AIM:  What is an Empirical Formula?

Empirical Formula: a chemical formula "reduced" to its simplest terms
For example:
        The Empirical Formula of N₂O₄ is NO₂ (2 divides into both subscript numbers)
        The Emprical Formula of C₆H₆ is CH  (6 divides into both subscript numbers)

• Since we can calculate % composition from chemical formulas, we can also determine chemical formulas from % compositions.
• However because a chemical formula is a molar ratio of the elements in it, we can only calculate empirical formulas.

Example 1:
        Calculate the empirical formula of a substance that is 50% sulfur and 50% oxygen by mass.

STEP 1: we have to assume a mass of the substance, to keep the math simple assume 100g
STEP 2: set-up mole calculations for each element in the compound
                #mols= mass/molar mass

mols S =  50g / 32g/mol = 1.56mol S
mols O = 50g / 16g/mol = 3.125mol O

• These are the relative mole ratios for the compound
• They need to be converted from decimals into whole numbers

STEP 3: turn mole ratio into whole number ratio by dividing all the elements by the smallest                           number of moles calculated.
mols S = 1.56mols S / 1.56  = 1mol S
mols O = 3.125mols O / 1.56 = 2mols O

The Empirical Formula is therefore SO₂

Example 2:
Calculate the empirical formula of a substance that is 80% carbon and 20% hydrogen by mass.

• First assume 100g of substance
• set-up mole calculations:

mols C = 80g / 12g/mol = 6.667mols C
mols H = 20g / 1g/mol = 20mols H

• divide by the smallest number of mols

mols C = 6.667_mols C / 6.667 = 1_mol C
mols H = 20mols H / 6.667 = 3mols H

The empirical formula is therefore CH₃

• We can also calculate empirical formula if we are given masses of the elements

Example 3:
0.3546g of Vanadium reacts with Oxygen and forms a product with a mass of 0.6330g.
What is the empirical formula of the product?

STEP 1: Find masses of the starting elements
        Mass V = 0.3546g
        Mass O = (mass product - mass V) = 0.6330g - 0.3546g = 0.2784g

STEP 2: set-up mole equations

mols V = 0.3546g / 50.94g/mol = 0.006961mols V

mols O = 0.2784g / 16g/mol = 0.0174mols O

STEP 3: calculate whole number ratios

mols V = 0.006961mols V / 0.006961 = 1mol V

mols O = 0.0174mols O / 0.006961 = 2.5mols O

• The ratio is 1mol V to 2.5 mols O but we cannot have fractions so we double both numbers and our new ratio becomes 2mols V to 5 moles O

Therefore the empirical formula is V₂O₅


HWK: Ch 4 ques# 6, 17, 19 & 26