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Comparing Empirical Formula 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}
Using the periodic table of the elements we can determine the empirical formula mass of this compound(CH_{2}): 12.01 + (2 × 1.008) = 14.026
If we know the molecular mass of the compound is 28.00 (M_{r} = 28.00) then we can find the value of "n" in the molecular formula C_{n}H_{2n}:
Write the general expression: 
M_{r}  =  n × empirical formula mass 
Substitute in the known values: 
28.00  =  n × 14.026 
Divide both sides of equation by 14.026 
28.00 14.026  =  n × 14.026
14.026 
Solve for n 
2  =  n 
Subsitute this value, n = 2, back into the general molecular formula C_{n}H_{2n} to get the molecular formula of this compound.
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}.
Examples 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^{(2)}) 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)}
Theory Behind Calculating Empirical Formula from Percentage Composition
We can use the percentage composition (percent composition) of a compound to determine an empirical formula for the compound.
Remember that the percentage composition gives us the percent by mass of each element present in the compound, for example
If a compound is made up of 10% by mass of element X and 90% by mass of element Z
Then 100 g of the compound will contain:
 10 g of element X
mass of X = 
10 % 
× 
100 g 
= 
10
100 
× 100 
= 10 g 
 90 g of element Z
mass of Z = 
90 % 
× 
100 g 
= 
90
100 
× 100 
= 90 g 
10 g of element X and 90 g of element Z is present in 100 g of the compound.
On the other hand, the empirical formula of the compound gives us the lowest whole number ratio of atoms of each element in the compound, for example
Empirical formula XZ tells us that for every 1 atom of X there is 1 atom of Z in the compound.
Empirical formula XZ_{2} tells us that for every 1 atom of X there are 2 atoms of Z, etc
Clearly, we will need to convert the "mass" of each element in the compound into a "number of atoms".
6.02 × 10^{23} atoms of element X = 1 mole of atoms of element X
Mass of 1 mole of atoms of element X = molecular mass expressed in grams (molar mass)
So we can use the "mass" and "molar mass" (or relative molecular mass) to calculate the moles of atoms of each element present,
moles of element X = 
mass of element X in grams molar mass of element X in grams per mole 
moles of element Z = 
mass of element Z in grams molar mass of element Z in grams per mole 
Next we need to write the relationship between moles of X and moles of Z in the compound:
moles of element X 
: 
moles of element Z 
and then to find the lowest whole number ratio of these moles
 If moles of Z < moles of X, divide both by moles of Z
moles of element X moles of element Z 
: 
moles of element Z
moles of element Z 
that is 
moles of element X moles of element Z 
: 1 
 If moles of X < moles of Z, divide both by moles of X
moles of element X
moles of element X 
: 
moles of element Z moles of element X 
that is 
1 : 
moles of element Z moles of element X 
Be aware that the calculation above to get the mole ratio of one element to another may not result in whole numbers.
A further calculation may be needed to get the lowest whole number ratio of moles of one element to another.
This is where a recognition of the decimal equivalent of common fractions, as shown in the table below, can be 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 
Once you have calculated the lowest whole number ratio of moles of one element to the other, you are ready to write the empirical formula, for example
 If the mole ratio of X to Z is 1:1, empirical formula is XZ
 If mole ratio of X to Z is 1:2, empirical formula is XZ_{2}
 If mole ratio of X to Z is 2:3, empirical formula is X_{2}Z_{3}
Steps for Calculating Empirical Formula from Percentage Composition
 Assume 100 g of sample
 Convert all percentages to a mass in grams
 Find the relative atomic mass of each element present using the Periodic Table
 Calculate the moles of each element present: n = mass ÷ relative atomic 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)
Worked Example: 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:
 Assume mass of sample of compound is 100 g
 Convert all percentages to a mass in grams

Elements 

Cu 
Cl 
% by mass 
47.25 
52.75 
(b) mass in 100 g of compound 


 Find the relative atomic mass (molar mass) of each element present using the Periodic Table

Elements 

Cu 
Cl 
% by mass 
47.25 
52.75 
(b) mass in 100 g of compound 


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

35.45

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

Elements 

Cu 
Cl 
% by mass 
47.25 
52.75 
(b) mass in 100 g of compound 


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

35.45

(d) moles / mol 


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

Elements 

Cu 
Cl 
% by mass 
47.25 
52.75 
(b) mass in 100 g of compound 


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

35.45

(d) moles / mol 


(e) divide moles by 0.7435 


 If the numbers in the mole ratio are all whole numbers (integers) convert this to an empirical formula:
mole ratio Cu : Cl is  1:2 (1 and 2 are both whole numbers) 
empirical formula is  CuCl_{2} 
 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)
This step is not required since we have already achieved the lowest ratio of whole numbers
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
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 
Part 2: Determine the molecular formula using the empirical formula and molar mass of the compound
 empirical formula is HO (calculated in part 1 above)
 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 HO:
= 1.008 + 16.00 = 17.008 g mol^{1}
 Calculate "n"
molar mass of compound = n × molar mass of empirical formula
34.0 = n × 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}
(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.
This empirical formula of an ionic compound therefore tells us the lowest whole number ratio of cations to anions.
(2) And yes, there are even more possibilities if you think about the structural isomers of butene such as: but1ene (1butene) and but2ene (2butene)