Answer:
The empirical formula for the compound is :
[tex]N_{1}H_{1}O_{3}[/tex]
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 :
[tex]moles =\frac{given\ mass}{Molar\ mass}[/tex]
[tex]moles = \frac{Number\ of\ atoms}{Avogadro\ number}[/tex]
Avogadro number is represented by N0 and contain =
[tex]N_{0} = 6.022\times 10^{23}[/tex] atoms
1.Calculate number of moles of O
Number of Atoms =
[tex]7.7\times 10^{24}[/tex]
[tex]N_{0} = 6.022\times 10^{23}[/tex]
Insert value in :
[tex]moles = \frac{Number\ of\ atoms}{Avogadro\ number}[/tex]
[tex]moles = \frac{7.7\times 10^{24}}{6.022\times 10^{23}}[/tex]
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
[tex]moles =\frac{given\ mass}{Molar\ mass}[/tex]
[tex]moles =\frac{56}{14}[/tex]
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
[tex]\frac{12}{4.0} = 3[/tex]
N = 1
[tex]\frac{4}{4.0} = 1[/tex]
H = 1
[tex]\frac{4}{4.0} = 1[/tex]
So the empirical formula should be :
[tex]N_{1}H_{1}O_{3}[/tex]
or
NHO3
