Download CHEMISTRY 2202 UNIT 1 STOICHIOMETRY

CHEMISTRY 2202
UNIT 1
STOICHIOMETRY
Introduction
Chemistry is both a qualitative (describing with words)
and a quantitative (describing with numbers) science.
Chemical equations are ‘recipes’ that show chemists
what we need and what will be produced in a reaction.
eg. H2SO4(aq) + 2 NaOH(aq)  Na2SO4(aq) + 2 H2O(l)
Introduction
This balanced equation helps us answer questions like:
●
How much sulfuric acid is needed to neutralize
30 mL of 5.00 M sodium hydroxide?
●
What volume of water will be produced when
30 mL of sodium hydroxide reacts with 10 mL of
sulfuric acid?
Chemical equations help us “count” molecules or
QUANTIFY chemical reactions.
Unit Outline
STOICHIOMETRY:
- the calculation of quantities USED IN or PRODUCED
by a chemical reaction.
3 Sections:
Part 1 - Mole calculations
(ch. 2 & 3 in text)
Part 2 - Using mole ratios in chemical equations (ch. 4)
Part 3 - Solution stoichiometry
(ch. 6)
PART 1 – MOLE CALCULATIONS
●
Isotopes and Atomic Mass
●
Average Atomic Mass
●
●
●
●
●
●
●
Molar Mass
Mole Calculations
Molar Volume
Percent Composition
Empirical Formulas
Molecular Formulas
Formula of a Hydrate
ISOTOPES and ATOMIC MASS
Atomic number:
▫The # of protons in an atom or ion.
+
▫In a neutral atom: #p = #e
-
eg.
●
Atomic number of C is _____
ISOTOPES and ATOMIC MASS
•Mass number:
▫The # located in the bottom of each box on the
periodic table (has decimal places).
▫Mass Number = #p+ + #no
ie. mass # - #p+ = #no
eg.
● mass number of C = _____
● # of no in C = _____
Your Turn!!!
Atom
Atomic #
Atomic Mass
#
Carbon (C)
6
12
Carbon (C)
6
13
Carbon (C)
6
14
+
#p
-
#e
o
#n
ISOTOPES AND ATOMIC MASS . . .
•ISOTOPE
▫Atoms that have the same number of protons
(ie. same element) but a different NUMBERS OF
NEUTRONS are called isotopes
ISOTOPES AND ATOMIC MASS . . .
▫Isotopes have different mass numbers.
eg. Carbon has 3 naturally occurring isotopes:
●
carbon-12 
98.9 % of all carbon in the world
●
carbon-13 
1.1 % of all carbon in the world
●
carbon-14 
0.0000000001 % of all carbon
▫When these isotope values are averaged, with their percent
abundance in nature, we get the number that appears on the
periodic table (12.01 g/mol).
ISOTOPES AND ATOMIC
MASS . . .
ISOTOPE NOTATION:
ISOTOPES and ATOMIC MASS
Example
Carbon-12
Carbon-13
Carbon-14
How many neutrons are in each isotope?
ISOTOPES AND ATOMIC
MASS . . .
Name each isotope of magnesium.
How many neutrons are in each isotope?
YOUR TURN!!!
A. hydrogen-2
B.
YOUR TURN!!!
Write the isotope notation for an isotope of:
+
A. Copper that has 29 p and 35 n
+
o
o
B. Vanadium that has 23 p and 28 n .
AVERAGE ATOMIC MASS (AAM)
ANALOGY:
●
●
●
●
The average mark you receive in this course is a
weighted average of all your scores.
Your final grade depends on how well you do in
different pieces of work and their % of the total
grading scheme.
Your final mark is NOT a straight average.
On the periodic table, the atomic mass numbers
are also an average, based on different types of
isotopes and their relative percent abundance.
AVERAGE ATOMIC MASS . . .
●
This average is based on the concept of the
ATOMIC MASS UNIT (AMU)
●
The symbol for the AMU is μ.
●
1 μ = 1/12 the mass of a carbon-12 atom
(from IUPAC conference, 1961)
ie. 1 carbon-12 atom has a mass of 12 μ.
●
On the periodic table, carbon’s atomic mass is
an average, 12.01, expressed in g/mol.
AVERAGE ATOMIC MASS . . .
To calculate average atomic mass (AAM):
1.Change percent abundance value to decimal
value (divide by 100).
2. Multiply the atomic mass by its % abundance
for each isotope.
3. Add these numbers together.
4. Use significant digits.
In other words:
AAM = (exact atomic mass x % abundance as a decimal) +
(exact atomic mass x % abundance as a decimal) +
(exact atomic mass x % abundance as a decimal) + .…
AVERAGE ATOMIC MASS . . .
Calculate the average atomic mass of lithium from the
given data:
ISOTOPE
ATOMIC MASS
% ABUNDANCE
Lithium-6
6.01513 u
7.420
Lithium-7
7.01601 u
92.58
AVERAGE ATOMIC MASS . . .
Calculate the average atomic mass of Mg given that it has
three isotopes:
●
●
●
magnesium-24 → 23.98 μ; 78.60% abundance
magnesium-25 → 24.99 μ; 10.11 % abundance
magnesium-26 → 25.98 μ; 11.29 % abundance
Naturally occurring magnesium exists as a mixture of three
isotopes. Mg-24 has an atomic mass of 23.985 amu and a
relative abundance of 78.70 %. Mg-25 has an atomic mass
of 24.985 amu and a relative abundance of 10.13%. The
average atomic mass of magnesium is 24.31 amu.
Calculate the atomic mass of the remaining isotope.
For you to do!
Page 45-#2
Page 46-#2
The Mole
•INTRODUCTION:
▫Chemists face the
difficult task of trying to
measure extremely small
particles, such as:
●
●
●
FORMULA UNITS and IONS for ionic compounds
MOLECULES for molecular compounds
ATOMS for elements
The Mole
eg.
23
1 teaspoon of water → 1.67 x 10 molecules
(5.0 mL)
•A device that could count molecules at a rate of 1
million per second would take over 5 billion years to
count the molecules in a teaspoon of water!!!
The mole
Names to represent some common quantities:
Item
Quantity
Amount
Gloves
Pair
2
Soft-drinks
six-pack
6
eggs
dozen
12
pens
Gross (12 dozen)
144
paper
ream
500
The Mole
IUPAC to the RESCUE!!
•Chemists use the concept of the MOLE as a
convenient way to deal with counting huge
numbers of particles.
One mole is the number of atoms contained
in exactly 12 g of carbon-12.
The Mole
1 mol = 6.02 x 1023 particles
(atoms, ions, molecules, formula units)
•This number is referred to as Avogadro’s Number
(in honour of scientist Amadeo Avogadro who
discovered it), and is written:
NA = 6.02 x 1023 particles/mol
The letter n is used to represent
# of moles
The Mole
•Just how big is this number?
•Put it this way . . .
23
▫If we had 6.02 x 10 marbles ( 1 mol of marbles), it
would be enough to cover all of Earth to a depth of 80
km!
23
▫If you won 6.02 x 10 dollars, it would be enough to
spend a billion dollars a day for over a trillion years
before you had to worry about running out!
▫(In book, how big? p. 49)
The Mole
•One mole always contains 6.02 x 10
23
particles.
ie.
●
●
●
1 mol C
=
6.02 x 10
atoms
23
1 mol O2 =
6.02 x 10 molecules of O2
1 mol CH4 =
=
●
23
23
6.02 x 10 molecules of CH4
24
2.41 x 10 atoms of H
23
1 mol NaCl = 6.02 x 10 formula units
23
+
= 6.02 x 10 Na ions
23
-
= 6.02 x 10 Cl ions
Mole Calculations – Mole to Particle
●
Where:
n = number of moles (mol)
N = number of particles
(formula units, ions, molecules, or atoms)
NA = Avogadro’s number
23
(6.02 x 10 particles/mol)
Mole Calculations – Mole to Particle
eg #1:
35
How many moles in 1.21 x 10 molecules of CO2?
Mole Calculations – Mole to Particle
eg #2: How many atoms in 0.35 mol of Fe?
Mole Calculations – Mole to Particle
eg #3:
How many formula units in 5.69 mol of Na2SO4?
Mole Calculations – Mole to Particle
eg #4:
+
How many Na ions in 5.69 mol of Na2SO4?
HOMEWORK!!!
Pg. 52 #'s 7,8,9,10
MOLAR MASS (M)
INTRODUCTION:
•Moles cannot be measured directly.
•Chemists measure chemical amounts using
mass for solids and volume for gases & liquids.
•To convert between moles and mass we need
the mass of one mole or the MOLAR MASS.
Molar Mass (M) of Elements
•The mass of 1 mol of a substance is referred to
as its molar mass (M) and is measured in g/mol.
eg.
1 mol C → 12.01 g/mol
1 mol H → 1.01 g/mol
1 mol Pb → 207.19 g/mol
1 mol Na → 22.99 g/mol
Molar Mass (M) of Compounds
•The molar mass of a compound is the sum of the
molar masses of each element in the compound.
eg.
Calculate the molar mass of:
●
●
●
H2O
Na2SO4
.
Ba(MnO4)2 8 H2O
For you to do!
●
Pg. 57 # 17 &19
Answers on page 77
Mole Calculations – Mole to Mass
Rearrange:
Where:
n = number of moles (mol)
m = mass (g)
M = molar mass (g/mol)
Mole Calculations – Mole to Mass
eg #1:
How many moles in 4.67 g of copper ?
Mole Calculations – Mole to Mass
eg # 2:
What is the mass of 0.487 mol of H2O ?
Mole Calculations – Mole to Mass
eg # 3:
Calculate the mass of 2.00 mol of H2(g).
Mole Calculations – Mole to Mass
eg # 4:
How many moles are in 80.0 g of NH4Cl ?
For you to do!!
Pg. 59 #20 & Pg. 60 #25
Answers on page 77
Note
●
●
●
Molar amount = number of moles of whatever
1 kg = 1000 g
1 g = 1000 mg
Mole Calculation – Mass to Particles
• We do not have a direct quantity that relates
mass (m) and number of particles (N), but they
are related through number of moles (n).
Mole Calculation – Mass to Particles
eg.
How many molecules are in 26.9 g of H 2O(l)?
m = 26.9 g
Mwater= 18.02 g/mol
23
NA = 6.02 x 10 molecules/mol
Find N
Mole Calculation – Mass to Particles
eg.
28
A sample of Sn contains 4.69 x 10 atoms. Calculate
its mass.
N = 4.69 x 10
28
atoms
23
NA = 6.022 x 10 molecules/mol
MSn = 118.69 g/mol
Find m
Mole Calculation – Mass to Particles
eg.
How many molecules are in 4.78 g of glucose?
m = 4.78 g
Mglucose= 180.18 g/mol
23
NA = 6.022 x 10 molecules/mol
Find N
Mole Calculation – Mass to Particles
xM
X NA
Moles
(n)
Particles
(N)
÷ NA
Mass
(m)
÷M
Mole Calculation – Mass to Particles
Practice:
p. 63 #’s 28 & 33
p. 64 #’s 34 & 37
MOLAR VOLUME (gases)
Pressure (P)
•The force exerted per unit of surface area.
•SI unit is the Pascal (Pa).
(Used to be the atmosphere - atm)
Temperature (T)
•measures the kinetic energy of the particles.
o
•SI unit is Celsius degrees ( C).
•Other unit is Kelvin (K).
MOLAR VOLUME (gases)
•The volume of a gas is dependent on P and T.
A ‘COOL’ balloon
Pressure & Volume
MOLAR VOLUME (gases)
Standard Conditions
(needed to measure gas volumes)
STP – Standard Temperature and Pressure
o
T = 0 C (273 K)
P = 101.3 kPa (1 atm)
SATP – Standard Ambient Temperature and Pressure
o
T = 25 C
P = 100 kPa
MOLAR VOLUME (gases)
•Experimental evidence shows that the volume of
one mole of ANY GAS at STP is 22.4 L/mol.
ie. MV = 22.4 L/mol OR VSTP = 22.4 L/mol
Mole Calculations – Molar Volume
Where:
n
=
number of moles, in mol
v
=
volume, in L
V
STP
=
molar volume of a gas at STP
22.4 L/mol
Mole Calculations – Molar Volume
eg 1:
What is the volume, in L, of 0.600 mol of SO 2 gas at
STP?
Mole Calculations – Molar Volume
eg 2:
What is the number of moles in 33.6 L of He gas at
STP?
Mole Calculations – Molar Volume
eg 3:
What is the mass of 44.8 L of methane at STP?
Mole Calculations – Molar Volume
eg 4:
A chemist isolates 2.99 g of a gas. The sample
occupies 800.0 mL at STP. Is the gas most likely to be
O2(g), Kr(g), Ne(g), or F2(g)?
(Hint: Determine the molar mass of the gas.)
HOMEWORK!!!
p. 73 #’s 38 - 43
SUMMARY
PERCENT COMPOSITION
The composition of many substances is
expressed in terms
of percentages.
The London 2012 Olympic medals weigh between
375 - 400g.
Gold medal - 92.5% silver, 1.34% gold, with the remainder
copper.
Silver medal - 92.5% silver, with the remainder copper.
Bronze medal - 97.0% copper, 2.5% zinc and 0.5% tin.
3.5 % of sea water is dissolved salt
EXCEPT
the Dead Sea which is 33% salt.
PERCENT COMPOSITION
Two types of calculations:
1.Percent Composition given mass
% element = mass of element
mass of compound
x 100
Percent Composition given mass (g)
eg. 1
8.20 g of Mg combines with 5.40 g of O to form MgO.
Determine the percent composition of this compound?
Percent Composition given mass (g)
eg. 2
29.0 g of Ag combines with 4.30 of S to
form Ag2S. Calculate the mass percent for S
in this compound?
Your Turn!
p. 82 #’s 2 & 4
PERCENT COMPOSITION
•Percent composition is the percent by mass of
each element in a compound
•Should add up to 100%.
•AKA: Mass Percent
eg.
C8H8O3 – the vanillin molecule
C-->63.1 %
H-->5.3 %
O-->31.6 %
TOTAL: 100 %
PERCENT COMPOSITION
2.Percent Composition given the formula
% element = M of element
M of compound
x 100
M = molar mass
Percent Composition given the
Formula
eg. 3
Calculate the percent composition of C 2H6.
79.85 %
20.2 %
Based on your answer, calculate the mass
of C in an 82 g sample of C2H6.
Percent Composition given the
Formula
eg. 4
Calculate the mass percent of NaHSO4.
M = 120.07 g/mol
19.15 % 0.841 % 26.71 % 53.30 %
Based on your answer, calculate the mass of H in a
20.2 g sample of NaHSO4.
For you to do!
Pg. 85 #'s 5 & 8
EMPIRICAL and MOLECULAR FORMULAS
●
●
●
●
An empirical formula gives the simplest ratio of
elements in a compound.
A molecular formula shows the actual number
of atoms in a molecule of a compound.
Ex: glucose: molecular formula-->
empirical formula-->
Ionic compounds are always wrote as empirical
formulas – simplest ratio
A molecular formula is only used for molecular
compounds
EMPIRICAL and MOLECULAR FORMULAS
Compound
Butane
Molecular Formula Empirical Formula
C H
CH
C H O
CH O
2 5
4 10
Glucose
6 12
6
6 12 6
Water
H O
2
H O
2
CH
Benzene
C H
6 6
EMPIRICAL FORMULAS
The empirical formula of a compound may be
determined from the % composition of a
compound.
Method:
1.Change % to mass out of 100 g.
2.Convert mass to moles.
3.Divide by the smallest number of moles to get the
ratio for the empirical formula.
4.Multiply to get a whole number ratio.
(if needed)
EMPIRICAL FORMULAS
Page 90 in book.
Helpful for EF
calculations.
A FUN WAY TO REMEMBER EF calculations
...
Percent to mass
Mass to Mole
Divide by small
Multiply till whole
By Joel Thompson
EMPIRICAL FORMULAS
eg. 1
A compound was analyzed and found to contain 87.4% N
and 12.6 % H by mass. Determine the empirical formula of
the compound.
EMPIRICAL FORMULAS
eg. 2
What is the empirical formula of a compound that is 25.9
% N and 74.1 % O ?
EMPIRICAL FORMULAS
eg. 3
A compound contains 89.91% C and 10.08 % H by mass.
Determine the empirical formula of the compound.
MOLECULAR FORMULA CALCULATIONS
•We can determine the molecular formula of a
compound using the molar mass and the empirical
formula of a compound.
METHOD:
1.Determine empirical formula (may be given).
2.Divide the molar mass by the empirical formula
mass.
3.Round off the result to the nearest whole number.
4.Multiply empirical formula by this whole number to
get the molecular formula.
MOLECULAR FORMULA CALCULATIONS
eg. 1
Determine the molecular formula of a compound with
a molar mass of 60.00 g/mol and an empirical formula
of CH4N.
MOLECULAR FORMULA CALCULATIONS
eg. 2
The empirical formula of hydrazine is NH 2. The molar
mass of hydrazine is 32.06 g/mol. What is the
molecular formula for hydrazine?
For You to Try
p. 97 #’s 17 – 20
Unit 1 Part 2
CALCULATIONS AND CHEMICAL EQUATIONS
I know the answer!
TEE HEE!
UNIT 1 – PART 2 (CH 4)
Topics:
●
●
●
●
Stoichiometry
●
definition – what it is
●
types of stoichiometry
●
mole ratios
●
calculations using n & m
●
other calculations using v & V
STP
Theoretical yield, actual yield and percent yield
Limiting and Excess Reagent
Core Lab 2 – % Yield in a Chemical Reaction
STOICHIOMETRY
●
●
Stoichiometry comes from the Greek words
stoikheion, meaning “element” and metron,
meaning “to measure”.
Stoichiometry is the determination of quantities
needed for, or quantities produced by, chemical
reactions.
STOICHIOMETRY
3 types of stoichiometry
1. Gravimetric stoichiometry
●
stoichiometry using mass (g) of solids(s).
2. Solution stoichiometry
●
stoichiometry using concentration & volume of
solutions.
3. Gas stoichiometry
●
stoichiometry using volume (L) of gases (g) at
STP.
STOICHIOMETRY
Q: How do we calculate quantities used
or quantities produced in chemical reactions ?
A:
Mole Ratios
•
The coefficients from a balanced chemical
equation provide the mole ratios of reactants
and products in a chemical reaction.
•
Coefficients from a balanced chemical
equation, work like a sandwich recipe . . .
MOLE RATIOS
Clubhouse sandwich recipe
MOLE RATIOS
Fill in the missing quantities
Slices of
Toast
Slices of
Turkey
Strips of
Bacon
# of
Sandwiches
12
27
66
100
MOLE RATIOS
●
Mole ratios from balanced chemical equations
work in the same way
eg: equation for an air bag exploding
sodium azide 
2 NaN3(s) 
2 UNITS
or
2 MOLES
sodium + nitrogen gas
2 Na(s) + 3 N2(g)
NOT 2 g
MOLE RATIOS
2 NaN3(s) 
2 Na(s)
+
3 N2(g)
6
10
150
10
MOLE RATIOS
●
●
A mole ratio is a mathematical expression that
shows the relative amounts of two species involved
in a chemical change.
A mole ratio
● comes from a balanced chemical equation
●
show the relative amounts of reactants &
products in moles
looks like ---> coefficient required
coefficient given
MOLE RATIOS
●
Mole ratios from balanced chemical equations
represent relative number of particles (atoms,
molecules, ions, formula units or moles).
C3H8(g) +
5 O2(g)  3 CO2(g) + 4 H2O(g)
15 mol
18
molecules
35
molecules
7.38 mol
MOLE RATIOS
NOTE:
Mole ratios
DO NOT REPRESENT
MASS
P4 + 5 O2 → 2 P2O5
1 g of phosphorus reacts with 5 g of oxygen
MOLE TO MOLE STOICHIOMETRY
STEPS:
1.Write a balanced chemical equation with
subscripts
(Identify given and required values).
2.Use MOLE RATIOS from balanced chemical
equation .
nrequired = ngiven x coefficient required
coefficient given
MOLE TO MOLE STOICHIOMETRY
eg.1
How many moles of ammonia gas are produced when
0.60 mol of nitrogen gas reacts with hydrogen gas?
(1.2mol)
MOLE TO MOLE STOICHIOMETRY
eg.2
How many moles of CO2 are produced when 31.5 mol
of C3H8 is burned? (94.5 mol)
MOLE TO MOLE STOICHIOMETRY
eg.3
Calculate the number of moles of magnesium needed
to produce 0.586 mol of Mg3N2. (1.76 mol)
MOLE TO MOLE STOICHIOMETRY
eg.4
How many moles of sulfur are produced when 4.25
mol of SO3 decomposes? (0.531 mol)
FOR YOU TO TRY
●
Pg. 114-117 #'s 2,5,9
MORE STOICHIOMETRY
CALULATIONS
●
Other types of Stoichiometry:
●
●
●
●
Mole to Mass (or Mass to Mole)
Mass to Mass
Volume to Volume (for gases at STP)
SolutionStoichiometry (Part 3 of Unit 1...)
MOLE TO MASS STOICHIOMETRY
●
●
Mole to Mass Calculations are the same as
Mole to Mole but also require conversion
between mass and moles:
Use
n=m
M
or m= n x M
MOLE TO MASS STOICHIOMETRY
STEPS:
1.Balanced
chemical - identify given and required
2.Convert mass to moles
3.Mole Ratio
nrequired = ngiven x coefficient required
coefficient given
MOLE TO MASS STOICHIOMETRY
eg. 1:
Calculate the number of moles of oxygen that will
react with 6.49 g of aluminum. (0.180 mol)
MOLE TO MASS STOICHIOMETRY
eg. 2:
How many moles of water are produced when 20.6 g
of CH4 burns? (2.57 mol)
MOLE TO MASS STOICHIOMETRY
eg. 3: Calculate the mass of aluminum oxide that can be
produced from the reaction between 0.1804 mol of oxygen
gas and excess aluminum. (18.39g)
MOLE TO MASS STOICHIOMETRY
eg. 4: What mass of Ca3P2 is produced when
4.38 mol of Ca reacts with phosphorus? (266g)
For you to Try
Pg. 149 #'s 6 & 7
Mass to Mass Stoichiometry
STEPS:
1.Balanced
chemical (given!! required!! subscripts!!)
2.Convert mass to moles.
3.Mole Ratio
nrequired = ngiven
x
coefficient required
coefficient given
4.Convert
moles to mass
Mass to Mass Stoichiometry
NOTE: The coefficients from balanced equations
represent MOLE RATIOS.
MASSES are NOT to be used
in the MOLE RATIO step.
Mass to Mass Stoichiometry
eg1:
How many grams of NH3 gas are produced when 5.40 g of
hydrogen gas reacts with nitrogen gas? (30.4g)
Mass to Mass Stoichiometry
eg 2:
What mass of nitrogen gas is needed to react completely
with hydrogen gas and produce 30.6 g of ammonia gas?
(24.72g)
Mass to Mass Stoichiometry
eg 3:
What mass of CaCl2(aq) is produced when Ca(NO3)2(aq)
reacts with 4.39 g of NaCl(s)? (4.17g)
Mass to Mass Stoichiometry
eg 4:
Given that 45 g of sulfur (S8) gas reacts with aluminum
solid, find the mass of product. (7.0x101g)
For you to try
Page 121 #'s 11-14
LAW OF CONSERVATION OF
MASS
●
Mass is conserved in chemical reaction.
OR
total mass of
the reactants
●
=
total mass of
the products.
We can verify this law by using the mole ratios
and molar masses of reactants and products
from balanced chemical equations.
Law of Conversation of Mass
Quantity
2 Mg(s) +
MgO(s)
O2(g) 
2
n (mol)
2
1
2
m (g)
2(24.31)
32.00
2(40.31)
The total mass of the reactants is 2(24.31) + 32.00 = 80.62g
The total mass of the products is 2(24.31+16.00) = 2(40.31) =
80.62g
The mass of the products and reactants are the same. So mass
is conserved!!
Law of Conservation of Mass
Ex:
The decomposition of 500.00 g of Na N produces
3
323.20 g of N . How much Na is produced in this
2
decomposition?
Mole to Volume Stoichiometry
If gas volume at STP is GIVEN or REQUIRED you
must use VSTP in your calculations.
eg 1:
What volume of oxygen at STP is needed to react
with 2.34 g of NO2 ?
2 N2O5(s) 
4 NO2(g) + O2(g)
Mole to Volume Stoichiometry
eg 2:
Calculate the mass of N2H4(g) needed to
produce 157 L of nitrogen gas at STP in the
reaction below.
2 N2H4(l) + N2O4(l) 
3 N2(g) + 4 H2O(l)
For you to try
p. 123 # 16
LIMITING AND EXCESS
REAGENTS
Sandwich Time Again!
LIMITING AND EXCESS
REAGENTS
Here’s what we find in our kitchen:
◦How many sandwiches can we make??
6 Slices of Toast
12 Pieces of Turkey
20 slices of bacon
What limits the # of sandwiches??
LIMITING AND EXCESS
REAGENTS
Chemical reactions with 2 or more reactants, stop
when 1 reactant is completely used up.
Limiting reagent
● AKA limiting reactant
● The reactant that is completely used in a
chemical reaction, thereby limiting the amount
of product that can form.
LIMITING AND EXCESS
REAGENTS
●
Excess reagent
AKA excess reactant
● the reactant that is left over after the reaction
is complete.
● there is more than is needed for complete
reaction.
We can predict which reactant is the LR by
using mole ratios from balanced chemical
equations.
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DETERMING THE LIMITING
REAGENT
How to Solve:
1.Write balanced chemical equation.
2.Calculate n(mols) for EACH reactant.
3.Use the mole ratio for each reactant.
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The LR is the reactant that produces less moles
of product
4.Calculate the amount of product using the LR.
DETERMINE THE LIMITING
REAGENT
eg 1:
Determine the Limiting Reagent and Excess Reagent if
14.8 g of C3H8(g) reacts with 2.14 g of O2(g).
NOTE:
When you are given the quantity (mass, volume or
moles) of 2 or more reactants in a stoichiometry
question, YOU MUST ALWAYS determine the
LR and use it to determine the amount of product.
Limiting Reagent
eg 2.
What mass of NaCl is produced when 6.70 mol of Na
reacts with 3.20 mol of Cl2?
Limiting Reagent
eg 3.
What mass of CaCO3 will be produced when 20.0 g of
Ca3(PO4)2 reacts with 15.0 g of Na2CO3 ?
Limiting Reagent
eg 4.
What volume of H2 at STP will be produced when 10.0 g of
Zn reacts with 20.0 g of HCl(aq)?
For you to try
p. 134
#’s 27ab & 30a
Percent Yield Calculations
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The term YIELD refers to the mass of product
formed in a chemical reaction.
eg. Burning 99.5 g of propane in sufficient oxygen
produces 298 g of carbon dioxide gas.
THEORETICAL YIELD
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what you are supposed to get
The amount of product predicted or calculated.
Percent Yield Calculations
●
A question that asks you to find the theoretical
yield is asking you to calculate the amount
expected.
Actual yield
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what you actually get in an experiment
The amount actually produced in a chemical
reaction in the LAB
Percent Yield Calculations
For most reactions, your actual yield is less than your
theoretical yield, for many reasons, including:





Reaction is slow and was incomplete.
Impurities are present.
Reaction goes to equilibrium (never really
finishes).
Sample is wet when weighed.
Experimental error.
Percent Yield Calculations
Percent Yield Calculations
eg 1.
Calculate the theoretical yield of copper if 1.87 g of Al
reacts with excess aqueous copper (II) sulfate?
2 Al + 3 CuSO4 → Al2(SO4)3 + 3 Cu
Percent Yield Calculations
eg 1.(cont’d)
If the reaction produces 3.74 g of copper, what is the
percent yield of copper?
Percent Yield Calculations
eg 2.
20.0 g of bromic acid, HBrO3 (aq) , is reacted with excess HBr.
HBrO3(aq) + 5 HBr(aq)  3 H2O(l) + 3 Br2(aq)
(a) What is the theoretical yield of Br2 for this reaction?
(74.4 g)
(b) If 47.3 g of Br2 are produced, what is the percentage
yield of Br2? (63.6%)
Percent Yield Calculations
eg 3.
Barium sulfate forms a precipitate in this reaction:
Ba(NO3)2 (aq) + Na2SO4 (aq)  BaSO4(s) + 2 NaNO3(aq)
When 35.0 g of Ba(NO3)2 reacts with excess Na2SO4,
29.8 g of BaSO4 are recovered by the chemist.
(a) Calculate the theoretical yield of BaSO4. (31.3 g)
(b) Calculate the percentage yield of BaSO4. (95.2 %)
Percent Yield Calculations
eg 4.
Yeasts can act on a sugar, such as glucose, C6H12O6, to produce
ethyl alcohol, C2H5OH, and carbon dioxide.
C6H12O6 (s)  2 C2H5OH(l) + 2CO2(g)
If 223 g of ethyl alcohol are recovered after 1.63 kg of
glucose react, what is the percentage yield of the
reaction? (834 g, 26.7 %)
Solutions Chemistry
Chapter 7
Solutions
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Terms
Molar Concentration (mol/L)
Dilutions
% Concentration (pp. 255 – 263)
Solution Process
Solution Preparation
Solution Stoichiometry
Dissociation
Terms
solution
solvent
solute
concentrated
dilute
aqueous
miscible
immiscible
alloy
solubility
molar solubility
saturated
unsaturated
supersaturated
dissociation
electrolyte
non-electrolyte
soluble
insoluble
limiting reagent
excess reagent
actual yield
theoretical yield
decanting
pipetting
filtrate
precipitate
dynamic equilibrium
Terms
solution - is a homogeneous mixture
solute - the substance that dissolves OR the
substance in lesser quantity
solvent - the substance which dissolves the
solute OR the substance in greater quantity
concentrated - a large amount of solute relative
to the amount of solvent
dilute - a small amount of solute relative to the
amount of solvent
Terms
saturated – contains the maximum amount of
dissolved solute at a given temperature and
pressure
unsaturated – contains less than the maximum
amount of dissolved solute at a given
temperature and pressure
supersaturated – contains more than the
maximum amount of dissolved solute for a given
temperature and pressure
Terms
miscible – liquids that dissolve in each other
immiscible – liquids that do not dissolve in each
other
aqueous - the solvent is water
alloy - a solid solution of two or more metals
Terms
Solubility - the maximum amount of solute that
can be dissolved under specific temperature
and pressure conditions
eg. the solubility of HCl at 25 °C is 12.4 mol/L
eg. 100.0 mL of water at 25°C dissolves 36.2 g of
sodium chloride
Terms
soluble – solubility is greater than 1 g per 100 mL
of solvent.
insoluble - solubility is less than 0.1 g per 100
mL of solvent.
9 types of solution
Rate of dissolving
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for most solids, the rate of dissolving is greater
at higher temperatures
stirring a mixture or by shaking the container
increases the rate of dissolving.
decreasing the size of the particles increases
the rate of dissolving.
Solubility
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small molecules are often more soluble than
larger molecules.
solubility of solids increases with temperature.
the solubility of most liquids is not affected by
temperature.
the solubility of gases decreases as
temperature increases
an increase in pressure increases the solubility
of a gas in a liquid.
“Like dissolves like”
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ionic solutes and polar covalent solutes both
dissolve in polar solvents
non-polar solutes dissolve in non-polar
solvents.
compounds with very strong ionic bonds, such
as AgCl, tend to be less soluble in water than
compounds with weak ionic bonds, such as
NaCl.
Surprise Review
●
Find the molar mass of Ca(OH)2
●
How many moles in 45.67 g of Ca(OH)2?
●
Find the mass of 0.987 mol of Ca(OH)2.
Molar Concentration
The terms concentrated and dilute are
qualitative descriptions of solubility.
A quantitative measure of solubility uses
numbers to describe how much solute is
dissolved or the concentration of a
solution.
Molar Concentration
The MOLAR CONCENTRATION of a
solution is the number of moles of solute
(n) per litre of solution (v).
Molar Concentration
FORMULA:
Molar Concentration = number of moles
volume in litres
C= n
V
Rearranged Formulae
mol/L OR M (molar)
n CxV
or
n
V 
C
Molar Concentration
Calculate the molar concentration of:
1). 4.65 mol of NaOH is dissolved to
prepare 2.83 L of solution. (1.64 mol/L)
2). 15.50 g of NaOH is dissolved to
prepare 475 mL of solution.
( 0.3875 mol → 0.816 mol/L)
Molar Concentration
3).
the volume of 6.00 mol/L HCl(aq) that can be
made using 0.500 mol of HCl. (0.083L)
4).
the volume of 1.60 mol/L HCl(aq) that can be
made using 20.0 g of HCl. (0.343L)
Dilution (p. 272)
When a solution is diluted:
- The concentration decreases
- The volume increases
-
The number of moles remains the
same
ni = nf
Number of moles
before dilution
Number of moles
after dilution
Dilution (p. 272)
ni = n f
Ci Vi = Cf Vf
eg. Calculate the molar concentration of a
vinegar solution prepared by diluting 10.0 mL
of a 17.4 mol/L solution to a final volume of
3.50 L. (0.0497 mol/L)
p. 273 #’s 25 – 27
p. 276
#’s 1, 2, 4, & 5
I will actually be checking these
questions. So please do them!!
June 2010 # 12.
A lab technician prepares a dilute solution of
hydrochloric acid. If 50.0 mL of 2.50 mol/L
hydrochloric acid is added to 450.0 mL of
water, what is the new concentration?
(A) 0.250 mol/L
(C) 3.60 mol/L
(B) 0.278 mol/L
(D) 4.00 mol/L
Answer: (B)
Percent Concentration
Concentration may also be given as a %.
●
The amount of solute is a percentage of the
total volume or mass of solution.
 liquids in liquids - % v/v
 solids in liquids - % m/v
 solids in solids - % m/m
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Percent Concentration
mass of solute (g)
Percent (m/v) 
x 100
volume of solution (mL)
mass of solute (g)
Percent (m/m) 
x 100
mass of solution (g)
volume of solute (mL)
Percent (v/v) 
x 100
volume of solution (mL)
Concentration in ppm and ppb
Parts per million (ppm) and parts per billion
(ppb) are used for extremely small
concentrations
ppm x msolution
msolute 
6
10
ppb x msolution
msolute 
9
10
eg. 5.00 mg of NaF is dissolved in 100.0
kg of solution. Calculate the
concentration in:
a) ppm
b) ppb
ppm = 0.005 g x 106
100,000 g
= 0.05 ppm
ppb = 0.005 g x 109
100,000 g
= 50.0 ppb
For you to do as homework!!
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p. 265 #’s 15 – 17
Solution Preparation & Dilution
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standard solution – a solution of known
concentration
volumetric flask – a flat-bottomed glass vessel that
is used to prepare a standard solution
delivery pipet – a pipet that accurately measures
one volume
graduated pipet – pipets that have a series of lines
that can be used to measure many different volumes
To prepare a standard solution:
1. calculate the mass of solute needed
2. weigh out the desired mass
3. dissolve the solute in a beaker using less than
the final volume
4. transfer the solution to a volumetric flask
(rinse the beaker into the flask)
5. add water until the bottom of the meniscus is
at the etched line
To dilute a standard solution:
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1. Rinse the pipet several times with deionized
water.
2. Rinse the pipet twice with the standard
solution.
3. Use the pipet to transfer the required
volume.
4. Add enough water to bring the solution to
its final volume.
Solution Stoichiometry
1. Write a balanced equation
2. Calculate moles given
n=m/M
OR
n = CV
3. Mole ratios
4. Calculate required quantity
m=nxM
OR
n
V
C
OR
n
C
V
Solution Stoichiometry
eg. 45.0 mL of a HCl(aq) solution is used to
neutralize 30.0 mL of a 2.48 mol/L NaOH
solution.
Calculate the molar concentration of the
HCl(aq) solution.
Sample Problems
1. What mass of copper metal is needed to
react with 250.0 mL of 0.100 mol/L silver
nitrate solution? (0.794g)
2. Calculate the volume of 2.00 M HCl(aq)
needed to neutralize 1.20 g of dissolved
NaOH. (0.0150L)
3. What volume of 3.00 mol/L HNO3(aq) is
needed to neutralize 450.0 mL of 0.100 mol/L
Sr(OH)2(aq)? (30.0mL)
Sample Problem Solutions
Cu(s) + 2 AgNO3(aq) → 2 Ag(s) + Cu(NO3)2(aq)
Step 2
n = 0.02500 mol AgNO3
Step 3
n = 0.01250 mol Cu
Step 4
m = 0.794 g Cu
Sample Problem Solutions
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Step 2
n = 0.0300 mol NaOH
Step 3
n = 0.0300 mol HCl
Step 4
V = 0.0150 L HCl
Sample Problem Solutions
2 HNO3(aq) + Sr(OH)2(aq) →
2 H2O(l) + Sr(NO3)2(aq)
Step 2
n = 0.04500 mol Sr(OH)2
Step 3
n = 0.0900 mol HNO3
Step 4
V = 0.0300 mol/L HNO3
The Solution Process (p. 299)
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Dissociation occurs when an ionic compound
breaks into ions as it dissolves in water.
A dissociation equation shows what happens to an
ionic compound in water.
eg. NaCl(s) → Na+(aq) + Cl-(aq)
K2SO4(s) → 2 K+(aq) + SO42-(aq)
Ca(NO3)2(s) → Ca2+(aq) + 2 NO3-(aq)
The Solution Process (p. 299)
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Solutions of ionic compounds conduct
electric current.
A solute that conducts an electric current
in an aqueous solution is called an
electrolyte.
The Solution Process (p. 299)
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Acids are also electrolytes.
Acids form ions when dissolved in water.
eg. H2SO4(aq) → 2 H+(aq) + SO42-(aq)
HCl(s) → H+(aq) + Cl-(aq)
The Solution Process (p. 299)
Molecular Compounds DO NOT dissociate in
water.
eg. C12H22O11(s) → C12H22O11(aq)
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Because they DO NOT conduct electric
current in solution, molecular compounds are
non-electrolytes.
The Solution Process (p. 299)
The molar concentration of any dissolved ion is
calculated using the ratio from the
dissociation equation.
eq. What is the molar concentration of each ion
in a 5.00 mol/L MgCl2(aq) solution:
5.00 mol/L
5.00 mol/L
10.00 mol/L
What mass of calcium chloride is required to
prepare 2.00 L of a solution in which the Cl(aq) concentration is 0.120 mol/L ?