Chapter 12
The Mole Concept
12.1 Determining Formula and Molar Masses
Lesson Objectives
The student will:
• calculate the formula mass of a compound given its name and a periodic table.
Vocabulary
• formula mass
• molecular mass
Introduction
When atoms of one element chemically combine with atoms of another element, a compound is formed.
Compounds have names and formulas associated with them. The formula of a compound contains chemical
symbols that tell us what elements are in the compound. The subscripts in the formula tell us the ratios
of the elements present. For example, the formula MgCl2 tells us that this compound is composed of the
elements magnesium and chlorine, which combine in the ratio of two atoms of chlorine for each atom of
magnesium. Using the formula of the compound and the relative masses of elements from the periodic
table, we can then calculate the relative formula mass for the compound.
Formula Mass
The periodic table tells us the relative masses of all the elements. Looking at the squares for carbon and
helium, we can see that a carbon atom has about three times the mass of a helium atom. In this way, we
can compare the relative masses of any two atoms in the table. By looking at the chemical formulas and
the relative atomic masses, we can also compare the masses of different compounds. The formula mass
of a compound is the sum of all the atomic masses in the formula. The formula for water, for example,
is H2 O. This formula tells us that water is composed of hydrogen and oxygen and has a ratio of two
237
www.ck12.org
hydrogen atoms for each oxygen atom. We can determine the formula mass for water by adding up the
atomic masses of its components.
Example:
What is the formula mass of H2 O?
Solution:
Element Atomic Mass Number of Atoms
per Formula
H
1.0 daltons
2
O
16.0 daltons
1
Product
2.0 daltons
16.0 daltons
18.0 daltons
The formula mass of H2 O = 18.0 daltons.
Example:
What is the formula mass of Ca(NO3 )2 ?
Solution:
Element Atomic Mass Number of Atoms
per Formula
Ca
40.0 daltons
1
N
14.0 daltons
2
O
16.0 daltons
6
Product
40.0 daltons
28.0 daltons
96.0 daltons
164.0 daltons
The formula mass of Ca(NO3 )2 = 164.0 daltons.
These formula masses are in the same units as atomic masses and therefore are exactly comparable. For
example, the atomic mass of an oxygen atom is 16 daltons, the atomic mass of a fluorine atom is 19 daltons,
and the formula mass of a water molecule is 18 daltons. This means that a water molecule is slightly more
massive than an oxygen atom and slightly less massive than a fluorine atom.
Terminology
Since ionic compounds do not form individual molecules, the term formula mass is the only proper term
to describe this relative mass. In comparison, the formula mass for covalent compounds may also be called
the molecular mass because covalent compounds do form molecules. While it is important to understand
this distinction, no professional chemists have had their degrees in chemistry recalled when they referred
to the “molecular mass” of NaCl due to a slip of the tongue.
Lesson Summary
• The molecular mass of a molecule is found by adding the atomic masses of all the atoms in one
molecule.
• Not all substances exist as molecules, so the term molecular mass is not used for all substances. The
masses of ionic compounds and empirical formulas are called formula mass.
Further Reading / Supplemental Links
This website has lessons, worksheets, and quizzes on various high school chemistry topics. Lesson 5-6 is
on molecular and formula mass.
www.ck12.org
238
• http://www.fordhamprep.org/gcurran/sho/sho/lessons/lesson56.htm
Review Questions
1. Calculate the formula mass for each of the following.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
K2 SO4
CuO
Mg3 (AsO4 )2
Ca3 (PO4 )2
Fe2 O3
Al(OH)3
(NH4 )2 S
C12 H22 O11
2. On average, how many times heavier are bromine atoms than neon atoms?
3. An unknown element, M, combines with oxygen to form a compound with a formula of MO2 . If
25.0 grams of the unknown element combines with 4.50 grams of oxygen, what is the atomic mass of
M?
12.2 The Mole
Lesson Objectives
The student will:
• express the value of Avogadro’s number and explain its relevance to moles.
• use Avogadro’s number to convert to moles given the number of particles of a substance, and vice
versa.
• use the molar mass to convert to grams given the number of moles of a substance, and vice versa.
Vocabulary
• Avogadro’s number
• molar mass
• mole
Introduction
When objects are very small, it is often inconvenient, inefficient, or even impossible to deal with one object
at a time. For these reasons, we often deal with groups of small objects and even have names for these
groups. The most common of these is dozen, which refers to 12 objects. We frequently buy objects in
groups of twelve, like doughnuts or pencils. Even smaller objects, such as straight pins or staples, are
usually sold in boxes of 144, or a dozen dozen. A group of 144 objects is called a gross.
The problem of trying to deal with very small things individually also occurs in chemistry. Atoms and
molecules are too small to see, let alone to count or measure. Chemists needed to select a group of atoms
or molecules that would be convenient to operate with.
239
www.ck12.org
This video introduces the concept of the mole (3b, 3c): http://www.youtube.com/watch?v=AsqEkF7hcII
(9:44).
Figure 12.1: (Watch Youtube Video)
http://www.ck12.org/flexbook/embed/view/381
Avogadro’s Number
In chemistry, it is nearly impossible to deal with a single atom or molecule. Instead, chemists have defined a
group of particles that is convenient to work with. Since molecules are extremely small, you might suspect
that the number of particles in this group is going to be very large. In fact, the number of particles in this
group is 6.02 × 1023 particles, and the name of this group is the mole (abbreviated as mol). The number
6.02 × 1023 is called Avogadro’s number and is symbolized as the capital letter N. Although Italian scientist
Amedeo Avogadro did not determine this number, but the number was named in honor of him. One mole
of any object contains an Avogadro’s number, or 6.02 × 1023 , of those objects.
The number of objects in a mole was not chosen arbitrarily. Instead, there is a very particular reason
that the number 6.02 × 1023 was chosen. When chemists carry out chemical reactions, it is important to
understand the relationship between the numbers of particles of each element involved in the reaction.
Chemists realized that they obtained equal numbers of particles when they used one atomic or molecular
mass in grams of the substances. By looking at the atomic masses on the periodic table, chemists knew
that the mass ratio of one carbon atom to one sulfur atom was 12 amu to 32 amu. If they measured out
one atomic mass in grams of both substances (in other words, 12 grams of carbon and 32 grams of sulfur),
they would have the same number of atoms of each element. They didn’t know how many atoms were in
each pile, but they knew the number in each pile had to be the same. This logic is the same as knowing
that if a basketball has twice the mass of a soccer ball, then 100 lbs of basketballs and 50 lbs of soccer
balls both contain the same number of balls.
This amount of substance (its molecular mass in grams) became known as a gram-molecular mass. One
gram-molecular mass of any substance had the same number of particles in it. Years later, when it became
possible to count particles using electrochemical reactions, the number of particles in a gram-molecular
mass was counted. That number turned out to be 6.02 × 1023 particles. This number of particles continued
to be called a gram-molecular mass for many years, but eventually the name was changed to the mole.
The mole is defined so that 1.00 mole of carbon-12 atoms has a mass of 12.0 grams and contains 6.02 ×
1023 atoms. Likewise, 1.00 mole of water has a mass of 18.0 grams and contains 6.02 × 1023 molecules. One
mole of any element or compound has a mass equal to its molecular mass in grams and contains 6.02 × 1023
particles. The mass in grams of 6.02 × 1023 particles of a substance is now called the molar mass (mass
of 1.00 mole).
www.ck12.org
240
Converting Molecules to Moles and Vice Versa
We now know that because the mass of a single molecule of H2 SO4 is 98 daltons, the mass of an Avogadro’s
number of H2 SO4 molecules is 98 grams. We can use this information to find the mass in grams of
a single H2 SO4 molecule because we know that 98 grams contains 6.02 × 1023 molecules. If we divide
6.02 × 1023 molecules into 98 grams, we will get the mass of a single H2 SO4 molecule in grams. After
performing this calculation, we would obtain an answer of 1.6 × 10−22 grams/molecule – tiny, indeed. If we
are given a number of molecules of a substance, we can convert it into moles by dividing by Avogadro’s
number, and vice versa.
Example:
How many moles are present in 1, 000, 000, 000 (1 billion or 1 × 109 ) molecules of water?
Solution:
moles = (1, 000, 000, 000 molecules) ·
(
1 mole
6.02×1023 molecules
)
= 1.7 × 10−15 moles
You should note that this amount of water is too small for even our most delicate balances to determine
the mass. A very large number of molecules must be present before the mass is large enough to be detected
with our balances.
Example:
How many molecules are present in 0.00100 mole?
Solution:
molecules = (0.00100 mole) ·
(
6.02×1023 molecules
1 mole
)
= 6.02 × 1020 molecules
Unlike the previous example, the mass of 602, 000, 000, 000, 000, 000, 000 can be measured with a balance.
Converting Grams to Moles and Vice Versa
We can also convert back and forth between grams of substance and moles. The conversion factor for this
is the molar mass of the substance. To convert the grams of a substance into moles, we divide by the molar
mass. To convert the moles of a substance into grams, we multiply by the molar mass.
Example:
How many moles are present in 108 grams of water?
Solution:
moles =
grams
molar mass
=
108 grams
18.0 grams/mole
= 6.00 moles
Example:
What is the mass of 7.50 moles of CaO?
Solution:
grams = (moles) · (molar mass) = (7.50 moles) · (56.0 grams/mole) = 420. grams
241
www.ck12.org
This video tells a story to help students remember the easy way to perform mole conversions (3d): http:
//www.youtube.com/watch?v=Kg-zaG0ckVg (2:52).
Figure 12.2: (Watch Youtube Video)
http://www.ck12.org/flexbook/embed/view/382
Lesson Summary
• There are 6.02 × 1023 particles in 1.00 mole. The number 6.02 × 1023 is called Avogadro’s number.
• The number of moles in a given number of molecules of a substance can be found by dividing the
number of molecules by Avogadro’s number.
• The number of moles in a given mass of substance can be found by dividing the mass by the formula
mass expressed in grams/mole.
Further Reading / Supplemental Links
The learner.org website allows users to view streaming videos of the Annenberg series of chemistry videos.
You are required to register before you can watch the videos, but there is no charge to register. The website
has a video that apply to this lesson called “The Mole.”
• http://learner.org/resources/series61.html
This website has lessons, worksheets, and quizzes on various high school chemistry topics. Lesson 9-1 is
on the mole.
• http://www.fordhamprep.org/gcurran/sho/sho/lessons/lesson91.htm
The website below reviews how to calculate the molar mass of chemical compounds.
• http://misterguch.brinkster.net/molarmass.html
Review Questions
1. Convert the following to moles.
(a) 60.0 grams of NaOH
(b) 2.73 grams of NH4 Cl
(c) 5.70 grams of ZrF4
www.ck12.org
242
(d) 10.0 grams of PbO2
2. Convert the following to grams.
(a)
(b)
(c)
(d)
0.100 moles of CO2
0.437 moles of NaOH
0.500 moles of (NH4 )2 CO3
3.00 moles of ZnCl2
3. How many molecules (formula units for ionic compounds) are present in the following quantities?
(a) 0.250 mole of H2 O
(b) 6.00 moles of H2 SO4
(c) 0.00450 mole of Al2 (CO3 )3
4. How many moles are present in the following quantities?
(a) 1.00 × 1020 molecules of H2 O
(b) 1.00 × 1025 molecules of H2
(c) 5, 000, 000, 000, 000 atoms of carbon
5. How many molecules, atoms, or formula units are present in the following masses?
(a) 1.00 gram of Na2 CO3
(b) 8.00 grams of helium
(c) 1000. grams of H2 O
6. Convert the following to grams.
(a) 1.00 × 1023 molecules of H2
(b) 1.00 × 1024 formula units of AlPO4
(c) 1.00 × 1022 formula units of NaOH
7. What is the mass of a single atom of silver, Ag, in grams?
8. If you determined that the volume of a silver bar is 100. mL, how many atoms of silver would be in
the bar? The density of silver is 10.5 g/mL.
12.3 Percent Composition
Lesson Objectives
The student will:
• calculate the percent composition by mass given the masses of elements in a compound.
• calculate the percent composition by mass given the formula or name of a compound.
Vocabulary
• percent composition
Introduction
Metals useful to man are typically extracted from ore. The ore is removed from the mine and partially
purified by washing away dirt and other materials not chemically bound to the metal of interest. This
partially purified ore is then treated chemically in smelters or other purifying processes to separate pure
243
www.ck12.org
metal from the other elements. The value of the original ore is very dependent on how much pure metal can
eventually be separated from it. High-grade ore and low-grade ore command significantly different prices.
The ore can be evaluated before it is mined or smelted to determine what percent of the ore can eventually
become pure metal. This process involves determining what percentage of the ore is metal compounds and
what percentage of the metal compounds is pure metal.
Percent Composition from Masses
Compounds are made up of two or more elements. The law of definite proportions tells us that the
proportion, by mass, of elements in a compound is always the same. Water, for example, is always 11%
hydrogen and 89% oxygen by mass. The percentage composition of a compound is the percentage by
mass of each element in the compound.
Percentage composition can be determined experimentally. To do this, a known quantity of a compound is
decomposed in the laboratory. The mass of each element is measured and then divided by the total mass
of the original compound. This tells us what fraction of the compound is made up of that element. The
fraction can then be multiplied by 100% to convert it into a percent.
Example:
Laboratory procedures show that 50.0 grams of ammonia, NH3 , yields 41.0 grams of nitrogen and 9.00 grams
of hydrogen upon decomposition. What is the percent composition of ammonia?
Solution:
% nitrogen =
( 41.0
% hydrogen =
)
grams
grams )· (100%) = 82%
(50.0
9.00 grams
50.0 grams · (100%) = 18%
Example:
The decomposition of 25.0 grams of Ca(OH)2 in the lab produces 13.5 grams of calcium, 10.8 grams of
oxygen, and 0.68 grams of hydrogen. What is the percent composition of calcium hydroxide?
Solution:
(
)
grams
% calcium = 13.5
· (100%) = 54.0%
25.0
grams
( 10.8 grams )
% oxygen = 25.0 grams · (100%) = 43.2%
(
)
grams
% hydrogen = 0.68
25.0 grams · (100%) = 2.8%
You should note that the sum of the percentages always adds to 100%. Sometimes, the sum may total to
99% or 101% due to rounding, but if it totals to 96% or 103%, you have made an error.
Percent Composition from the Formula
Percent composition can also be calculated from the formula of a compound. Consider the formula for
the compound iron(III) oxide, Fe2 O3 . The percent composition of the elements in this compound can be
calculated by dividing the total atomic mass of the atoms of each element in the formula by the formula
mass.
Example:
What is the percent composition of iron(III) oxide, Fe2 O3 ?
Solution:
www.ck12.org
244
Element Atomic Mass Number of Atoms
Product
per Formula
Fe
55.8 daltons
2
111.6 daltons
O
16.0 daltons
3
48.0 daltons
Formula mass = 159.6 daltons
(
)
daltons
· (100%) = 69.9%
% iron = 111.6
159.6
( daltons )
48.0 daltons
% oxygen = 159.6 daltons · (100%) = 30.1%
Example:
What is the percent composition of aluminum sulfate, Al2 (SO4 )3 ?
Solution:
The formula mass of Al2 (SO4 )3 is: 2 · (27.0 daltons) + 3 · (32.0 daltons) + 12 · (16.0 daltons) = 342.0 daltons
)
(
daltons
· (100%) = 15.8%
% aluminum = 54.0
342
daltons
(
)
daltons
% sulfur = 96.0
· (100%) = 28.1%
(342 daltons )
192 daltons
% oxygen = 342 daltons · (100%) = 56.1%
Lesson Summary
• The percent composition of a compound is the percent of the total mass contributed by each element
in the compound.
• Percent composition can be determined either from the masses of each element in the compound or
from the formula of the compound.
Further Reading / Supplemental Links
This website has solved example problems for a number of topics covered in this lesson, including the
calculation of percent composition by mass.
• http://www.sciencejoywagon.com/chemzone/05chemical-reactions/
This website has lessons, worksheets, and quizzes on various high school chemistry topics. Lesson 5-8 is
on percent composition.
• http://www.fordhamprep.org/gcurran/sho/sho/lessons/lesson58.htm
The website below reviews how to calculate percent composition.
• http://www.ausetute.com.au/percentc.html
Review Questions
Determine the percent composition of the following compounds.
1.
2.
3.
4.
5.
BF3
Ca(C2 H3 O2 )2
FeF3
CrCl3
(NH4 )3 PO4
245
www.ck12.org
12.4 Empirical and Molecular Formulas
Lesson Objectives
The student will:
• reduce molecular formulas to empirical formulas.
• determine the empirical formula of a compound given either percent composition or masses.
• determine the molecular formula of a compound given either percent composition and molar mass or
masses.
Vocabulary
• molecular formula
Introduction
The empirical formula is the simplest ratio of atoms in a compound. Formulas for ionic compounds are
always empirical formulas, but for covalent compounds, the empirical formula is not always the actual
formula for the molecule. Molecules such as benzene, C6 H6 , would have an empirical formula of CH.
Finding Empirical Formula from Experimental Data
Empirical formulas can be determined from experimental data or from percent composition. Consider the
following example.
Example:
We find that a 2.50 gram sample of a compound contains 0.900 grams of calcium and 1.60 grams of chlorine.
The compound contains only these two elements. We can calculate the number of moles of calcium and
chlorine atoms in the compound. We can then find the molar ratio of calcium atoms to chlorine atoms.
From this, we can determine the empirical formula.
Solution:
First, we convert the mass of each element into moles.
0.900 g
= 0.0224 mol Ca
40.1 g/mol
1.60 g
atoms = 35.5
= 0.0451 mol
g/mol
moles of Ca =
moles of Cl
Cl
At this point, we have the correct ratio for the atoms in the compound. The formula Ca0.0224 Cl0.0451 ,
however, isn’t acceptable. We need to find the simplest whole number ratio. To find a simple whole
number ratio for these numbers, we divide each of them by the smaller number.
Ca =
0.0224
= 1.00
0.0224
Cl =
0.0451
= 2.01
0.0224
Now, we can see the correct empirical formula for this compound is CaCl2 .
www.ck12.org
246
Finding Empirical Formula from Percent Composition
When finding the empirical formula from percent composition, the first thing to do is to convert the
percentages into masses. For example, suppose we are given that the percent composition of a compound
as 40.0% carbon, 6.71% hydrogen, and 53.3% oxygen. Since every sample of this compound will have the
same composition in terms of the ratio of atoms, we could choose a sample of any size. Suppose we choose
a sample size of 100. grams. The masses of each of the elements in this sample will be 40.0 grams of
carbon, 6.71 grams of hydrogen, and 53.3 grams of oxygen. These masses can then be used to find the
empirical formula. You could use any size sample, but choosing a sample size of 100. grams is usually most
convenient because it makes the arithmetic simple.
Example:
Find the empirical formula of a compound whose percent composition is 40.0% carbon, 6.71% hydrogen,
and 53.3% oxygen.
Solution:
We choose a sample size of 100. grams and multiply this 100. gram sample by each of the percentages to get
masses for each element. This would yield 40.0 grams of carbon, 6.71 grams of hydrogen, and 53.3 grams
of oxygen. The next step is to convert the mass of each element into moles.
moles of C =
moles of H =
moles of O =
40.0 g
12.0 g/mol
6.71 g
1.01 g/mol
53.3 g
16.0 g/mol
= 3.33 moles C
= 6.64 moles H
= 3.33 mole Ca
Then, we divide all three numbers by the smallest one to get simple whole number ratios:
C=
3.33
=1
3.33
6.64
=2
3.33
H=
O=
3.33
=1
3.33
Finally, we can write the empirical formula CH2 O.
Sometimes, this technique of dividing each of the moles by the smallest number does not yield whole
numbers. Whenever the subscript for any element in the empirical formula is 1, dividing each of the moles
by the smallest will yield a simple whole number ratio, but if none of the elements in the empirical formula
has a subscript of 1, then this technique will not yield a simple whole number ratio. In those cases, a little
more work is required.
Example:
Determine the empirical formula for a compound that is 66.0% calcium and 34.0% phosphorus.
Solution:
We choose a sample size of 100. grams and multiply the 100. grams by the percentage of each element to
get masses. This yields 66.0 grams of calcium and 34.0 grams of phosphorus. We then divide each of these
masses by their molar mass to convert the masses into moles:
66.0 g
= 1.65 moles Ca
40.1 g/mol
34.0 g
= 1.10 moles P
31.0 g/mol
moles of Ca =
moles of P =
We then divide each of these moles by the smallest.
Ca =
1.65
= 1.50
1.10
P=
247
1.10
= 1.00
1.10
www.ck12.org
In this case, dividing each of the numbers by the smallest one does not yield a simple whole number ratio.
In such a case, we must multiply both numbers by some factor that will produce a whole number ratio.
If we multiply each of these by 2, we get a whole number ratio of 3 Ca to 2 P. Therefore, the empirical
formula is Ca3 P2 .
Finding Molecular Formulas
Empirical formulas show the simplest whole number ratio of the atoms in a compound. Molecular formulas
show the actual number of atoms of each element in a compound. When you find a empirical formula from
either masses of elements or from percent composition, you are finding the empirical formula. For the
compound N2 H4 , you will get an empirical formula of NH2 , and for C3 H6 , you will get CH2 . If we want
to determine the actual molecular formula, we need one additional piece of information. The molecular
formula is always a whole number multiple of the empirical formula. In order to get the molecular formula
for N2 H4 , you must double each of the subscripts in the empirical formula. Since the molecular formula
is a whole number multiple of the empirical formula, the molecular mass will be the same whole number
multiple of the formula mass. The formula mass for NH2 is 16 g/mol, and the molecular mass for N2 H4
is 32 g/mol. When we have the empirical formula and the molecular mass for a compound, we can divide
the formula mass into the molecular mass and find the whole number that we need to multiply each of the
subscripts in the empirical formula.
Example:
Find the molecular formula for a compound with percent composition of 40.0% carbon, 67.1% hydrogen,
and 53.3% oxygen. The molecular mass of the compound is 180 g/mol.
Solution:
This is the same as an earlier example, except now we also have the molecular mass of the compound.
Earlier, we determined the empirical formula of this compound to be CH2 O. The empirical formula has
a formula mass of 30.0 g/mol. In order to find the molecular formula for this compound, we divide the
formula mass into the molecular mass (180 divided by 30) and find the multiplier for the empirical formula
to be 6. As a result, the molecular formula for this compound will be C6 H12 O6 .
Example:
Find the molecular formula for a compound with percent composition of 85.6% carbon and 14.5% hydrogen.
The molecular mass of the compound is 42.1 g/mol.
Solution:
We choose a sample size of 100. g and multiply each element percentage to get masses for the elements in
this sample. This yields 85.6 g of C and 14.5 g of H. Dividing each of these by their atomic mass yields
7.13 moles of C and 14.4 moles of H. Dividing each of these by the smallest yields a whole number ratio of
1 carbon to 2 hydrogen. Thus, the empirical formula will be CH2 .
The formula mass of CH2 is 14 g/mol. Dividing 14 g/mol into the molecular mass of 42.1 g/mol yields a
multiplier of 3. The molecular formula will be C3 H6 .
Lesson Summary
• The empirical formula of a compound indicates the simplest whole number ratio of atoms present in
the compound.
• The empirical formula of a compound can be calculate from the masses of the elements in the
compound or from the percent composition.
www.ck12.org
248
• The molecular formula of a compound is some whole number multiple of the empirical formula.
Further Reading / Supplemental Links
This website has solved example problems for a number of topics covered in this lesson, including the
determination of empirical and molecular formulas.
• http://www.sciencejoywagon.com/chemzone/05chemical-reactions/
This website has lessons, worksheets, and quizzes on various high school chemistry topics. Lesson 5-5 is
on empirical formulas.
• http://www.fordhamprep.org/gcurran/sho/sho/lessons/lesson55.htm
The website below reviews ow to calculate empirical formulas.
• http://www.chem.lsu.edu/lucid/tutorials/empiricalform.html
Review Questions
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
What is the empirical formula for C8 H18 ?
What is the empirical formula for C6 H6 ?
What is the empirical formula for WO2 ?
A compound has the empirical formula C2 H8 N and a molar mass of 46 g/mol. What is the molecular
formula of this compound?
A compound has the empirical formula C2 H4 NO. If its molar mass is 116.1 g/mol, what is the
molecular formula of the compound?
A sample of pure indium chloride with a mass of 0.5000 grams is found to contain 0.2404 grams of
chlorine. What is the empirical formula of this compound?
Determine the empirical formula of a compound that contains 63.0 grams of rubidium and 5.90 grams
of oxygen.
Determine the empirical formula of a compound that contains 58.0% Rb, 9.50% N, and 32.5% O.
Determine the empirical formula of a compound that contains 33.3% Ca, 40.0% O, and 26.7% S.
Find the molecular formula of a compound with percent composition 26.7% P, 12.1% N, and 61.2%
Cl and with a molecular mass of 695 g/mol.
Image Sources
All images, unless otherwise stated, are created by the CK-12 Foundation and are under the Creative
Commons license CC-BY-NC-SA.
249
www.ck12.org