Respuesta :
Answer:
OsCO or COOs
Explanation:
Data given
Carbon = 15.89 %
Oxygen = 21.18 %
Osmium = 62.93%
Empirical formula = ?
Solution:
First find the masses of each component
Consider total compound is 100g
As we now
mass of element = % of component
So,
15.89 g of C = 15.89 % Carbon
21.18 g of O = 21.18 % Oxygen
62.93 g of Os = 62.93% Osmium
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 = 15.89 g/ 12 g/mol
no. of mole = 1.3242
mole of C = 1.3242
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 = 21.18 g/ 16 g/mol
no. of mole =
mole of O = 1.3238
For Os
Molar mass of Os = 190 g/mol
no. of mole = mass in g / molar mass
Put value in above formula
no. of mole = 62.93 g/ 190 g/mol
no. of mole =
mole of Os = 1.3312
Now we have values in moles as below
C = 1.3242
O = 1.3238
Os = 1.3312
Divide the all values on the smalest values to get whole number ratio
C = 1.3242 /1.3238 = 1.0003
O = 1.3238 /1.3238 = 1
Os = 1.3312 /1.3238 = 1.0056
So all have round value 1 mole
C = 1
O = 1
Os = 1
So the empirical formula will be (OsCO) i.e. all 3 atoms in simplest small ratio
