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when 0.422g of phosphorus is burned, 0.967g of a white oxide is obtained.

ADetermine the empirical formula of the oxide
b. write a balanced equation for the reaction of phosphorus and molecular oxygen on the basis of this empirical formula

Verified Answer :

moles = mass / molar mass 

moles P = 0.422 g / 30.97 g/mol = 0.01363 mol 
moles O = (0.967 g - 0.422g) / 16.00 g/mol = 0.03406 moles 

So ratio moles P : moles O 
= 0.01363 mol : 0.03406 mol 

Divide each number in the ratio by the smallest number 

(0.01363 / 0.01363) : (0.03406 / 0.01363) 
= 1 : 2.5 

The empirical formula needs to be the smallest whole number ratio of atoms in the molecules. Since you have a non-whole number, multiply the ratio by the smallest number needed to make both number whole numbers. In this case x 2 

2 x (1 : 2.5) 
= 2 : 5 

Answer :

(a) The empirical formula of a compound is, \(P_2O_5\)

(b) The balanced chemical equation will be:

\(P_4+5O_2\rightarrow 2P_2O_5\)

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 = \(\frac{\text{ given mass of P}}{\text{ molar mass of P}}= \frac{0.422g}{31g/mole}=0.0136moles\)

Moles of O = \(\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{0.545g}{16g/mole}=0.0341moles\)

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

For P = \(\frac{0.0136}{0.0136}=1\)

For O = \(\frac{0.0341}{0.0136}=2.5\)

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 = \(P_2O_5\)

The balanced chemical equation will be:

\(P_4+5O_2\rightarrow 2P_2O_5\)

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