Download Chemical Reaction Equations

CLASSIFYING
CHEMICAL
REACTIONS
Review of Writing and Balancing
Pg. 58-64
Review:
When counting elements, don’t forget to look at both the subscript
and the coefficient.
For example:
P2O5 = has 2 phosphorus atoms and 5 oxygen atoms
2P2O5 = has 4 phosphorus atoms and 10 oxygen atoms
Because there are 2 molecules (indicated by the coefficient) and 2
atoms in each molecule (indicated by the subscript) –
So you multiply!!
Remember:
Never change a subscript to balance an equation!
O2(g) + H2(g)  H2O(l) Is unbalanced – but you can’t
change it the following way!
O2(g) + H2(g)  H2O2(l) X
Make sure the coefficients are the lowest whole-number ratio :
2O2(g) + 4H2(g)  4H2O(l)
This is a balanced formula but these are not the lowest
numbers you could use:
O2(g) + 2H2(g)  2H2O(l)
Balancing Chemical Equations
1) Write the chemical formulas for the reactants and products including the states
 Cu(s) + AgNO3(aq)  Ag(s) + Cu(NO3)2(aq)
2) Balance the element (atom or ion) present in the greatest number by
multiplying by the lowest coefficient possible
 (NO3)2(aq) = 2 present (lowest coefficient possible to balance = 2)
 Cu(s) +
2AgNO3(aq)  Ag(s) + Cu(NO3)2(aq)
3) Repeat step 2 for the rest of the elements
 Now we have 2 Ag, so balance the other side
 Cu(s) + 2AgNO3(aq)  2Ag(s) + Cu(NO3)2(aq)
4) Count elements on each side of the final equation to ensure they balance:
 1 Cu(s) = 1 Cu(aq) ; 2Ag = 2Ag(s) ; 2 NO3 = (NO3)2(aq)
What do you remember?
You learned about five reaction types, can you match them up
Composition (Formation)
CH4(g) + O2(g)  CO2(g) + H2O(g)
Decomposition
Mg(s) + O2(g)  MgO(s)
Combustion
Cu(s) + AgNO3(aq)  Ag(s) + Cu(NO3)2(aq)
Single Replacement
CaCl2(aq) + Na2CO3(aq)  CaCO3(s) + NaCl(aq)
Double Replacement
H2O(l)  O2(g) + H2(g)
Classifying Chemical Reactions:
Composition (Formation)
2Mg(s) + O2(g)  2MgO(s)
element + element  compound
Predict and balance the following:
Li(s) + Cl2(g) 
Na(s) + F2(g) 
Ba(s) + N2(g) 
Classifying Chemical Reactions:
Decomposition
2H2O(l)  O2(g) + 2H2(g)
compound  element + element
Predict and balance the following:
NaCl(s) 
Sr3P2(s) 
Cs2O(s) 
Classifying Chemical Reactions:
Combustion
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
fuel + oxygen  carbon dioxide + water
Predict and balance the following:
C3H8(g) + O2(g) 
C2H6(g) + O2(g) 
C4H10(g) + O2(g) 
Classifying Chemical Reactions: Single
Replacement
2AgNO3(aq) + Cu(s)  Cu(NO3)2(aq) + 2Ag(s)
compound + element  compound + element
2NaI(aq) + Cl2 (g)  I2 (s) + 2NaCl(aq)
Predict and balance the following:
NaBr(aq) + O2(g) 
CuCl2(aq) + Al(s) 
Li2CO3(aq) + K(s) 
Classifying Chemical Reactions: Double
Replacement
2AgNO3(aq) + CuCl2(aq)  Cu(NO3)2(aq) + 2AgCl(aq)
compound + compound  compound + compound
CaCI2(aq) + Na2CO3 (aq)  2NaCl (aq) + CaCO3(s)
Predict and balance the following:
NaBr(aq) + MgO(aq) 
CuCl2(aq) + AlF3(aq) 
Li2CO3(aq) + K2O(aq) 
Using the solubility
table:
Chemical Reaction
Equations
Section 7.1 pg. 278-285
Reaction Assumptions
Reactions are spontaneous – reactions will occur when all
the reactants are mixed together
Reactions are fast – the reaction must occur within a
reasonable time (see pg. 280)
Reactions are quantitative – one that is more than 99%
complete; in other words, at least one reactant is
completely used up
Reactions are stoichiometric – means that there is a
simple whole-number ratio of chemical amounts of
reactants and products (the coefficients for a balanced
equation do not change)
Net Ionic Equations
A chemical reaction equation that includes only reacting entities
(molecules, atoms and/or ions) and omits any that do not change
Writing Net Ionic Equations:
1) Write a complete balanced chemical equation
2) Dissociate all high-solubility ionic compounds, and ionize all
strong acids to show the complete ionic equation
3) Cancel identical entities that appear on both the reactant and
product sides
4) Write the net ionic equation, reducing coefficients if
neccessary
When cancelling spectator ions,
they must be identical in every way:
chemical amount, form (atom, ion,
molecule) and state of matter
Practice
Write the net ionic equation for the reaction of aqueous barium chloride and
aqueous sodium sulfate. (Refer to the solubility table)
1) BaCl2(aq) + Na2SO4(aq)  BaSO4(s) + 2NaCl(aq)
2) Ba2+(aq) + 2Cl-(aq) +2Na+(aq) + SO42-(aq)  BaSO4(s) + 2Na+(aq) + 2Cl-(aq)
(Complete ionic equation)
3) Ba2+(aq) + 2Cl-(aq) +2Na+(aq) + SO42-(aq)  BaSO4(s) + 2Na+(aq) + 2Cl-(aq)
4) Ba2+(aq)) + SO42-(aq)  BaSO4(s) (Net ionic equation)
Ions that are present but do not take part in (change during) a reaction are
called spectator ions (like spectators at a sports game: they are present but
do not take part in the game)
Practice
Write the net ionic equation for the reaction of zinc metal and aqueous
copper (II) sulfate (Refer to the solubility table)
1) Zn(s) + CuSO4(aq)  Cu(s) + ZnSO4(aq)
2) Zn(s) + Cu2+(aq) + SO42-(aq)  Cu(s) + Zn2+(aq) + SO42-(aq)
(Complete ionic equation)
3) Zn(s) + Cu2+(aq) + SO42-(aq)  Cu(s) + Zn2+(aq) + SO42-(aq)
4) Zn(s) + Cu2+(aq)  Cu(s) + Zn2+(aq) (Net ionic equation)
For entities involving strong
acids, H+(aq) is used as a matter
of convenience over H3O+(aq)
Practice
Write the net ionic equation for the reaction of hydrochloric acid and
barium hydroxide
1) 2HCl(aq) + Ba(OH)2(aq)  BaCl2(aq) + 2HOH(l)
aka: 2H2O(l)
2) 2H+(aq) + 2Cl-(aq) + Ba2+(aq) + 2OH-(aq)  Ba2+(aq) + 2Cl-(aq) + 2H2O(l)
(Complete ionic equation)
3) 2H+(aq) + 2Cl-(aq) + Ba2+(aq) + 2OH-(aq)  Ba2+(aq) + 2Cl-(aq) + 2H2O(l)
4) H+(aq)) + OH-(aq)  H2O(l) (Net ionic equation) – coefficients reduced to 1
Limiting and Excess Reagents
Cu(s) + AgNO3(aq) 
What is in the container when the reaction is finished?
 Cu(NO3)2(aq) + Ag(s)
(Cu2+(aq) = blue)
But how do you know the reaction is done?? There is still copper left!
(a) Copper wire and a beaker with aqueous silver nitrate solution
(b) A few moments after the wire is immersed
(c) The beaker contents after 24 h
Limiting and Excess Reagents
When no further changes appear to be occurring, we assume that all of the
AgNO3(aq) that was initially present has now been completely reacted.
A limiting reagent is the reactant whose entities are completely consumed in a
reaction, meaning the reaction stops.
In order to make sure this happens, more of the other reactant must be
present than is required
An excess reagent is the reactant whose entities are present in surplus amounts,
so that some remain after the reaction ends..
In our reaction: much more copper was used than
needed (evidenced by the unreacted copper) so we
assume the reaction ended when no more silver ions
were left, so silver nitrate was the limiting reagent.
Homework:
WB pg 59 Q 10
WB pg 71 Q 42
WB pg 91 Q 1-3
* Classifying & Balancing Equations WS
Stoichiometry
Section 7.2 pg. 286-293
Stoichiometry
A method of problem-solving using known quantities of one
entity in a chemical reaction to find out unknown quantities
of another entity in the same reaction
Always requires a balanced chemical equation
Series of steps for measurement calculations
Always involves a mole ratio to “switch substances”
Writing chemical reactions, net ionic equations and
dissociation equations is essential
Converting grams to moles (and vice versa) is also essential
Gravimetric
Stoichiometry
Gravimetric Stoichiometry – method used to calculate the masses
of reactants or products in a chemical reaction
Gravimetric = mass measurement
Using gravimetric stoichiometry, we can apply our knowledge of
balanced chemical reactions and mass to mol conversions to:
Predict the mass of product we will get from a reaction
Estimate the amount of reactant we need for a reaction to produce a
certain mass of product
Gravimetric
Stoichiometry Steps:
Step 1: Write a balanced chemical reaction equation
List the measured mass (given), the unknown quantity (required) and
conversion factors (M = molar mass)
Step 2: Convert the mass of the measured substance to moles
Step 3: Calculate the moles of the unknown substance using the
mole ratio required
given
Step 4: Convert from moles of the required substance to its mass.
Stoichiometry
Calculations
(Measured quantity)
solids/liquids
m 
n
mole
ratio
solids/liquids
m 
(Required quantity)
n
How many grams of oxygen are required to completely burn 10.0 g of propane
(C3H8(g))?
The balanced chemical equation is:
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g)
The number of moles of propane reacting is: (MASS to MOLES)
10.0 g x 1 mol = 0.227 mol
44.11 g
The number of moles of oxygen produced is: (MOLE RATIO)
0.227 mol x 5 mol = 1.13 mol
1 mol
The mass of oxygen produced is: (MOLES TO MASS)
1.1335 mol x 32.00 g = 36.3 g
1 mol
Practice #2
20.0 g of butane (C4H10(g)) are completely burned in a lighter.
How many grams of CO2(g) are produced?
Start with a balanced chemical equation:
2 C4H10(g) + 13 O2(g) 
8 CO2(g) + 10 H2O (g)
m = 20.0g
m=?
M = 58.14 g/mol
M = 44.01 g/mol
Practice #2 (Team Unit
Analysis)
2 C4H10(g) + 13 O2(g) 
8 CO2(g) + 10 H2O (g)
m = 20.0g
m=?
M = 58.14 g/mol
M = 44.01 g/mol
2) Mass to moles, mole ratio, moles to mass
20.0 g x
1 mol x . 8 mol CO2
x
58.14 g
2 mol C4H10
44.01 g = 60.6 g
1 mol
Practice #3 (Unit
Analysis)
What mass of iron (III) oxide is required to produce 100.0 g of
iron?
Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g)
m=?
m = 100.0g
M = 159.70g/mol
M = 55.85 g/mol
m Fe2O3(s): 100.0 g x
1 mol x 1 mol x 159.70 g
55.85 g
2 mol
1 mol
= 143.0 g Fe2O3
Practice
Remember to keep the
unrounded values in your
calculator for further
calculation until the final
answer is reported.
WB pg 77-80 ALL
Pg. 290 #8-14
(Answers pg. 785)
Percent Yield for
Reactions
We can use stoichiometry to test experimental designs,
technological skills, purity of chemicals, etc. We evaluate these by
calculating a percent yield.
This is the ratio of the actual (experimental) quantity of product
obtained and the theoretical (predicted) quantity of product obtained
from a stoichiometry calculation
Percent yield = actual yield
predicted yield
x 100
Some forms of experimental uncertainties:
All measurements (limitations of equipment)
Purity of chemical used (80-99.9% purity)
Washing a precipitate (some mass is lost through filter paper)
Estimation of reaction completion (qualitative judgements i.e.
color)
Percent Yield Example
#1
Example: In a chemical analysis, 3.00 g of silver nitrate in solution
was reacted with excess sodium chromate to produced 2.81 g of
filtered, dried precipitate.
Predicted value: 2AgNO3(aq) + Na2CrO7(aq) Ag2CrO7(s) +2NaNO3(aq)
m = 3.00 g
M = 169.88g/mol
3.00g x
1 mol x
169.88g
1 mol x
2 mol
m=
M = 331.74 g/mol
331. 74 g = 2. 93 g
1 mol
Percent yield = actual yield x 100%
predicted yield
= 2.81g x 100%
2.93g
= 95.9%
Testing the
Stoichiometric Method
Stoichiometry is used to predict the mass of precipitate
actually produced in a reaction.
Filtration is used to separate the mass of precipitate actually
produced in a reaction
Testing the
Stoichiometric Method
What mass of lead is produced by the reaction of 2.13 g of
zinc with an excess of lead(II) nitrate solution?
Design: A known mass of zinc is place in a beaker with an excess of
lead(II) nitrate solution. The lead is produced in the reaction is separated
by filtration and dried. The mass of the lead is determined
Prediction: Zn(s) + Pb(NO3)2(aq)  Zn(NO3)2(aq) + Pb(s)
m = 2.13 g
M = 65.41 g/mol
2.13 g x 1 mol x 1
65.41 g
1
m=?
M = 207.2 g/mol
x 207.2 g =
1 mol
6.75 g
Testing the Stoichiometric Method
Prediction: 6.75 g of lead will be produced (Stoichiometric calculation)
Evidence: In the beaker, crystals of a shiny black solid were produced, and
all the zinc disappeared.
Mass of filter paper = 0.92 g
Mass of dried filter paper plus lead = 7.60g
Analysis:
mass of lead product = 7.60g – 0.92g = 6.68g
According to the evidence collected, 6.68 g of lead precipitated.
Evaluation: % difference = (experimental – predicted) x 100%
predicted
6.68 – 6.75 x 100%
6.75
= 1%
Given such a small difference, which can readily be accounted for by normal sources of error, the
prediction is clearly verified, and so the stoichiometric method for predicting the mass of lead
formed in this experiment is judged to be acceptable.
Testing the Stoichiometric Method
Prediction: 6.75 g of lead will be produced (Stoichiometric calculation)
Analysis:
mass of lead product = 7.60g – 0.92g = 6.68g
According to the evidence collected, 6.68 g of lead precipitated.
Evaluation:
Could you also calculate the % yield ?
Percent yield = actual yield
predicted yield
x 100%
= 6.68 g x 100%
6.75g
= 99.0%
Homework
Pg 293 Q 1,2 & 6-10
WB pg 85-87 Q 3, 5, 6, 9, 10, 11
Gas Stoichiometry
Section 7.3 pg.294-298
Molar Volume
STP = 22.4L/mol
SATP = 24.8 L/mol
Molar volume is the same for all gases at the
same temperature and pressure (remember, all
gases have the same physical properties)
At STP, molar volume = 22.4 L/mol (101.325 kPa and 0oC)
At SATP, molar volume = 24.8 L/mol (100 kPa and 25oC)
This can be used as a conversion factor just like molar mass!
At STP, one mole of gas has a volume
of 22.4 L, which is approximately the
volume of 11 “empty” 2 L pop bottles.
Comparing Kelvin and
Celsius Scales
To convert degrees Celsius to
Kelvin , you add 273.
K = oC + 273
To convert Kelvin to degrees
Celsius, you subtract 273.
oC
= K - 273
Examples:
What is 254 K in oC ?
-19oC
What is -34oC in K ?
239K
Gas Stoichiometry
Many chemical reactions involve gases as a reactant or a product
Gas Stoichiometry – the procedure for calculating the volume of
gases as products or reactants
Gases also have a molar volume (L/mol) rather than
concentration.
This is the conversion factor used to convert (litres of gas) to
(moles of gas)
The Ideal Gas Law (PV = nRT) may also be required to:
A) find the number of moles of reactant
B) Find the V, P, or T of the product
Example #1
If 300g of propane burns in a gas barbecue, what volume
of oxygen measured at SATP is required for the reaction?
Remember: 24.8L/mol for SATP
C3H8(g)
+
m = 300g
44.11g/mol
300 g
x 1 mol
44.11 g
5O2(g)  3CO2(g) + 4H2O(g)
V=?
24.8L/mol
x 5 mol
1 mol
x
24.8 L = 843 L O2(g)
1 mol
**Remember – molar volume is the conversion factor for gases just like
molar mass is the conversion factor in gravimetric stoichiometry
Example #2
Hydrogen gas is produced when sodium metal is added
to water. What mass of sodium is necessary to produce
20.0L of hydrogen at STP?
Remember: 22.4L/mol for STP
2Na(s)
+
2H2O (l)  2NaOH(aq) + H2(g)
m=?
22.99g/mol
20.0L
x 1 mol
22.4 L
V = 20.0L
22.4L/mol
x 2mol
1 mol
x
22.99g = 41.1 g Na(s)
1 mol
**Remember – molar volume is the conversion factor for gases just like
molar mass is the conversion factor in gravimetric stoichiometry
Example #3
If the conditions are not STP or SATP, the
molar volume cannot be used! You must use
the ideal gas law to find the gas values using
moles determined from stoichiometry
What volume of ammonia at 450kPa and 80oC can be obtained from the
complete reaction of 7.5kg of hydrogen with nitrogen?
2N2(g) +
3H2(g) 
m = 7500g
M = 2.02 g/mol
2NH3(g)
m=?
P = 450kPA
T = 353.13K
7500 g x 1 mol x 2 = 2475.2475 mol NH3(g)
2.02 g
3
PV = nRT
 V = nRT
P
= (2475.2475 mol)(8.314kpa•L/mol•K)(353.15K)
(450kPa)
= 16150.10L  1.6 x 104 L of NH3(g)
1.
Gas Stoichiometry
Summary
Write a balanced chemical equation and list the
measurements, unknown quantity symbol, and conversion
factors for the measured and required substances.
2. Convert the measured quantity to a chemical amount using
the appropriate conversion factor
3. Calculate the chemical amount of the required substance
using the mole ratio from the balanced chemical equation.
4. Convert the calculated chemical amount to the final
quantity requested using the appropriate conversion factor.
Stoichiometry
Calculations
(Measured quantity)
solids/liquids
gases
gases
solids/liquids
(Required quantity)
m
V, T, P
V,T,P
m

n
mole
ratio

n
Homework
Pg. 299 #1-6
WB pg 85-87 Q 2, 4 & 8
Solution Stoichiometry
Section 7.4 pg. 300-303
Solution Stoichiometry
The majority of work in research and industry involves
solutions. Recall that solutions are easy to handle and are
usually easier to control in reactions.
The major difference compared to gravimetric and gas
stoichiometry is that we use molar concentration (mol/L) as a
conversion factor rather than molar mass or molar volume
Solution stoichiometry – the procedure for calculating the
molar concentration or volume of solution products or
reactants
Example #1
Ammonia and phosphoric acid solutions are used to produce
ammonium hydrogen phosphate fertilizer. What volume of
14.8mol/L NH3(aq) is needed to react with 1.00kL of 12.9mol/L of
H3PO4(aq)?
2NH3(aq) +
V=?
14.8mol/L
1.00kL
H3PO4(aq)

(NH4)2HPO4(aq)
V = 1.00kL
12.9 mol/L
x 12.9 mol
1L
x 2mol x 1 L
= 1.74 kL
1mol
14.8 mol = 1.74 x 103 L
Example #2
In an experiment, a 10.00 mL sample of sulfuric acid solution reacts
completely with 15.9 mL of 0.150 mol/L potassium hydroxide. Calculate
the amount concentration of the sulfuric acid.
H2SO4(aq) + 2KOH(aq)
V = 10.00mL
c=?

2H2O(l) + K2SO4(aq)
V = 15.9 mL
0.150 mol/L
15.9mL x 0.150 mol x 1mol
1L
2 mol
x 1
= 0.119 mol/L
10.0 mL
Gravimetric, Gas and
Solution Stoichiometry
Summary
1. Write a balanced chemical equation and list the quantities
and conversion factors for the given substance and the one
to be calculated.
2. Convert the given measurement to its chemical amount
using the appropriate conversion factor
3. Calculate the amount of the other substance using the mole
ratio from the balanced chemical equation.
4. Convert the calculated chemical amount to the final
quantity requested using the appropriate conversion factor.
Remember to use the Ideal Gas Law for all gases not at STP or SATP
Stoichiometry
Calculations
(Measured quantity)
solids/liquids
gases
solutions
solutions
gases
solids/liquids
(Required quantity)
m
V, T, P
c, V
c, V
V,T,P
m

n
mole
ratio

n
Guided Practice
WB pg 85-87 Q 1, 7, 12-13
Introduction to Chemical
Analysis
Using Stoichiometry
Section 8.1-8.2 (Pg. 314-319)
Chemical Analysis
Analysis of an unknown chemical sample can include both
qualitative and quantitative analysis
Qualitative – the identification of a specific substance present
Quantitative – the determination of the quantity of a substance
present
Three methods commonly used to do this:
Colorimetry – analysis by color, which used light emitted, absorbed or
transmitted by a chemical (Section 8.1)
Gravimetric Analysis – stoichiometric calculations from a measured
mass of a reagent (Section 8.2)
Titration Analysis – uses stoichiometric calculations from a measured
solution volume of a reagent (Section 8.4)
Colorimetry
Ion
Solution colour
Group 1, 2, 17
Colourless
Cr2+(aq)
Blue
Cr3+(aq)
Green
Co2+(aq)
Pink
Cu+(aq)
Green
Cu2+(aq)
Blue
Fe2+(aq)
Pale-green
Fe3+(aq)
Yellow-brown
Mn2+(aq)
Pale pink
Ni2+(aq)
Green
CrO4
2-
(aq)
Which solution is which?
potassium dichromate
sodium chloride
sodium chromate
Yellow
Cr2O72-(aq)
Orange
MnO4-(aq)
Purple
potassium permanganate
nickel (II) nitrate
copper (II) sulfate
6
2
4
1
3
5
Colorimetry
Aqueous ions can sometimes be identifies qualitatively by
eye, (like the previous example) but for a more precise
identification, technology must be used
Spectrophotometer – is a device that measures
the quantity of light absorbed at any desired
wavelength when a light beam is passed
through a solution.
Can measure the concentration of any desired
coloured ion, even in a solution that has
several different mixed colours.
It can be adjusted to only “see” one precise
wavelength (colour) selected
Colorimetry
We can also use flame test to detect
the presence of several metal ions.
The technology has become more
sophisticated now, instead we use
atomic absorption
spectrophotometers, which analyze
the light absorbed by samples
vaporized in a flame
Gravimetric Analysis
Lab technicians sometimes perform the same chemical
analysis on hundreds of samples every day.
For example, in a medical laboratory, blood and urine samples
are routinely analyzed for specific chemicals such as cholesterol
and sugar.
In many industrial and commercial laboratories, technicians
read the required quantity of a chemical from a graph that
has been prepared in advance.
This saves the time and trouble of doing a separate
stoichiometric calculation for each analysis performed.
By completing the Analysis of the following investigation
report, you will be illustrating this practice. (see pg. 317)
Gravimetric Analysis
Purpose: The purpose of this lab exercise is to use a graph of a
precipitation reaction’s stoichiometric relationship to determine the
mass of lead(II) nitrate present in a sample solution.
Problem: What mass of lead(II) nitrate is in 20.0 mL of a solution?
Design: Samples of two different lead(II) nitrate solutions are used. Each
sample is reacted with an excess quantity of a potassium iodide solution,
producing lead(II) iodide, which has a low solubility and settles to the
bottom of the beaker (Figure 1).
After the contents of the beaker are filtered and dried, the mass of
lead(II) iodide is determined. Use the Evidence pg. 317
Analysis: The analysis is completed by graphing Table 1 and then reading
the mass of lead(II) nitrate present in each solution from the graph.
Gravimetric Analysis
Analysis: Read the mass of lead(II) nitrate present in each solution
from the graph.
According to the
evidence collected
and the graph:
The mass of
lead(II) nitrate in
solution 1 = 3.15 g
solution 2 = 5.40 g.
Precipitation Completeness
1. Precisely measure a sample volume of the solution containing the limiting
reagent.
2. Add (while stirring) an approximately equal volume of the excess reagent
solution.
3. Allow the precipitate that forms to settle, until the top layer
of solution is clear.
4. With a medicine dropper, add a few more drops of excess reagent
solution. Allow the drops to run down the side of the container,
and watch for any cloudiness that may appear when the drops mix
with the clear surface layer
5. If any new cloudiness is visible, the reaction of the limiting reagent
sample is not yet complete. Repeat steps 2 to 4 of this procedure as
many times as necessary, until no new precipitate forms during the
test in step 4.
6. When no new cloudiness is visible (the test does not form any
Colorimetry Analysis
Pg. 316 #1-4
Gravimetric Analysis
Pg. 319 #1, 2, 4
Section 8.3
pg. 320-324
Limiting
and
Excess
Reagents
In any chemical reaction, it is easy to run out of one or
another reactant – which has an impact on the amount of
products that can result from a reaction
To solve these problems you must identify which of the
reactants is going to run out first.
This is the “limiting reagent”
The other is the “excess reagent”
Example: LEGO – If you had three yellow blocks and four red blocks,
you could only make three yellow/red combinations – because
there is not enough yellow blocks to make four. The yellow block
is the limiting reagent and the red block is the excess reagent
+
=
Identifying Limiting and
Excess Reagents
WHY DO WE CARE??
It is often desirable to know how much excess reagent is
required to ensure that a reaction goes to completion.
The general rule is to assume that a reasonable quantity of
excess reagent is to use 10% more than the quantity required
(Not the case in commercial chemistry)
When you know the quantity of more than one reagent, you
may need to know which one will limit the reaction.
1) You want to test the stoichiometric method using the
reaction of 2.00 g of copper(II) sulfate in solution with an
excess of sodium hydroxide in solution. What would be a
reasonable mass of sodium hydroxide to use?
 To answer this question, you need to calculate the minimum
mass required and then add 10%.
CuSO4(aq) + 2 NaOH(aq)  Cu(OH)2(s) + Na2SO4(aq)
2.00g
m=?
159.62g/mol 40.00g/mol
2.00 g x
1 mol x 2
159.62 g
1
1.00 g x 1.10%= 1.10g
x
40.0g
1 mol
= 1.00 g
10% = 0.10g
Practice pg. 321 #2
3) If 10.0 mol of methane and 10.0 mol of oxygen react,
which is the limiting reagent.
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
10.0 mol
10.0 mol
n CH4(g): 10.0 mol x 2
1
= 20.o mol
n O2(g): 10.0 mol x 1 = 5.00 mol
2
20.0 mol of oxygen
would be required
to react with 10.0
mol of methane
5.00 mol of methane
would be required to
react with 10.0 mol
of oxygen
Since 20.0 mol of oxygen is required to react with 10.0 mol of methane,
but only 10.0 mol is available, oxygen is the limiting reagent.
How much methane would be left? 5.0 mol – 5.00 mol = 5.00 mol
Practice pg. 324 #3b-d
2) If 10.0g of copper is placed in solution of 20.0g of silver
nitrate, which reagent will be the limiting reagent?
 All reactants must be converted to moles, then using the
mole ratio, determine which reactant will run out first.
Cu(s)
+ 2 AgNO3(aq)  2 Ag(s) + Cu(NO3)2(aq)
10.0g
63.55 g/mol
20.0g
169.88g/mol
n Cu(s): 10.0g x 1 mol
= 0.157 mol x 2 = 0.315 mol
63.55 g
1
n AgNO3: 20.0g x 1 mol
= 0.118 mol x 1
169.88 g
= 0.0589 mol
2
You must test one of the values using the mole ratio.
Assume one chemical is completely used up and see
if enough of the second chemical is present.
That much silver
nitrate is not
available so
copper is not the
limiting reagent
More copper than
that is available so
silver nitrate is the
limiting reagent
3) From the previous example, where 10.0 g of copper reacts
with 20.0 g of silver nitrate, what mass of copper will be in
excess? (leftover when the reaction is complete)
ER
Cu(s)
LR
+ 2 AgNO3(aq)  2 Ag(s) + Cu(NO3)2(aq)
0.157 mol
0.118 mol
n Cu(s): 0.118 mol x 1 = 0.0589 mol x 63.55 g =
2
1 mol
3.74 g of copper
will be required
in this reaction
10 g mass
– 3.74 gof= silver
6.3 g ofwill
copper
be leftover
4) What
be will
produced?
n Ag(s): 0.118 mol x 2 = 0.118 mol x 107.87 g =
2
1 mol
Practice pg. 324 #4
12.7 g of silver
will be produced
in this reaction
Putting it all together…
In an experiment, 26.8g of iron (III) chloride in solution is combined with
21.5g of sodium hydroxide. Which reactant is in excess, and by how
much? What mass of precipitate will be obtained?
FeCl3(aq) + 3NaOH(aq)  Fe(OH)3(s) + 3NaCl(aq)
26.8g
21.5g
m=?
162.20g/mol
40.00g/mol 106.88g/mol
nFeCl3 = 26.8g x 1 mol = 0.165 mol x 3 = 0.496 mol
162.20g
1
nNaOH = 21.5 g x 1 mol = 0.538 mol
40.0g
There is more
NaOH than
this so FeCl3
is the LR
0.538mol – 0.496 mol = 0.042 mol x 40.00 g = 1.7 g NaOH (excess)
1 mol
0.165 mol x 1 x 106.88 g = 17.7 g of Fe(OH)3(s)
1
mol
Limiting and Excess Reagents
Summary
Identify the limiting reagent by choosing either reagent
amount, and use the mole ratio to compare the required
amount with the amount actually present.
The quantity in excess is the difference between the
amount of excess reagent present and the amount
required for complete reaction.
A reasonable reagent excess to use to ensure complete
reaction is 10%.
Homework
Pg. 327 #4, 6-8
Titration Analysis
Section 8.4 – pg. 328-332
Experimental Design for
Analysis
Experimental designs discussed so far have been
QUALitative (flame test, solution colour, litmus test,
conductivity, solubility)
QUANtitative experimental designs allow measures
amounts to be obtained.
The experimental design chosen for analysis depends on
equipment, time available, and degree of accuracy
required.
Experimental Design for
Analysis
There are four types of analytical experimental designs:
1. Crystallization – the solvent is vaporized from a solution, with or without
heating, leaving a solid behind which is measured for mass.
2. Filtration – a low solubility solid, produced from a single or double
replacement reaction, is separated with a filter then dried so mass can be
measured (filtrate = solution that goes through)
3. Gas Collection – a gas, formed as a product of a reaction, is collected and
volume, pressure and temperature are measured.
4. Titration – a solution in a burette (called the “titrant”) is progressively
added to a measured volume of another solution (called the “sample”) in
an Erlenmeyer flask. The volume of the titrant at the endpoint is measured.
Chemical Analysis by
Titration
Titration – is a common experimental
design used to determine the amount
concentration of substances in solution.
The solution of known concentration
may be either the titrant or the sample;
it makes no difference to the analysis
Titration breakdown:
Carefully adding a solution (titrant)
from a burette into a measured, fixed
volume of another solution (sample) in
an Erlenmeyer flask until the reaction is
judged to be complete
Chemical Analysis by
Titration
Burette – precisely marked glass cylinder with a
stopcock at one end. Allows precise, accurate
measurement and control of the volume of reacting
solution.
When doing a titration, there will be a point at which the
reaction is complete; when chemically equivalent amounts
of reactants have combined. This is called the equivalence
point:
Equivalence point – the point during a titration at which
the exact theoretical chemical amount of titrant has
been added to the sample. (QUANTITATIVE)
To measure this equivalence point experimentally, we
look for a sudden change in an observable property, such
as color, pH, or conductivity. This is called the endpoint.
(QUALITATIVE)
Chemical Analysis by
Titration
An initial reading of the
burette is made before any
titrant is added to the sample.
Then the titrant is added until
the reaction is complete; when
a final drop of titrant
permanently changes the
colour of the sample.
The final burette reading is
then taken.
The difference between the
readings is the volume of
titrant added.
Near the endpoint, continuous gentle swirling of the solution is important
Chemical Analysis by
Titration
A titration should involve
several trials, to improve
reliability of the answer.
A typical requirement is to
repeat titrations until
three trials result in
volumes within a range of
0.2mL.
These three results are
then averaged before
carrying out the solution
stoichiometry calculation;
disregard any trial
volumes that do not fall in
the range.
Titration Tips
Any property of a solution such as colour, conductivity, or pH
that changes abruptly can be used as an endpoint. However,
some changes may not be very sharp or may be difficult to
measure accurately.
This may introduce error into the experiment. Any difference
between the titrant volumes at the empirical (observed)
endpoint and the theoretical equivalence point is known as
the titration error.
How to read a burette:
**Initial readings will always be
lower (1 mL) than final readings
(13.2 mL) because 0 is at the top
of the burette. You will see this
in the example problem next.
Sample Problem
Determine the concentration of hydrochloric acid in a
commercial solution.
A 1.59g mass of sodium carbonate, Na2CO3(s), was dissolved to
make 100.0mL of solution. Samples (10.00mL) of this standard
solution were then taken and titrated with the hydrochloric acid
solution.
Trial
1
2
3
4
Final burette reading (mL)
13.3
26.0
38.8
13.4
Volume of HCl(aq) added
13.1
12.7
12.8
12.8
Indicator colour
Red
The titration evidence collected is below. Methyl orange
Initial burette reading (mL)
0.2
13.3
26.0
0.6
indicator was used.
Orange Orange Orange
TIP: In titration analysis, the first trial is typically done very quickly. It is just for practice, to learn
what the endpoint looks like and to learn the approximate volume of titrant needed to get to the
endpoint. Greater care is taken with subsequent trials
Sample Problem
First: Determine the concentration of the 100.0 mL sodium carbonate
solution.
1.59 g Na2CO3(s) x 1 mol
105.99g
x
1
0.100L
= 0.150 mol/L
Second: Write a balanced chemical equation:
2HCl(aq) + Na2CO3(aq)  H2CO3(aq) + 2NaCl(aq)
*12.8 mL
c=?
10.0 mL
c = 0.150 mol/L
c HCl (aq): 10.0 mL x 0.150 mol x 2 x
1
L
1
12.8mL
* The volume of HCl(aq)
used is an average of
trials 2, 3, and 4.
= 0.234 mol/L
Using titration analysis, the concentration of the commercial hydrochloric
acid solution is 0.234 mol/L
Titration Analysis Summary
Titration is the technique of carefully controlling the addition of a
volume of solution (the titrant) from a burette into a measured fixed
volume of a sample solution until the reaction is complete.
The concentration of one reactant must be accurately known.
The equivalence point is the point at which the exact theoretical
(stoichiometric) reacting amount of titrant has been added to the
sample.
The endpoint is the point during the titration at which the sudden
change of an observable property indicates that the reaction is
complete.
Several trials must be completed. When at least three trials result in
values that are all within a range of 0.2 mL, those values are averaged.
The average value is used for the stoichiometry calculation.
Homework
Pg. 331 #1-4 – go through answers as a class
Volumetric Stoichiometry – Titration Worksheet homework
Acid–Base Titration Curves
and Indicators
Section 8.5 (The last one!)
Pg. 333-339
Titration Curves
A titration curve is a graph of the pH (vertical axis) versus
the amount of the reagent progressively added to the
original sample.
As the equivalence point is approached, there is a rapid
change in the pH.
When a titration is done to create a pH curve, the addition
of titrant is not stopped at the endpoint, but is continued
until a large excess has been added.
So what is happening?
The initial addition of the titrant (in the burette) to the acid does not produce large
changes. This relatively flat region of the pH curve is where a buffering action
occurs.
As the titration proceeds, and base is added, some of the acid is reacted with the
added base, but anywhere before the equivalence point some excess acid will
remain, so the pH stays relatively low.
• Very near the equivalence
point, a small excess of acid
becomes a small excess of
base with the addition of a
few more drops, so the pH
abruptly changes.
• The equivalence point is the
centre of this change, where
the curve is the most
vertical.
What reaction is
happening?
HCl(aq) + NaOH(aq)  HOH(l) + NaCl(aq)
Net-ionic equation? Do you remember how?
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)  H2O(l) + Na+(aq) + Cl-(aq)
H+(aq) + OH-(aq)  H2O(l) (net ionic equation)
For convenience, hydrogen
ions are written in their
simplest (Arrhenius) form
Why was the equivalence point 7?
Remember water has a neutral pH of 7, and the spectator ions are neutral, so a strong monoprotic
acid-strong monoprotic base titration must have a pH of 7 at the equivalence point.
Equivalence points
It is important to note, that the
equivalence point pH is 7 ONLY for
strong acid-strong base reactions.
• For every other acid-base reaction, the equivalence
point solution will contain ions or molecules that are
not spectators – so titration curves must be done
empirically to determine the equivalence point .
Gener
al
Rule
Strong Acid to
Weak Base:
pH at
equivalence
point is always
lower than 7
Strong Base to
Weak Acid:
pH at
equivalence
point is always
higher than 7
Strong Acid (HCl)– Weak Diprotic Base (Na2CO3)
The equivalence point is still below 7 but do you notice anything else?
If you observe the curve closely, you see that there are two places where the curve
steepens as the titration proceeds.
This happens because the base is diprotic, meaning that it will react with two
hydrogen ions; so the hydrogen ions attach to the carbonate one at a time.
We use the second reaction equivalence point, because we want the pH value
when the reaction is complete.
Why does the curve
start at the top?
Because now a
strong acid is being
added to the weak
base, not vice versa
Strong Acid (HCl)– Weak Diprotic Base (Na2CO3)
CO32-(aq) + H+(aq)  HCO3-(aq) – first equivalence point
HCO3-(aq)+ H+(aq)  H2CO3(aq) – second equivalence point
Net reaction:
CO32-(aq) + 2H+(aq)  H2CO3(aq)
You will want to
choose an
indicator that
changes color at
the second
equivalence point
Strong Acid (HCl)– Weak Diprotic Base (Na2CO3)
Why is the equivalence point less than 7???
Balanced reaction: Na2CO3(aq) + 2HCl(aq)  2NaCl(aq) + H2CO3(aq)
Net ionic equation:
2Na+ (aq) + CO32-(aq) + 2H+(aq) + 2Cl-(aq)
So at the equivalence point, there is
neutral water, neutral spectator ions, and
 2Na+ (aq) + 2Cl-(aq) + H2CO3(aq)
some weak acid H2CO3(aq). What do you
think keeps the pH below 7?
Why do we care about titration curves?
Acid base reaction pH curves provide a wealth of
information:
Initial pH levels
Equivalence point volume of titrant
Number of reaction steps
Equivalence point pH for indicator selection; so the endpoint
observed for the indicator chosen will closely match the
equivalence point of the reaction
• Thymol blue is an
unsuitable indicator
for this titration
because it changes
colour before the
equivalence point
(pH 7).
• Alizarin yellow is
also unsuitable
because it changes
colour after the
equivalence point.
Simulation
• Bromothymol blue is suitable because its endpoint pH of 6.8
(assume the middle of its pH range) closely matches the reaction
equivalence point pH of 7, and the colour change is completely on the
vertical portion of the pH curve.
Final
Tips
• It is also critical
that the amount of
indicator used be
extremely small.
• Some of the titrant volume is used to react with the indicator to
make it change color. But if the amount of indicator is small, the
volume of titrant used this way will be very small, and the
accuracy of the titration will not be affected.
Summary
An indicator for an acid–base titration analysis must be
chosen to have an endpoint (change of colour) at very
nearly the same pH as the pH at the equivalence point of
the reaction solution.
The pH of the solution at the equivalence point for a
strong monoprotic acid–strong monoprotic base reaction
will be 7.
The pH of the solution at the equivalence point for any
other acid–base reaction must be determined
Homework
Pg. 336 #1-4
Pg. 339 #1-3, 5-7, 9
Unit 4 Review
Exam Next Week – chapter 7-8