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Resources
Chapter Presentation
Bellringer
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Sample Problems
Visual Concepts
Standardized Test Prep
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Chapter 7
The Mole and Chemical
Composition
Table of Contents
Section 1 Avogadro’s Number and Molar Conversions
Section 2 Relative Atomic Mass and Chemical
Formulas
Section 3 Formulas and Percentage Composition
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Chapter 7
Section 1 Avogadro’s Number and
Molar Conversions
Bellringer
• List as many common counting units as you can.
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Chapter 7
Section 1 Avogadro’s Number and
Molar Conversions
Objectives
• Identify the mole as the unit used to count particles,
whether atoms, ions, or molecules.
• Use Avogadro’s number to convert between amount
in moles and number of particles.
• Solve problems converting between mass, amount in
moles, and number of particles using Avogadro’s
number and molar mass.
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Chapter 7
Section 1 Avogadro’s Number and
Molar Conversions
Avogadro’s Number and the Mole
• The SI unit for amount is called the mole (mol). A
mole is the number of atoms in exactly 12 grams of
carbon-12.
• Scientists use the mole to make counting large
numbers of particles easier.
• The number of particles in a mole is called
Avogadro’s Number. Avogadro’s number is
6.02214199  1023 units/mole.
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Chapter 7
Section 1 Avogadro’s Number and
Molar Conversions
Avogadro’s Number and the Mole, continued
The Mole Is a Counting Unit
• The mole is used to count out a given number of
particles, whether they are atoms, molecules, formula
units, ions, or electrons.
• The mole is just one kind of counting unit:
• 1 dozen = 12 objects
• 1 hour = 3600 seconds
• 1 mole = 6.022  1023 particles
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Chapter 7
Visual Concepts
The Mole
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Chapter 7
Section 1 Avogadro’s Number and
Molar Conversions
Avogadro’s Number and the Mole, continued
Amount in Moles Can Be Converted to Number of
Particles
• Counting units are used to make conversion factors.
• The definition of one mole is
6.022  1023 particles = 1 mol
• The conversion factor is
6.022 × 1023 particles
 1
1 mol
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Chapter 7
Section 1 Avogadro’s Number and
Molar Conversions
Avogadro’s Number and the Mole, continued
Choose the Conversion Factor That Cancels the
Given Units
• All conversion factors are equal to 1, so you can use
them to convert among different units.
• You can tell which conversion factor to use, because
the needed conversion factor should cancel the units
of the given quantity to give you the units of the
answer or the unknown quantity.
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Chapter 7
Section 1 Avogadro’s Number and
Molar Conversions
Converting Between Amount in Moles and
Number of Particles
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Chapter 7
Section 1 Avogadro’s Number and
Molar Conversions
Converting Amount in Moles to Number of
Particles
Sample Problem A
Find the number of molecules in 2.5 mol of sulfur
dioxide.
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Chapter 7
Section 1 Avogadro’s Number and
Molar Conversions
Converting Amount in Moles to Number of
Particles
Sample Problem A Solution
2.5 mol SO2  ? = ? molecules SO2
You are converting from the unit mol to the unit
molecules. The conversion factor must have the units
of molecules/mol.
You use 6.022  1023 molecules/1 mol.
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Chapter 7
Section 1 Avogadro’s Number and
Molar Conversions
Converting Amount in Moles to Number of
Particles
Sample Problem A Solution, continued
2.5 mol SO2  ? = ? molecules SO2
6.022 × 1023 molecules SO2
2.5 mol SO2 ×
 ? molecules SO2
1 mol SO2
2.5 mol SO2 = 1.5  1024 molecules SO2
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Chapter 7
HW. 3,4
Sample Problem A, practice pg. 228
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Section 1 Avogadro’s Number and
Molar Conversions
Chapter 7
Avogadro’s Number and the Mole, continued
Number of Particles Can Be Converted to Amount
in Moles
• The reverse calculation is similar to that in Sample
Problem A but the conversion factor is inverted to get
the correct units in the answer.
• example: How many moles are 2.54  1022 iron(III)
ions?
2.54 × 10
22
ions Fe
3
1 mol Fe3
3
×

0.0422
mol
Fe
6.022  1023 ions Fe3
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Chapter 7
Section 1 Avogadro’s Number and
Molar Conversions
Converting Number of Particles to Amount
in Moles
Sample Problem B
A sample contains 3.01  1023 molecules of sulfur
dioxide, SO2. Determine the amount in moles.
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Chapter 7
Section 1 Avogadro’s Number and
Molar Conversions
Converting Number of Particles to Amount
in Moles
Sample Problem B Solution
3.01 × 1023 molecules SO2  ? = ? mol SO2
You are converting from the unit molecules to the unit
mol. The conversion factor must have the units of
mol/molecules.
You use 1 mol/6.022  1023 molecules.
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Chapter 7
Section 1 Avogadro’s Number and
Molar Conversions
Converting Number of Particles to Amount
in Moles
Sample Problem B Solution, continued
3.01  1023 molecules SO2  ? = ? mol SO2
3.01 1023 molecules SO2 ×
1 mol SO2
 ? mol SO2
6.022  1023 molecules SO2
3.01  1023 molecules SO2 = 0.500 mol SO2
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Chapter 7
HW. 5
Sample Problem B, practice pg. 229
3. A biologist estimates that there are 2.7  1017 termites
on Earth. How many moles of termites is this?
4. How many moles do 5.66  1025 lithium ions, Li , equal?

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Chapter 7
Section 1 Avogadro’s Number and
Molar Conversions
Molar Mass Relates Moles to Grams
Amount in Moles Can Be Converted to Mass
• The molar mass is the mass in grams of one mole of
an element or compound.
• Molar mass is numerically equal to the atomic mass
of monatomic elements and the formula mass of
compounds and diatomic elements.
• The units for molar mass are g/mol.
• Molar mass can be used as a conversion factor in
problems converting between mass and amount.
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Chapter 7
Visual Concepts
Molar Mass
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Chapter 7
Visual Concepts
Molar Mass as a Conversion Factor
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Chapter 7
Section 1 Avogadro’s Number and
Molar Conversions
Molar Mass Relates Moles to Grams, continued
The Mole Plays a Central Part in Chemical
Conversions
• To convert from number of particles to mass, you
must use a two-part process:
• First, convert number of particles to amount in moles.
• Second, convert amount in moles to mass in grams.
• One step common to many problems in chemistry is
converting to amount in moles.
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Chapter 7
Section 1 Avogadro’s Number and
Molar Conversions
Converting Between Mass, Amount, and
Number of Particles
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Chapter 7
Section 1 Avogadro’s Number and
Molar Conversions
Converting Number of Particles to Mass
Sample Problem C
Find the mass in grams of 2.44  1024 atoms of carbon,
whose molar mass is 12.01 g/mol.
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Chapter 7
Section 1 Avogadro’s Number and
Molar Conversions
Converting Number of Particles to Mass
Sample Problem C Solution
First part:
2.44  1024 atoms  ? = ? mol
Select the conversion factor that will take you from
number of atoms to amount in moles.
You use 1 mol/6.022  1023 atoms.
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Chapter 7
Section 1 Avogadro’s Number and
Molar Conversions
Converting Number of Particles to Mass
Sample Problem C Solution, continued
Second part:
? mol  ? = ? g
Select the conversion factor that will take you from
amount in moles to mass in grams.
You use the molar mass of carbon, 12.01 g C/1 mol.
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Chapter 7
Section 1 Avogadro’s Number and
Molar Conversions
Converting Number of Particles to Mass
Sample Problem C Solution, continued
2.44  10
24
1 mol
12.01 g C
atoms ×
×
23
6.022  10 atoms
1 mol
= 48.7 g C
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Chapter 7
HW. 3,4
Sample Problem C , Practice pg. 231
Find the mass in grams of
1. 2.11 . 1024 atoms of copper (molar mass of Cu  63.55
g/mol)
2. 3.01  1023 formula units of NaCl (molar mass of NaCl 
58.44 g/mol)
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Chapter 7
Section 1 Avogadro’s Number and
Molar Conversions
Molar Mass Relates Moles to Grams, continued
Mass Can Be Converted to Amount in Moles
• Converting from mass to number of particles is the
reverse of the operation in the previous problem.
• To convert from mass to number of particles, you
must use a two-part process:
• First, convert mass in grams to amount in
moles.
• Second, convert amount in moles to number
of particles.
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Chapter 7
Section 1 Avogadro’s Number and
Molar Conversions
Converting Mass to Number of Particles
Sample Problem D
Find the number of molecules present in
47.5 g of glycerol, C3H8O3. The molar mass of glycerol
is 92.11 g/mol.
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Chapter 7
Section 1 Avogadro’s Number and
Molar Conversions
Converting Mass to Number of Particles
Sample Problem D Solution
First part:
47.5 g  ? = ? mol
Select the conversion factor that will take you from
mass in grams to amount in moles.
You use the inverse of the molar mass of glycerol:
1 mol
92.11 g C3H8O3
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Chapter 7
Section 1 Avogadro’s Number and
Molar Conversions
Converting Mass to Number of Particles
Sample Problem D Solution, continued
Second part:
? mol  ? = ? molecules
Select the conversion factor that will take you from
amount in moles to number of particles.
You use
6.022  1023 molecules
.
1 mol
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Chapter 7
Section 1 Avogadro’s Number and
Molar Conversions
Converting Mass to Number of Particles
Sample Problem D Solution, continued
1 mol
6.022  1023 molecules
47.5 g C3H8O3 

92.11 g C3H8O3
1 mol
= 3.11  1023 molecules
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Chapter 7
Sample Problem D, practice pg. 233
1. Find the number of atoms in 237 g Cu (molar mass
of Cu  63.55 g/mol).
2. Find the number of ions in 20.0 g Ca (molar mass
of Ca  40.08 g/mol).
2
2
3. Find the number of atoms in 155 mol of arsenic.
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Chapter 7
Section 1 Avogadro’s Number and
Molar Conversions
Molar Mass Relates Moles to Grams, continued
Remember to Round Consistently
• Remember that an answer must never be given to
more significant figures than is appropriate.
• Round molar masses from the periodic table to two
significant figures to the right of the decimal point.
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Homework
Section Review, pg.
233, Qs. 1-13
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Section 7.1 Review pg. 233
1. What is the definition of a mole?
2. How many particles are there in one mole?
3. Explain how Avogadro’s number can give
two conversion factors.
4. Which will have the greater number of ions,
1 mol of nickel(II) or 1 mol of copper(I)?
5. Without making a calculation, is 1.11 mol Pt
more or less than 6.022 x10 atoms?
23
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Section 7.1 Review pg. 233
6. Find the number of molecules or ions.
a. 2.00 mol Fe3
7. Find the number of sodium ions, Na.
b. 3.00 mol Na4P2O7
8. Find the number of moles.
c. 5.610 . 1022 ions Na
9. Find the mass in grams.
a. 4.30  1016 atoms He, 4.00 g/mol
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Section 7.1 Review pg. 233
10. Find the number of molecules or ions
b. 3.5 g Cu2, 63.55 g/mol
11. What is the mass of 6.022  1023 molecules of ibuprofen
(molar mass of 206.31 g/mol)?
12. Find the mass in grams.
c. 1.842  1019 ions Na, 22.99 g/mol
13. Find the number of molecules.
a. 2.000 mol H2, 2.02 g/mol
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1) There are exactly 1000 mg in 1
g. A cup of hot chocolate has 35.0
mg of sodium ions, Na+. One cup
of milk has 290 mg of calcium
ions, Ca2+.
a. How many moles of sodium ions
are in the cup of hot chocolate?
b. How many moles of calcium
ions are in the milk?
3) How many moles of NaNO2
are there in a beaker that
contains 0.500 kg of NaNO2
(molar mass of NaNO2 = 69.00
g/mol)?
2) Cyclopentane has
the molecular formula
C5H10. How many
moles of hydrogen
atoms are there in 4
moles of
cyclopentane?
____________________
How many atoms of Fe
are in the formula
Fe3C? How many
moles of Fe are in one
mole of Fe3C
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Chapter 7
Section 2 Relative Atomic Mass and
Chemical Formulas
Bellringer
• Compare the masses of a roll of pennies and a roll of
dimes. Both contain 50 coins.
• Why are the masses of the rolls different when both
rolls contain the same number of coins?
• If given a roll of mixed coins, what information would
you need to determine the mass of the roll?
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Chapter 7
Section 2 Relative Atomic Mass and
Chemical Formulas
Objectives
• Use a periodic table or isotopic composition data to
determine the average atomic masses of elements.
• Infer information about a compound from its
chemical formula.
• Determine the molar mass of a compound from its
formula.
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Chapter 7
Section 2 Relative Atomic Mass and
Chemical Formulas
Average Atomic Mass and the Periodic Table
Most Elements are a Mixture of Isotopes
• Isotopes are atoms that have different numbers of
neutrons than other atoms of the same element do.
12C, 14C
potassium-39, potassium-41
chlorine-35, chlorine-37
234Np, 235Np, 236Np, 237Np, 238Np, 239Np
• Average atomic mass is a weighted average of the
atomic mass of an element’s isotopes.
If you know the abundance of each isotope, you can
calculate the average atomic mass of an element.
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Chapter 7
Visual Concepts
Average Atomic Mass
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Chapter 7
Section 2 Relative Atomic Mass and
Chemical Formulas
Calculating Average Atomic Mass
Sample Problem E
The mass of a Cu-63 atom is 62.94 amu, and that of a
Cu-65 atom is 64.93 amu. Using the data below, find
the average atomic mass of copper.
• abundance of Cu-63 = 69.17%
• abundance of Cu-65 = 30.83%
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Chapter 7
Section 2 Relative Atomic Mass and
Chemical Formulas
Calculating Average Atomic Mass
Sample Problem E Solution
The contribution of each isotope is equal to its atomic mass
multiplied by the fraction of that isotope.
• contribution of Cu-63:
• contribution of Cu-65:
62.94 amu × 0.6917
64.93 amu × 0.3083
Average atomic mass is the sum of the individual contributions:
(62.94 amu × 0.6917) + (64.93 amu × 0.3083)
= 63.55 amu
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Chapter 7
Sample Problem E
Practice p.236
1) Calculate the average atomic mass for gallium if
60.00% of its atoms have a mass of 68.926 amu and
40.00% have a mass of 70.925 amu.
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Chapter 7
Sample Problem E
Practice p.236
2)Calculate the average atomic mass of oxygen.
Its composition is 99.76% of atoms with a mass
of 15.99 amu, 0.038% with a mass of 17.00 amu,
and 0.20% with a mass 18.00 amu.
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HW.
SectionReview,p.240 # 8,9
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Chapter 7
Section 2 Relative Atomic Mass and
Chemical Formulas
Chemical Formulas and Moles
Formulas Express Composition
• A compound’s chemical formula tells you which
elements, as well as how much of each, are present
in a compound.
• Formulas for covalent compounds show the elements
and the number of atoms of each element in a
molecule.
• Formulas for ionic compounds show the simplest ratio
of cations and anions in any pure sample.
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Chapter 7
Section 2 Relative Atomic Mass and
Chemical Formulas
Chemical Formulas and Moles, continued
Formulas Express Composition, continued
• Any sample of compound has many atoms and ions,
and the formula gives a ratio of those atoms or ions.
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Chapter 7
Section 2 Relative Atomic Mass and
Chemical Formulas
Chemical Formulas and Moles, continued
Formulas Give Ratios of Polyatomic Ions
• Formulas for polyatomic ions show the simplest ratio
of cations and anions.
• They also show the elements and the number of
atoms of each element in each ion.
• For example, the formula KNO3 indicates a ratio of

one K+ cation to one NO3 anion.
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Chapter 7
Section 2 Relative Atomic Mass and
Chemical Formulas
Understanding Formulas for Polyatomic Ionic
Compounds
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Chapter 7
Section 2 Relative Atomic Mass and
Chemical Formulas
Chemical Formulas and Moles, continued
Formulas Are Used to Calculate Molar Masses
• The molar mass of a molecular compound is the sum
of the masses of all the atoms in the formula
expressed in g/mol.
• The molar mass of an ionic compound is the sum of
the masses of all the atoms in the formula expressed
in g/mol.
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Chapter 7
Visual Concepts
Formula Mass
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Chapter 7
Section 2 Relative Atomic Mass and
Chemical Formulas
Calculating Molar Mass for Ionic Compounds
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Chapter 7
Section 2 Relative Atomic Mass and
Chemical Formulas
Calculating Molar Mass of Compounds
Sample Problem F
Find the molar mass of barium nitrate, Ba(NO3)2.
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Chapter 7
Section 2 Relative Atomic Mass and
Chemical Formulas
Calculating Molar Mass of Compounds
Sample Problem F Solution
Find the number of moles of each element in 1 mol
Ba(NO3)2:
•
Each mole has 1 mol Ba, 2 mol N, and 6 mol O.
Use the periodic table to find the molar mass of each
element in the formula:
• molar mass of Ba: 137.33 g/mol
• molar mass of N: 14.01 g/mol
• molar mass of O: 16.00 g/mol
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Chapter 7
Section 2 Relative Atomic Mass and
Chemical Formulas
Calculating Molar Mass of Compounds
Sample Problem F Solution, continued
Multiply the molar mass of each element by the
number of moles of each element. Add these masses
to get the total molar mass of Ba(NO3)2.
mass of 1 mol Ba = 1  137.33 g/mol = 137.33 g/mol
mass of 2 mol N = 2  14.01 g/mol = 28.02 g/mol
+ mass of 6 mol O = 6  16.00 g/mol = 96.00 g/mol
molar mass of Ba(NO3)2 = 261.35 g/mol
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Chapter 7
HW. #3,4
Sample Problem F
Practice p.239
2) Write the formula and then find the molar mass.
a. sodium hydrogen carbonate
b. cerium hexaboride
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c. magnesium perchlorate
d. aluminum sulfate
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e. iron(III) hydroxide
f. tin(II) chloride
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g. tetraphosphorus decoxide
h. iodine monochloride
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Chapter 7
Section 2 Relative Atomic
Mass and Chemical Formulas
HW.
SectionReview,p.240
# (1-7),10( d, e),
11( b, d), 12(a, b)
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Section 7.2 Review, pg. 240
1. What is a weighted average?
It is an average that takes into account both the value and the
frequency of each number involved.
2. On
the periodic table, the average atomic mass of
carbon is 12.01 g. Why is it not exactly 12.00?
Some atoms of carbon have a mass greater than 12.00, so they
raise the average atomic mass.
3. What
Cs2CO3
is the simplest formula for cesium carbonate?
4. What
ions are present in cesium carbonate?
Two Cs+ ions for every Co3 2- ion
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5. What
is the ratio of N and H atoms in NH ?
3
One nitrogen atom: 3 hydrogen atoms
6. What
is the ratio of calcium and chloride ions in
CaCl ?
2
One calcium ion : 2 chloride ions
7. Why
is the simplest formula used to determine
the molar mass for ionic compounds
An ionic compound is composed of positive and negative
ions in an infinite crystal. The simplest formula reflects the
specific ratio of positive and negative ions
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Section 7.2 Review, pg. 240
8) Calculate the average atomic mass of
chromium. Its composition is: 83.79% with
a mass of 51.94 amu; 9.50% with a mass of
52.94 amu; 4.35% with a mass of 49.95 amu;
2.36% with a mass of 53.94 amu. (51.99 amu)
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Section 7.2 Review, pg. 240
9) Element X has two isotopes. One has a
mass of 10.0 amu and an abundance of
20.0%.The other has a mass of 11.0 amu and
an abundance of 80.0%. Estimate the average
atomic mass. What element is it? (10.80, boron)
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Section 7.2 Review, pg. 240
10. Find the molar mass.
d. (NH4)2HPO4
e. C2H5NO2
(132.08g/mol)
( 75.08g/mol)
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Section 7.2 Review, pg. 240
11. Determine the formula, the molar mass, and the
number of moles in 2.11 g of each of the following
compounds.
b. phosphorus trifluoride
d. mercury(II) bromate
( PF3, 87.97g/mol, 2.40E-2 mol)
( Hg(BrO3)2, 456.39g/mol, 4.62E-3mol)
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12. Find the molar mass and the mass of 5.0000 mol of
each of the following compounds.
a. calcium acetate, Ca(C2H3O2)2 (158.18g/mol, 790.90 g)
b. iron(II) phosphate, Fe3(PO4)2
(357.49g/mol, 1787.4g)
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Chapter 7
Section 3 Formulas and Percentage
Composition
Bellringer
• Brainstorm a list of what you know about
percentages.
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Chapter 7
Section 3 Formulas and Percentage
Composition
Objectives
• Determine a compound’s empirical formula from its
percentage composition.
• Determine the molecular formula or formula unit of a
compound from its empirical formula and its formula
mass.
• Calculate percentage composition of a compound
from its molecular formula or formula unit.
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Chapter 7
Section 3 Formulas and Percentage
Composition
Using Analytical Data
• The percentage composition is the percentage by
mass of each element in a compound.
• Percentage composition helps verify a substance’s
identity.
• Percentage composition also can be used to compare
the ratio of masses contributed by the elements in two
different substances.
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Chapter 7
Percentage Composition of Iron Oxides
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Chapter 7
Section 3 Formulas and Percentage
Composition
Using Analytical Data, continued
Determining Empirical Formulas
• An empirical formula is a chemical formula that
shows the simplest ratio for the relative numbers and
kinds of atoms in a compound.
• An actual formula shows the actual ratio of elements
or ions in a single unit of a compound.
• For example, the empirical formula for ammonium
nitrate is NH2O, while the actual formula is NH4NO2.
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Chapter 7
Empirical and
Actual Formulas
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Chapter 7
Section 3 Formulas and Percentage
Composition
Using Analytical Data, continued
Determining Empirical Formulas, continued
•
You can use the percentage composition for a
compound to determine its empirical formula.
1. Convert the percentage of each element to g.
2. Convert from g to mol using the molar mass of
each element as a conversion factor.
3. Compare these amounts in mol to find the
simplest whole-number ratio among the
elements.
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Chapter 7
Visual Concepts
Percentage Composition
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Chapter 7
Section 3 Formulas and Percentage
Composition
Determining an Empirical Formula from
Percentage Composition
Sample Problem G
Chemical analysis of a liquid shows that it is 60.0% C,
13.4% H, and 26.6% O by mass. Calculate the
empirical formula of this substance.
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Chapter 7
Section 3 Formulas and Percentage
Composition
Determining an Empirical Formula from
Percentage Composition
Sample Problem G Solution
Assume that you have a 100.0 g sample, and convert
the percentages to grams.
for C:
60.0%  100.0 g = 60.0 g C
for H:
13.4%  100.0 g = 13.4 g H
for O:
26.6%  100.0 g = 26.6 g O
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Chapter 7
Section 3 Formulas and Percentage
Composition
Determining an Empirical Formula from
Percentage Composition
Sample Problem G Solution, continued
Convert the mass of each element into the amount in
moles, using the reciprocal of the molar mass.
1 mol C
60.0 g C ×
 5.00 mol C
12.01 g C
1 mol H
13.4 g H ×
 13.3 mol H
1.01 g H
26.6 g O ×
1 mol O
 1.66 mol O
16.00 g O
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Chapter 7
Section 3 Formulas and Percentage
Composition
Determining an Empirical Formula from
Percentage Composition
Sample Problem G Solution, continued
The formula can be written as C5H13.3O1.66, but you
divide by the smallest subscript to get whole numbers.
5.00 mol C
1.66
 3.01 mol C
13.3 mol H
1.66
 8.01 mol H
1.66 mol O
1.66
 1.00 mol O
The empirical formula
is C3H8O.
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Chapter 7
Section 3 Formulas and
Percentage Composition
Sample Problem G , practice p. 243
Determine the empirical formula for each substance.
1) A dead alkaline battery is found to contain a
compound of Mn and O. Its analysis gives 69.6%
Mn and 30.4% O.
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Chapter 7
Sample Problem G , practice p. 243
2) A compound is 38.77% Cl and 61.23% O.
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Chapter 7
Section 3 Formulas and
Percentage Composition
HW.
Practice 3,4;
p.243
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Chapter 7
Section 3 Formulas and Percentage
Composition
Using Analytical Data, continued
Molecular Formulas Are Multiples of Empirical
Formulas
• The formula for an ionic compound shows the
simplest whole-number ratio of the large numbers of
ions in a crystal of the compound.
• A molecular formula is a whole-number multiple of
the empirical formula.
• The molar mass of any compound is equal to the
molar mass of the empirical formula times a whole
number, n.
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Chapter 7
Visual Concepts
Comparing Molecular and Empirical Formulas
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Chapter 7
Section 3 Formulas and Percentage
Composition
Determining a Molecular Formula from an
Empirical Formula
Sample Problem H
The empirical formula for a compound is P2O5. Its
experimental molar mass is 284 g/mol. Determine the
molecular formula of the compound.
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Chapter 7
Section 3 Formulas and Percentage
Composition
Determining a Molecular Formula from an
Empirical Formula
Sample Problem H Solution
Find the molar mass of the empirical formula P2O5.
+
2  molar mass of P = 61.94 g/mol
5  molar mass of O = 80.00 g/mol
molar mass of P2O5 = 141.94 g/mol
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Chapter 7
Section 3 Formulas and Percentage
Composition
Determining a Molecular Formula from an
Empirical Formula
Sample Problem H Solution, continued
exp erimental molar mass of compound
n 
molar mass of empirical formula
n 
284 g / mol
 2.00
141.94 g / mol
n (empirical formula) = 2 (P2O5) = P4O10
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Chapter 7
Section 3 Formulas and
Percentage Composition
Sample Problem H , practice p. 245
1 ) A compound has an experimental molar mass of
78 g/mol. Its empirical formula is CH. What is its
molecular formula?
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HW.
Practice 2,3; p.245
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Sample Problem H , practice p. 245
2) A compound has the empirical formula CH O. Its
experimental molar mass is 90.0 g/mol. What is its
molecular formula?
2
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Sample Problem H , practice p. 245
3) A brown gas has the empirical formula NO . Its
experimental molar mass is 46 g/mol. What is its
molecular formula?
2
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Chapter 7
Section 3 Formulas and Percentage
Composition
Using Analytical Data, continued
Chemical Formulas Can Give Percentage Composition
• If you know the chemical formula of any compound,
then you can calculate the percentage composition.
• From the subscripts, determine the mass contributed
by each element and add these to get molar mass.
• Divide the mass of each element by the molar mass.
• Multiply by 100 to find the percentage composition of
that element.
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Chapter 7
Section 3 Formulas and Percentage
Composition
Using Analytical Data, continued
Chemical Formulas Can Give Percentage Composition
• CO and CO2 are both made up of C and O, but they
have different percentage compositions.
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Chapter 7
Section 3 Formulas and Percentage
Composition
Using a Chemical Formula to Determine
Percentage Composition
Sample Problem I
Calculate the percentage composition of copper(I)
sulfide, Cu2S, a copper ore called chalcocite.
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Chapter 7
Section 3 Formulas and Percentage
Composition
Using a Chemical Formula to Determine
Percentage Composition
Sample Problem I Solution
Find the molar mass of Cu2S.
+
2 mol  63.55 g Cu/mol = 127.10 g Cu
1 mol  32.07 g S/mol = 32.07 g S
molar mass of Cu2S = 159.17 g/mol
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Chapter 7
Section 3 Formulas and Percentage
Composition
Using a Chemical Formula to Determine
Percentage Composition
Sample Problem I Solution, continued
Calculate the fraction that each element contributes to
the total mass by substituting the masses into the
equations below and rounding correctly.
mass % Cu 
mass of 2 mol Cu
× 100 
molar mass of Cu2S
127.10 g Cu
× 100 
159.17 Cu2S
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79.852% Cu
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Chapter 7
Section 3 Formulas and Percentage
Composition
Using a Chemical Formula to Determine
Percentage Composition
Sample Problem I Solution, continued
mass of 1 mol S
mass % S 
× 100 
molar mass of Cu2S
32.07 g S
× 100 
159.17 g Cu2S
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20.15% S
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Chapter 7
Section 3 Formulas and
Percentage Composition
Sample Problem I, practice p.248
1) Calculate the percentage composition of Fe3C, a
compound in cast iron.
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Chapter 7
Sample Problem I, practice p.248
2) Calculate the percentage of both elements in sulfur
dioxide.
3) Calculate the percentage composition of
ammonium nitrate, NH4NO3
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Sample Problem I, practice p.248
4) Calculate the percentage composition of each of
the following:
a. SrBr2
c. Mg(CN)2
b. CaSO4
d. Pb(CH3COO)2
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Chapter 7
HW.
Section 7.3 Review,
p.248
#(1-3), 4c, 6c, 7c
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Section 7.3 Review, pg. 248
1. a. Suppose you know that a compound is 11.2% H
and 88.8% O. What information do you need to
determine the empirical formula?
The molar masses of H and O
b. What additional information do you need to
determine the molecular formula?
The compound’s experimental molar mass
2. Isooctane has the molecular formula C8H18.
What is its empirical formula?
C4H9
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3. What information do you need to calculate the
percentage composition of CF4
The molar masses of carbon and fluorine
4. Determine the empirical formula.
c. The analysis of a compound shows that it
is 27.0% Na, 16.5% N, and 56.5% O.
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6. Determine the formula, and then calculate the
percentage composition.
c. silver nitrate
7. Calculate the percentage composition.
c. iron(III) sulfate, Fe2(SO4)3
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Chapter 7
Standardized Test Preparation
Understanding Concepts
1. Element A has two isotopes. One has an atomic
mass of 120 and constitutes 60%; the other has an
atomic mass of 122 and constitutes 40%. Which
range below includes the average atomic mass of
Element A?
A.
B.
C.
D.
less than 120
between 120 and 121
between 121 and 122
greater than 122
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Chapter 7
Standardized Test Preparation
Understanding Concepts
1. Element A has two isotopes. One has an atomic
mass of 120 and constitutes 60%; the other has an
atomic mass of 122 and constitutes 40%. Which
range below includes the average atomic mass of
Element A?
A.
B.
C.
D.
less than 120
between 120 and 121
between 121 and 122
greater than 122
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Chapter 7
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Understanding Concepts
2. Which of the following can be determined from the
empirical formula of a compound alone?
F.
G.
H.
I.
the true formula of the compound
the molecular mass of the compound
the percentage composition of the compound
the arrangement of atoms within a molecule of the
compound
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Chapter 7
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Understanding Concepts
2. Which of the following can be determined from the
empirical formula of a compound alone?
F.
G.
H.
I.
the true formula of the compound
the molecular mass of the compound
the percentage composition of the compound
the arrangement of atoms within a molecule of the
compound
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Chapter 7
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Understanding Concepts
3. How many ions are in 0.5 moles of NaCl?
A.
B.
C.
D.
1.204  1023
3.011  1023
6.022  1023
9.033  1023
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Chapter 7
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Understanding Concepts
3. How many ions are in 0.5 moles of NaCl?
A.
B.
C.
D.
1.204  1023
3.011  1023
6.022  1023
9.033  1023
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Understanding Concepts
4. How many moles of calcium (mass = 40.1) are in a
serving of milk containing 290 mg of calcium?
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Chapter 7
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Understanding Concepts
4. How many moles of calcium (mass = 40.1) are in a
serving of milk containing 290 mg of calcium?
Answer: 7.23  10–3 mol
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Chapter 7
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Understanding Concepts
5. How is Avogadro's number related to moles?
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Understanding Concepts
5. How is Avogadro's number related to moles?
Answer: Avogadro's number is the number of particles in
a mole.
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Chapter 7
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Understanding Concepts
6. Antimony has two isotopes. One, amounting to 57.3%
of the atoms, has a mass of 120.9 amu. The other,
42.7% of the atoms, has a mass of 122.9 amu. What
is the average atomic mass of antimony?
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Chapter 7
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Understanding Concepts
6. Antimony has two isotopes. One, amounting to 57.3%
of the atoms, has a mass of 120.9 amu. The other,
42.7% of the atoms, has a mass of 122.9 amu. What
is the average atomic mass of antimony?
Answer: 121.8 amu
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Chapter 7
Standardized Test Preparation
Reading Skills
Read the passage below. Then answer the questions.
In 1800 two English chemists, Nicholson and Carlisle,
discovered that when an electric current is passed through water,
hydrogen and oxygen were produced in a 2:1 volume ratio and a
1:8 mass ratio. This evidence helped to support John Dalton's
theory that matter consisted of atoms, demonstrating that water
consists of the two elements in a constant proportion. If the same
number of moles of each gas occupy the same volume, then
each molecule of water must consist of twice as much hydrogen
as oxygen, even though the mass of hydrogen is only one-eighth
that of oxygen.
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Chapter 7
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Reading Skills
7. Based on this experiment, what is the empirical
formula of water?
F.
G.
H.
I.
HO
H2O
H2O8
O 8H
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Reading Skills
7. Based on this experiment, what is the empirical
formula of water?
F.
G.
H.
I.
HO
H2O
H2O8
O 8H
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Reading Skills
8. How would the experimental result have been
different if hydrogen gas existed as individual atoms
while oxygen formed molecules with two atoms
bound by a covalent bond?
A.
B.
C.
D.
The result would be the same.
The ratio of hydrogen to oxygen would be 1:1.
The ratio of hydrogen to oxygen would be 1:4.
The ratio of hydrogen to oxygen would be 4:1.
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Reading Skills
8. How would the experimental result have been
different if hydrogen gas existed as individual atoms
while oxygen formed molecules with two atoms
bound by a covalent bond?
A.
B.
C.
D.
The result would be the same.
The ratio of hydrogen to oxygen would be 1:1.
The ratio of hydrogen to oxygen would be 1:4.
The ratio of hydrogen to oxygen would be 4:1.
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Chapter 7
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Reading Skills
9. How does the empirical formula for water compare to
its molecular formula?
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Reading Skills
9. How does the empirical formula for water compare to
its molecular formula?
Answer: Both formulas are the same, H2O.
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Chapter 7
Standardized Test Preparation
Interpreting Graphics
Use the diagram below to answer questions 10–13.
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Chapter 7
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Interpreting Graphics
10. How many moles of oxygen atoms are there in 100.0
moles of carbon dioxide?
F.
G.
H.
I.
66.7
72.7
100.0
200.0
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Interpreting Graphics
10. How many moles of oxygen atoms are there in 100.0
moles of carbon dioxide?
F.
G.
H.
I.
66.7
72.7
100.0
200.0
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Interpreting Graphics
11. Explain why the percentage of oxygen in carbon
dioxide is not twice the percentage of oxygen in
carbon monoxide, if there are twice as many oxygen
atoms.
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Interpreting Graphics
11. Explain why the percentage of oxygen in carbon
dioxide is not twice the percentage of oxygen in
carbon monoxide, if there are twice as many oxygen
atoms.
Answer: Although there are twice as many oxygen
atoms in carbon dioxide, the percentage composition
is based on the mass, not the number of atoms.
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Chapter 7
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Interpreting Graphics
12. If you did not know the true formulas for carbon
monoxide and carbon dioxide, what information would
you need beyond what is provided in the illustration in
order to calculate them?
A.
B.
C.
D.
the percentage compositions
the atomic masses of carbon and oxygen
the melting and boiling points of each compound
the number of atoms of each element in the
compound
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Chapter 7
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Interpreting Graphics
12. If you did not know the true formulas for carbon
monoxide and carbon dioxide, what information would
you need beyond what is provided in the illustration in
order to calculate them?
A.
B.
C.
D.
the percentage compositions
the atomic masses of carbon and oxygen
the melting and boiling points of each compound
the number of atoms of each element in the
compound
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Interpreting Graphics
13. How many grams of carbon are contained in 200.0
grams of carbon dioxide?
F.
G.
H.
I.
27.29
42.88
54.58
85.76
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Chapter 7
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Interpreting Graphics
13. How many grams of carbon are contained in 200.0
grams of carbon dioxide?
F.
G.
H.
I.
27.29
42.88
54.58
85.76
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