Key Concepts
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 CH_{2}.
The molecular formula for the same compound will equal to n × (CH_{2}), in other words, the molecular formula for this compound will be C_{n}H_{2n}
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 C_{n}H_{2n}:
M_{r}  =  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 C_{2}H_{4}
There are many compounds that can have the empirical formula CH_{2} and therefore a molecular formula of the form C_{n}H_{2n}.
These include:
 C_{2}H_{4} (ethene or ethylene) molecular mass=28.0 and n = 2, that is, C_{(1 × 2)}H_{(2 × 2)}
 C_{3}H_{6} (propene or propylene) molecular mass=42.0 and n = 3, that is, C_{(1 × 3)}H_{(2 × 3)}
 C_{3}H_{6} (cyclopropane) molecular mass=42.0 and n = 3, that is, C_{(1 × 3)}H_{(2 × 3)}
 C_{4}H_{8} (butene or butylene) molecular mass=56.0 and n = 4, that is, C_{(1 × 4)}H_{(2 × 4)}
 C_{4}H_{8} (cyclobutane) molecular mass=56.0 and n = 4, that is, C_{(1 × 4)}H_{(2 × 4)}
Calculating Empirical Formula from Percentage Composition
 Assume you have 100 g of sample.
 Convert all percentages to a mass in grams:
mass in grams = 
% composition given
100 
× 100 
= % given 
 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 
 Calculate the moles of each element present:
moles of element = mass in grams ÷ molar mass 
 Divide the moles of each element by the smallest of these to get a mole ratio
 If the numbers in the mole ratio are all whole numbers (integers) convert this to an empirical formula
 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 
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 


(c) molar mass / g mol^{1} 
63.55

35.45

(d) moles = mass ÷ molar mass 


(e) divide moles by 0.7435 


(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 CuCl_{2}
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 


(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 
H_{1}O_{1} is HO 
Solution Part 2: Determine the molecular formula using the empirical formula and molar mass of the compound
 empirical formula is HO
 molecular formula = whole number × empirical formula
molecular formula = n × empirical formula
molecular formula is H_{n}O_{n}
 Calculate the molar mass of the empirical formula:
= 1.008 + 16.00 = 17.008 g mol^{1}
 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
 Write the molecular formula for the compound
molecular formula is H_{n}O_{n}
n = 2
substitute this value for n into the molecular formula H_{n}O_{n}:
molecular formula of the compound is therefore H_{2}O_{2}

What would you like to do now? 

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.
