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Empirical and Molecular Formula

Key Concepts

  • Empirical Formula of a compound shows the lowest whole number 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.1

  • 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 × empirical formula mass
        where Mr = molecular mass
        and n = a whole number (1 ,2 3, etc)

    Alternatively, you can relate the molar mass of the molecular formula and the empirical formula in the same way:
        molar mass of molecular formula = n × molar mass of empirical formula

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

Comparing 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 molecular formula for the same compound will equal to n × (CH2), in other words, the molecular formula for this compound will be CnH2n

The empirical formula mass of this compound is: 12.01 + (2 x 1.008) = 14.026

If we know the molecular mass of the compound is 28.00 then we can find the value of "n" in the molecular formula CnH2n:

Mr = n × empirical formula mass
28.00 = n × 14.026
28.00
14.026
= n × 14.026
14.026
2 = n

So the molecular formula for the compound is:
C(1 × 2)H(2 × 2) which is C2H4

There are many compounds that can have the empirical formula CH2 and therefore a molecular formula of the form CnH2n.
These include:

  • C2H4 (ethene or ethylene) molecular mass=28.0 and n = 2, that is, C(1 × 2)H(2 × 2)

  • C3H6 (propene or propylene) molecular mass=42.0 and n = 3, that is, C(1 × 3)H(2 × 3)

  • C3H6 (cyclopropane) molecular mass=42.0 and n = 3, that is, C(1 × 3)H(2 × 3)

  • C4H8 (butene or butylene) molecular mass=56.0 and n = 4, that is, C(1 × 4)H(2 × 4)

  • C4H8 (cyclobutane) molecular mass=56.0 and n = 4, that is, C(1 × 4)H(2 × 4)

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Calculating Empirical Formula from Percentage Composition

  1. Assume you have 100 g of sample.

  2. Convert all percentages to a mass in grams:

    mass in grams = % composition given
    100
    × 100 = % given

  3. Find the molar mass (relative atomic mass in grams) of each element present using the Periodic Table

      Common Elements
    carbon hydrogen oxygen
    molar mass
    g mol-1
    12.01 1.008 16.00

  4. Calculate the moles of each element present:

    moles of element = mass in grams ÷ molar 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
0.667 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

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Worked Example of Determining Empirical Formula from Percentage Composition

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

Solution:

  Elements
Cu Cl
% by mass
(given in question)
47.25 52.75
(a) assume you have 100 g of compound    
(b) mass of element in 100 g of compound
47.25
100
× 100
= 47.25 g
52.75
100
× 100
= 52.75 g
(c) molar mass / g mol-1 63.55 35.45
(d) moles = mass ÷ molar mass
47.25
63.55
= 0.7435
52.75
35.45
= 1.488
(e) divide moles by 0.7435
0.7435
0.7435
= 1
1.488
0.7435
= 2
(f) If the numbers in the mole ratio are all whole numbers (integers) convert this to an empirical formula: Cu :
1:
Cl
2

Empirical formula for this compound is CuCl2

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Worked Example: Molecular Formula from Percentage Composition and Molar Mass

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

The solution to this problem is in two parts:

  • Part 1: determine the empirical formula for the compound based on the percentage composition

  • Part 2: determine the molecular formula for the compouund based on the empirical formula and molar mass

Solution Part 1: Determine the empirical formula using the percentage composition of the compound

element H O
% by mass
(from question)
5.88 94.12
(a) assume 100 g of compound    
(b) mass in grams
5.88
100
× 100 = 5.88
94.12
100
× 100 = 94.12
(c) molar mass / mol g-1
(relative atomic mass in g)
1.008 16.00
(d) moles = mass ÷ molar mass 5.88 ÷ 1.008 = 5.83 94.12 ÷ 16.00 = 5.88
(e) divide throughout by
the smallest number
5.88 ÷ 5.83 = 1 5.88 ÷ 5.83 = 1
(f) Convert mole ratio
to an empirical formula
H1O1 is HO

Solution Part 2: Determine the molecular formula using the empirical formula and molar mass of the compound

  1. empirical formula is HO

  2. molecular formula = whole number × empirical formula
    molecular formula = n × empirical formula
    molecular formula is HnOn

  3. Calculate the molar mass of the empirical formula:
    = 1.008 + 16.00 = 17.008 g mol-1

  4. Calculate "n"
    molar mass of compound = n x molar mass of empirical formula
    34.0 = n x 17.008
    n = 34.0 ÷ 17.008 = 2

  5. Write the molecular formula for the compound
    molecular formula is HnOn
    n = 2
    substitute this value for n into the molecular formula HnOn:
    molecular formula of the compound is therefore H2O2
See also Experimental determination of empirical formula of magnesium oxide.


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1. Only a molecule, that is a substance made up of 2 or more atoms covalently bonded in a discrete unit, can have a molecular formula.
An ionic substance consists of positive and negative ions arranged in a lattice so that there are no discrete molecules.
We should not speak of the molecular formula for an ionic substance, and we should only use the empirical formula to describe the lattice.

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