15.89 percent carbon, 21.18 percent oxygen, and 62.93 percent osmium. what is the empirical formula

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