Solving Empirical Formula Problems There are two common types of empirical formula problems. Luckily, the steps to solve either are almost exactly the same. Example #1: Given mass % of elements in a compound.
A compound was found to contain 32.65% Sulfur, 65.3% Oxygen and 2.04% Hydrogen. What is the empirical formula of the compound?
1) The first step in this problem is to change the % to grams.
32.65%→32.65g
of S
65.3%→65.3g of O
2.04%→2.04g of H
2) Next divide all the given masses by their molar mass.
32.65g of S/ 32gm-1 = 1.0203 moles of S
65.3g of O/ 16gm-1 = 4.08 moles of O
2.04g of H/ 1.008gm-1 = 2.024 moles of H
3) Then, pick the smallest answer in moles from the previous step and divide all the answers by that. Remember that if you calculate a number that is x0.9 round to the nearest whole number
1.0203 moles of S/ 1.0203 = 1
4.08 moles of O/1.0203 =
3.998 ≈ 4
2.024 moles of H/1.0203 = 1.984 ≈ 2
4) Lastly, the coefficients calculated in the previous step will become the subscripts in the chemical formula.
S = 1
O = 4
H = 2
H2SO4
Example#2: Given the mass of a reactant before a chemical reaction and the mass of a product after a reaction.
When 0.273g of Mg is heated in a Nitrogen (N2) environment a chemical reaction occurs. The product of the reaction is 0.378g . Calculate the empirical formula.
1) In any empirical formula problem you must first find the mass % of the elements in the compound. Since the total mass of the final product was 0.378 we find that:
0.378g total-0.273g magnesium = 0.105g nitrogen
0.105g nitrogen/0.378g total (100) = 27.77%
0.273g magnesium/0.378g total (100) =72.23%
2) Then change the % to grams
27.77%→27.77g of N
72.23%→72.23g of Mg
3) Next, divide all the masses by their respective molar masses.
27.77g/14gm-1
= 1.98 moles N
72.23g/24.31gm-1 = 2.97 moles Mg
4) Pick the smallest answer of moles and divide all figures by that.
1.98 moles N/ 1.98 = 1
2.97 moles Mg/ 1.98 = 1.5
Since our answer for magnesium is not close enough to round to the nearest whole number we must choose a factor to multiply all the figures by that will yield us whole numbers
1*2= 2 N
1.5*2= 3 Mg
5) Lastly, the coefficients calculated in the previous step will become the subscripts in the
chemical formula.
1.98 moles N/ 1.98 = 1
1.98 moles N/ 1.98 = 1
N = 2
1.98 moles N/ 1.98 = 1
Mg = 3
Mg3N2
EMPIRICAL AND MOLECULAR FORMULAE
EMRIRICAL FORMULA – This represents the simplest whole number ratio of
atoms within a molecule
MOLECULAR FORMULA – This represents the exact numbers of
each atom in a molecule
Empirical formulae can be calculated from either % mass compostion data or
mass composition data. Molecular formula can then be calculated if the molecular
mass of the compound is known.
EXAMPLE 1 – from % composition
A substance containing carbon and hydrogen and oxygen only contains 73.47%
carbon and 10.20% hydrogen by mass, its relative molecular mass is 98. Calculate its
molecular formula.
Step 1 calculate empirical formula
ATOMSCARBONHYDROGENOXYGEN
PERCENTAGE MASS73.4710.20(100 – (73.47 + 10.20) =
16.33
ATOMIC MASS12116
MASS/ATOMIC MASS6.122510.201.020
RATIO(divide by
smallest)
6101
Empirical formula = C6H10O
Step 2 use MR to find molecular formula
MR = mass (C6H10O) x n
98 = (98) x nn = 1molecular formula =
C6H10O
EXAMPLE 2 – from mass data
A substance contains 12.28g nitrogen, 3.51g hydrogen, 28.07g sulphur and
56.10g oxygen, its MR is 228 .Calculate its molecular formula.