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General Chemistry 103 (Dr. Amah)
Determination of Chemical Formulas: Empirical and Molecular
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Chemical formulas can be determined by measuring the mass of each element present in
a sample of the compound. The mass of each element (grams) is converted to number of moles,
or molecules of each element presenting the compound. You will need to do such calculations in
order to determine the amounts of elements present in samples of compounds needed to
produce various materials.
Learning Objectives
 Identify the formula of a chemical compound from its composition.
 Determine the mass composition of the chemical compound.
Models: Chemical Analysis of Acetic Acid and Methane Gas
Acetic acid is the active ingredient in vinegar. A chemical analysis of 157.5 g of acetic
acid provided the following information.
Element
Mass of
Element(g)
carbon
63.00
oxygen
83.93
hydrogen 10.57
Mass Percent Moles of
Element
40.00
5.246
53.29
5.246
6.714
10.486
Simplest Ratio of
Moles of Element**
1.00
1.00
2.00
Mass percent = mass x 100/total mass
Moles = mass/molar mass
The 1:1:2 ratio in the last column means that the empirical formula of acetic acid is C 1O1H2
which is the same as: CHO2
Methane is the major gas ingredient used in gas stoves. A chemical analysis of 100.00 g
of this gas provided the following information.
Element
carbon
hydrogen
Mass of
Element (g)
75.00
25.00
Mass Percent
75.00
25.00
Moles of
Element
6.25
25.00
Simplest ratio of
Mole of Element**
1.00
4.00
Mass percent = mass x 100/total mass
Moles = mass/molar mass
The 1:1: ratio in the last column means that the empirical formula of methane is C 1H1
2
Key Questions
1. How was the mass percent in the above tables calculated from the mass for each
element?
2. How was the number of moles of each element calculated?
3. How is the empirical formula determined from the data in the tables?
4. What information does the empirical formula provide?
5. What feature, related to composition, do all compounds with the same mass percent
composition have?
6. If the molecular mass of acetic acid is 60.00 g/mol, and methane gas is 16.00 g/mol,
what is the?:
a) molecular formula of acetic acid; and
b) molecular formula of methane gas.
7. A sample of a liquid with mass of 8.657 g was decomposed into its element and gave
5.217 g of carbon, 0.9620 g of hydrogen, and 2.478 g of oxygen. Determine the percent
composition of this compound.
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8. One of the compounds of iron and oxygen, “black iron oxide”, occurs in the mineral
magnetite. When a 2.448 g sample was analyzed it was found to have 1.771 g of Fe.
Determine the empirical formula of this compound.
*Occasionally, mole ratios my be encountered that are fractions such as:
1.25
1.33
1.40
1.50
1.67
2.25
In such situations, one must multiply these mole ratios by a number in order to obtain whole mole
numbers ratio as shown in the following table:
Mole ratio Multiply by Whole Number
Mole Ratio
1.25
4
5
1.33
3
4.
1.4
5
7
1.5
2
3
1.67
3
5
2.25
4
10
2.5
2
5
Examples are shown in the tables A and B shown below:
Table A:
Element
Mass Percent
Mole of
Element
Mole Ratio
Of Element
Fe
O
58.266
41.734
1.04335
2.6085
1.000
2.50
Whole Number
Mole Ratio of
Element
2.00
5.00
Therefore, empirical formula is Fe2O5
Table B:
Element
Mass Percent
Mole of
Element
Mole Ratio
Of Element
Fe
O
69.942
30.058
1.2524
1.8787
1.000
1.5
Therefore, empirical formula is Fe2O3
Whole Number
Mole Ratio of
Element
2.00
3.00
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9. Complete the following table C and determine the empirical formula of the iron oxide
compound.
Table C:
Element Mass Percent Mole of Mole Ratio Whole Number
Element Of Element Mole Ratio of
Element
X: ?
Fe
72.36
1.2957
1.000
?
?
Y: ?
O
1.7275
Therefore, empirical formula is FexOy