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

Answer:

The empirical formula for the compound is :

$N_{1}H_{1}O_{3}$

or , NHO3

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 :

$moles =\frac{given\ mass}{Molar\ mass}$

$moles = \frac{Number\ of\ atoms}{Avogadro\ number}$

Avogadro number is represented by N0 and contain =

$N_{0} = 6.022\times 10^{23}$ atoms

1.Calculate number of moles of O

Number of Atoms =

$7.7\times 10^{24}$

$N_{0} = 6.022\times 10^{23}$

Insert value in :

$moles = \frac{Number\ of\ atoms}{Avogadro\ number}$

$moles = \frac{7.7\times 10^{24}}{6.022\times 10^{23}}$

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

$moles =\frac{given\ mass}{Molar\ mass}$

$moles =\frac{56}{14}$

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

$\frac{12}{4.0} = 3$

N = 1

$\frac{4}{4.0} = 1$

H = 1

$\frac{4}{4.0} = 1$

So the empirical formula should be :

$N_{1}H_{1}O_{3}$

or

NHO3