Question

A particular compound in the chemistry laboratory is found to contain 7.2x10^24 atoms of oxygen, 56.0g of nitrogen, and 4.0 mol of hydrogen. What is it’s empirical formula?

Answer 1

The empirical formula for the compound is :

`N_(1)H_(1)O_(3)`

or , NHO3

Step-by-step explanation:

Empirical formula : It is the simplest ratio of atoms present in a compound.

Calculate number of moles of H , N and O using formula :

`moles =(given\ mass)/(Molar\ mass)`

`moles = (Number\ of\ atoms)/(Avogadro\ number)`

Avogadro number is represented by N0 and contain =

`N_(0) = 6.022* 10^(23)` atoms

1.Calculate number of moles of O

Number of Atoms =

`7.7* 10^(24)`

`N_(0) = 6.022* 10^(23)`

Insert value in :

`moles = (Number\ of\ atoms)/(Avogadro\ number)`

`moles = (7.7* 10^(24))/(6.022* 10^(23))`

O = 11.95 mole = 12.0 mole

(11.95 is almost equal to 12 )

2.Calculate number of moles of N

Given mass = 56 .0 g

Molar mass of N = 14 g/mol

`moles =(given\ mass)/(Molar\ mass)`

`moles =(56)/(14)`

N = 4.0 mole

3.Calculate number of moles of H

Already given in moles

H = 4.0 moles

3.Calculate Simple ratio : Divide the moles of each H ,O ,N by the 4.0 mole ( It is the smallest number)

Ratios Are :

O = 3

`(12)/(4.0) = 3`

N = 1

`(4)/(4.0) = 1`

H = 1

`(4)/(4.0) = 1`

So the empirical formula should be :

`N_(1)H_(1)O_(3)`

or

NHO3