Chemical Formulas
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Printed: November 24, 2013
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C ONCEPT
Concept 1. Chemical Formulas
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Chemical Formulas
Lesson Objectives
•
•
•
•
Use a chemical formula or mass data to calculate the percent composition of a compound.
Use the percent composition of a compound to calculate the mass of an element in a given sample.
Be able to calculate the empirical formula for a compound when given percent composition data.
Be able to calculate the molecular formula for a compound when you know its molar mass and its empirical
formula.
Lesson Vocabulary
• percent composition: The percent by mass of each element in a compound.
• empirical formula: An elemental formula showing the lowest whole-number ratio of the elements in a
compound.
• molecular formula: A formula showing how many atoms of each element are present in one molecule of a
molecular compound.
Check Your Understanding
• How can you calculate the amount of a substance in moles from its mass and its molar mass?
Introduction
Packaged foods that you eat typically have nutritional information provided on the label. The label of a popular
brand of peanut butter (Figure 1.1) reveals that one serving size is considered to be 32 g. The label also gives the
masses of various types of compounds that are present in each serving. One serving contains 7 g of protein, 15 g of
fat, and 3 g of sugar. This information can be used to determine the composition of the peanut butter on a percent
by mass basis. For example, to calculate the percent of protein in the peanut butter, we could perform the following
calculation:
7 g protein
× 100% = 22% protein
32 g
In a similar way, chemists often need to know what elements are present in a compound and in what percentages.
The percent composition is the percent by mass of each element in a compound. It is calculated in a way that is
similar to what we just saw for the peanut butter.
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FIGURE 1.1
Foods like peanut butter provide nutritional information on the label in the form of masses of different types of
compounds present per serving.
Percent Composition and Mass
Determining Percent Composition from Mass Data
We can calculate percent composition based on the following formula:
mass of element × 100%
% by mass = mass
of compound
The example below shows how the percent composition of a compound can be calculated based on mass data:
Example 10.13
A certain newly synthesized compound is known to contain the elements zinc and oxygen. When a 20.00 g sample
of the compound is decomposed, 16.07 g of pure zinc remains. Determine the percent composition of the compound.
Answer:
If the compound contained only zinc and oxygen, and 16.07 grams was due to the zinc, we can subtract to determine
the mass of oxygen in the original sample:
Mass of oxygen = 20.00 g – 16.07 g = 3.93 g O
Then, we divide the individual masses of each element by the total mass of the sample to determine the percent (by
mass) of each element in the compound:
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Concept 1. Chemical Formulas
16.07 g Zn
× 100% = 80.35% Zn
20.00 g
3.93 g O
× 100% = 19.65% O
%O=
20.00 g
% Zn =
The calculations make sense because the sum of the two percentages adds up to 100%. By mass, the compound is
mostly zinc.
Determining Masses from Percent Composition Data
We can also perform the reverse calculation, determining the mass of an element in a given sample, if we know the
total mass of the sample and its percent composition.
Example 10.14
You have a 10.0 g sample of a metal alloy that contains only aluminum and zinc. If the sample is 36% aluminum by
mass, what masses of Al and Zn are present?
Answer:
We are told that the sample is 36% aluminum by mass. Because the only other component is zinc, it must make up
the remaining 64% of the mass. We can multiply each of these percentages by 10.0 grams to find the masses of each
element.
10.0 g sample × 0.36 = 3.6 g Al
10.0 g sample × 0.64 = 6.4 g Zn
Empirical Formulas
Recall that an empirical formula is one that shows the lowest whole-number ratio of the elements in a compound.
Because the structure of ionic compounds is an extended three-dimensional network of positive and negative ions,
only empirical formulas are used to describe ionic compounds. However, we can also consider the empirical formula
of a molecular compound. Ethene is a small hydrocarbon compound with the formula C2 H4 (Figure 1.2). While
C2 H4 is its molecular formula and represents its true molecular structure, it has an empirical formula of CH2 . The
simplest ratio of carbon to hydrogen in ethene is 1:2. In each molecule of ethene, there is 1 carbon atom for every
2 atoms of hydrogen. Similarly, we can also say that in one mole of ethene, there is 1 mole of carbon for every 2
moles of hydrogen. The subscripts in a formula represent the molar ratio of the elements in that compound.
FIGURE 1.2
Ball-and-stick model of ethene, C2 H4 .
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Determining Percent Composition from a Chemical Formula
The percent composition of a compound can also be determined from its chemical formula. The subscripts in the
formula are first used to calculate the mass of each element found in one mole of the compound. That value is then
divided by the molar mass of the compound and multiplied by 100%.
% by mass =
mass of element in 1 mol of compound
× 100%
molar mass of compound
The percent composition of a given compound is always the same as long as the compound is pure.
Example 10.15
Dichlorine heptoxide (Cl2 O7 ) is a highly reactive compound used in some synthesis reactions. Calculate the percent
composition of dichlorine heptoxide.
Answer:
Determine the mass of each element in one mole of the compound, and find the total molar mass of the compound:
• mass of Cl in 1 mol Cl2 O7 = 2 x molar mass of Cl = 70.90 g
• mass of O in 1 mol Cl2 O7 = 7 x molar mass of O = 112.00 g
• molar mass of Cl2 O7 = 70.90 g/mol + 112.00 g/mol = 182.90 g/mol
Now, calculate the percent by mass of each element by dividing the mass of that element in 1 mole of the compound
by the molar mass of the compound and multiplying by 100%.
70.90 g Cl
× 100% = 38.76% Cl
182.90 g
112.00 g O
× 100% = 61.24% O
%O=
182.90 g
% Cl =
As expected, the percentages add up to 100%.
Determining Empirical Formulas from Percent Composition
A procedure called elemental analysis allows us to determine the empirical formula of an unknown compound.
Percent composition data can be directly obtained with this technique, and these values can be used to find the molar
ratios of the elements, which gives us the empirical formula. The steps to be taken are outlined below.
1. Assume a 100 g sample of the compound so that the given percentages can be directly converted into grams.
2. Use each element’s molar mass to convert the grams of each element to moles.
3. In order to find a whole-number ratio, divide the moles of each element by the smallest value obtained in step
2.
4. If all the values at this point are whole numbers (or very close), each number is equal to the subscript of the
corresponding element in the empirical formula.
5. In some cases, one or more of the values calculated in step 3 will not be whole numbers. Multiply each of
them by the smallest number that will convert all values into whole numbers (or very close to whole numbers).
Note that all values must be multiplied by the same number so that the relative ratios are not changed. These
values can then be used to write the empirical formula.
Example 10.16
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Concept 1. Chemical Formulas
A compound of iron and oxygen is analyzed and found to contain 69.94% iron and 30.06% oxygen by mass. Find
the empirical formula of the compound.
Answer:
Follow the steps outlined in the text.
1. Assume a 100 g sample. In 100 grams of the compound, there would be 69.94 g Fe and 30.06 g O.
2. Convert to moles.
1 mol Fe
= 1.252 mol Fe
55.85 g Fe
1 mol O
30.06 g O ×
= 1.879 mol O
16.00 g O
69.94 g Fe ×
3. Divide both values by the smallest of the results.
1.252 mol Fe
= 1 mol Fe
1.252
1.879 mol O
= 1.501 mol O
1.252
4. Since the moles of O is still not a whole number, both numbers can be multiplied by 2. The results are now close
enough to be rounded to the nearest whole number.
1 mol Fe × 2 = 2 mol Fe
1.501 mol O × 2 = 3 mol O
The empirical formula of the compound is Fe2 O3 .
Molecular Formulas
Molecular formulas tell us how many atoms of each element are present in one molecule of a molecular compound.
In many cases, the molecular formula is the same as the empirical formula. For example, the molecular formula of
methane is CH4 , and because 1:4 is the smallest whole-number ratio that can be written for this compound, that is
also its empirical formula. Sometimes, however, the molecular formula is a simple whole-number multiple of the
empirical formula. Acetic acid is an organic acid that gives vinegar its distinctive taste and smell. Its molecular
formula is C2 H4 O2 . Glucose is a simple sugar that cells use as their primary source of energy. Its molecular formula
is C6 H12 O6 . The structures of both molecules are shown in Figure 1.3. They are very different compounds, yet both
have the same empirical formula, CH2 O.
The following table shows a few other compounds with their empirical and molecular formulas:
TABLE 1.1: Empirical and Molecular Formulas
Compound
water
hydrogen peroxide
methane
butane
Empirical Formula
H2 O
HO
CH4
C2 H5
Molecular Formula
H2 O
H2 O2
CH4
C4 H10
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FIGURE 1.3
Acetic acid (left) has a molecular formula
of C2 H4 O2 , while glucose (right) has a
molecular formula of C6 H12 O6 . Both have
the empirical formula CH2 O.
Empirical formulas can be determined from the percent composition of a compound. In order to determine its
molecular formula, it is necessary to also know the molar mass of the compound. Chemists have various methods
to determine the molar mass of an unknown compound. In order to go from the empirical formula to the molecular
formula, follow these steps:
1. Calculate the empirical formula mass (EFM), which is simply the molar mass represented by the empirical
formula.
2. Divide the molar mass of the compound by the empirical formula mass. The result should be a whole number
or very close to a whole number.
3. Multiply all of the subscripts in the empirical formula by the whole number found in step 2. The result is the
molecular formula.
Example 10.17
The empirical formula of a compound that contains boron and hydrogen is BH3 . Its molar mass is 27.7 g/mol.
Determine the molecular formula of the compound.
Answer:
Follow the steps outlined above.
1. The empirical formula mass (EFM) = 13.84 g/mol
1.
molar mass
27.7
=
=2
EFM
13.84
1. BH3 × 2 = B2 H6
The molecular formula of the compound is B2 H6 . The molar mass of the molecular formula matches the molar mass
of the compound.
You can watch a video lecture about molecular and empirical formulas at http://www.khanacademy.org/science/c
hemistry/chemical-reactions-stoichiometry/v/molecular-and-empirical-formulas.
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Concept 1. Chemical Formulas
You can watch a video lecture about determining molecular and empirical formulas from percent composition
at http://www.khanacademy.org/science/physics/thermodynamics/v/molecular-and-empirical-forumlas-from-percen
t-composition.
Lesson Summary
• The percent composition of a compound is the percent by mass of each of the elements in the compound. It
can be calculated from mass data or from the chemical formula.
• Percent composition data can be used to determine a compound’s empirical formula, which is the molar ratio
between the elements in the compound.
• The empirical formula and the molar mass of a substance can be used to determine its molecular formula,
which is the number of each kind of atom in a single molecule of the compound.
Lesson Review Questions
1. Calculate the percent carbon in C2 H6 .
2. Calculate the percent nitrogen in CH3 CH2 CH2 NH2 .
3. A sample of a given compound contains 13.18 g of carbon and 3.32 g of hydrogen. What is the percent
composition of this compound?
4. 5.00 g of aluminum is reacted with 7.00 g of fluorine to form a compound. When the compound is isolated, its
mass is found to be 10.31 g, with 1.69 g of aluminum (and no fluorine) left unreacted. Determine the percent
composition of the compound.
5. Calculate the percent by mass of each element present in sodium sulfate (Na2 SO4 ).
6. Vitamin C contains carbon (40.9%), hydrogen (4.6%), and oxygen (54.5%). Calculate the empirical formula
for vitamin C. The molecular mass is about 180. Determine the molecular formula for vitamin C.
7. Calculate the empirical formula of each compound from the percentages listed:
a. 63.65% N, 36.35% O
b. 81.68% C, 18.32% H
8. A compound was analyzed and found to contain 13.5 g Ca, 10.8 g O, and 0.675 g H. What is the empirical
formula of the compound?
9. Calculate the percent composition of the following compounds:
a. magnesium fluoride, MgF2
b. silver nitrate, AgNO3
10. A compound with the empirical formula CH has a molar mass of 78 g/mol. Determine its molecular formula.
11. A compound is found to consist of 43.64% phosphorus and 56.36% oxygen. The molar mass of the compound
is 284 g/mol. Find the molecular formula of the compound.
Further Reading/Supplementary Links
• Dalton and relative weights: http://dl.clackamas.edu/ch104-03/relative.htm
• Calculating formulas and composition: http://library.thinkquest.org/10429/low/chemcomp/chemcomp.htm
• Finding empirical formulas given percent composition: http://www.chemteam.info/Mole/Emp-formula-givenpercent-comp.html
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Points to Consider
In the next chapter we will be able to use this information to determine how much material we need for a chemical
reaction and how much product we can produce as a result of a reaction.
References
1. CK-12 Foundation - Joy Sheng. . CC-BY-NC-SA 3.0
2. Ben Mills (Wikimedia: Benjah-bmm27). http://commons.wikimedia.org/wiki/File:Ethylene-CRC-MW-3Dballs.png. Public Domain
3. (left) Ben Mills (Wikimedia: Benjah-bmm27); (right) Ben Mills (Wikimedia: Benjah-bmm27), User:Yikrazuul/Wikimedia
Commons. (left) http://commons.wikimedia.org/wiki/File:Acetic-acid-2D-flat.png; (right) http://commons.wi
kimedia.org/wiki/File:D-glucose-chain-2D-Fischer.png. Public Domain
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