Chapter Two, Reading Assignment

• Atoms, Molecules and Ions

1. Atomic structure and symbolism
1. Chemical Symbols and Isotopes
2. Atomic mass unit and average atomic mass
2. Chemical Formulas
3. The Periodic Table
1. Historical development of the periodic table
2. Classification of elements in the periodic table
4. Ionic and Molecular Compounds
1. Formation of Ionic Compounds
2. Formation of Molecular Compounds
5. Chemical Nomenclature
1. Ionic compounds
2. Molecular Compounds 1
 Chapter 3

Composition of Substances and Solutions

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Learning Objectives of the Chapter:
At the end of this chapter you will be able to
Define the amount unit mole and the related quantity Avogadro’s
number
Explain the relation between mass, moles, and numbers of atoms
or molecules, and perform calculations deriving these quantities
from one another
Compute the percent composition of a compound
Determine the empirical and molecular formula of a compound
Calculate solution concentrations using molarity
Perform dilution calculations using the dilution equation
Define the concentration units of mass percentage, volume
percentage, mass-volume
percentage, parts-per-million (ppm), and parts-per-billion (ppb)

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Atomic mass is the mass of an atom in
atomic mass units (amu)

By definition:
1 atom 12C “weighs” 12 amu

On this scale
1
H = 1.008 amu
16
O = 16.00 amu

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The average atomic mass is the weighted
average of all of the naturally occurring
isotopes of the element.

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 Naturally occurring lithium is:
7.42% 6Li (6.015 amu)
92.58% 7Li (7.016 amu)

Average atomic mass of lithium:

7.42 x 6.015 + 92.58 x 7.016

= 6.941 amu
100

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Average atomic mass (6.941)

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Molecular mass (or molecular weight) is the sum of
the atomic masses (in amu) in a molecule.

1S 32.07 amu
2O + 2 x 16.00 amu
SO2 SO2 64.07 amu

For any molecule

molecular mass (amu) = molar mass (grams)

1 molecule SO2 = 64.07 amu

1 mole SO2 = 64.07 g SO2 8
Formula mass is the sum of the atomic masses
(in amu) in a formula unit of an ionic compound.

1Na 22.99 amu

NaCl 1Cl + 35.45 amu
NaCl 58.44 amu

For any ionic compound

formula mass (amu) = molar mass (grams)

1 formula unit NaCl = 58.44 amu

1 mole NaCl = 58.44 g NaCl 9
What is the formula mass of Ca3(PO4)2 ?

1 formula unit of Ca3(PO4)2

3 Ca 3 x 40.08
2P 2 x 30.97
8O + 8 x 16.00
310.18 amu

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The mole is an amount unit similar to familiar units like pair, dozen,
gross, etc. It provides a specific measure of the number of atoms or
molecules in a sample of matter.

Dozen = 12

Pair = 2

The mole (mol) is the amount of a substance that contains as

many elementary entities as there are atoms in exactly 12.00
grams of 12C

1 mol = NA = 6.0221367 x 1023

Avogadro’s number (NA) 11
Molar mass is the mass of 1 mole of atoms in grams

1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g

1 12C atom = 12.00 amu

1 mole 12C atoms = 12.00 g 12C

1 mole lithium atoms = 6.941 g of Li

For any element

atomic mass (amu) = molar mass (grams)
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One Mole of:

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Deriving Moles from Grams for an Element
According to nutritional guidelines from the US Department of
Agriculture, the estimated average requirement for dietary
potassium is 4.7 g. What is the estimated average requirement of
potassium in moles?

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Eg. Beryllium is a light metal used to fabricate transparent X-ray
windows for medical imaging instruments. How many moles of
Be are in a thin-foil window weighing 3.24 g?

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 Deriving Number of Atoms from Mass for an Element

M= molar mass in g/mol, NA = Avogadro’s number

How many atoms are in 0.551 g of potassium (K) ?
1 mol K = 39.10 g K
1 mol K = 6.022 x 1023 atoms K
1 mol K 6.022 x 1023 atoms K
0.551 g K x x =
39.10 g K 1 mol K

8.49 x 1021 atoms K

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Check Your Learning
1. A prospector panning for gold in a river collects 15.00 g of pure
gold. How many Au atoms are in this quantity of gold?
2. How many moles of sucrose, C12H22O11, are in a 25-g sample of
sucrose?
3. What is the mass of 0.443 mol of hydrazine, N2H4?

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Deriving the Number of Atoms and Molecules from the Mass of
a Compound
A packet of an artificial sweetener contains 40.0 mg of saccharin
(C7H5NO3S), which has the structural formula:

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19
How many H atoms are in 72.5 g of C3H8O ?

1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O

1 mol C3H8O molecules = 8 mol H atoms
1 mol H = 6.022 x 1023 atoms H

1 mol C3H8O 8 mol H atoms 6.022 x 1023 H atoms

72.5 g C3H8O x x x =
60 g C3H8O 1 mol C3H8O 1 mol H atoms

5.82 x 1024 atoms H

Check Your Learning
How many C4H10 molecules are contained in 9.213 g of this
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compound? How many hydrogen atoms?
 Determining empirical and molecular formulas
Percent composition of an element from molecular mass or
molar mass of elements =
n x molar mass of element
x 100%
molar mass of compound

n is the number of moles of the element in 1 mole of the

compound

2 x (12.01 g)
%C = x 100% = 52.14%
46.07 g
6 x (1.008 g)
%H = x 100% = 13.13%
46.07 g
1 x (16.00 g)
%O = x 100% = 34.73%
46.07 g
C2H6O 52.14% + 13.13% + 34.73% = 100.0%
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Calculation of Percent Composition From given mass
Analysis of a 12.04-g sample of a liquid compound composed of
carbon, hydrogen, and nitrogen showed it to contain 7.34 g C, 1.85
g H, and 2.85 g N. What is the percent composition of this
compound?
Solution
To calculate percent composition, divide the experimentally derived
mass of each element by the overall mass of the compound, and
then convert to a percentage:

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Ex. A 24.81-g sample of a gaseous compound containing only
carbon, oxygen, and chlorine is determined to contain 3.01 g C,
4.00 g O, and 17.81 g Cl. What is this compound’s percent
composition?

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Percent Composition and Empirical Formulas
Determine the empirical formula of a
compound that has the following
percent composition by mass:
K 24.75, Mn 34.77, O 40.51 percent.

1 mol K
nK = 24.75 g K x = 0.6330 mol K
39.10 g K
1 mol Mn
nMn = 34.77 g Mn x = 0.6329 mol Mn
54.94 g Mn
1 mol O
nO = 40.51 g O x = 2.532 mol O
16.00 g O

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Percent Composition and Empirical Formulas

nK = 0.6330, nMn = 0.6329, nO = 2.532

0.6330 ~
K: ~ 1.0
0.6329
0.6329
Mn : = 1.0
0.6329
2.532
O: ~ 4.0
~
0.6329

KMnO4

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Determining an Empirical Formula from Percent Composition
Eg. The bacterial fermentation of grain to produce ethanol forms a
gas with a percent composition of 27.29% C and 72.71% O. What
is the empirical formula for this gas?

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 Determination of molecular formulas
• Molecular formulas are derived by comparing the compound’s
molecular or molar mass to its empirical formula mass.
• As the name suggests, an empirical formula mass is the sum of
the average atomic masses of all the atoms represented in an
empirical formula.
• If the molecular (or molar) mass of the substance is known, it
may be divided by the empirical formula mass to yield
• the number of empirical formula units per molecule (n):

• The molecular formula is then obtained by multiplying each

subscript in the empirical formula by n, as shown by the generic
empirical formula

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Determination of the Molecular Formula for Nicotine
Nicotine, an alkaloid in the nightshade family of plants that is
mainly responsible for the addictive nature of cigarettes, contains
74.02% C, 8.710% H, and 17.27% N. If 40.57 g of nicotine
contains 0.2500 mol nicotine, what is the molecular formula?

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29
Ex. What is the molecular formula of a compound with a percent
composition of 49.47% C, 5.201% H, 28.84% N, and 16.48% O,
and a molecular mass of 194.2 amu?

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 Molarity and Other Concentration Units
• The relative amount of a given solution component is known as
its concentration.
• A solution in which water is the solvent is called an aqueous
solution.
• A solute is a component of a solution that is typically present at
a much lower concentration than the solvent.
• Solute concentrations are often described with qualitative terms
such as dilute (of relatively low concentration) and concentrated
(of relatively high concentration).
• Molarity (M) is a useful concentration unit for many
applications in chemistry.
• Molarity is defined as the number of moles of solute in exactly
1 liter (1 L) of the solution:

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Calculating Molar Concentrations
A 355-mL soft drink sample contains 0.133 mol of sucrose (table
sugar). What is the molar concentration of sucrose in the beverage?
Solution
Since the molar amount of solute and the volume of solution are
both given, the molarity can be calculated using the definition of
molarity.
Per this definition, the solution volume must be converted from mL
to L:

Eg. A teaspoon of table sugar contains about 0.01 mol sucrose.

What is the molarity of sucrose if a teaspoon of sugar has been
dissolved in a cup of tea with a volume of 200 mL?
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Calculating Molar Concentrations from the Mass of Solute
Distilled white vinegar is a solution of acetic acid, CH3CO2H, in
water. A 0.500-L vinegar solution contains 25.2 g of acetic acid.
What is the concentration of the acetic acid solution in units of
molarity?

Exercise
1.Calculate the molarity of 6.52 g of CoCl2 (128.9 g/mol) dissolved
in an aqueous solution with a total volume of 75.0 mL.
2. How many grams of NaCl are contained in 0.250 L of a 5.30-M
solution?
3. The concentration of acetic acid in white vinegar was determined
to be 0.839 M. What volume of vinegar contains 75.6 g of acetic
acid?
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 Dilution of Solutions
Dilution is the process whereby the concentration of a solution is
lessened by the addition of solvent. For example, a glass of iced tea
becomes increasingly diluted as the ice melts.

Dilution is also a common means of preparing solutions of a

desired concentration. By adding solvent to a measured portion of a
more concentrated stock solution, a solution of lesser concentration
may be prepared.
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• For example, commercial pesticides are typically sold as solutions
in which the active ingredients are far more concentrated than is
appropriate for their application.
• Before they can be used on crops, the pesticides must be diluted.
This is also a very common practice for the preparation of a
number of common laboratory reagents.
• A simple mathematical relationship can be used to relate the
volumes and concentrations of a solution before and after the
dilution process.
• According to the definition of molarity, the number of moles of
solute in a solution (n) is equal to the product of the solution’s
molarity (M) and its volume in liters (L):

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the dilution equation is often written in the more general form:
C1V1=C2V2
Where, C and V are concentration and volume, respectively.

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Determining the Concentration of a Diluted Solution
If 0.850 L of a 5.00-M solution of copper nitrate, Cu(NO3)2, is
diluted to a volume of 1.80 L by the addition of water, what is the
molarity of the diluted solution?
Solution
The stock concentration, C1, and volume, V1, are provided as well
as the volume of the diluted solution, V2.
Rearrange the dilution equation to isolate the unknown property, the
concentration of the diluted solution, C2:
C1V1=C2V2
C2=C1V1/ V2

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 Check Your Learning
1. What is the concentration of the solution that results from
diluting 25.0 mL of a 2.04-M solution of CH3OH to 500.0 mL?
2. What volume of 0.12 M HBr can be prepared from 11 mL (0.011
L) of 0.45 M HBr?
3. A laboratory experiment calls for 0.125 M HNO3. What volume
of 0.125 M HNO3 can be prepared from 0.250 L of 1.88 M HNO3?
4. What volume of 1.59 M KOH is required to prepare 5.00 L of
0.100 M KOH?
5. What volume of a 0.575-M solution of glucose, C6H12O6, can
be prepared from 50.00 mL of a 3.00-M glucose solution?

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 Percentage (W/W, W/V and V/V)
Mass Percentage
• The mass percentage of a solution component is defined as the
ratio of the component’s mass to the solution’s mass, expressed as a
percentage:

• Mass percentage is also referred to by similar names such as

percent mass, percent weight, weight/weight percent
• The most common symbol for mass percentage is simply the
percent sign, %, although more detailed symbols are often used
including %mass, %weight, and (w/w)%.

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Calculation of Percent by Mass
•A 5.0-g sample of spinal fluid contains 3.75 mg (0.00375 g) of
glucose. What is the percent by mass of glucose in spinal fluid?
Solution
•The spinal fluid sample contains roughly 4 mg of glucose in 5000
mg of fluid, so the mass fraction of glucose should be a bit less
than one part in 1000, or about 0.1%.
•Substituting the given masses into the equation defining mass
percentage yields:

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Calculations using Mass Percentage
“Concentrated” hydrochloric acid is an aqueous solution of 37.2%
HCl that is commonly used as a laboratory reagent. The density of
this solution is 1.19 g/mL. What mass of HCl is contained in 0.500 L
of this solution?
Solution

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 Check Your Learning
1. A bottle of a tile cleanser contains 135 g of HCl and 775 g of
water. What is the percent by mass of HCl in this cleanser?
2. What volume of concentrated HCl solution contains 125 g of
HCl?
Volume Percentage
•Liquid volumes over a wide range of magnitudes are conveniently
measured using common and relatively inexpensive laboratory
equipment.
•The concentration of a solution formed by dissolving a liquid
solute in a liquid solvent is therefore often expressed as a volume
percentage, %vol or (v/v)%:

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Calculations using Volume Percentage
Rubbing alcohol (isopropanol) is usually sold as a 70%vol aqueous
solution. If the density of isopropyl alcohol is 0.785 g/mL, how
many grams of isopropyl alcohol are present in a 355 mL bottle of
rubbing alcohol?
Solution
Per the definition of volume percentage, the isopropanol volume is
70% of the total solution volume. Multiplying the isopropanol
volume by its density yields the requested mass:

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Check Your Learning
Wine is approximately 12% ethanol (CH3CH2OH) by volume.
Ethanol has a molar mass of 46.06 g/mol and a density 0.789 g/mL.
How many moles of ethanol are present in a 750-mL bottle of wine?
Mass-Volume Percentage
• For example, physiological saline solution, used to prepare
intravenous fluids, has a concentration of 0.9% mass/volume
(m/v), indicating that the composition is 0.9 g of solute per 100
mL of solution.
• The concentration of glucose in blood (commonly referred to as
“blood sugar”) is also typically expressed in terms of a mass-
volume ratio.

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(a) The NaCl concentration of physiological saline is 0.9% (m/v).
(b) This device measures glucose levels in a sample of blood. The
normal range for glucose concentration in blood (fasting) is around
70–100 mg/dL.

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Parts per million (ppm) and Part per billion (ppb)
Very low solute concentrations are often expressed using
appropriately small units such as parts per million (ppm) or parts
per billion (ppb). Like percentage (“part per hundred”) units, ppm
and ppb may be defined in terms of masses, volumes, or mixed
mass-volume units.
There are also ppm and ppb units defined with respect to numbers
of atoms and molecules.

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• Both ppm and ppb are convenient units for reporting the
concentrations of pollutants and other trace contaminants in
water.
• Concentrations of these contaminants are typically very low in
treated and natural waters, and their levels cannot exceed
relatively low concentration thresholds without causing adverse
effects on health and wildlife.
• For example, the Environmental Protection Agency (EPA) has
identified the maximum safe level of fluoride ion in tap water to
be 4 ppm.
• Inline water filters are designed to reduce the concentration of
fluoride and several other trace-level contaminants in tap water.

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Calculation of Parts per Million and Parts per Billion
Concentrations
•According to the EPA, when the concentration of lead in tap water
reaches 15 ppb, certain remedial actions must be taken. What is this
concentration in ppm? At this concentration, what mass of lead (μg)
would be contained in a typical glass of water (300 mL)?
Solution
•The definitions of the ppm and ppb units may be used to convert the
given concentration from ppb to ppm.
•Comparing these two unit definitions shows that ppm is 1000 times
greater than ppb
•(1 ppm = 103ppb). Thus:

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• The definition of the ppb unit may be used to calculate the
requested mass if the mass of the solution is provided.
• Since the volume of solution (300 mL) is given, its density must
be used to derive the corresponding mass.
• Assume the density of tap water to be roughly the same as that of
pure water (~1.00 g/mL), since the concentrations of any
dissolved substances should not be very large.
• Rearranging the equation defining the ppb unit and substituting
the given quantities yields:

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Check Your Learning
A 50.0-g sample of industrial wastewater was determined to
contain 0.48 mg of mercury.
Express the mercury concentration of the wastewater in ppm and
ppb units.
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