Answer :
(a) The empirical formula of a compound is, 
(b) The balanced chemical equation will be:

Solution : Given,
Mass of phosphorus = 0.422 g
Mass of white oxide = 0.967 g
Molar mass of phosphorus = 31 g/mole
Molar mass of oxygen = 16 g/mole
First we have to calculate the mass of oxygen.
Mass of oxygen = Mass of white oxide - Mass of phosphorus
Mass of oxygen = 0.967 - 0.422
Mass of oxygen = 0.545 g
Now convert given masses into moles.
Moles of P = 
Moles of O = 
For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For P = 
For O = 
The ratio of P : O = 1 : 2.5
To make this ratio in a whole number we multiple ratio by 2, we get:
The ratio of P : O = 2 : 5
The mole ratio of the element is represented by subscripts in empirical formula.
The Empirical formula = 
The balanced chemical equation will be:
