Question

When a 30.98-g sample of phosphorus reacts with oxygen, a 71.00-g sample of phosphorus oxide is formed. What is the percent composition of the compound? What is the empirical formula for this compound?

Answer 1

Answer : The percent composition of P = 43.63 % and O = 53.37 % .

Empirical formula of Phosphorous oxide = P₂O₅

Part A : Percent composition :

It is percent of each element present in compound . It is given by formula :

`Percent composition = (mass of element)/( total mass of compound ) * 100`

Given : Mass of Phosphorous (P ) = 30.98 g

Mass of Compound Phosphorous oxide = 71.00 g

Mass of Oxygen (O) = mass of compound - mass of P

= 71.00 g - 30.98 g = 40.02 g

`Percent composition of P = (mass of P )/( mass of compound)`
`= (30.98 g )/(71.00 g) * 100`

Percent composition of P = 43.63 %

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Percent composition of O = 53.37 %

Part B : Empirical formula of Phosphorous Oxide.

Empirical formula is formula which shows the proportion of element present in a compound . Following are the steps to calculate empirical formula of an compound :

Step 1 : Find masses of each element .

Mass of P = 30.98 g

Mass of O = 40.02 g

Step 2 : Conversion of masses of element to its mole .

`Mole = (mass )/(molar mass )`

Given : Mass of P = 30.98 g
`Molar mass of P = 30.97 (g)/( mol)`

`Mole of <strong>P</strong> = (30.98 g )/(30.97(g)/(mol) ) = <strong>1 mol</strong>`

Given: Mass of O = 40.02 g
`Molar mass of O = 15.99(g)/(mol)`

`Mole of <strong>O</strong> = (40.02 g)/(15.99(g)/(mol)) = <strong>2.5 mol </strong>`

Step 3 : Finding Ratio of mole .

In this step ratio is found by dividing each mole by smallest mole .Since mole of P is smaller , so this will be used for division.

`Ratio of mole of P= (mole of P )/(mole of P ) = (1 mol)/(1 mol) = 1`

`Ratio of mole of O = ( mole of O )/(mole of P ) = (2.5 )/(1 ) = 2.5`

Hence , Ratio of P : O = 1 : 2.5

Since the ratio is in fraction , it need to be converted to whole number . So we multiply the ratio by such a minimum number which gives us a whole number ratio. This step is skipped if the ratio already comes in whole number.

On multiplication the ratio by 2 :

Ratio of P : O = ( 1 : 2.5 ) * 2 = 1 * 2 : 2.5 * 2

Ratio of P : O = 2 : 5

Step 4 : Writing the empirical formula

The ratio of P and O is 2: 5 , which gives the empirical formula of compound as P₂O₅ .

Answer 2

Answer:- P = 43.63% and O = 56.37% and the empirical formula is
`P_2O_5`

Solution:- mass of Phosphorous = 30.98 g

mass of product(phosphorous oxide) = 71.00 g

mass of oxygen = 71.00 - 30.98 = 40.02 g

percentage of phosphorous in the compound = (30.98/71.00)100 = 43.63%

percentage of oxygen in the compound = (40.02/71.00)100 = 56.37%

moles of P = 30.98 g x ( 1mol/30.97 g) = 1.00 mol

moles of O = 40.02 g x (1 mol/ 16.00 g) = 2.50 mol

Ratio of moles of P to O is 1.00:2.50

The whole number ratio is 2:5

So, the empirical formula of the phosphorous oxide formed is
`P_2O_5`