Chemical
Composition
COUNTING ATOMS BY
WEIGHING
Problem:
The “lead” in my pencil is graphite-an
almost pure carbon compound.
I wrote a really long English essay and
used up most of a pencil, how many
carbon atoms did I use?
Can I figure this out.
What is the mass of 1 carbon
atom?

1.99 * 10-23 g
◦ Is this a practical number to use for calculations?
◦ Scientists have defined atomic mass units or amu’s to avoid this.
◦ 1 amu = 1.66 * 10-24 g

The average atomic mass of 1 carbon atom is 12.01 amu.
◦ Any sample of carbon from nature can be treated as though it were
composed of identical carbon atoms, each with a mass of 12.01
amu.
Conversion Relationship #1

1 amu = 1.66 * 10-24 g

This equivalency allows conversion between these two
measurements of mass.
◦ Its just like using 2.2046 lbs = 1 Kg
Practice: g  amu

What is the mass of 5.00 g Carbon in
atomic mass units?

What is the mass of 5 million amus of
Silver in grams?
Conversion Relationship #2

The mass of a single atom of an element
equals its mass on the Periodic Table in
atomic mass units.
◦ For Carbon: 1 C atom = 12.0107 amu
◦ For Hydrogen:1 H atom = 1.00795 amu
◦ For Chlorine: 1 Cl atom = 35.453 amu
Conversion Relationship #2

1 atom = mass on PT in amu

The periodic table contains equivalency
statements that can be used to convert
between mass in amu and number of atoms.

For example, What is the mass of 12 Carbon
atoms?
◦ 1 C atom = 12.01 amu
◦ 12 C atoms (12.01 amu / 1 C atom) = 144.1 amu

Practice: # atoms  amu
Calculate the mass, in amu, of a sample
of aluminum that contains 75 aluminum
atoms.
1 Al atom = 26.981538 amu
Practice: # atoms  amu

Calculate the mass of a sample that
contains 23 nitrogen atoms.
1 N atom = 14.0067 amu

Practice: # atoms  amu
Calculate the number of sodium atoms
present in a sample that has a mass of
1172.49 amu.
1 Na Atom = 22.98977 amu

Practice: # atoms  amu
Calculate the number of oxygen atoms
in a sample that has a mass of 288 amu.
1 O Atom = 15.9994 amu
Try these…

Calculate the mass of a sample that contains 2493 Lithium atoms.

Which weighs more, 25 Platinum atoms, or 5 iron atoms?

How many atoms are in 10462.692 amu of oxygen?
THE MOLE
What is a Dozen?

A Dozen is a word that represents a number.
◦ 1 Dozen = 12

1 dozen apples = ?
◦ = 12 apples

2 dozen cookies = ?
◦ = 24 cookies

3.2 dozen eggs = ?
◦ = 38.4 eggs (38 eggs)
What is a Mole?

A Mole is a word that represents a number.
◦ 1 mole = 1 mol = 6.022 x 1023

This number is called Avogadro’s number.
◦ Avogadro’s Number = 1 mol = 6.022 x 1023

A Mole is a lot bigger than a Dozen!
◦ 1 mol apples = ?
 = 6.022 x 1023 apples
◦ 2 mol cookies = ?
 = 1.2044 x 1024 cookies
◦ 3.2 mol eggs = ?
 = 1.92704 x 1024 eggs
 = 1,927,040,000,000,000,000,000,000 eggs
Conversion Relationship #3

1 mol = 6.022 x 1023 atoms

How many calcium atoms are present in a
sample that contains 3.12 mol Ca?

How many moles are present in a sample
of gold that contains 2.6 x 1052 atoms?
Why is the mole necessary?

Comparing equal masses of chemicals does not compare equal numbers of
particles.

Consider the synthesis of barium sulfide from its elements: Ba(s) + S(s) 
BaS
◦ If there are 100 amu of each element, how many molecules of BaS will form?
◦ Calculate the number of atoms in each 100 amu sample.
◦ Are they the same?

It is much more efficient to use moles to compare quantities of chemicals.
◦ If there are 100 moles of Ba and 100 moles of S, how many moles of BaS will form?
◦ 100 moles is the same number of atoms no matter what kind of atom it is. It is a basis
for an equal comparison.
◦ That is a much easier way of thinking of things.

The Mole is the Most
Important Quantity in
Chemistry!!!!

Comparing quantities in chemistry using
moles allows an equal basis of
comparison.

Comparing quantities in chemistry using
grams does not.
Conversion Relationship #4
The Mass of a Mole

The Periodic Table of Elements tells us the mass of 1
mole of each of the atoms.
◦ 1 mole of an element = mass on Periodic Table in grams.

To Clarify between the mass of 1 atom and the mass
of one mole of atoms:
◦ 1 atom = atomic mass in amu
 The mass of 1 carbon atom is 12.0107 amu
◦ 1 mole of atoms = atomic mass in grams
 The mass of 1 mole of carbon atoms is 12.0107 g
Practice: g-mol & mol-#atoms

How many moles Al is contained in
10.0 g Al?

How many atoms does this sample
contain?
Practice: grams-moles-#atoms

A silicon chip used in an integrated
circuit of a microcomputer has a mass
of 5.68 mg. How many Si atoms are
present in this chip
Practice: #atoms-moles-grams

Calculate both the number of moles in,
and the mass of, a sample of chromium
containing 5.00 x 1020 atoms.
8.4 MOLAR MASS
Goals:
1.
2.
3.
To define Molar Mass.
To determine the molar mass of a polyatomic
substance.
To convert between moles and mass of a substance.
The Molar Mass of Atoms

The molar mass of an atom is its atomic
mass from the Periodic Table in grams.

What is the molar mass of an atom of
Oxygen?
◦ 15.9994 g/mol

What is the molar mass of elemental
oxygen? O2(g)
◦ 2 x 15.9994 g/mol = 31.9988 g/mol
The Molar Mass of
Polyatomic Substances

A polyatomic substance is a collection of atoms.
◦ The molar mass of a polyatomic substance is simply the sum of
the molar masses of all the atoms of which the substance is
made.

Example: Find the molar mass of methane (CH4)
1. Determine the chemical composition:
 CH4 = 1 C + 4 H
2. Then add up the masses of all the atoms.
 CH4 = 1(12.0107) + 4(1.00794) = 16.04246
 The molar mass is the mass of 6.022 x 1023 molecules of
CH4, in grams.
 The molar mass of methane = 16.0425 g/mol CH4
Summary of Molar Mass
 The
Molar Mass of a substance is the
mass (in grams) of 1 mole of the
substance.
1.
2.
For Elements, the molar mass is equal to the atomic mass on
the periodic table (in grams).
Diatomic & Polyatomic Substances, the
For
molar mass is equal to the sum of the atomic weights of all the
atoms in the substance (in grams).
Practice: Calculating Molar Mass

Calculate the molar mass of sulfur
dioxide.
1. Chemical Composition:

SO2 = S + 2O
2. Add the molar mass of all the atoms:

SO2 = 1(32.065) + 2(15.9994)
 Answer:
 64.0638 = 64.064 g/mol SO2
Practice: Calculating Molar Mass

Polyvinyl chloride (called PVC) which is widely used
for floor coverings (“vinyl”) and for plastic pipes in
plumbing systems, is made from a molecule with the
formula C2H3Cl. Calculate the molar mass of this
substance.
1. Chemical Composition:
 C2H3Cl = 2C + 3H + 1Cl
2. Sum of molar masses:
 C2H3Cl = 2(12.0107) + 3(1.00794) + 1(35.453)
Answer: 62.49822 = 62.498 g/mol C2H3Cl
Conversion Relationship #4
The Mass of a Mole (Revisited)

Once the molar mass of a substance is known, it can be used to convert
between moles and masses of the substance.
◦ The molar mass of methane (CH4) is 16.04246 g/mol.
◦ This gives an equivalency statement that can be used as a conversion factor:
1 mol CH4 = 16.04246 g CH4
1. What is the mass of 3.2 moles of methane?
2. How many moles are in 5.93 g methane?
Practice: grams-moles (with
molecules this time)

Calcium Carbonate, CaCO3 (also called
calcite), is the principal mineral found in
limestone, marble, chalk, pearls, and the
shells of marine animals such as clams.
a. Calculate the molar mass of calcium
carbonate.
b. What is the mass in grams of 4.86 mol
CaCO3?
Practice: grams-moles

Calculate the molar mass for sodium
sulfate, Na2SO4. A sample of sodium
sulfate with a mass of 300.0 g represents
what number of moles of sodium sulfate?
a. Calculate the molar mass of Sodium Sulfate.
b. Convert mass Na2SO4  moles Na2SO4

Practice: moles-grams
Juglone, a dye known for centuries, is produced from
the husks of black walnuts. It is also a natural
herbicide (weed killer) that kills off competitive
plants around the black walnut tree but does not
affect grass and other noncompetitive plants. The
formula for juglone is C10H6O3.
a. Calculate the molar mass of Juglone:
b. How many moles are contained in a 1.56 g sample of
juglone?

Conversion Relationship #3
(Revisited)
Remember Avogadro’s Number?
◦ 6.022 x 1023

Avogadro’s number represents 1 mole of anything.
◦ 1 mole of any substance contains 6.022 x 1023 units of that
substance.

This relationship is an equivalency factor that can be used
to make conversions between moles and # of molecules.
◦ 1 mol = 6.022 x 1023 molecules
Practice: mass-moles-#molecules

Isopentyl Acetate, C7H14O2, the compound responsible for the scent of bananas, can be produced
commercially. Interestingly, bees release about 1μg (1x10-6 g) of this compound when they sting.
This attracts other bees, which then join the attack. How many moles and how many molecules of
isopentyl acetate are released in a typical bee sting?
a. Determine the molar mass of C7H14O2:
b. Convert mass C7H14O2  moles C7H14O2  # of
molecules C7H14O2:
Practice: mass-moles-#molecules

The substance Teflon, the slippery coating on
many frying pans, is made from the C2F4
molecule. Calculate the number of C2F4 units
present in 135 g of Teflon.
a. Calculate the molar mass of C2F4 :
b. Convert mass  moles  # molecules
MOLAR
RELATIONSHIPS
RELATIONSHIP #1
PARTICLES IN A MOLE
Particles in a
Mole
 A chemical mole is defined as the number
of atoms present in a sample of Carbon12 that weighs exactly 12.0107g.

A Mole is a word that represents a
Number, and that number is also known
as Avogadro’s Number.

1 mole = 6.022 x 1023 particles
◦ Particles are either atoms or molecules.
Practice: Moles - # Particles

How many carbon atoms are in a sample that
contains 5.16 moles C?

How many moles are present in a sample of nitric
acid that contains 5 billion molecules?
RELATIONSHIP #2
ATOMS IN A MOLECULE
Atoms in Molecules

Consider a single water molecule:
◦ What is the composition by type of atom?

Using moles, 1 mole H2O contains 2 moles
H and 1 mole O.

The following equivalencies can be made:
◦ 1 mol H2O = 2 mol H
◦ 1 mol H2O = 1 mol O
Practice: Atoms in
Molecules

How many moles of carbon are in a 6.57
mol sample of Aluminum Acetate?
8.5 Pages 226 - 228
PERCENT
COMPOSITION BY
MASS
Mass Percent

The Mass Percent of an element in a
compound is the percent composition by
mass of the element in the compound.
𝐓𝐨𝐭𝐚𝐥 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐭𝐡𝐞 𝐞𝐥𝐞𝐦𝐞𝐧𝐭 𝐢𝐧 𝐭𝐡𝐞 𝐜𝐨𝐦𝐩𝐨𝐮𝐧𝐝
(
𝐓𝐨𝐭𝐚𝐥 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐭𝐡𝐞 𝐜𝐨𝐦𝐩𝐨𝐮𝐧𝐝

𝐌𝐚𝐬𝐬 𝐏𝐞𝐫𝐜𝐞𝐧𝐭 =

m%element = n(molar masselement) / (molar masscompound) x 100
 Where n is the number of atoms of the element in the compound.
)
x 100
Calculating Mass Percent
Consider the compound Ethanol (C2H5OH)
 Follow these steps to determine the mass
percent of Carbon in Ethanol:

1. Determine the number of moles of C in one mole
of ethanol:
 There are 2 moles of Carbon, so n = 2.
2. Determine the mass of C in ethanol:
 2 mol C x (12.0107 g C / 1 mol C) = 24.0214 g C
3. Determine the molar mass of ethanol:
 2(12.0107) + 6(1.00794) + 1(15.9994) = 46.0684 g
C2H5OH
Practice: Calculating Mass Percent

Carvone is a substance that occurs in two forms, both of which have the same
molecular formula (C10H14O) and molar mass. One type of carvone gives caraway
seeds their characteristic smell; the other is responsible for the smell of spearmint
oil. Compute the mass percent of each element in Carvone.
Practice: Calculating Mass Percent

Penicillin F has the formula C14H20N2SO4. Calculate the mass
percent of each element in Penicillin F.
Using Mass Percent

A sample of Calcium Carbonate has a
mass of 5.69 g. What is the mass of
Oxygen present in the sample?
Using Mass Percent

What mass of carbon is present in 15.39
mol C6H12O6?
8.6 Pages 228 - 230
THE EMPIRICAL
FORMULA
Combustion Analysis

When a compound containing only C, H,
& O is burned in the presence of excess
oxygen, the composition by mass of C, H,
& O can be determined by measuring the
mass of CO2, and H2O produced. 0.2015
g of a substance containing C, H, & O is
found to contain 0.0806 g C, 0.01353 g H,
and 0.1074 g O. What is the ratio of
C:H:O atoms in the compound?
Step 1: Convert mass to moles

0.0806 g C (1 mol C / 12.0107 g C)
= 0.006710683 mol C = .00671 mol C

0.01353 g H (1 mol H / 1.00794 g H)
= 0.0134234181 mol H = 0.01342 mol H

0.1074 g O (1 mol O / 15.9994 g O)
= 0.0067127517 mol O = 0.006713 mol
O
Step 2: Convert Moles to Atoms

0.00671 mol C (6.022 x 1023 atoms C / 1 mol C)
= 4.04 x 1021 atoms C

0.01342 mol H (6.022 x 1023 atoms H / 1 mol H)
= 8.084 x 1021 atoms H

0.006713 mol O ((6.022 x 1023 atoms H / 1 mol H)
= 4.042 x 1021 atoms O
Step 3: Determine the Ratio of C, H, & O
C:H:O
(4.04 x 1021 atoms C) : (8.084 x 1021 atoms H) : (4.042 x 1021 atoms O)
1:2:1
Notice that the mole ratio is the same…
(0.00671 mol C) : (0.01342 mol H) : (0.006713 mol O)
Is this ratio the molecular formula?

The ratio of C:H:O is 1:2:1

Does this mean the molecular formula is
CH2O?

What about the molecular formulas of
the following compounds…?
◦ C2H4O2, C3H6O3, C5H10O5

All of the above have a ratio of C:H:O of
Empirical Formulas

The Empirical Formula of a compound is
the smallest whole-number (integer) ratio
of atoms.

All four of the compounds CH2O,
C2H4O2, C3H6O3, & C5H10O5 have the
same empirical formula; CH2O.

The Empirical Formula of a compound is
the most simplified formula.
Determining Empirical Formulas

Determine the empirical formula for each
compound:
1. C6H6
2. C12H4Cl4O2
3. C6H16N2

Answers:
1. CH
2. C6H2Cl2O
3. C3H8N
8.7 Pages 230 - 235
CALCULATING EMPIRICAL
FORMULAS
Calculating Empirical Formulas
1.
Determine the number of moles of each
element present in the compound. (This can be
done 1 of 2 ways)
a) If the amounts are given in grams, simply convert
them to moles using the molar masses of the
elements from the periodic table of elements.
b) If the amounts are give in percent composition by
mass, simply base your calculations on a sample of
mass = 100. g of the compound. The percent mass
of each element will represent the mass in grams
of that element.
2.
Divide each value of the number of moles by
the smallest of the values. If each resulting
Practice: Calculating Empirical Formulas
An oxide of aluminum is formed by the reaction of
4.151 g Al with 3.692 g O. Calculate the empirical
formula of the compound.
1. Determine the mass of each element: Given
2. Determine the moles of each:

4.151 g Al (1 mol Al / 26.981538 g Al) = 0.1538 mol Al
3.692 g O (1mol O / 15.9994 g O) = 0.2308 mol O
Divide each by the least number of moles:
Al: 0.1538 / 0.1538 = 1
O: 0.2308 / 0.1538 = 1.5
4. Multiply to get whole numbers to determine the
3.
Practice: Calculating Empirical Formulas

1.
When a 0.3546 g sample of vanadium metal is
heated in air, it reacts with oxygen to achieve a final
mass of 0.6330 g. Calculate the empirical formula of
this vanadium oxide.
Determine the mass of each element present:
◦ mV = 0.3546 g
◦ mO = 0.6330 – 0.3546 = 0.2784 g
2.
Determine the moles of each element present:
◦ molV = 0.3546 g (1 mol / 50.9415 g) = 0.0069609258 mol V
◦ mol O = 0.2784 g (1 mol / 15.9994 g) = 0.0174006525 mol O
3.
Determine the mole:mole ratio of elements:
◦ V: 0.0069609258 / 0.0069609258 = 1
◦ O: 0.0174006525 / 0.0069609258 = 2.499761242 ≈ 2.5
Practice: Calculating Empirical Formulas
A sample of lead arsenate, an insecticide used against the
potato beetle, contains 1.3813 g of lead, 0.00672 g of
hydrogen, 0.4995 g arsenic, and 0.4267 g oxygen.
Calculate the empirical formula for lead arsenate.
1. Determine the mass of each element: given
2. Determine moles of each element:

◦
◦
◦
◦
3.
◦
◦
◦
◦
1.3813 g Pb (1 mol / 207.2 g) = 0.0066665058 mol Pb
0.00672 g H (1 mol / 1.00794 g) = 0.0066670635 mol H
0.4995 g As (1 mol / 74.9216 g) = 0.0066669692 mol As
0.4267 g O (1 mol / 15.9994 g) = 0.0266697501 mol O
Determine the mole ratios:
Pb: 0.0066665058 / 0.0066665058 = 1
H: 0.0066670635 / 0.0066665058 = 1.000083657 ≈ 1
As: 0.0066669692 / 0.0066665058 = 1.000069512 ≈ 1
O: 0.0266697501 / 0.0066665058 = 4.000559048 ≈ 4
Practice: Calculating Empirical Formulas
Analysis of the chemical carbamic acid finds that it is
composed of 0.8007 g C, 0.9333 g N, 0.2016 g H, and
2.133 g O. Determine the empirical formula of carbamic
acid.
1. Determine the mass of each element: given
2. Determine moles of each element:

◦
◦
◦
◦
3.
◦
◦
◦
◦
0.8007 g C (1 mol / 12.0107 g) = 0.0666655565 mol C
0.9333 g N (1 mol / 14.0067 g) = 0.0666323974 mol N
0.2016 g H (1 mol / 1.00794 g) = 0.2000119055 mol H
2.133 g O (1 mol / 15.9994 g) = 0.1333174994 mol O
Determine the mole ratios:
C: 0.0666655565 / 0.0666323974 = 1.00497642 ≈ 1
N: 0.0666323974 / 0.0666323974 = 1
H: 0.2000119055 / 0.0666323974 = 3.001721584 ≈ 3
O: 0.1333174994 / 0.0666323974 = 2.000790976 ≈ 2
Practice: Calculating Empirical Formulas

Cisplatin, the common name for a platinum compound that is
used to treat cancerous tumors, has the composition (mass
percent) 65.02 % Pt, 9.34 % N, 2.02 % H, and 23.63 %Cl.
Calculate the empirical formula for cisplatin.
Determine the mass of each element: given in % so use a
100.g sample:
1.
◦
Determine moles of each element:
2.
◦
◦
◦
◦
3.
65.02 g Pt + 9.34 g N + 2.02 g H + 23.63 g Cl = 100.00 g sample
65.02 g Pt (1 mol / 195.08 g) = 0.3332991593 mol Pt
9.34 g N (1 mol / 14.0067 g) = 0.6668237344 mol N
2.02 g H (1 mol / 1.00794 g) = 2.004087545 mol H
23.63 g Cl (1 mol / 35.453 g) = 0.6665162328 mol Cl
Determine the mole ratios:
◦ Pt: 0.3332991593 / 0.3332991593 = 1
◦ N: 0.6668237344 / 0.3332991593 = 2.000676317 ≈ 2
Practice: Calculating Empirical Formulas

The most common form of nylon (nylon 6) is 63.68% C,
12.38% N, 9.80% H, and 14.14% O. Calculate the
empirical formula of nylon 6.
1.
◦
2.
◦
◦
◦
◦
3.
Determine the mass of each element: given in % so use a
100.g sample:
63.68 g C + 12.38 g N + 9.80 g H + 14.14 g O = 100.00 g
sample
Determine moles of each element:
63.68 g C (1 mol / 12.0107 g) = 5.301939104 mol C
12.38 g N (1 mol / 14.0067 g) = 0.8838627228 mol N
9.80 g H (1 mol / 1.00794 g) = 9.7228096 mol H
14.14 g O (1 mol / 15.9994 g) = 0.883731419 mol O
Determine the mole ratios:
◦ C: 5.301939104 / 0.883731419 = 5.999491463 ≈ 6
8.8 CALCULATION OF
MOLECULAR FORMULAS
Pages 235 – 237
Calculating Molecular
Formulas

Knowing the percent composition or
knowing the mass ratios of a compound
allows us to calculate the empirical
formula of the compound.
◦ But this does not determine the identity of
the compound…

In order to calculate the molecular
formula of a compound, the molar mass
must also be known.
Practice: Calculating Molecular
Formulas

A white powder is analyzed and found to
have an empirical formula of P2O5. The
compound has a molar mass of 283.88 g.
What is the compound’s molecular
formula?
1.
Determine molar mass of the empirical
formula:
◦ P2O5 = 141.94452 g/mol
2.
Determine the value of n:
◦ n = 283.88 / 141.94452 = 1.999936313 ≈ 2
Practice: Calculating Molecular
Formulas

A compound used as an additive for gasoline to help
prevent engine knock shows the following percentage
composition: 71.65% Cl, 24.27% C, 4.07% H. The
molar mass is known to be 98.96 g. Determine the
empirical and the molecular formula for this
compound.

Determine the empirical formula:
◦ Mass of each, moles of each, mole ratio, lowest whole
number mole ratio = empirical formula
◦ The empirical formula is ClCH2

Determine the molar mass of the empirical formula:
◦ ClCH2 = 49.47958 g/mol

Determine the value of n:
◦ 98.96 g / 49.47958 g = 2.000016977 ≈ 2

Determine the molecular formula: