Question

A given compound has 52.00% of Chromium and 48.00% of Oxygen . Determine its empirical formula.

Answer 1

Given the percentage of chromium in compound = 52.00%

 

Given the percentage of oxygen in compound = 48.00%

 

• Step 1:

Find the molar mass of chromium and oxygen

 

For chromium = 51.996.

 

For oxygen = 15.999.

 

• Step 2

Conver the Molar mass into moles

 

For chromium - 51.996 g has 1.000075 moles.

 

For Oxygen - 15.99 g has 3.001125 moles.

 

• Step 3

Divide the moles of both Cr and O by the smallest mole value which is 1.000075.

 

Therefore Cr = 1.000075/1.000075 = 1

 

For O = 3.001125/1.000075 = 2.999887.

 

• Step 4

 

Divide by the fractional components of each mole value we get

 

For Chromium we will get - 1.0001125

 

For Oxygen we will get - 3.0002250

 

Round off to the closest whole number we will get 1 and 3

 

Therefore empirical formula of the compound is CrO3.

The empirical formula of given compound is CrO3