Determination of Empirical Formula From the Mass of each Element
We can determine the empirical (simpliest) formula by using mass of each
element in the compound data. A compound is composed of 7.20g of
carbon, 1.20g of hydrogen, and 9.60g of oxygen. Find the empirical
formula for this compound.
1. Convert the grams of each element to moles knowing that 1 mole of
any element = formula mass in grams
7.20 / 12.01 = 0.600 mol C
1.20 / 1.01 = 1.19 mol H
9.60 / 16.00 = 0.600 mol O
2. Divide each of the three mole figures by the lowest of the three in
order to simplify the mole ratio since the definition of simple
formula is the SIMPLEST whole number mole ratio.
0.600 mol C / 0.600 = 1.00 mol C
1.19 mol H / 0.600 = 1.98 mol H
0.600 mol O / 0.600 = 1.00 mol O
3. If these mole figures are not whole numbers or within .2 of being to
the next whole number then find the lowest common factor (ie 2,3,4)
that when you multiply by that factor each of the mole figures
determined in step 2 that will result in whole numbers since the
definition mentioned in step 2 talks about WHOLE numbers.
4. Use these mole figures arrived at in either step 2 or 3 (whichever are
whole number ratio) as the subscripts and use them after the
respective symbol.
C1H2O1 would be the Empirical Formula
Determination of Empirical Formula From Mass % Composition
One variation of the previous example is if the composition is given in mass
percent. If this is the case then one assumes a 100 gram sample and then
convert all percentage figures to gram figures. Then you would proceed
with step 1. Assuming 100 gram sample allows us to use the same numbers
as gram figures instead of percentage figures. The numerical percentages
will remain fixed according to the Law of Constant Composition. Let's
take the problem above where the combustion of the sample gave the mass
% The compound had the following mass % composition:
38.71% Carbon
9.71% Hydrogen
51.58% Oxygen
Determine the Empirical Formula for the compound.
1. Assume 100 grams of sample: so in a 100 gram sample there will be
38.71 grams of Carbon, 9.71 grams Hydrogen, and 51.58 grams
Oxygen
2. Convert grams of each element in that 100 gram sample to mol of
the element dividing by the atomic mass of the element:
38.71 grams C X 1 mol C / 12.01 grams C = 3.223 mol C
9.71 grams H X 1 mol H / 1.01 grams H = 9.61 mol H
51.58 grams O X 1 mol O / 16.00 grams O = 3.223 mol O
3. Get a simple ratio of mol by dividing each mol figure by the lowest
of the three:
3.223 mol C / 3.223 = 1.00 mol C
9.61 mol H / 3.223 = 2.98 mol H
3.22 mol O / 3.223 = 1.00 mol O
4. If these are whole numbers they represent the subscripts in the
Empirical Formula:
CH3O
Determining The Molecular Formula
Suppose we had a compound whose empirical formula was
CH2O
and whose molecular formula was
C6H12O6
If we compared the mass based on the empirical formula (Simple mass)
with the mass based on the molecular formula (molecular mass) we would
find that the molecular mass was six times the mass of the empirical
formula. Therefore it is easy to understand why the molecular formula has
six times as many atoms of each element in its formula. If we know the
molecular mass and the empirical formula we can then determine that
factor difference by dividing the molecular mass by the simple mass, and
then multiply each subscript in the simple formula by that factor to get the
subscripts in the molecular formula. Let's take an example to illustrate
how this is done.
Combustion Analysis gives the following mass %:
26.7% Carbon
2.2% Hydrogen
71.1% Oxygen
If a separate analysis determined the molecular mass of the compound to
be 90. Determine the Empirical Formula and the Molecular Formula of
the compound.
1. Assume a 100 gram sample which will convert the given percentages
to gram amounts
2. Convert the grams of each element to mols by dividing the mass by
the atomic mass of the element:
26.7 grams C X 1 mol C / 12.01 grams C = 2.223 mol C
2.2 grams H X 1 mol H / 1.01 grams H = 2.2 mol H
71.1 grams O X 1 mol O / 16.00 grams O = 4.44 mol O
3. Determine the simpliest ratio of mols by dividing each of the mol
figures by the lowest of the three.
2.223 mol C / 2.2 = 1.0 mol C
2.2 mol H / 2.2 = 1.0 mol H
4.44 mol O / 2.2 = 2.0 mol O
4. Identify the empirical formula:
CHO2
5. Determine the mass based on the Empirical Formula.
For CH2 the mass would be:
(1)(12.0) + 1(1.0) + 2(16.0) = 45.0 = Empirical mass
6. Determine the common factor that defines how much more massive
the molecular formula is compared to the Empirical Formula.
molecular mass / Empirical mass = 90 / 45 = 2 = factor
7. Take the determined whole number factor determined in the above
step and multiply each subscript in the empirical formula and that
will be the new subscripts in the molecular formula.
(CHO2)2 = C2H2O4