Determination of Molecular Formulas

Activity 6
Determination of Molecular Formulas
Why?
Molecular formulas are also called chemical formulas. They tell you how many atoms of each
element are present in a molecule. Molecular formulas can be determined by measuring the mass
of each element present in a sample of the compound. This conversion of macroscopic quantities
of material (grams) to the microscopic composition (number of atoms of each element present in a
molecule) is used by chemists, biochemists, pharmacologists and others who work in the production
of new materials for research, technology, and medicine.
Learning Objective
• Understand the relationship between the mass percent composition of a chemical compound
and its molecular formula
Success Criteria
• Quickly calculate mass percent composition from a molecular formula
• Determine the molecular formula from the mass percent composition
Prerequisite
• Activity 3: Molecular Representations
Information
The molecular or chemical formula of a molecule tells you the number of atoms of each element
that comprise the molecule. It also tells you the number of moles of each element needed to make
1 mole of the compound.
Molecular formulas can be determined from the mass percent of each element present in a
compound.
The mass percent of an element in a sample of a pure compound is calculated in the following
way:
mass of the element present in a sample
mass % of an element = × 100%
total mass of the sample
Activity 6 —Determination of Molecular Formulas 33

The mass % of an element in a compound can also be calculated if the molecular formula is known.
This calculation is done in the following way.
x 100%
mass % of an element =
mass of the element in one mole of the compound
molar mass of the compound
× 100%
For example, to calculate the percent C in propene, C 3 H 6 : the mass of carbon in 1 mol propene
is 3 × 12.01 g/mol = 36.03 g/m. The molar mass of propene is 42.08 g/mol, so dividing 36.03 by
42.12 and multiplying by 100% produces 85.62%.
Model 1: What Do these Different Compounds Have in
Common?
Table 6.1
Name
Line Structure
Molecular
Formula
Mass % Composition
C
H
ethene
propene C 3 H 6 85.64 14.38
1-butene
2-butene
1-pentene
2-pentene
Task
Complete Table 6.1 for Model 1 by filling in the molecular formula and the mass percent composition
information.
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Foundations of Chemistry

Key Questions
1. What are common features of the six compounds in Model 1 that you can identify from
Table 6.1?
2. Is it possible to determine the molecular formula of a compound from the mass percent
composition? Explain why or why not.
3. How are the molecular formulas of the six compounds in Model 1 similar, and how are they
different?
Information
An empirical formula for a compound contains integer subscripts that provide the ratio of elements
in the compound by using the smallest whole numbers. The molecular formula is some multiple of
the empirical formula. Additional information, for example the molar mass, is needed to determine
the molecular formula from the empirical formula.
Key Questions
4. What is the empirical formula for the compounds in Model 1?
Model 2: Chemical Analysis of Acetic Acid
Acetic acid is the active ingredient in vinegar. A chemical analysis of 157.5 g of acetic acid provided
the following information:
Table 6.2
Element Mass of Element (g) Moles of Element
Whole Number
Ratio
carbon 63.00 5.246 5.246/5.246 = 1
oxygen 83.93 5.246 5.246/5.246 = 1
hydrogen 10.57 10.486 10.486/5.246 = 2
moles = mass of element / molar mass of element
The 1:1:2 ratio in the last column means that the empirical formula is (COH 2 ).
Activity 6 —Determination of Molecular Formulas 35

Key Questions
5. How was the number of moles of each element calculated?
6. How were the ratios of the elements in acetic acid determined from the moles of each element
present in the sample?
7. How is the empirical formula determined from the ratios of the elements present in the
sample?
8. What information does the empirical formula provide?
9. What is the relationship between the ratio of moles of each element present in the sample, and
the ratio of the number of atoms of each element present in each molecule of acetic acid?
10. Why is the atomic mass percent composition of an unknown chemical compound an important
quantity to determine in a chemical analysis?
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Foundations of Chemistry

Exercises
1. Sodium carbonate has the molecular formula Na 2 CO 3 . Calculate the molar mass of this
compound, the mass percent composition of each element in the compound, and the mass of
each element present in a 73.6 g sample.
Molar mass calculation:
(2 × 22.99) + (12.01) + (3 × 16.00) = 105.99 g/mol
Mass percent calculations:
(2 × 22.99)
105.99
× 100% = 43.38% Na
12.01
100% 11.33% C
105.99 × =
(3 × 16.00)
105.99
× 100% = 45.29% O
Amount in 73.6 g sample: multiply the percent by 73.6 g to get:
31.91 g Na
8.339 g C
33.33 g O
2. A sample of sodium bicarbonate (baking soda) was found to consist of 9.122 g Na, 0.4000 g
H, 4.766 g C, and 19.04 g O. Calculate the moles of each element present, and determine the
empirical formula of sodium bicarbonate.
9.122 g Na
22.99 g Na/mol Na =
0.3968 mol Na
0.4000 g H
1.008 g H/mol H =
4.766 g C
12.01 g C/mol C =
19.04 g O
16.00 g O/mol O =
0.3968 mol H
0.3968 mol C
1.19 mol O
Divide all by the smallest:
0.3968 1.19
= 1 and = 3
0.3968 0.3968
So mol ratio is 1:1:1:3, so the empirical formula is NaHCO 3
Activity 6 —Determination of Molecular Formulas 37

Problems
1. Show how you can determine the molecular formula for acetic acid, which is found in vinegar,
given the empirical formula CH 2 O and the added information that the molar mass of acetic acid
is 60.05 g/mole.
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Foundations of Chemistry

2. One of the chlorofluorocarbons (a freon), which is used in refrigerator compressors and
contributes to destruction of ozone in the upper atmosphere, has a molar mass of 132.9 g/mole
and a percent composition of 53.34% Cl, 28.59% F, and 18.07% C. Hint: Whenever the data
give the percent of the elements present rather than the mass, you can assume any sample size
and calculate the mass. Why is it particularly convenient to assume the data came from a 100 g
sample?
a) How many grams of chlorine are there in a 100 g sample of the freon?
b) How many moles of chlorine are there in a 100 g sample of the freon?
53.34 g
35.45 g /mol
=
1.50 moles chlorine
c) What is the ratio of the moles of chlorine to the moles of carbon in this freon?
moles of C
18.07 g
= =
12.01 g/mol
1.5 mol carbon
so 1.5:1.5 = 1:1 ratio of chlorine to carbon in this freon.
d) Determine the empirical and molecular formulas of this freon.
mol of F
28.59 g
= =
19.0 g/mol
1.50 mol F
so CClF is the empirical formula.
the empirical formula molar mass is 66.45 g/mol,
which is half the molar mass of the compound,
so Cl 2 F 2 C 2 is the molecular formula.
Activity 6 —Determination of Molecular Formulas 39

3. Combustion of 10.68 mg of a compound containing only C, H, and O produces 16.01 mg CO 2
and 4.37 mg H 2 O. The molar mass of the compound is 176.1 g/mol. What is the empirical
formula of the compound? What is the molecular formula of the compound?
Hints
The empirical formula gives the proportions in terms of mole ratios of each element present in the
compound. To find the empirical formula, you first need to determine how many moles of each
element are present in a sample of the compound.
1. You can determine the mass and moles of carbon in the compound from the mass of carbon
dioxide that was produced in the combustion. Can you explain why the fraction of the carbon
dioxide mass that is due to carbon is 12.01/44.01?
12.01 (16.01 mg) = 4.369 mg
44.01
4.369 mg
moles of C = = 0.3638 mmol
-1
12.01 g mol
2. You can determine the mass and moles of hydrogen in the compound from the mass of water that
was produced in the combustion by first determining the fraction of the mass of water that is due to
hydrogen.
2.016 (4.37 mg) = 0.489 mg
18.016
0.489 mg
moles of H = = 0.485 mmol
-1
1.008 g mol
3. You can determine the mass and moles of oxygen because the sum of O, C, and H masses
must add up to 10.68 mg, which is the mass of the sample.
mass of oxygen = (10.68 - 4.369 - 0.489) mg = 5.822 mg
moles of oxygen
5.822 mg
= = 0.3639 mmol
-1
16.00 g mol
4. From the moles of C, H, and O, you can determine the mole ratios, which leads to the empirical
formula.
0.3638 0.485 0.3639
C:H:O = : : = 1.000 : 1.33 : 1.000
0.3638 0.3638 0.3638
5. Given the empirical formula and the molar mass of 176.1 g/mol, you can identify the molecular
formula.
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Foundations of Chemistry