Answer :
Let suppose the compound weights 100 g,
Nitrogen = 25.9 % = 25.9 g
Oxygen = 74.1 % = 74.1 g
Compound = Nitrogen + Oxygen
Compound = 25.9 g + 74.1 g = 100g
Calculate moles of Nitrogen and Oxygen,
As,
Mole = mass / M.mass
For Nitrogen,
Mole = 25.9 g / 14.0 g/mol
Mole = 1.85 moles of Nitrogen
For Oxygen,
Mole = 74.1 g / 15.999 g/mol
Mole = 4.631 moles of Oxygen
Now finding Ratio of Nitrogen to Oxygen,
Nitrogen : Oxygen
1.85 : 4.631
Now,
Dividing this ratio with 1.85 as it is the lowest number,
So,
(1.85 ÷ 1.85) : (4.631 ÷ 1.85)
1 : 2.50
Multiply this ratio with 2 so that a whole numbers are obtained
(1 : 2.50) × 2
Result:
2 : 5
There are 2 Nitrogen atoms and 5 Oxygen atoms in this compound.
N₂O₅ (Empirical Formula)
Nitrogen = 25.9 % = 25.9 g
Oxygen = 74.1 % = 74.1 g
Compound = Nitrogen + Oxygen
Compound = 25.9 g + 74.1 g = 100g
Calculate moles of Nitrogen and Oxygen,
As,
Mole = mass / M.mass
For Nitrogen,
Mole = 25.9 g / 14.0 g/mol
Mole = 1.85 moles of Nitrogen
For Oxygen,
Mole = 74.1 g / 15.999 g/mol
Mole = 4.631 moles of Oxygen
Now finding Ratio of Nitrogen to Oxygen,
Nitrogen : Oxygen
1.85 : 4.631
Now,
Dividing this ratio with 1.85 as it is the lowest number,
So,
(1.85 ÷ 1.85) : (4.631 ÷ 1.85)
1 : 2.50
Multiply this ratio with 2 so that a whole numbers are obtained
(1 : 2.50) × 2
Result:
2 : 5
There are 2 Nitrogen atoms and 5 Oxygen atoms in this compound.
N₂O₅ (Empirical Formula)