# Calculating Empirical Formula

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```					                                              Calculating Empirical Formula
Steps for determining Empirical Formula of a compound.
Step 1
From the mass of each element convert this to a number of moles of each element.
Step2
Divide the number of moles of each element by the smallest number of moles to convert the smallest number to 1. If all these
numbers are whole numbers these are the subscripts in the empirical formula. If not go to step 3
Step 3
Multiply the numbers you derived in the above step by the smallest inter that will convert all of them to whole numbers.

Worked Example 1
A sample of an oxide of copper is found to contain 4.0 g of Cu and 0.50 g of oxygen. Calculate its empirical formula.

Step 1 – Find the number of moles of each element
Cu = 4.0 g x 1 mole
63.546 g
= 0.0625 mole
O = 0.50 g x 1 mole
15.999 g
= 0.03125 mole
Step 2 – Divide by lowest number to find ratio of moles
Cu: O = 2:1              Empirical Formula = Cu2O

Worked Example 2:
Find the empirical formula of an iron oxide that contains 78% iron.
The iron oxide must also contain 22% oxygen
Mole of Fe =                    78 g x 1 mol
55.8g               = 1.4 mol Fe
Mole of O = 22 g x 1 mol
16.0 g                              = 1.4 mol O

Simple ratio 1: 1         Empirical Formula = FeO

Worked Example 3:
An oxide of aluminium is formed by the reaction of 4.151 g of aluminium with 3.692 g of oxygen. Calculate the empirical formula
for this compound.
Step 1
Mass Al= 4.151 g
Mole Al = 4.151/26.89 g = 0.1539 mol
Mass O = 3.692 g
Mole O = 3.692/16.0 = 0.2308 mol
Step 2
Divide by 0.1539 (smallest number)
Ratio mole Al:O= 1: 1.5
Step 3
Multiply to give whole numbers (multiply by 2)
Ratio mole Al:O = 2:3

Empirical Formula     Al2O3

Example 4
A sample of lead arsenate contains 1.3813 g of lead, 0.00672 g of hydrogen, 0.4995 g of arsenic and 0.4267 g of oxygen. Calculate
the empirical formula.