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Empirical and Molecular Formula

Key Concepts

  • Empirical Formula of a compound shows the ratio of elements present in a compound.

  • Molecular Formula of a compound shows how many atoms of each element are present in a molecule of the compound.

  • The empirical formula mass of a compound refers to the sum of the atomic masses of the elements present in the empirical formula.

  • The Molecular Mass (formula mass, formula weight or molecular weight) of a compound is a multiple of the empirical formula mass.
        Mr = n x empirical formula mass
        where Mr = molecular mass
        and n = a whole number (1 ,2 3, etc)

  • Empirical Formula can be calculated from the percentage (or percent) composition of a compound.

Examples of Empirical and Molecular Formula

If carbon and hydrogen are present in a compound in a ratio of 1:2, the empirical formula for the compound is CH2.

The empirical formula mass of this compound is: 12.0 + (2 x 1.0) = 14.0 g mol-1

If we know the molecular mass of the compound is 28.0 g mol-1 then we can find the molecular formula for the compound.
Mr = n x empirical formula mass
28.0 = n x 14.0
n = 2
So the molecular formula for the compound is 2 x empirical formula, ie, 2 x (CH2) which is C2H4

There are many compounds that can have the empirical formula CH2.
These include:

  • C2H4 (ethene or ethylene) molecular mass=28.0g mol-1 and n=2

  • C3H6 (propene or propylene) molecular mass=42.0g mol-1 and n=3

  • C3H6 (cyclopropane) molecular mass=42.0g mol-1 and n=3

  • C4H8 (butene or butylene) molecular mass=56.0g mol-1 and n=4

  • C4H8 (cyclobutane) molecular mass=56.0g mol-1 and n=4

Calculating Empirical Formula from Percentage Composition

  1. Assume 100g of sample

  2. Convert all percentages to a mass in grams, eg, 21% = 21g, 9% = 9g

  3. Find the relative atomic mass of each element present using the Periodic Table

  4. Calculate the moles of each element present: n = mass ÷ relative atomic mass

  5. Divide the moles of each element by the smallest of these to get a mole ratio

  6. If the numbers in the mole ratio are all whole numbers (integers) convert this to an empirical formula

  7. If the numbers in the mole ratio are NOT whole numbers, you will need to further manipulate these until the mole ratio is a ratio of whole numbers (integers)

    Recognising the decimal equivalent of common fractions is very helpful!

    Common Decimal Equivalent Fraction Mole Ratio Example
    0.125 1/8 1 : 1.125 converts to 1 : 9/8
    multiply throughout by 8 to give 8 : 9
    0.25 1/4 1 : 0.25 converts to 1 : 1/4
    multiply throughout by 4 to give 4 : 1
    0.33 1/3 1 : 1.33 converts to 1 : 4/3
    multiple throughout by 3 to give 3 : 4
    0.375 3/8 1 : 1.375 converts to 1 : 11/8
    multiply throughout by 8 to give 8 : 11
    0.5 1/2 2 : 1.5 converts to 2 : 3/2
    multiply throughout by 2 to give 4 : 3
    0.625 5/8 1 : 1.625 converts to 1 : 13/8
    multiply throughout by 8 to give 8 : 13
    0.66 2/3 2 : 1.66 converts to 2 : 5/3
    multiply throughout by 3 to give 6 : 5
    0.875 7/8 1 : 0.875 converts to 1 : 7/8
    multiply throughout by 8 to give 8 : 7

Example 1

A compound is found to contain 47.25% copper and 52.75% chlorine.
Find the empirical formula for this compound.

element Cu Cl
mass in grams 47.25 52.75
relative atomic mass 63.6 35.5
moles = mass ÷ relative atomic mass 47.25 ÷ 63.6 = 0.74 52.75 ÷ 35.5 = 1.49
divide throughout by lowest number 0.74 ÷ 0.74 = 1 1.49 ÷ 0.74 = 2.01 = 2

Empirical formula for this compound is CuCl2

Example 2

A compound with a molecular mass of 34.0g mol-1 is known to contain 5.88% hydrogen and 94.12% oxygen.
Find the molecular formula for this compound.

First, find the empirical formula of the compound.

element H O
mass in grams 5.88 94.12
relative atomic mass 1.0 16.0
moles = mass ÷ relative atomic mass 5.88 ÷ 1.0 = 5.88 94.12 ÷ 16.0 = 5.88
divide throughout by the smallest number 5.88 ÷ 5.88 = 1 5.88 ÷ 5.88 = 1

Empirical formula is HO

Calculate the empirical formula mass: 1.0 + 16.0 = 17.0 g mol-1

Molecular Mass = n x empirical formula mass
34.0 = n x 17.0
n = 34.0 ÷ 17.0 = 2

Molecular Formula is 2 x (HO) which is H2O2


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