Answer:
C₂HO₂Br₃
Step-by-step explanation:
Data given
Carbon = 8.09%
Hydrogen = 0.34%
Oxygen = 10.78%
bromine = 80.78%
Empirical formula = ?
Solution:
First find the masses of each component
Consider total compound is 100g
As we now
mass of element = % of component
So,
8.09 g of C = 8.09% of Carbon
0.34 g og H = 0.34% of Hydrogen
10.78 g of O = 10.78% of Oxygen
80.78 g of bromine = 80.78% of bromine
Now convert the masses to moles
For Carbon
Molar mass of C = 12 g/mol
no. of mole = mass in g / molar mass
Put value in above formula
no. of mole = 8.09 g/ 12 g/mol
no. of mole = 0.674
mole of C = 0.6742
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 = 0.34 g/ 1 g/mol
no. of mole = 0.34
mole of H = 0.34 mole
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 = 10.78 g/ 16 g/mol
no. of mole = 0.674
mole of O = 0.674 mole
For Br
Molar mass of Br = 80 g/mol
no. of mole = mass in g / molar mass
Put value in above formula
no. of mole = 80.78 g/ 80 g/mol
no. of mole = 1.0098
mole of Br = 1.0098 moles
Now we have values in moles as below
C = 0.6742
H = 0.34
O = 0.674
Br = 1.0098
Divide the all values on the smallest values to get whole number ratio
C = 0.6742 / 0.34 = 1.983 ≅ 2
H = 0.34 / 0.34 = 1
O = 0.674 /0.34 = 1.983 ≅ 2
Br = 1.0098 / 0.34 = 2.97 ≅ 3
So all have round value 1 mole
C = 2
H = 1
O = 2
Br = 3
So the empirical formula will be C₂HO₂Br₃ i.e. all 3 atoms in simplest small ratio