Answer:
HNO₃
Explanation:
Data given
Nitrogen = 9.8 g
Hydrogen = 0.70 g
Oxygen = 33.6 g
Empirical formula = ?
Solution:
Convert the masses to moles
For Nitrogen
Molar mass of N = 14 g/mol
no. of mole = mass in g / molar mass
Put value in above formula
no. of mole = 9.8 g/ 14 g/mol
no. of mole = 0.7
mole of N = 0.7 mol
For Hydrogen
Molar mass of H = 1 g/mol
no. of mole = mass in g / molar mass
Put value in above formula
no. of mole = 0.70 g/ 1 g/mol
no. of mole = 0.7
mole of H = 0.7 mol
For Oxygen
Molar mass of O = 16 g/mol
no. of mole = mass in g / molar mass
Put value in above formula
no. of mole = 33.6 g / 16 g/mol
no. of mole = 2.1
mole of O = 2.1 mol
Now we have values in moles as below
N = 0.7
H = 0.7
O = 2.1
Divide the all values on the smallest values to get whole number ratio
N = 0.7 / 0.7 = 1
H = 0.7 / 0.7 = 1
O = 2.1 / 0.7 = 3
So all have following values
N = 1
H = 1
O = 3
So the empirical formula will be HNO₃ i.e. all three atoms in simplest small ratio.
