The Mole
Kevin Pyatt, Ph.D.
Donald Calbreath, Ph.D.
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AUTHORS
Kevin Pyatt, Ph.D.
Donald Calbreath, Ph.D.
EDITORS
Donald Calbreath, Ph.D.
Max Helix
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Chapter 1. The Mole
C HAPTER
1
The Mole
C HAPTER O UTLINE
1.1
The Mole Concept
1.2
Mass, Volume, and the Mole
1.3
Chemical Formulas
1.4
References
Humans have long been interested in how much of something they have. Ancient commerce was often conducted
by trading goods for a certain weight of gold or silver. Both religious and civil statutes forbade the use of false
weights in conducting business. In chemistry, we routinely weigh materials to determine how much of a chemical
we are adding to a reaction. The amount of product is often weighed to determine reaction yield. Chemical solutions
are prepared by weighing a specified amount of material and dissolving it in a defined volume of solvent. Weight
and volume are two different ways to measure how much of a substance is present. In this chapter, we will explore
how to express amounts of materials in a way that indicates the number of atoms or molecules contained in a given
sample.
User:T helmadatter/Wikimedia Commons. commons.wikimedia.org/wiki/File:BalanceMineralPachuca.JPG. Public Domain.
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1.1. The Mole Concept
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1.1 The Mole Concept
Lesson Objectives
•
•
•
•
•
Describe three methods of measuring matter.
Define Avogadro’s number.
Define mole.
Perform calculations to convert between the number of moles and the number of particles in a sample.
Calculate the molar mass of a substance.
Lesson Vocabulary
•
•
•
•
mole: The name given to a group of 6.022 × 1023 items.
Avogadro’s number: 6.022 × 1023 .
representative particle: The smallest unit of a substance that retains its chemical identity.
molar mass: The mass of one mole of representative particles of a substance.
Check Your Understanding
• What units are used to describe the atomic mass (or weight) of an element?
Introduction
As you have learned, matter comprises most everything around us. The atoms and molecules that make up this matter
are so tiny that you cannot see them with the naked eye. Imagine you had a bag of apples. You could measure, rather
easily, the weight of the apples, the volume of the apples, and the number of apples in your bag. But what if you
wanted to know the number of atoms that were in those apples? How could you possibly count something you
cannot even see? In this lesson you will learn about the mole concept, which enables scientist to quantify things as
tiny as atoms.
Determining the Amount of Material
Determining by Weight
One common method of seeing how much material is present is to weigh it. This practice goes back many centuries.
The Code of Hammurabi (established around 1760 B.C.) spelled out methods for assuring fair weights in business
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Chapter 1. The Mole
and trade dealings. Many societies had harsh penalties for using false weights. Today we have state and federal
agencies that make sure the scales used for business are accurate.
Shopping in a grocery store provides many opportunities for purchases by weight. Most fresh meat and fresh
vegetables are marketed by weight, as are flour, sugar, and many other packaged goods.
Determining by Volume
Volume is another way of measuring material. Gases cannot be easily weighed, but their volume can be determined
quickly and accurately. Liquid chemicals as well as foods and beverages are also sold by volume. Gasoline for our
cars and trucks is sold by the gallon. At the grocery store, we can buy a gallon of milk or a two-liter bottle of soda
pop.
Determining by Counting
The third way to indicate how much of something is present is to count the objects. At the grocery store, we could
buy a dozen eggs, six donuts, or two loaves of bread. We might buy movie tickets or t-shirts, based on price per item.
The weight or volume for many of items could be determined, but counting is generally faster for small numbers of
objects and does not require any additional measurement tools.
Converting between Different Measurements
Sometimes one method of measurement is easiest but does not provide us with the information that we need. For
example, say you had a huge sack of pennies and wanted to know how much it was worth. You could count them, but
it might be much faster to weigh the sack. Say you find out that you have 4.801 pounds (2177.5 grams) of pennies. If
you know that each penny weighs 2.5 grams (on average), then you could divide 2177.5 by 2.5, and you would know
that you have about 871 pennies, or $8.71. Such conversions become even more important when dealing with much
more miniscule objects, such as atoms or molecules. Because the numbers involved when talking about amounts of
atoms are so large, a new unit was developed to make discussing such amounts easier. This unit is referred to as the
mole.
The Mole
As we will see in the chapter on gases, the Italian scientist Amadeo Avogadro (1776-1856) proposed that the volume
of a gas is directly proportional to the number of atoms or molecules in a given sample. Due to the vast quantities
of atoms and molecules in an easily observable sample of any material, chemists needed a name that can stand for a
very large number of items. A mole (abbreviated "mol") is the name given to a group of 6.022 × 1023 items. This
value is named Avogadro’s number as a tribute to his influential work.
The word "mole" is analogous to the word "dozen", in that it is a name that stands for a number. A dozen donuts
is simply another way of saying 12 donuts. Similarly, a mole of donuts would be another way of saying 6.022 ×
1023 donuts. Obviously, this would be an excessive amount of donuts. However, it becomes a more useful quantity
when talking about objects like molecules. One mole of water molecules is equivalent to 6.022 × 1023 molecules of
water. This amount of liquid water has a volume of a little over one tablespoon, which is a very normal amount to
be dealing with.
Technically speaking, the mole is defined as the amount of carbon atoms in exactly 12 grams of carbon-12 (the
isotope of carbon that contains 6 protons and 6 neutrons). This value cannot be determined by literally counting
atoms, but it has been calculated in a variety of ways. When written out to 8 significant figures, the currently
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FIGURE 1.1
Amadeo Avogadro (1776-1856)
accepted value is 6.0221413 × 1023 . However, because we don’t usually deal with other measurements that are this
precise, the rounded value of 6.022 × 1023 is sufficient for any situations that you are likely to encounter.
The mole is such an important parameter in chemistry that Mole Day was established in the U.S. on October 23,
where it is celebrated from 6:02 AM to 6:02 PM. The day was originally established to generate interest in chemistry.
The scale of Avogadro’s number is nearly incomprehensible in terms of everyday objects. This can be illustrated
with examples like the following problem, in which we use the dimensional analysis technique from the chapter
Measurement to convert between units.
Example 10.1
If you were given one million dollars every second, how long would it take to accumulate a mole of dollars?
Answer:
23
1 day
1 year
dollars
1 sec
1 min
1 hour
10
1 mol of dollars × ( 6.022×10
1 mol of dollars )( 1,000,000 dollars )( 60 sec )( 60 min )( 24 hours )( 365 days ) = 1.91 × 10 years
Even at this very fast rate, it would take over 19 billion years to obtain a mole of dollars. To put this into perspective,
the universe is believed to be about 14 billion years old, and our Sun will run out of fuel and burn out about 6 billion
years from now.
Representative Particles
When we talk about one mole of a particular chemical substance, it is important to know exactly how much material
is being referenced. For example, the phrase "a mole of water molecules" is often shortened to "a mole of water."
When we talk about moles of a molecular substance, it is assumed that we are referring to moles of molecules.
Similarly, the phrase "five moles of carbon dioxide" would refer to five moles of CO2 molecules. In these cases, the
molecule is sometimes called a representative particle, which is simply the smallest unit of a substance that retains
its chemical identity.
What about when we are discussing non-molecular substances? Recall that an ionic compound does not exist as
discrete molecules, but rather as an extended three-dimensional network of ions called a crystal lattice. The empirical
formula of an ionic compound tells us the ratio of the ions in the crystal. In these cases, one mole of the substance
is assumed to mean one mole of each ion in the empirical formula. For example, one mole of sodium oxide (Na2 O)
refers to two moles of sodium ions (Na+ ) and one mole of oxide ions (O2− ). The empirical formula gives you the
representative particle for ionic substances. Unlike molecules, these representative particles do not exist as isolated
units (sodium oxide does not exist as small clusters of three atoms each).
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Chapter 1. The Mole
In the case of pure metals, the representative particle is simply the atom. One mole of zinc refers to a mole of Zn
atoms.
Example 10.2
Calculate the number of moles of iron atoms that are present in 2 moles of Fe2 O3 .
mol Fe
2 mol Fe2 O3 × ( 1 2mol
Fe2 O3 ) = 4 mol Fe
Example 10.3
Calculate the number of atoms that must be present in 3.05 moles of Pb.
23
atoms Pb
) = 1.84 × 1024 atoms Pb
3.05 mol Pb × ( 6.022×10
1 mol Pb
Molar Mass
Because we are not able to count individual atoms, it is important to have a way to convert between amounts, which
are expressed in moles, and a unit of quantity that we can more easily measure, such as mass. We begin by looking
at the periodic table, which tells us the relative masses of various elements.
Molar Masses of the Elements
As you learned previously, the average atomic masses found on the periodic table are in atomic mass units (amu).
For example, one atom of the most abundant isotope of hydrogen has a mass of approximately 1 amu, and one atom
of helium has a mass of about 4 amu. Atomic masses are relative masses; they are based on the definition that one
amu is equal to 1/12th of the mass of a single atom of carbon-12. Therefore, one atom of carbon-12 has a mass of
12 amu, which is three times heavier than an atom of helium. This ratio would hold for any number of carbon and
helium atoms. One hundred carbon-12 atoms would have three times the mass of one hundred helium atoms. By
extension, 12.00 g of carbon-12 would contain the same number of atoms as 4.00 g of helium.
The relative scale of atomic masses in amu is also a relative scale of masses in grams. We said before that the mole
is officially equal to the number of carbon atoms in exactly 12 g of carbon-12. In other words, one carbon-12 atom
has a mass of exactly 12 amu, and one mole of carbon atoms has a mass of exactly 12 grams. This relationship is
true for all substances. If one atom of helium has a mass of 4.00 amu, one mole of helium atoms has a mass of 4.00
g. One molecule of water has a mass of 18.0 amu, so one mole of water molecules has a mass of 18.0 grams. Molar
mass is defined as the mass of one mole of representative particles of a substance. It is expressed in units of grams
per mole (g/mol).
The values on the periodic table can be read either as average atomic masses or as molar masses. For example,
the mass associated with chlorine is 35.45, even though no atoms of chlorine actually have a mass of 35.45 amu.
However, in naturally occurring samples of chlorine atoms, about 75% will be chlorine-35 atoms (which have a mass
of 35.0 amu), and about 25% will be chlorine-37 atoms (which have a mass of 37.0 amu). Since any usable quantity
of chlorine contains a very, very large number of atoms, one mole of chlorine atoms will contain a mixture of these
two isotopes in this ratio, resulting in a sample with a mass of 35.45 grams. Therefore chlorine has a molar mass of
35.45 g/mol.
Molar Masses of Compounds
The molecular formula of carbon dioxide is CO2 . One molecule of carbon dioxide consists of 1 atom of carbon and
2 atoms of oxygen. We can calculate the molar mass of carbon dioxide by adding together the molar masses of each
atom present in the compound.
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1.1. The Mole Concept
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12.01 g/mol + 2(16.00 g/mol) = 44.01 g/mol
One mole of carbon dioxide molecules would have a mass of 44.01 grams. What about for an ionic compound? The
molar mass of sodium sulfide (Na2 S) can be calculated as follows:
2(22.99 g/mol) + 32.06 g/mol = 78.04 g/mol
Because Na2 S is the empirical formula for this substance, the representative particle consists of two sodium atoms
and one sulfur atom.
Example 10.4
What is the molar mass of hydrogen peroxide (H2 O2 ).
Answer:
H - 2 × 1.0 g/mol = 2.0 g/mol
O - 2 × 16.0 g/mol = 32.0 g/mol
H2 O2 = 34.0 g/mol
Example 10.5
How many moles of carbon are present in a 10.00 gram sample?
Answer:
23
atoms C
) = 6.022 × 1024 atoms C
10.00 g C × ( 6.022×10
1 mol C
Lesson Summary
• One mole of any chemical substance contains 6.022 x 1023 representative particles.
• The masses on the periodic table can be read as average atomic masses or as molar masses.
• Molar mass is the mass of one mole of any given substance.
Lesson Review Questions
1. Imagine that you have 1 mole of coins, each of which is 1.5 mm thick. If they were placed in a single stack,
how tall would the stack be (in km)? If the closest distance between the earth and the moon is 356,400 km,
would the coins reach the moon? (Hint: Use the technique of dimensional analysis.)
2. Calculate the number of moles of N that must be present in 3 moles of NH3 .
3. Calculate the number of moles of oxygen that must be present in 2 moles of C6 H12 O6 .
4. How many moles of gold are present in a sample containing 1.81 x 1024 gold atoms?
5. How many atoms are in 0.065 moles of Hg?
6. Calculate the number of Br− ions in 0.0038 moles of CaBr2 .
7. Calculate the molar mass of K2 Cr2 O7 .
8. Calculate the molar mass of dinitrogen tetroxide (N2 O4 ).
Further Reading / Supplemental Links
• Avogadro biography: http://www.bulldog.u-net.com/avogadro/avoga.html
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Chapter 1. The Mole
• The mole –its history and use: http://www.visionlearning.com/library/module_viewer.php?mid=53
• Mole Day: http://chemistry.about.com/od/historyofchemistry/a/mole-day.htm
• National Mole Day Foundation: http://www.moleday.org/
Points to Consider
• How can we use this information to determine the number of atoms or molecules in a given amount of a
material?
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1.2. Mass, Volume, and the Mole
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1.2 Mass, Volume, and the Mole
Lesson Objectives
•
•
•
•
•
Be able to convert between mass and moles of a substance.
Explain how Avogadro’s hypothesis relates to volumes of gases at standard temperature and pressure.
Convert between moles and volume of a gas at STP.
Calculate the mass of a gas at STP when given its volume.
Make conversions between mass, number of particles, and gas volumes.
Lesson Vocabulary
• standard temperature and pressure (STP): A pressure of one atmosphere and a temperature of 0°C.
Check Your Understanding
• What is the value of Avogadro’s number, and what does it represent?
Introduction
As we discussed in the previous lesson, there are many different ways to measure how much of something you
have. Usually, there is a particular unit of measurement that is easiest to use, depending on what you are trying to
quantify. In this lesson you will learn the significance of using moles in converting between measurements and in
understanding how much of something you have.
Mass and Moles
When we need materials for a chemical reaction, counting out a certain number of atoms or molecules is obviously
impractical, so we weigh out a certain mass of each substance instead. As we will see in a later chapter, chemical
equations tell us the molar ratios in which chemicals react with one another. This information can be used to
determine how much of one chemical is needed to fully react with a set amount of another substance.
Example 10.6
In a certain reaction, we want to use two moles of silver nitrate, AgNO3 . We need to know how many grams of
silver nitrate will be needed. First, we determine the molecular weight of the chemical:
Ag: 1 x 107.9 g/mol = 107.9 g/mol
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Chapter 1. The Mole
N: 1 x 14.0 g/mol = 14.0 g/mol
O: 3 x 16.0 g/mol = 48.0 g/mol
AgNO3 - 169.9 g/mol
Because 169.9 grams of AgNO3 is equivalent to one mole of AgNO3 , we can use that as a conversion factor. We
start with the amount that we need (2.0 moles), and convert to the desired units (grams of AgNO3 ):
grams AgNO3
2.0 moles AgNO3 × 169.9
= 339.8 grams AgNO3
1 mole AgNO3
The units of moles cancel, and we are left with grams.
Example 10.7
In another reaction, we determine that we need 0.45 moles of NaBr. We check our chemical supplies and find we
have 37.2 grams NaBr on hand. Do we have enough for this reaction?
Na: 1 x 23.0 g/mol = 23.0 g/mol
Br: 1 x 79.9 g/mol = 79.9 g/mol
NaBr - 102.9 g/mol
grams NaBr
= 46.3 grams NaBr
0.45 moles NaBr × 102.9
1 mole NaBr
We need 0.45 moles, which our calculation tells us is equivalent to 46.3 grams. We only have 37.2 grams available,
so we do not have enough for this experiment.
Example 10.8
We run an experiment that gives us 65.4 grams of Rb2 O as a product. How many moles did we obtain?
First, we need to calculate the molar mass of the compound:
Rb: 2 x 85.5 g/mol = 171.0 g/mol
O: 1 x 16.0 g/mol = 16.0 g/mol
Rb2 O - 187.0 g/mol
1 mol Rb2 O
65.4 g Rb2 O × 187.0
g Rb2 O = 0.350 mol Rb2 O
We obtained 0.350 moles of Rb2 O. Note that the conversion factor of molar mass (187.0 grams Rb2 O = 1 mole
Rb2 O) was written "upside-down," with the grams on the bottom and the moles on top. This is so that the unwanted
units (grams) cancel, leaving only the desired units (moles) in our final answer. Whenever we have this type of
conversion factor, the choice of which quantity to put in the numerator and denominator depends on the units that
we wish to cancel.
Volume and Moles
Avogadro’s Hypothesis
Avogadro proposed that equal volumes of gases at the same temperature and pressure contain the same number of
particles, and therefore the same combining ratios. This means that at a given temperature and pressure, one mole
of any gas will take up the same volume, regardless of its identity. This is very different than the case for solids and
liquids. For example, a mole of water takes up significantly more space than a mole of gold, which is quite dense.
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Because different gases have different molar masses, the mass contained in a set volume of gas will depend on the
identity of the gas. For example, at a certain temperature and pressure, one liter of hydrogen gas has a mass of
0.0899 grams, while a liter of oxygen gas has a mass of 1.43 grams. This ratio is approximately equal to 1:16, which
is the ratio of the molar masses for these two elements. The table below shows the amount of mass contained in a
given volume of various gases.
FIGURE 1.2
Relative masses of various gases.
The implications of Avogadro’s work were extensive and helped in developing the kinetic theory of gases in the
last half of the 19th century. They also were useful as practical guidelines for how to run reactions with gases.
For example, water has a formula of H2 O, which means that for each mole of oxygen atoms, there are two moles of
hydrogen atoms. A complete reaction can occur with no leftover hydrogen or oxygen only if the volume of hydrogen
used is twice the volume of oxygen used. Because the volume of a gas is easier to measure than its mass, this is a
useful experimental tool.
Example 10.9
Let’s say you have three balloons. One is filled with hydrogen gas, one with oxygen gas, and one with nitrogen gas.
The molar masses of these gases are 2 g/mol for H2 , 32 g/mol for O2 , and 28 g/mol for N2 . If all three balloons are
the same volume, which contains the most mass, and which contains the least?
Answer:
Based on their molar masses, hydrogen is the lightest molecule, and oxygen is the heaviest. Because all three
volumes are the same, each balloon contains the same number of gas molecules. Therefore, the hydrogen balloon
will have the lowest mass, and the oxygen balloon will have the highest.
Calculations Involving Gases
Because the volume of a gas is dependent on the pressure and temperature, scientists found it useful to collect
data at fixed pressures and temperatures so that they could be compared between different gases. A pressure of
one atmosphere and a temperature of 0°C is known as standard temperature and pressure (STP). Under these
conditions, one mole of any gas takes up a volume of 22.4 liters. This information allows us to convert between
liters and moles for gases at STP.
Example 10.10
At STP, we have 46.2 liters of helium. How many moles of helium do we have?
1 mol He
(46.2 L He)( 22.4
L He ) = 2.06 mol He
Example 10.11
What volume does 4.96 moles of O2 occupy at STP?
L O2
(4.96 mol O2 )( 22.4
1 mol O2 ) = 111 L O2
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Chapter 1. The Mole
If we know the volume of a gas at STP, we can combine this information with its molar mass to determine how much
mass is present in the sample.
Example 10.12
What is the mass of 86.7 liters of N2 at STP?
The molar mass of N2 is 28.0 grams/mole, and because one mole of a gas at STP takes up 22.4 L of space, we can
perform the following calculation.
28.0 g N2
1 mol N2
(86.7 L N2 )( 22.4
L N2 )( 1 mol N2 ) = 108 g N2
The conversion factors are arranged so that all units cancel except for grams, which are the units of our final answer.
When studying chemical reactions, we frequently need to convert back and forth between mass, volume, moles,
and number of particles. This will be expanded upon in our chapter on stoichiometry. Figure 1.3 summarizes the
relationships that we have studies so far between these different quantities.
FIGURE 1.3
Mole relations for mass, volume, and particles.
Lesson Summary
• Molar masses can be used to determine the mass of a given quantity of material.
• At standard temperature and pressure (1 atmosphere of pressure and 0°C), one mole of a gas occupies 22.4
liters.
• The mass of a sample can be calculated for a given volume of a known gas at STP using Avogadro’s number.
Review Questions
1. Calculate the number of moles in 56.3 grams of CaCO3 .
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2. What is the mass (in grams) of 3.2 moles of glucose (C6 H12 O6 )?
3. You are running an experiment that needs three moles of NaHCO3 , and you only have 150 grams of material.
Calculate the number of moles you have, and determine if this is enough to run the experiment.
4. A balloon contains 96 grams of CO2 gas at STP. How many moles are present?
5. How many atoms are there in 12.2 grams of Zn?
6. A container holds 68.5 liters of HBr gas. What is the mass of HBr in the container?
Further Reading / Supplemental Links
• Mole-mass calculations: http://www.ausetute.com.au/massmole.html
• Molar mass calculations: http://misterguch.brinkster.net/molarmass.html
• Gas calculations: http://www.sciencegeek.net/Chemistry/taters/Unit5MolarVolume.htm
Points to Consider
• How can you use information about the composition of a substance to determine its chemical formula?
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Chapter 1. The Mole
1.3 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 1.4) 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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1.3. Chemical Formulas
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FIGURE 1.4
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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Chapter 1. The Mole
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 1.5). 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 1.5
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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Chapter 1. The Mole
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 1.6. They are very different compounds, yet both
have the same empirical formula, CH2 O.
The following Table 1.1 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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1.3. Chemical Formulas
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FIGURE 1.6
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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Chapter 1. The Mole
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.
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Chapter 1. The Mole
1.4 References
1.
2.
3.
4.
5.
C. Sentier. http://commons.wikimedia.org/wiki/File:Avogadro_Amedeo.jpg . Public Domain
Jodi So. CK-12 Foundation . CC BY-NC 3.0
Christopher Auyeung. CK-12 Foundation . CC BY-NC 3.0
Joy Sheng. CK-12 Foundation . CC BY-NC 3.0
Ben Mills (User:Benjah-bmm27/Wikimedia Commons). http://commons.wikimedia.org/wiki/File:EthyleneCRC-MW-3D-balls.png . Public Domain
6. (”left”) Ben Mills (User:Benjah-bmm27/Wikimedia Commons); (”right”) Ben Mills (User:Benjah-bmm27/Wikimedia
Commons), User:Yikrazuul/Wikimedia Commons. http://commons.wikimedia.org/wiki/File:Acetic-acid-2Dflat.png; http://commons.wikimedia.org/wiki/File:D-glucose-chain-2D-Fischer.png . Public Domain
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