Topic 1
Quantitative Chemistry
Chapter 4 page 97
The Mole
Mole is a word which is used to describe an exact quantity.
It is comparable to other words we use to describe
quantities:
dozen = 12
gross = 144
ream = 500
score = 20
mole = 6.02 X 1023
Definition
 A mole is the measure of the amount of a substance
containing the same number of particles as there are atoms in
exactly 12 g of C-12.
 Its unit is mol and its symbol is n.
 One mole of a substance contains:
6.02 × 1023 of atoms, ions or molecules (particles).
 The number is called Avogadro’s number.
 In 12g of carbon-12 there are 6.02 × 1023 atoms. The mass of all
other atoms is relative to C-12.
A review of standards
 A relative scale is one in which all measurements are
compared to one reference or standard.
 Why carbon?
o Cheap and widely available
o Easy to isolate and purify this isotope
o Not toxic
602000000000000000000000
The mole is a pretty big
number…
This helps us to work with very small atoms.
Or peas which are very small vegetables.
Assigned Work
 10 min on side one of Molar Mass of a Substance
Review of Relative Atomic Mass (Ar)
 The relative atomic mass (Ar) of an element is defined as the
weighted mean of all the naturally occurring isotopes of an
element on a scale on which the mass of carbon-12 is taken to
be 12 exactly.
 We calculated these in Chapter 1:
Relative Mass (Ar)= ∑(Ir × % abundance)
 Relative atomic masses are found in the periodic table. The
units are u or amu (atomic mass unit).
 One amu = 1/12 the mass of a C-12 atom.
Relative Molecular mass (Mr) and Relative
Formula mass
 Just as relative atomic mass refers to atoms, relative
molecular mass (Mr) refers to molecules.
 The relative molecular mass (Mr) of a molecule is the sum
of the relative atomic masses of the elements as given in
the molecular formula.
 The relative formula mass (Mr) is identical to relative
molecular mass, except it is used for ionic compounds
instead of molecular compounds.
 Units are often not given for relative masses (some texts
use u or amu for atomic mass unit).
Molar Mass (M)
 The mass of one mole of any substance is known as molar
mass (M). It is equal to the relative atomic (molecular or
formula) mass but it is measured in units of grams per
mole.
 From this point on in the course, this is the unit that you
will need to know.
 Units (gram per mole) are written several ways:
g mol-1
or g
mol
or
g/mol
Different atoms have different masses. There is 1 mole
of atoms in each of these substances.
Therefore, different substances will have different
masses for 1 mole.
Formulas and moles
Mg(OH)2
 This is an ionic compound
 It is represented by a formula unit.
 The formula tells us how many moles of each are present in 1
mole of the substance.
Moles of atoms
Mg -1 mol
O – 2 mol
H – 2 mol
Calculating Molar Mass of Compounds
METHOD: Find the sum of the number of moles of each
element times its molar mass.
Example:
Find the molar mass (M) of Ca(HCO3)2
Element # moles
atomic mass (Ar)
Ca
1
×
40.01
=40.01
H
2
×
1.01
=2.02
C
2
×
12.0
=24.00
O
6
×
16.0
=96.00
MCa(HCO3)2 = 162.03 g mol ¯1
 The molar mass of Ca(HCO3)2 is 162.03 g mol ¯1 or 162.03g
1 mol
Assigned Work
 Complete worksheet Molar Mass of Substances.
Converting between mass and moles in the lab.
 In chemistry you are quite often asked to consider the number
of moles of a substance. Yet, we do not have a instrument to
measure moles.
 We do have electronic balances which measure mass in grams.
 We can convert moles → mass and mass to moles using the
molar mass determined from the periodic table as a
conversion factor.
 Molar Mass (M) Conversion factor:
x grams
1 mol
 Note: A conversion factor is always equal to the number 1 because the top
quantity = the bottom quantity (e.g. 12 = 1 dozen). Because of this, we can
flip it upside down and the value is not changed (e.g. 1 dozen/12).
Converting:
mol (n) →mass (g)
amount mol (n)
× grams
1 mol
mass (g)
(M, molar mass)
Example. What is the mass (grams) of 0.750 mol CO2
Step 1. Find molar mass(Mr) of CO2
C: 1 × 12.01 = 12.01 g mol¯1
O: 2 × 16.00 = 32.00 g mol¯1
molar mass of CO2 = 44.01 g mol¯1
Step 2. convert mol→ mass
0.750 mol CO2 × 44.01g = 33.0 g CO2 (3 sig fig)
1 mol CO2
Converting:
mass (g)→ mol (n)
mass (g)
amount mol (n)
×
1 mol
gram)
(inverse of M)
Example. How many moles are in 23.6 grams of CH3COOH?
Step 1. Find molar mass of CH3COOH
(12.01 × 2) + (16.00 × 2) + (1.01 × 4) = 60.06 g
Step 2. convert mass → moles
23.6g
× 1mol = 0.393 mol CH3COOH
60.06 g
Using the triangle to help solve these problems.
 If you have any two of three variables, you can use the triangle
to help you rearrange the formula.
(g mol –1)
(mol)
Example 6 on page 105
(g mol –1)
(mol)
2.117g of a shiny metallic element is 0.04071 moles.
Calculate the molar mass of this substance and determine
its identity by referring to the periodic table.
M= m
n
=
2.117g
= 52.00 g mol -1
0.04071 mol
The element is Cr
Assigned Work
 Exercises 4.2 page 106-7 Complete problems
1-11. Where there are several parts to a
questions just select a few. Work to Mastery
Significant Digits
 DEFINITION: Significant digits are measurements that include
all the certain digits for that measuring instrument, plus one
more uncertain digit.
 Certain digits are ones represented by a line (division) on the
instrument. The uncertain digit is one taken from between the
lines of the smallest division.
4.86
.
ONES
(certain)
TENTHS
(certain)
HUNDREDTHS
(uncertain)
Significant Digit Rules
 Every non-zero digit and zeros between these
digits are significant.
 How many?
24.7
0.743
all have 3 sig. figures
714
10.1
Determining the number of Sig Figs when there is
a decimal in number
 If there is a decimal, start on the right and count to the left until you
reach the last nonzero digit.
100.10
54 3 21
5 sig fig
0.0010
2 1
2 sig fig
0.001
1
1 sig fig
Determining the number of sig figs when there is
no decimal in the number
 If the number doesn’t have a decimal, start on the left and
count to the right until you reach the last nonzero digit.
10010
1234
4 sig fig
1000
1
1 sig fig (unless it is an actual count;
then it is infinite)
Scientific (Exponential) Notation
Used in science because we have very large numbers...
mass of the Earth is
6 000 000 000 000 000 000 000 000 kg
and very small numbers...
mass of the atom...
0.000 000 000 000 000 000 000 001 67 g
Using scientific notation
 This is standard notation: 234 000 000
 This same number written in scientific notation is:
2.34 x 108
A number
between
1 and 10
multiplied by
a power of
10
Converting from Scientific Notation back to
Expanded Notation (everyday numbers)
1. If you have a positive exponent, move the decimal to
the right→ (number becomes larger).
Ex. 23.546 x 102 = 2354.6
2. If you have a negative exponent, move the decimal to
the ←left (number becomes smaller).
Ex. 23.546 x 10 -2 = 0.23546
Examples
Rounding Numbers
If the first digit to be dropped is 4 or less, the last kept digit is not
changed.
78.9432 (rounded to 4 digits) → 78.94
If the first digit to be dropped is 5 or greater, the last kept digit is
increased by 1.
57.66 (rounded to 3 digits) → 56.7
When rounding, use only the first digit to the right of the kept
digit to decide on rounding. i.e. ROUND ONLY ONCE
CORRECT: 6.248 (rounded to 2 digits) → 6.2
INCORRECT: 6.248 rounds to 6.25 then to 6.3
Sig Fig and Scientific Notation
How many sig fig?
Write with 3 sig figs in
scientific notation
 300
(1)
 3.00 × 102
 184 480
(5)
 1.84 × 105
 0.15
(2)
 1.50 × 10¯1
 0.0010
(2)
 1.00 × 10¯3
Calculating with Sig Figs
Addition and Subtraction
 The final answer must have the same number of
decimal places as the number with the least
decimal places. The number of significant digits
can change.
 Example: 12.52 + 349.0 + 8.24 = 369.76
round to 369.8 (1 number after the decimal)
Try These:
1. 2.365 + 5.8
1. 8.165
=
8.2
2. 6.98 – 5.7124
2. 1.2676
=
1.27
3. 1.3333 + 5.4215
3. 6.7548
4. 5 – 2.6123
4. 2.3877
=
2
5. 8.23 + 0.65
5. 8.88
Calculating with Sig Figs
Multiplication and Division
 The final answer must have the same number of
significant digits as the number with the least
significant digits.
Example: 2.8723 × 1.6 = 4.59568
→round to 4.6 (2 significant digits)
Try These:
1. 2.365 x 5.8
13.717 = 14
2. 6.18 x 2.714
16.77252 = 16.8
3. 1.3433 x 5.421
7.2820293 = 7.282
4. 5 ÷ 2.6123
1.914022126 = 2
5. 8.23 ÷ 0.65
12.66153846 = 13
Two of the Greatest Sins with Sig Figs
1. Writing too many digits in an answer
(intermediate or final); more than justified by
the number of digits in the data.
2. Rounding-off too soon. Carry at least one more
sig. fig. in intermediate steps than should be in
the final answer or carry all digits in calculator
and round at end.
Assigned Work
Section 5.1 Exercises, page 170-1 questions 6-9
All we gotta
do is…..
Avogadro’s Constant
•6.02 X 1023 is an experimentally determined
number, called Avogadro’s Constant (L) after an
Italian, Amedeo Avogadro.
•Avogadro’s constant can be used to convert the
number of moles (n) to number of particles (N) using
the equation:
n = N
L
Rearranged: N = n x L
N
÷
n
L
6.02 x 10 23
×
Converting # of moles → particles
Page 101
mol
number of
particles
× 6.02 x 10 23 particles
1 mol
But what is
a particle?
Types of particles represented by a formula:
 Atoms→ Na, Mg, Cl
 Ions → Na+, Mg2+, Cl¯, PO43¯
 Molecules groups of atoms bonded together by covalent bonds; no
charge
(non-metals + non-metals) → CH4, PCl5, C6H12O6
 Formula Units →the empirical formula of an ionic compound
(metals + non-metals (or polyatomic ions) → AlCl3, Na2O, MgSO4,
Ba(NO3)2
mol of
particles
×
number of
particles
6.02 x 10 23 particles
1 mol
Example 1: How many atoms in 1.5 mol of Au?
1.5 mol Au × 6.02 × 1023 atoms = 9.0 × 1023 Au atoms
1 mol Au
Example 2: How many molecules in 3.2 mol CH3?
3.2 mol CH3 × 6.02 × 1023 molecules = 1.9 × 1024 CH3 molecules
1 mol CH3
number of
particles
mol of
particle
×
1 mol
6.02 x 10 23 particles
See example on page 102
How many moles in 3.1 x 1023 molecules of H2 gas?
3.1 x 1023 H2 molecules H2 ×
1mol
= 0.50 mol H2
6.02 × 1023 H2 molecules
When you have to convert the type of particle…
 How many atoms in 1.5 mol of Au?
 You are given Au and asked to find (atoms of ) Au.
 How many molecules in 3.2 mol CH3?
 You are given CH3 and asked to find (molecules of) CH3.
 How many hydroxide ions (OH-) are in 0.250 mol
Mg(OH)2?
 You are given Mg(OH)2 and asked to find OH_.
 For these situations you have to perform an extra preliminary
step to convert from mol of particle A to mol of particle B.
More on Formulas and moles
Mg(OH)2
 This is an ionic compound
 It is represented by a formula unit.
 The formula tells us how many moles of each atom and each
type of ion are present in 1 mole of the substance.
Moles of atoms
Moles of ions
Mg -1 mol
O – 2 mol
H – 2 mol
Mg2+ - 1 mol
OH- - 2 mol
total mol ions = 3 mol
More on Formulas and moles
Mg(OH)2
 We can make several conversion units using the
information in the formula unit.
1mol Mg atoms or 2 mol OH- or 3mol ions
1 mol Mg(OH)2
1 mol Mg(OH)2
1 mol Mg(OH)2
Moles of atoms
Moles of ions
Mg -1 mol
O – 2 mol
H – 2 mol
Mg2+ - 1 mol
OH- - 2 mol
total mol ions = 3 mol
Molecular Formulas and moles
CH3O
 This is a covalent or molecular compound.
 The molecular formula represents one molecule or one mole
of molecules. It tells how many moles of each atom are
present in 1 mole of the substance. It does not contain ions
but other than that, molecular formulas are like formula
units.
Moles of atoms
Moles of ions
C -1 mol
H – 4 mol
O – 1 mol
0 mol (no ions in molecules)
Back to the problem...
How many hydroxide ions (OH-) are in 0.250 mol Mg(OH)2?
Step 1. Find moles of OH0.250 mol Mg(OH)2 × 2 OH- = 0.500 mol OH1 Mg(OH)2
Step 2. Convert to particles hydroxide ions (OH-)
0.500 mol OH- × 6.02 × 1023 OH- = 3.01 × 1023 OH-ions
1 mol OH-
moles of
molecule
mol of
atoms
× # atoms
1 molecule
# atoms
x 6.02 x 10 23
1 mol
Example 3: How many atoms of oxygen in 0.71 mol SO3?
This two step problem can be written out on one line and
entered in your calculator all at once.
0.71 mol SO3 × 3 O × 6.02 × 1023 O atoms = 1.3 × 1024 O atoms
1 SO3
1 mol O
Assigned Work
 Worksheet, Finding the Number of Moles and Ions in
a Compound
 Section 4.1 Exercises, page 102 questions 1-5
 Avogadro’s Number Worksheet
Finding the mass of atoms
number of
atoms
x
mol of atoms
1 mol
6.02 x 10 23 atoms
mass of
atoms
× grams
1 mol
How many grams are in 100 atoms of copper?
Step 1. ALWAYS CONVERT TO MOLES FIRST.
100 Cu atoms x 1 mol atoms
= 1.661 x 10-22 mol
(given)
6.02 x 10 23 atoms
Step 2. Covert mol to mass.
1.661 x 10-22 mol x 63.55 g Cu = 1.06 x 10-20 g Cu
1 mol Cu
(find)
Molar Mass
Assigned Work
 Questions 12-14 on page 107
The mole is the heart of chemistry
Assigned Work
 Problems 12-14 on page 107
Stoichiometry
 Stoichiometry is the study of the relationship between
reactants and products in a chemical reaction.
 In other words, how much of one substance will react or be
produced in a chemical reaction relative to the amount of
another substance in the reaction.
Al2S3(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2S(g)
Coefficients represent
moles of substance
6mol of HCl reacts with 1 mol of aluminum sulfide to produce 2 mol
of aluminum chloride and 3 mol of hydrosulfuric acid
Why is stoichiometry important?
 Chemists need to know how much reactant is necessary to
produce a required amount of product.
 If too little of one reactant is used, then the maximum
amount of product will not be produced.
 If too much of a reactant is used then you will be wasting
it. They must use the reactants in the proper way.
 In order to do this, chemists follow the recipe. They follow
the recipe (the chemical equation) by looking at the mole
and mass ratios to make sure they get the “right mix”.
Chemical Conversions in Stoichiometry
1. mol A → mol B
2. mixed: mass A (g) ↔ mol B
3. Mass A (g) → mass B (g)
mass (g) A
× 1 mol A
gA
molar mass
mol A
mol B
× mol B
mol A
Coefficients from
balanced equation
mass (g) B
× gB
1 mol B
molar mass
Example: How many mol of HCl are required to
produce 0.45 mol of H2S?
1
mass (g) A
mol A
× 1 mol A
gA
mol B
× mol B
mol A
Balance the equation:
Al2S3(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2S(g)
Step
1
Convert mol of H2S to mol of HCl
0.45 mol of H2S × 6 mol HCl = 0.90 mol HCl
3 mol H2S
mass (g) B
× gB
1 mol B
You Try.
mass (g) A
× 1 mol A
gA
mol A
mol B
× mol B
mol A
mass (g) B
× gB
1 mol B
 Nitrogen dioxide will decompose into nitrogen and oxygen. Calculate the
number of moles of nitrogen gas produced when 0.20 mol of nitrogen
dioxide is decomposed.
You Try.
mass (g) A
× 1 mol A
gA
mol A
1
mol B
× mol B
mol A
mass (g) B
× gB
1 mol B
 Nitrogen dioxide will decompose into nitrogen and oxygen.
Calculate the number of moles of nitrogen gas produced when
0.20 mol of nitrogen dioxide is decomposed.
Balanced equation: 2NO2 → 2O2 + N2
Step
1
Convert mol of NO2 to mol of N2
0.20 mol of NO2 × 1 mol N2 = 0.10 mol N2
2 mol NO2
Example: How many mol of Al2S3 are required to produce
156.34g of H2S?
1
mass (g) A
× 1 mol A
gA
2
mol A
mol B
× mol B
mol A
mass (g) B
× gB
1 mol B
Balanced equation: Al2S3(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2S(g)
Step
1
Convert grams of H2S to mol of Al2S3
156.34g H2S × 1 mol H2S = 4.58744 mol H2S
34.08g H2S
Step
2
(carry at least one extra s.f. than ans requires)
Convert mol of H2S to mol of Al2S3
4.58744 mol H2S × 1 mol Al2S3 = 1.52914 mol Al2S3 =
3 mol H2S
1.5291 mol Al2S3
You Try.
mass (g) A
× 1 mol A
gA
mol A
mol B
× mol B
mol A
mass (g) B
× gB
1 mol B
 How many grams of calcium hydroxide are required to react with 0.063 mol
of iron (III) chloride?
You Try.
mass (g) A
mol A
× 1 mol A
gA
1
2
mol B
× mol B
mol A
mass (g) B
× gB
1 mol B
 How many grams of calcium hydroxide are required to react with 0.063 mol
of iron (III) chloride?
Balanced equation: 3Ca(OH)2 + 2FeCl3 → 3CaCl2 + 2Fe(OH)3
Step
1
Convert mol of FeCl3 to mol of Ca(OH)2
0.063 mol of FeCl3 × 3 mol Ca(OH)2
2 mol FeCl3
Step
2
= 0.0945 mol Ca(OH)2
Convert mol of Ca(OH)2 to grams of FeCl3
0.0945 mol Ca(OH)2 × 74.1 g Ca(OH)2 = 7.002 g Ca(OH)2
1 mol Ca(OH)2
7.0 g Ca(OH)2
Example: What mass of Al2S3 is required to produce
35.00g of AlCl3?
1
2
mass (g) A
mol A
× 1 mol A
gA
Balanced equation:
Step
1
3
mol B
× mol B
mol A
mass (g) B
× gB
1 mol B
Al2S3(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2S(g)
Convert grams of AlCl3 to mol
35.00 gAlCl3 × 1 mol AlCl3 = 0.262506 mol AlCl3
133.33 g AlCl3
Step
2
Convert mol of AlCl3 to mol of Al2S3
0.262506 mol of AlCl3 × 1 mol Al2S3 = 0.13125 mol Al2S3
2 mol AlCl3
CONT’D What mass of Al2S3 is required to produce
35.00g of AlCl3?
1
mass (g) A
× 1 mol A
gA
Balanced equation:
Step
3
2
mol A
3
mol B
× mol B
mol A
mass (g) B
× gB
1 mol B
Al2S3(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2S(g)
Convert moles of Al2S3 to grams
0.13125 mol Al2S3 × 150.14 gAl2S3 = 19.7063 g Al2S3
1 mol Al2S3
19.71 g Al2S3
You Try.
mass (g) A
× 1 mol A
gA
mol A
mol B
× mol B
mol A
mass (g) B
× gB
1 mol B
Aluminum reacts with sulfuric acid to produce aluminum sulfate. What mass
of aluminum sulfate can be produced from 23.45g of Al?
You Try.
1
mass (g) A
× 1 mol A
gA
2
3
mol A
mol B
× mol B
mol A
mass (g) B
× gB
1 mol B
Aluminum reacts with sulfuric acid (H2SO4) to produce aluminum sulfate. What mass
of aluminum sulfate can be produced from 23.45 g of Al?
Balanced equation: 2Al + 3H2SO4 → Al2(SO4 )3 + 3H2
Step 1 Convert grams of Al to mol Al
23.45 g Al × 1 mol Al = 0.86916 mol Al
26.98g Al
Step 2 Convert mol of Al to mol of H2SO4
0.86916 mol Al × 3 mol H2SO4 = 1.3037 mol H2SO4
2 mol Al
Step 3
Convert moles of H2SO4 to grams
1.3037 mol H2SO4 × 98.08 g H2SO4 =
1 mol H2SO4
127.9 g H2SO4
1
2
mass (g) A
3
mol A
mol B
× mol B
mol A
× 1 mol A
gA
mass (g) B
× gB
1 mol B
Aluminum reacts with sulfuric acid (H2SO4) to produce aluminum sulfate. What mass
of aluminum sulfate can be produced from 23.45 g of Al?
Balanced equation: 2Al + 3H2SO4 → Al2(SO4 )3 + 3H2
Given
23.45 g Al ×
Step
1
Step
2
1 mol Al × 3 mol H2SO4
2 mol Al
26.98g Al
Step
3
× 98.08 g H2SO4
1 mol H2SO4
?
= 127.9 g H2SO4
Assigned Work
 Complete Stoichiometry worksheet.
 There is an extra worksheet on the blog.
Percent Composition (use MASS) – page 108
 When a new compound is created in the laboratory, the
formula is determined by calculating the percent
composition. This is the percent by mass of each element
in a compound.
% mass =
grams of element × 100%
grams of compound
% Composition. Three types of problems.
You will have to determine:
1. the percent for each element, given a formula of a
compound. (given: C6H6)
2. the amount of an element, mass and amount of the
compound (given: 15.5g of C6H6)
3. the percent of each element, given masses of each
element AND mass of the compound. (given: 10.5 gram
of a compound was found to contain 1.5g of hydrogen,
7.0g of oxygen and 2.0g of nitrogen.)
Type 1. Determining % composition from a formula
Example: Calculate the percent composition of ethane C2H6
Step 1
Determine the number of grams of each atom type in 1 mole of the compound and
calculate the molar mass of the compound:
M(C2H6) = 2(12.01) + 6(1.01) = 30.08
24.02 g C
6.06 g H
Step 2
Find % masses by dividing the number of gram the elements by M(C2H6):
C = 24.02 g × 100% = 80%
30.08 g
H = 6.06 g × 100% = 20%
30.08 g
C2H6 is composed of 80.0% C and 20.0%
Note: The % should add up to 100%. Round to whole numbers or 1 decimal place,
whichever allows you to obtain 100%.
Type 2. Determining the mass of an element given the mass and
formula of the compound
Example : Calculate the mass of carbon in 82.0 g of C2H6.
Step 1. Calculate the % of carbon present in C2H6.
From the last problem it was determined that C2H6 is composed
of 80.0% C and 20.0% H.
Step 2. Find the % of the given amount that is carbon.
80% × 82.0g = grams of carbon
80 × 82.0g= 65.6 g carbon
100
Type 3. Determining % composition from given masses of both
the elements and compounds –the formula is not given
Example: A compound of 2.225 g, known to be a carbohydrate (CHO),
was analyzed and found to contain 0.890 g of carbon, and 0.148 g of
hydrogen. Determine the % composition by mass of the compound.
Step 1. Since the amount of the third element is not given find this first.
Mass of oxygen = 2.225 g -(0.890 + 0.148) g = 1.187g
Step 2. Find % of the total mass of the compound for each element.
% C = 0.890 × 100
2.225
% C = 40.0%
% H = 0.148 × 100
% O = 1.187 × 100
2.225
2.225
% H = 6.7%
% O = 53.3%
Summary of % Composition Problems (use grams)
Given:
1. a formula


Change moles (subscripts) to grams for each element.
Divide the grams of each element by the molar mass of the
compound and multiply by 100.
the mass of an element and the mass and formula of the
compound
2.


Use the formula to find the % composition as above.
Multiply the % / 100 (i.e. change to a decimal) by the given amount
of the compound.
the mass of each element and mass of sample
3.

Make sure that you have the mass of ALL the elements first. You may
have to calculate one using subtraction. Divide the mass of each
element by the total mass of the sample and multiply by100.
Assigned Work
 Complete worksheet Percent Composition
Empirical Formula (use moles)
 The empirical formula provides data about the ratio in
which atoms are bonded, without indicating the actual
numbers of atoms present.
 It can be calculated from the percent composition data or
from experimental data which uses grams.
Empirical Formula vs. Molecular Formula
 The empirical formula is the simplest whole number ratio of
atoms of different elements in a compound.
 The molecular formula is the actual number of atoms of
different elements covalently bonded in a molecule.
PENTENE
Molecular Formula: C5H10
Empirical Formula: CH2
Calculating Empirical Formula- from lab data
A compound of carbon and hydrogen is analyzed and found to
consist of 75.0% by mass carbon and 25.0% by mass of hydrogen.
Determine the empirical formula of the hydrocarbon.
Find the mass
Change to moles
Divide by the smallest
Multiply ‘til whole
Calculating Empirical Formula- from lab data
A compound of carbon and hydrogen is analyzed and found to consist of
75.0% by mass carbon and 25.0% by mass of hydrogen. Determine the
empirical formula of the hydrocarbon.
Assume that
the percents
are grams
C = 75.0 g
H = 25.0g
75.0
12.01
25.0
1.01
6.245
6.245
24.75
6.245
1
1
3.96
4
The empirical formula is CH4
Find the mass
Change to moles
Divide by the smallest
Multiply ‘til whole (or
(round)
Calculating Empirical Formula- page 110
A bright orange compound of potassium, chromium, and oxygen is analyzed to
determine its percentage composition by mass. A 1.308 g sample of the
compound was found to have 0.348 g potassium, 0.498 g of oxygen. Determine
the empirical formula of the compound.
Find mass of chromium: 1.308 – (0.348 + 0.498) = 0.462 g
K = 0.348g
0.348
39.10
Cr = 0.462 g
O = 0.498 g
0.462
52.00
0.496
16.00
0.00890
0.00888
0.00888
0.00888
0.0311
0.00888
1
1
3.5
2
2
7
Find the mass
Change to moles
Divide by the smallest
2×
The empirical formula is K2Cr2O7
Multiply ‘til whole
Summary of empirical formula calculations
(find moles)
Find the mass (by assuming the % = # grams, or using masses given)
Change to moles (divide by atomic mass of each element)
Divide by the smallest (smallest mole)
Multiply ‘til whole (if necessary- usually × 2 if dividing by the smallest
gives a 0.5 number; if the number is 4.98, round)
Determining Molecular Formula
 The molecular formula is a multiple of the empirical formula.
 To find molecular formula, you must have
The empirical formula CxHy and the molar mass of this
formula (you calculate)
2. molar mass (molecular mass, M) (given in question)
1.
Step 1. To find the multiple: a = Molecular Mass
Empirical Formula Mass
Step 2. To find the molecular formula →
Remember M.E.
a(CxHy)
Example
 Given that the empirical formula of a hydrocarbon is CH and its
molar mass is 26.02 g mol -1 determine its molecular formula.
Empirical Formula Mass: 12.01 + 1.01 = 13.02
Molecular Mass:
26.02
Step 1
a = Molar Mass
Empirical Formula Mass
Step 2
molecular formula →
2(CH)
The molecular formula of the hydrocarbon is C2H2
26.02 = 2
13.02
Assigned Work
 Section 4.3, page 113-114 Questions 1(c,d,e), 2-8
Warm-up
 Page 126-7, # 1 and 8
Solutions to warm up
mol I2 → mol HI (repeat for other 2 reactants)
mol Al2S3 → mol H2 → grams H2S
Limiting Reagents
 All of these problem and those to date assume that the
amount of both reactants is sufficient to keep the
reaction going until both are completely used up. But
what if there isn’t enough of one the reactants?
Limiting Reagent vs. Excess Reagent
Limiting Reagent/Reactant
 The reactant in a chemical reaction that limits, or determines,
the amount of product that can be formed.
 The reaction will stop when all of the limiting reactant is used up
or “runs out”.
Excess Reagent/Reactant
 The reactant in a chemical reaction that remains or is “left over”
when a reaction stops (when the limiting reactant is completely
consumed).
 The excess reactant remains because there is nothing with which
it can react.
You will be asked to:
1. Identify the limiting reagent
2. Identify the excess reagent
3. Calculate the product
4. Calculate how much excess remains (is left over).
excess
amount
limiting
reagent
excess
reagent
product
Car Stoichiometry
1
4
1
reactant
reactant
product
You have 15 car bodies and 72 tires.
a) What will you run out of first? (i.e. what is limiting reagent?)
b) What will be left over and how many? (i.e. How much excess
reagent is there?)
Do you have 60
tires? YES.
a) 15 car bodies x 4 tires = 60 tires are needed Therefore this is
your excess reagent
1 car body
The car bodies are
limiting .
b) 72 tires – 60 tires = 12 tires left over Amount of
excess
Example 1
You have 6.70 mol of Na reacting with 3.20 mol Cl2.
Identify the limiting reagent
b. Determine the moles of NaCl produced.
c. Calculate the moles of excess reagent remaining after the
reaction.
Write the balanced reaction and given moles:
2Na
+ Cl2 → 2NaCl
a.
6.70 mol
3 .20mol
Example 1 cont’d –Identify the limiting reagent
This is really like several stoichiometry problems:
2Na
+ Cl2
→ 2NaCl
Initial
6.70
3.20
0
Ratio
2
1
2
Coefficients from
equation
a) IF Na is the limiting reagent it will all react. How much Cl2 is needed for this?
6.70 mol Na × 1 mol Cl2 = 3.35 mol Cl2 needed.
2 mol Na
Do you have 3.35
mol Cl2?
NO. You only have 3.20, so Cl2 is the limiting reagent. (Na is in excess)
IF the answer had been YES then Na would be the limiting reagent.
Example 1 cont’d –Determine the moles of NaCl produced
E.R.
2Na
L.R.
+ Cl2
→ 2NaCl
Initial
6.70
3.20
0
Ratio
2
1
2
b) How much product is produced?
Coefficients from
equation
The limiting reagent
determines this
product
L.R.
3.20 mol Cl2 × 2 mol NaCl
1 mol Cl2
= 6.40 mol NaCl is produced.
Example 1 cont’d – Calculate the moles of excess reagent
E.R.
L.R.
2Na
+ Cl2
→ 2NaCl
Initial
6.70
3.20
0
Ratio
2
1
2
c) How much excess reagent (i.e. Na) is left-over?
3.20 mol Cl2 × 2 mol Na = 6.40 mol Na reacts.
1 mol Cl2
Amount left over = initial – amount that reacts
= 6.70 mol - 6.40 mol
= 0.30 mol Na left over
Find the amount of
excess reagent
needed
Subtract from
initial to find left
over
Example 2
Consider the reaction:
4P
7.0 mol
+ 5O2
→ 2P2O5
8.0mol
Identify the limiting reagent
b. Determine the grams of P2O5 produced.
c. Calculate the moles of excess reagent remaining after the
reaction.
a.
Example 2 cont’d -Identify the limiting reagent
E.R.
4P
L.R.
+ 5O2
→ 2P2O5
Initial
7.0
8.0
0
Ratio
4
5
2
Coefficients from
equation
a) IF P is the limiting reagent it will all react. How much O2 is needed for this?
7.0 mol P × 5 mol O2 = 8.8 mol O2 needed/used.
4 mol P
Do you have 8.8 mol
of O2?
NO, only have 8.0 so
this is L.R.
If you had started with O2 ...
8.0 mol O2 × 4 mol P = 6.4 mol P needed/used. Do you have 6.4 mol
of P?
5 mol O2
YES., have 7.0 mol
so P is excess
reagent.
Example 2 cont’d
E.R.
4P
- How many grams of P2O5 is produced?
L.R.
+ 5O2
→ 2P2O5
Initial
7.0
8.0
0
Ratio
4
5
2
L.R.
8.0 mol O2 × 2 mol P2O5 =
5 mol O2
Coefficients from
equation
3.20 mol P2O5 is produced.
2-P = 2(30.97) = 61.94
5-O = 5(16.00) = 80.00
141.94
g/mol
3.20 mol P2O5 × 141.94 g P2O5 = 454.208gP2O5
1 mol
=
450g P2O5
Example 2 cont’d –Calculate the moles of excess reagent
E.R.
4P
L.R.
+ 5O2
→ 2P2O5
Initial
7.0
8.0
0
Ratio
4
5
2
8.0 mol O2 × 4 mol P = 6.4 mol P needed/used.
5 mol O2
The amount of
excess reagent
needed was done in
step 1
Subtract the
amount need from
initial to find left
over.
Amount left over = initial – amount that reacts
= 7.0 mol - 6.40 mol
= 0.60 mol P left over at end of reaction
Example 2 cont’d – USING AN IRA TABLE
E.R.
4P
Initial
7.0
Reacts
Ratio
-4x4
-4x=6.4
After
0.60
L.R.
+ 5O2
8.0
5
-5x =8.0
-5x
0
→ 2P2O5
0
+2x=
+2x
2 3.2
3.2
8.0 mol O2 × 4 mol P = 6.4 mol P needed/used.
5 mol O2
Amount left over = initial – amount that reacts
= 7.0 mol - 6.40 mol
= 0.60 mol P left over at end of reaction
8.0 mol O2 × 2 mol P2O5 = 3.2 mol P2O5 produced.
5 mol O2
Assume that all
of O2 gets used.
5x=8.0
x= 8.0
5
x= 1.6
4x = 4(1.6) = 6.4
2x = 2(1.6) = 3.2
Example 2 cont’d – Let’s assume that P is L.R.
L.R.
4P
Initial
7.0
Reacts
Ratio
-4x=7.0
-4x4
After
0
E.R.
+ 5O2
8.0
-5x
-5x=8.75
5
-0.75
→ 2P2O5
0
+2x=
+2x
2 3.2
3.5
Since you can not have a negative number, our assumption
that P is limiting reagent is wrong.
Assume that all
of P gets used.
4x=7.0
x= 7.0
4
x= 1.75
5x = 5(1.75) = 8.75
2x = 2(1.75) = 3.5
Summary of Limiting Reagent Calculations
WORK IN MOLES- WRITE A BALANCED EQUATION

Identify the limiting reagent-the one that you don’t have enough of in initial
amounts
1.
2.

Start with initial amount of either reactant
Find amount of the other reactant that is required using coefficients from the
equation. If this amount is greater than what you started with, it is the LR.
Identify the excess reagent -the one that you have enough of
1.

Determined in last step
Calculate the theoretical yield (how much product)
1.

Start with LR, convert to desired product using coefficients from the equation
Calculate how much excess remains (is left over).
1.
2.
Start with L.R. to find how much E.R. reacts
Subtract this amount from initial amount of E.R.
Assigned Work
 Section 4.5 Exercises 10-13 on page 127
 REMEMBER Mass must be converted to moles first.
Percent Yield
 When we calculated the amount of product produced in previous
questions:
5.4 mol of Al reacts with 8.0 mol Cl2 to form aluminum chloride.
a. Identify the limiting reagent
b. Determine the moles of AlCl3 produced.
 We were considering the theoretical yield.
 In reality, the experimental yield (actual amount) that you get during
an experiment is significantly less.
 One way to express how successful the preparation of the product
was is to calculate the percentage yield.
Percent yield = experimental (actual) yield × 100
theoretical yield
Example 1
2HCl + Ca(OH)2 → 2H2O + CaCl2
a)
Two students did an experiment where they added 6.95 g of Ca(OH)2 to
an excess of HCl. Their chemistry teacher told them that they would get
the same mark as their percent yield. Since they knew that Ca(OH)2 was
the limiting reagent (HOW?), they were able to calculate the expected
amount of CaCl2 (they expected 10.4g). But to their dismay they were
only able to produce 8.76 g of CaCl2. What mark did they get?
Experimental
% yield = 8.76 g × 100 = 84.2 %
yield
10.4
-what they got
Theoretical yield
-what they should have
got
Example 2
 In the Haber Process, ammonia is made from nitrogen gas and
hydrogen gas.
N2(g) + 3H2(g) → 2NH3(g)
If 45.00g of ammonia is produced when 42.03 g of nitrogen reacts with
excess hydrogen gas, what is the % yield of the experiment?
Solution:
1. Find the theoretical yield
1.
2.
3.
4.
Identify limiting reagent (given in directly)
Convert mass to moles
Find the theoretical yield using the L.R.
Convert product moles to mass
2. Calculate the % yield.
Example 2 cont’d Find the theoretical yield using L.R.
L.R.
N2(g) + 3H2(g) → 2NH3(g)
1
3
2
2. Convert mass of L.R. to moles
42.03g N2 × 1mol =
28.02 g
1.500 mol N2
3. Find the theoretical yield using the L.R.
1.500 mol N2 × 2NH3 =
1 N2
3.000 mol NH3
4. Convert product moles to mass
3.000 mol NH3 × 17.03 g =
1 mol
51.09g NH3
Theoretical
Yield
Example 2 cont’d Calculate the percentage yield
L.R.
N2(g) + 3H2(g) → 2NH3(g)
The theoretical yield of NH3 is 51.09g.
The experimental (actual) amount was 45.00g
Percent yield = experimental (actual) yield × 100
theoretical yield
% yield = 45.00g × 100 = 88.08%
51.09g
Given in
question
Assigned Work
 Limiting Reagent Worksheet
 Section 4.5 Exercises, page 127-8 # 14, 15,17
Matter
Impure
Substances
Pure substances
(Mixtures)
Homogenous
Elements
Compounds
Heterogenous
Mixtures
Mixtures
(SOLUTIONS)
Metals
Nonmetals
Ionic
Molecular
Solutions (page142)
 Solutions are homogeneous mixtures; impure substances.
 Solutions are made up of solvents and solutes.
 The substance in greater quantity is the solvent. For our
course, we will study aqueous solutions (aq)– solutions in
which water is the solvent.
 The substance in less abundance is the solute. The solute is
dissolved in the solvent (i.e. the solvent molecules surround
the solute particles and keep it suspended).
 Solutions are clear but not colourless.
 For our course, the majority of the solutes
will be ionic compounds and the solvent will
be water.
Dissolving at a Molecular Level
The Dissolving Process
 The ionic bond (electrostatic attraction) between the
positively and negatively charged ions, and the hydrogen
bonds (also attractive forces) between the water
molecules, must be weaker than the attractive forces
(ion-dipole) between the water molecules and the ions of
the ionic compound.
 The water molecule is polar; the hydrogen ends being δ+
and the oxygen end being δ-.
 The hydrogen ends of the water molecules are attracted
to and surround the anions Cl-, while the oxygen end of
each water molecule is attracted to the cations (Na+).
A dissociation equation is a balanced equation that shows all the
ions that are produced when an ionic compound dissolves.
Equations must be balanced, include charges for the ions and
state subscripts.
NaCl(s)→ Na+(aq) + Cl-(aq)
Mg(NO3)2(s) → Mg2+(aq) + 2NO3-(aq)
Other solution types: liquids + liquids
Miscible vs. immiscible
 Oil and water don't mix. Pouring 10 mL of olive oil into 10
mL of water results in two distinct layers, clearly
separated by a curved meniscus.
 Each layer has the same volume and essentially the same
composition as the original liquids. Because very little
mixing has apparently occurred, the liquids are called
immiscible or unmixable .
 Pouring alcohol into water results in a single liquid phase;
a solution. No meniscus forms between the alcohol and
the water, and the two liquids are considered miscible.
Other solution types: liquids + liquids
Alloys: solids + solids




Alloys can be considered
mixtures of metals with metals.
Alloying metals together alters
the melting point and improves
properties such as luster,
conductivity, malleability,
ductility and strength.
Iron, the most important metal,
is only used when it is alloyed.
Brass and bronze are not
elements but alloys.
Saturation and Solubility
 Some solutes are more soluble
than others (i.e. you can
dissolve more KNO3 than KCl in
the same amount of water at
600 C).
 The amount of substance that
can be dissolved is dependent
on temperature and the type
of compound.
 Plotting the amount of solute
that can be dissolved in 100cm3
of water at varying
temperatures produces a graph
called a solubility curve.
Unsaturated, Saturated, Supersaturated
 When a solution can not
hold any more solute, it is
said to be saturated. The
graphed line represents
saturated.
 Quantities of solute below
the line would result in an
unsaturated solution.
 Quantities of solute above
the line would precipitate
out of solution or under
special conditions result in a
super-saturated solution.
Concentration (c)
 The concentration of a solution is a quantitative expression
describing how much solute is dissolved in the solution.
 Concentration is the amount of solute in a given volume of
solution, expressed in g dm-3, or mol dm-3 (or mol L-1).
 There are many ways to express concentration:
 Molarity: most important-the number of moles of solute
dissolved in 1L of solution (mol dm-3 or mol L-1)
 Concentration can also be g cm-3
 For low concentrations ppm (parts per million; mg dm-3) or
ppb (μg dm-3)
Concentration (C)
Molarity (C) = mol of solute (n) =
volume of solution (V)
mol = mol = mol/L = mol L-1 = mol dm-3
L
dm3
 You will be asked to find any one of the three variables, given the other
two.
convert gram →mol and ml →L before applying equation
It is easiest to work in L and change to dm-3 after
mol = mol/L = mol L-1 = mol dm-3
L
1000mL = 1 L
1000cm3 = 1dm3
1mL = 1 cm3
grams
Example 1
Find the concentration of :
a) 2.0 mol of NaCl in 8.0 dm3 of water.
mL
C = n = 2.0 mol = o.25mol dm-3
V
8.0 dm3
b) A solution of 6.5g of NaCl dissolved in 35.5 mL of water.
6.5 g NaCl × 1 mol = 0.1112 mol NaCl
58.44g
35.5 mL × 1 L = 0.0355 L
C = 0.1112 mol = 3.1 mol L-1
1000mL
0.0355 L
grams
Example 2
How many grams of AgNO3 (169.88g mol-1) are required
to make 100.0 cm3 of a 0.20 mol dm-3 solution?
You are given volume and conc, so find moles (CV) and convert to grams.
C= 0.20 mol dm-3
V = 100.0 cm3 →0.100dm3
0.100 dm3 × 0.20 mol = 0.0200 mol
1 dm3
0.0200 mol × 169.88g = 3.3976 g AgNO3 = 3.4 g AgNO3
1 mol
cm3
Concentration of Ions in Solution
 Often the concentration of the ion is what we are concerned with,
rather than the overall solution concentration . Once a compound
dissolves, the ions act independently of each other.
Example: Calculation of Ion Concentration
What is the concentration of Al3+ and Cl- in a 2.4mol L-1 solution of AlCl3?
Write the dissociation equation (breaks compound into two ions)
AlCl3 (s) → Al3+(aq) + 3 Cl-(aq)
molar ratio:
1
1
3
conc mol L-1
2.4
2.4
3 × (2.4) = 7.2
Dimensional Analysis Method: Beginning with solution molarity, calculate the ion
conc.
2.4M AlCl3 × 3 mol Cl- = 7.2M
1 mol AlCl3
ANSWER:
[Cl-] = 7.2 mol L-1
2.4M AlCl3 × 1 mol Al3+ = 2.4M
1 mol AlCl3
[Al3+] = 2.4 mol L-1
[ ] means conc
grams
IF YOU ARE GIVEN GRAMS YOU MUST FIND
THE MOLARITY FIRST.
Example: What is the concentration of lead in a solution made by
dissolving 3.54g Pb(NO3)2 in 500.0mL?
Step 1. Find molarity of the solution:
3.54g Pb(NO3)2 × 1 mol = 0.01069mol Pb(NO3)2
331.21g
0.01069mol = 0.0214 mol L-1 Pb(NO3)2
0.500L
Step 2. Write the dissociation equation and calculate [ion]:
Pb(NO3)2 →
Pb2+
+ 2NO3⁻
[Pb2+ ] = 0.0214 mol L1
1
2
0.0214
0.0214
0.0428
Note: You may be given the ion conc and asked to find the solution molarity.
Assigned Work
 Worksheet Concentration of Solutions Part A and C
Preparing Solutions cont’d
Reading the Meniscus
Measurement
is 50.0mL
Why one
decimal place?
4
Volumetric mark
3
2
1
beaker
Erlenmeyer
flask
graduated
cylinder
volumetric
flask
Pipette
(or pipet)
When selecting glassware...
 % uncertainty changes with the amount of material being
measured in labware such as a graduated cylinder.
 In order to minimize percentage uncertainty, choose the
smallest measuring device that will hold the required volume
being measured.
 Erlenmeyer flasks, and beakers are NOT measuring devices
except for rough estimates.
Using a Pipette Bulb
Steps to Making a Solution
Determine the number of grams solute needed.
Select the volumetric flask of the correct size.
Add the solute to the flask.
Add distilled water until the meniscus of the solution
reaches the volumetric mark of the flask (if you go past the
mark the solution will be too dilute and you need to start
over).
5) Stopper the flask and invert 7-10 times to thoroughly mix
the solution.
6) Label with molarity and necessary WHMIS safety
information if being stored.
1)
2)
3)
4)
DILUTIONS
(page 145)
 Many solutions are made from stock solutions.
 A stock solution is a highly concentrated solution.
 When you make a dilution you are reducing the concentration
of a solution by adding water to it.
 The number of moles of solute does not change when water is
added.
Dilution Formula
C1 × V1
C1
V1
C2
V2
=
C2 × V2
= molarity of stock or concentrated solution
(this is always the most concentrated solution or the highest
molarity)
= volume of stock solution
= molarity of final solution or “dilution”
= final volume after water is added (this will always be
greatest volume)
 Note: You may be asked for the amount of water that is to be added
to make the dilution.
Amount of water = V2 - V1
Dilution Calculations
Example: How would you prepare 100.0ml of 0.40 mol L-1 MgSO4
solution using a stock solution of 2.0mol L-1 MgSO4?
C1 = 2.0M
V1 = ?
C2 = 0.40M
V2 = 100mL
2.0M × V1 = 0.40M × 100mL
V1 = (0.40M × 100mL)
2.0M
V1 = 0.05 mL of stock solution required
Note that you do NOT have
to convert to L in this
formula since units cancel.
The same units MUST be
used on both sides
Steps to Making a Dilution
 Calculate the volume of stock solution required.
 Pick the appropriate sized volumetric flask.
 Measure the stock solution using a graduated cylinder or a
pipette.
 Rinse the cylinder with distilled water into the flask.
 Reading the meniscus, fill up to volumetric mark with distilled
water.
 Stopper the flask and invert 7-10 times to mix well.
Steps to Making a Dilution
Solution Stoichiometry page 146-150
Types of Problem:
1. Using concentration and volume (find moles)
2. Using grams and molar mass (find moles)
3. Limiting Reagents and Products.
Assigned Work: Section 4.8 exercises page 151
Gas – A state of matter
KINETIC THEORY
 The state of a substance is determined
by the space between the particles. The
more energy, the further apart the
particles.
 Solids particles are close together, have
the least energy and are very organized
and hold their shape.
 Liquids particles have more energy and
can flow past each other, but they will
take the shape of the container.
 Gases particles are the most energetic
with the greatest amount of space
between them. They conform to the
shape of the container and will fill the
container.
Avogadro’s Law
Equal volumes of gases at the same temperature and pressure
contain equal number of particles V1 = V2
n n
If you have 100cm3 of three gases at the same temp and press
they will have the same number of molecules AND the same
number of moles.
H2
HCl
O2
Using Avogadro’s Law in Stoichiometry
 Ratio of moles of products and reactants are the same as
the volume ratios. This means that you can use volumes
instead of moles in stoichiometry and other limiting
reagent problems.
28cm3
14cm3
28cm3
Example
N2(g) + 3H2(g) → 2NH3(g)
10.0 cm3
How many cm3 of H2 and N2 gas are required to
produce 10.0 cm3 of NH3 gas?
10.0 cm3 NH3 × 3 H2 = 15.0 cm3 H2
2 NH3
ratio from balanced equation;
VOLUME WORKS LIKE MOLES.
Standard Molar Volume (Vm)
 At standard conditions (STP) there are 22.4 L mol -1
 This applies for all gases. It doesn’t matter if it is H2(g),
CO2(g) , O2(g).
Converting: Litres→mol
Example 1.
15.3 L
How many moles are in 15.3L of CO2 at STP?
× 1mol
22.4 L
= 0.683 mol CO2
YOU TRY How many moles are in 0.45L of He gas at STP?
0.45 L × 1mol = 0.020 mol He
22.4 L
The type of gas or formula does NOT matter. Always use 22.4
L/mol for STP.
Converting: mol→Litres
Example 1.
at STP?
How many litres are in 0.450mol of CH4
0.450mol × 22.4 L = 10.1L CH4
1mol
YOU TRY. How many litres are in 3.5mol of H2 gas at
STP?
3.5 mol × 22.4 L
1mol
= 78L H2
Assigned Work
 Worksheet on Avogadro’s Number and Molar Volume
 Page 130-1; # 1-4
 Use conversion units on previous pages of text.
Important Variables for Gases
 pressure (P)
SI unit is Pascals (Pa) or kilopascals (kP)
 volume (V)
unit is dm3
 temperature (t)
unit is Kelvin (K)
 moles (n)
What is pressure?
•Pressure is the force exerted upon
an object per unit of surface area.
(P=F/A)
•As they collide with the inner
surface of a basketball, gas
particles exert pressure. The more
particles that collide, the greater
the pressure.
1kPa = 1000Pa
Standard Temperature and Pressure (STP) and units
 Standard conditions are used to measure molar volume for gases since
gases expand depending on the temperature and pressure.
 STP – 00C and 101.3kPa (1.013 × 105 Pa )
Pressure Conversion Units
1kPa = 1000Pa
101.3kPa = 1.013 × 105 Pa = 1 atm
=760mm Hg
Volume Conversion Units
1L = 1dm3 = 0.001 m3 = 1000mL = 1000 cm3
1cm3 = 1mL
1000L = 1000 dm3 =1 m3
Temperature Conversions
K = 0C + 273
Video of lungs
Interactive animation
Relationship of Pressure ↑ and Volume↓
 The pressure exerted by a given mass of gas at a constant
temperature is inversely proportional (xV = 1/x P) to the
volume occupied by the gas.
If you double (×2) the volume, you ½ the pressure.
 2V = 1/2P
 If you triple the volume, you have 1/3 of the pressure, etc.
 3V = 1/3P
Plot 1/P
(inverse of P)
Now a straight line
that intercepts zero
Relationship of Volume↑ and Temp ↑
 At a constant pressure, the volume of a given mass of gas
directly proportional to its temperature.
 If you double (× 2) the temperature,
you double (× 2) the volume.
 2T = 2V
 3T = 3V
Pearson Interactive Animation Hindenburg
Relationship between Pressure and Temperature
 At a constant volume, the pressure of a given mass of gas
directly proportional to its Kelvin temperature.
pressure, etc.
 2T = 2P
 3T = 3P
Pressure P
 If you double (×2) the temperature, you double (×2) the
Review of Gaseous Relationships
P1V1 = P2V2
T1
T2
Double Variable 1 (2x)
What happens to variable 2?
P1V1 = P2V2
2 × PRESSURE
1 × 1 = 2 × V2
1/2 VOLUME
V2 = ½
(assume that P1 and V1= 1)
V1 = V2
T1 T2
1 = V2
2
1
2 × TEMPERATURE
P1 = P2
1 = P2
T1 T2
1
2
2 × TEMPERATURE
V2 = 2
DOUBLE (2×) VOLUME
P2 = 2
DOUBLE (2×) PRESSURE
Example Problem 4
 If the volume on a 8.00 dm3 sample of gas is halved,
calculate the new pressure (P2) of the gas (assuming
constant temperature).
Pressure and volume are inversely proportional. If you
half the volume, you double the pressure.
½ V = 2×P
Check: (assume that P1 = 1)
P1V1 =P2V2
0.5 × 8.00 = 2P
1 × 8 = P2 × 4
4 = 2P
2=P
Example Problem 5
(# 3, page 141)
 During an experiment to study the Boyle’s Law the following results were
obtained.
232
85
234
232
232
a)
b)
State which two gas measures were kept constant in the experiments.
temperature and mass
State which reading of volume (1-5) was incorrect.
Since P1V1 = P2V1 all PV should be the same
#2 is incorrect since all P × V ≈ 232, except #2 (0.526 × 444 = 85)
The Combined Gas Law
USE TO CALCULATE
CHANGES IN
BEFORE AND AFTER
SITUATIONS of the
SAME gas sample
There are several gas laws that are combined to show the relationship of
all the variables affecting gases.
Avogadro’s Law: V1 = V2
n
n
Boyle’s Law P1V1 = P2V2 Charles' Law V1 = V2 Gay Lussac’s Law P1 = P2
T 1 T2
T1 T2
Can be combined into:
P1V1 = P2V2
nT1 nT2
 T in K
 n = moles (is often omitted if you are dealing with one sample)
 Other variables must be in the same unit on both sides but the unit can
vary.
Example 1
P1V1 = P2V2
T1
T2
 A cylinder of compressed oxygen has a volume of 30.0L and 100.0 atm
pressure at 27 0C. The cylinder is cooled until the pressure is 5.0 atm. What
is the new temperature in the cylinder?
P1 = P2
T1 T2
P1= 100.0
P2= 5.0
V1= 30.0 ←Doesn’t change→ V2=30.0
T1= 27+ 273
T2= ?
= 300
100 = 5.0
300 T2
100 x T2 = 5.0 × 300
T2 = 5.0 x 300
100
T2= 15 K
0C = 15- 273 = - 2580C
Example Problem 2
 A gas cylinder containing 50.0dm3 of hydrogen has a
pressure of 70.0 atm at 200C. What volume of hydrogen
would this produce at 5.0 atm?
P1= 70.0 atm
P2= 5.0 atm
V1= 50.0 dm3
V2 = ?
T1= 20.00C ←Doesn’t change→T2= 20.00C
P1V1 = P2V2
T1
T2
P1V1 = P2V2
70.0 × 50.0 = 5.0 × V2
70.0 × 50.0 = V2
5.0
V2 = 700 dm3
Example Problem 3
 An air balloon inflated in air-conditioned room at 270C is
heated to 570C. The final volume is 4.4L. What was the
volume at the time it was filled?
 T1 = 270C + 273 = 300K
 V1 = ?
 T2 = 570C + 273 = 330K
 V2 = 4.4L
Use Kelvin
P1V1
T1
= P2V2
T2
V1 = 4.4
300 330
V1 = 4.4 × 300
330
V1 = 4.0L
Assigned Work
 Complete Section 4.7 Exercises pages 141-142; #1, 4, 5, 6,
11, 12, 13, 15
The Ideal Gas Law (pg 137)
PV = nRT
 USE THESE UNITS (because of the constant R)!!
 P in kPa
 V in dm3 or L
 T in K
 n is mol
The constant R:
R = 8.314 J K-1 mol-1
USE WHEN TRYING
TO FIND MASS OR
MOLES IN PRESSURE
PROBLEMS THAT ARE
NOT BEFORE AND
AFTER
Example 1
What volume will 52.0 g of carbon dioxide gas occupy at a temperature
of 24°C and 2.034 atm?
 P = 2.034 atm x 101.3kPa = 206.04 kPa
1 atm




V= ?
R= 8.314 J K-1 mol-1
T= 240C K= 24 + 273 = 297K
n= 52.0 g x 1 mol = 1.182mol
44.01g
PV = nRT
(206.04 kPa)V = (1.182mol)(8.314 J K-1 mol-1 )(297K)
V = (1.182mol)(8.314 J K-1 mol-1 )(297K) = 14L
(206.04 kPa)
Example 2- Using gas laws to find Mr
 3.376g of a gas occupies 2.368dm3 at 17.60C and a pressure of 96.73kPa.
What is the molar mass of the gas?
P = 96.73kPa
V= 2.368dm3
R= 8.314 J K-1 mol-1
T= 17.60C K= 17.6 + 273 = 290.6K
n=?
PV = nRT → (96.73kPa)(2.368dm3)= n (8.314 J K-1 mol)( 290.6K)
n = (96.73kPa)(2.368dm3) = 0.0950mol
(8.314 J K-1 mol )(290.6K)
Mr = grams = 3.376g (given)
mol
0.0950mol
= 35.5g mol -1
Only ideal gases obey the gas laws.
 The Kinetic Theory describes “ideal” gases as ones whose
particles:
 Have no size; occupy zero volume
 Have no intermolecular forces; are not attracted to each
other
 Have elastic collisions; their kinetic energy does not change
when they collide.
 Real gases are not ideal; they do have some attractive forces
(otherwise they could not condense into liquids).
 Real gases behave most like an ideal gas (and follow the gas
laws) at high temperatures and low pressures.
WHY under these conditions?
Which one behaves ideally?
Assigned Work
Complete Section 4.7 Exercises pages 141-142; #2, 7, 8, 9, 10, 14
Page 157 #21, 22,
Also graphing problems 1-5 on page 181
That’s all for
this unit!!