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Empirical and Molecular Formulas
Empirical Formula – A chemical formula that gives the lowest whole number ratio of atoms in a molecule.
Ionic compounds are always written as empirical formulas (MgO instead of Mg2O2).
Molecular Formula – A chemical formula that is the same as the empirical formula for a given compound or is
some multiple of the empirical formula of that compound.
Examples
Example
1
2
Empirical Formula
CH3
CH3
Empirical Molar Mass
15 g/mol
15 g/mol
Molecular Formula
Molecular Molar Mass
C2H6
C3H9
30 g/mol
45 g/mol
Note that as the subscripts of the empirical formula are multiplied by a whole number, so is the empirical molar
mass. In Example 1, CH3 is multiplied by 2 to make C2H6 and the empirical molar mass of 15 g/mol is also
multiplied by 2 to get a molecular molar mass of 30 g/mol. Empirical formulas can be multiplied by any whole
number. In Example 2, the empirical formula and molar mass are both multiplied by 3.
Practice I
Directions: Write the empirical formula for each of the following
Molecular Formula
Empirical Formula
Molecular Formula
1) C2H4
4) C9H18O3
2) C6H6
5) P4O10
3) C6H12O6
6) P2H5
Empirical Formula
Practice II
Directions: Use the information given below to determine the molecular formula
Number
Empirical Formula
Empirical Molar
Mass
Molecular Mass
Factor used to get
Molecular Formula
Molecular
Formula
Example
1
2
3
4
5
CH3
As2O5
CH2
NO2
C2H3
CO2
15 g/mol
60 g/mol
690 g/mol
42 g/mol
92 g/mol
54 g/mol
44 g/mol
60/15 = 4
C4H12
Determining Empirical Formulas from Percent Composition Data
Scientists can determine the types of elements present in a sample (qualitative analysis) and also the
relative amounts of elements present in a sample (quantitative analysis.) Then the scientist can perform a
mathematical calculation that will determine the empirical formula of the compound for identification purposes.
For example, a scientist may be asked to find out information about a “mysterious” white powder.
Qualitative tests indicate that carbon, hydrogen and oxygen are present in the sample and quantitative tests
determine that 40.0 % is carbon, 6.7 % is hydrogen and 53.3 % is oxygen. From this information, the scientist
can determine the chemical formula for the substance. The poem below gives the steps for solving these types
of problems:
A Chemistry Poem:
“Percent to mass,
Mass to mole.
Divide by small,
Then multiply ‘til Whole”
Example 1:
Recall that the “mysterious white powder” contained 40.0 % carbon, 6.7 % hydrogen and 53.3 % oxygen. We
are trying to determine the empirical formula. We know that the chemical formula will be something like this:
C?H?O?. We are going to perform calculations to determine the “?” which are the subscripts or mole ratios of
the elements present in the compound.
Step 1: “Percent to Mass”  Simply change the “%” to “g”. Skip this step if data is already in grams.
Ex) 40.0 g carbon, 6.7 g hydrogen and 53.3 g oxygen
Step 2: “Mass to Mole”  Divide the mass of each element by its molar mass (no diatomic elements here.)
Ex) C = 40.0g C | 1 mole C = 3.33 mol
H = 6.7 g H | 1 mole H = 6.7 mol
O = 53.3 g O | 1 mole O = 3.33 mol
12 g C
1gH
16 g O
Step 3: “Divide by Small”  Select the smallest number calculated in Step 2 and divide all of the answers in
Step 2 by that number. You will always get a “1” as an answer for Step 3.
Ex) C = 40.0g C | 1 mole C = 3.33 mol H = 6.7 g H | 1 mole H = 6.7 mol
O = 53.3 g O | 1 mole O = 3.33 mol
12 g C
3.33
1gH
=1
3.33
16 g O
=2
3.33
=1
Step 4: “Multiply ‘til Whole”  Multiply the mole ratios calculated in Step 3 by a whole number if they are
decimals. In this example, the answers are whole numbers, so we can skip this step.
Answer: C1H2O1  CH2O
Since this is the empirical formula for sugar (C6H12O6) the mysterious white powder may be sugar.
Practice III
Directions: Determine the empirical formula for each of the following. SHOW ALL WORK.
1) A compound that is 34.8 % Fe and 65.2 % Cl
2) A compound that is 40.5 % Zn, 19.9 % S and 39.6 O
3) A compound that is 88.9 g Cu and 11.2 g O.
4) A compound that is 54.5 % C, 9.1 % H and 36.4 % O.
5) Determine the molecular formula of the compound in #4 if the molecular mass is 88 g/mol.