Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.

a) If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of P = 0.062 g

Mass of O= (0.142- 0.062) = 0.08 g

Step 1 : convert given masses into moles

Moles of P=$\frac{\text{ given mass of P}}{\text{ molar mass of P}}= \frac{0.062g}{31g/mole}=0.002moles$

Moles of O=$\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{0.08g}{16g/mole}=0.005moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For P = $\frac{0.002}{0.002}=1$

For O =$\frac{0.005}{0.002}=2.5$

The ratio of P: O = 1: 2.5

Converting them into whole number ratios by multiplying with 2:

Hence the empirical formula is $P_2O_5$