Stoichiometric
Calculations
A Directed Learning Activity
for Hartnell College
Chemistry 1
Funded by the Title V – STEM Grant
#P031S090007 through Hartnell College
For information contact lyee@hartnell.edu
Start
Student Learning Objectives
This tutorial will help you to:
1.
2.
3.
Determine limiting and/or excess reagents
for a balanced equation
Use data from gravimetric analysis to predict
mass of product and determine percent
yield
Use data from volumetric analysis to
calculate titration results
Next
Getting Started



This set of Power Point slides will lead you through a
series of short lessons and quizzes on the topics
covered by this Directed Learning Activity tutorial.
Move through the slideshow at your own pace.
There are several hyperlinks you can click on to
take you to additional information, take quizzes,
get answers to quizzes, and to go to other lessons.
You can end this slide show at any time by hitting
the “ESC” key on your computer keyboard.
Next
Table of Topics
 What
You Need to Know
 What is “Stoichiometry”?
 Balanced Chemical Equations
 Limiting Reagents
 Percent Yield
 Titrations
Next slide
What You Need to Know




How to write chemical formulas and
balanced chemical equations to describe
chemical reactions
Systems of expressing concentration of solutes
in solution (molarity, normality, percent)
Conversion between mass and moles and
different units of concentration
The Ideal Gas Laws
Please check with your textbook or other tutorials in this
series for help on these subjects.
Next slide
What is “Stoichiometry”?
“Stoichiometry” is a term that comes from
two Greek words that mean “measuring
elements”. Stoichiometric calculations
make use of the quantitative relationships
between the substances in a chemical
reaction to convert from the amount of one
substance in the chemical reaction to the
amount of another substance in that same
reaction. In order to do so, you need to
know the balanced chemical equation.
Next Slide
Balanced Chemical
Equations
How to Use Balanced Chemical Equations to
Convert Quantities
Next Slide
What a Balanced Chemical
Equation Can Tell You
If you have a balanced chemical equation,
it can tell you many things:
1. The formulas and symbols of the
reactants and products
2. The physical state of a substance (solid,
liquid, gas, or dissolved in solution)
3. If special conditions (e.g., heat, light,
catalyst) are needed, and
4. The relationships between the molecules
or moles of the reactants and products.
Next slide
A balanced chemical equation must obey the
Law of Conservation of Mass – i.e., the sum of
the masses of all the reactants must equal the
sum of the masses of all the products. Another
way of looking at a balanced chemical
equation is that the number and kind of atoms
on one side of the reaction arrow must equal
the number and kind of atoms on the other side
of the arrow.
We can use these facts to our advantage when
we are trying to calculate how much of a
product or reactant is present.
Next Slide
Mole-to-Mole Conversions
The coefficients of the balanced equation give
us the relationship between the moles of each
chemical species. If we look at this reaction:
𝐶𝐶 𝑠 + 2 𝐴𝐴𝐴𝐴3 𝑎𝑎 → 𝐶𝐶 𝑁𝑁3
2
𝑎𝑎 + 2 𝐴𝐴(𝑠)
We see these molar relationships:
1 mole Cu: 2 moles of AgNO3 : 1 mole Cu(NO3)2:
2 moles Ag, so if we know the quantity of one of
the reactants or products, we can calculate the
amounts of the other chemical species.
Next Slide
Mole-to-Mole cont’d
You can set up a simple equation using a conversion
factor like this:
𝑚𝑚𝑚𝑚𝑚 𝑜𝑜 𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑜𝑜 𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝑠𝑠𝑠𝑠𝑠𝑠𝑠
𝑚𝑚𝑚𝑚𝑚 𝑜𝑜 𝐾𝐾𝐾𝐾𝐾 𝑠𝑠𝑠𝑠𝑠𝑠𝑠
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑜𝑜 𝐾𝐾𝐾𝐾𝐾 𝑠𝑠𝑠𝑠𝑠𝑠𝑠
So if we were given a problem to calculate how much
Ag(s) would be formed if 0.295 mole of Cu(s) is reacted,
Ag (s) would be the DESIRED species and Cu(s) would
be the known species, which would lead to
2 𝑚𝑚𝑚𝑚 𝐴𝐴
𝑚𝑚𝑚𝑚𝑚 𝐴𝐴 = 0.295 𝑚𝑚𝑚𝑚 𝐶𝐶
= 0. 590 𝑚𝑚𝑚𝑚 𝐴𝐴
(1 𝑚𝑚𝑚𝑚 𝐶𝐶)
Next Slide
Mole-to-Mass and Mass-Mole
Conversions
So, continuing this process of examining
balanced equations, we can convert from
moles of one substance mass of another
substance, or vice-versa, as follows:
𝑚𝑚𝑚𝑚 𝑜𝑜 𝐾𝐾𝐾𝐾𝐾 → 𝑚𝑚𝑚𝑚𝑚 𝑜𝑜 𝐾𝐾𝐾𝐾𝐾 →
𝑚𝑚𝑚𝑚 𝑜𝑜 𝐷𝐷𝐷𝐷𝐷𝐷𝐷 or
𝑚𝑚𝑚𝑚 𝑜𝑜 𝐾𝐾𝐾𝐾𝐾 → 𝑚𝑚𝑚𝑚 𝑜𝑜 𝐷𝐷𝐷𝐷𝐷𝐷𝐷
→ 𝑚𝑚𝑚𝑚 𝑜𝑜 𝐷𝐷𝐷𝐷𝐷𝐷𝐷
Next Slide
Mass-to-Mass Conversions
We can also begin with a known mass of
one substance, and using the balanced
chemical equation, then calculate the
mass of another substance. This calculation
takes several steps:
1. Convert known mass of substance “A” to
moles of substance “A”
2. Convert from moles of substance “A” to
moles of some desired substance “B”
3. Convert from moles of “B” to mass of “B”
Next Lesson
Limiting Reagent
Calculations
How Much or How Little?
Next slide
The Limiting Reagent
In a reaction with two or more reagents, the
reagent which is completely consumed first is
referred to the “limiting reagent”. In other
words, this is the reagent that dictates how
much product it is possible to produce.
In order to determine the limiting reagent, we
are typically given the mass of the two
reagents. From this information and the
balanced equation, you can calculate the
amount of product the reagents will produce.
Let’s look at an example.
Next Slide
Limiting Reagent Example
If 1.0 mole of Cr is mixed with 1.0 mole O2
and the reaction is heated to produce
Cr2O3(s), what would be the amount of
product? The balanced equation is:
4 Cr(s) + 3 O2 (g) → 2 Cr2O3(s)
First calculate the number of moles of O2 (g)
needed to react with 1.0 mole Cr.
𝑚𝑚𝑚𝑚𝑚 𝑜𝑜 𝑂2 =
3 𝑚𝑚𝑚𝑚𝑚 𝑂2
1.0 𝑚𝑚𝑚𝑚 𝑜𝑜 𝐶𝐶 = 0.75 𝑚𝑚𝑚𝑚𝑚 𝑂2
4 𝑚𝑚𝑚𝑚𝑚 𝐶𝐶
Next Slide
Example cont’d
So, there is more than enough oxygen to
react with all the Cr. Therefore, what limits
the amount of product is Cr, which is the
limiting reagent. Note that you could have
started with the oxygen instead and found
that there was not enough Cr to react with
the available oxygen. You would reach the
same conclusion.
Now what?
Well, you know that only 0.75 mole Cr is
consumed, and also know the ratio
between Cr and Cr2O3(s).
Next Slide
Example cont’d
2𝑚𝑚𝑚𝑚𝑚𝑚𝑚2𝑂3
𝑚𝑚𝑚𝑚𝑚 𝐶𝐶2𝑂3 =
1 𝑚𝑚𝑚𝑚 𝐶𝐶
4 𝑚𝑚𝑚𝑚𝑚 𝐶𝐶
= 0.500 𝑚𝑚𝑚𝑚𝑚 𝐶𝐶2𝑂3
Note that if you had a gas as a product,
you could use the ideal gas laws to
calculate the volume of gas produced.
Quiz Questions
Quiz Questions
Ammonia reacts with oxygen to produce
nitrogen monoxide and liquid water
NH3(g) + O2(g) → NO(g) + H2O(l)
1. Balance the chemical equation
2. Determine the limiting reagent if 100g
ammonia and 100g oxygen are present
when the reaction begins.
3. Which is the excess reagent and how
much of it will be left when the reaction
is complete?
Review Examples
Check Answer
Answer to Quiz
The balanced equation is
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l)
Remember that you need to have the
same kind and number of atoms on each
side of the equation.
1.
Oxygen is the limiting reagent.
You can start the limiting reagent
calculations with either of the starting
reagents.
2.
Next Slide
Quiz Answer cont’d
Starting with ammonia, this is how to
calculate the amount of oxygen
consumed:
1𝑚𝑚𝑚𝑚𝑚3
100𝑔𝑔𝑔3
17.0𝑔𝑔𝑔3
5𝑚𝑚𝑚𝑚2
4𝑚𝑚𝑚𝑚𝑚3
32.0𝑔𝑔2
1𝑚𝑚𝑚𝑚2
So oxygen is the limiting reagent.
3.
= 235𝑔𝑔2
Ammonia is in excess and 57.5 g is left.
1𝑚𝑚𝑚𝑚2 4𝑚𝑚𝑚𝑚𝑚3 17.0𝑔𝑔𝑔3
100𝑔𝑔2
32.0𝑔𝑔2
5𝑚𝑚𝑚𝑚2
1𝑚𝑚𝑚𝑚𝑚3
= 42.5𝑔𝑔𝑔3
Next Slide
Quiz Answer cont’d
That tells us how much ammonia is
consumed. The excess would be difference
between the 100g you started with and the
amount consumed, or 57.5g NH3.
Return to Examples
Next Lesson
Percent Yield
Reality Compared to Theory
Next slide
Using More of What You Have
Learned
In the previous lesson, you saw how to predict
how much product would result from given
quantities of starting material. That amount
produced is referred to as the “theoretical
yield”. In other words, all things being perfect,
that amount would be the maximum quantity
of product you could create from the given
quantities of reactants. First, you had to
determine the limiting reagent and then use
that to calculate the (theoretical) yield.
Next Slide
Continued
𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝑌𝑌𝑌𝑌𝑌 =
𝐴𝐴𝐴𝐴𝐴𝐴 𝑌𝑌𝑌𝑌𝑌
𝑥
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑌𝑌𝑌𝑌𝑌
100%
Let’s look at an example –
Calculate the percent yield of sodium
sulfate if 32.18 g sulfuric acid reacts with
excess sodium hydroxide to give 37.91g
sodium sulfate.
Next Slide
Continued
32.18𝑔 𝐻2𝑆𝑆4
1𝑚𝑚𝑚 𝐻2𝑆𝑆4 1𝑚𝑚𝑚𝑚𝑚2𝑆𝑆4 142.0𝑔 𝑁𝑁2𝑆𝑆4
𝑥
𝑥
𝑥
98.08𝑔 𝐻2𝑆𝑆4 1𝑚𝑚𝑚 𝐻2𝑆𝑆4 1𝑚𝑚𝑚 𝑁𝑁2𝑆𝑆4
= 46.59𝑔 𝑁𝑁2𝑆𝑆4
However, the experimental yield was 37.91g
𝑁𝑁2𝑆𝑆4, so the percent yield is:
% yield = 100% 𝑥
Review Lesson
37.91𝑔
46.59𝑔
= 81.37%
Go to Quiz
Quiz Question
SO2(g) + H2O(l) → H2SO3(aq)
If this reaction yields 21.61g sulfurous acid
when 19.07g of sulfur dioxide reacts with
excess water, what is the percent yield?
Check Answer
Answer to Quiz
Solution: 88.46% yield.
The equation given is balanced. So, first you have to
calculate the theoretical yield based on the limiting
reagent, which is given as sulfur dioxide. The
theoretical yield is:
1𝑚𝑚𝑚𝑚𝑚3
1𝑚𝑚𝑚𝑚2𝑆𝑆3 82.07𝑔𝑔2𝑆𝑆3
19.07𝑔𝑔𝑔3 𝑥
64.05𝑔𝑔𝑔3
1𝑚𝑚𝑚𝑚𝑚3
1𝑚𝑚𝑚𝑚2𝑆𝑆3
= 24.43𝑔𝑔2𝑆𝑆3
21.61𝑔
% 𝑦𝑦𝑦𝑦𝑦 =
𝑥𝑥𝑥𝑥𝑥 = 88.46%
24.43𝑔
Review Lesson
Next Slide
Titrations
Using Liquid Concentrations
Next slide
Using Molarity, Normality and
Percent Concentrations
Sometimes we are asked to do quantitative
calculations of reactions based on volumes and
concentrations of solutions.
Suppose you have two different solutions containing
solutes that can react with each other to form a
product. A simple example would be reaction of an
acid solution and a base solution. If you know the
concentration of one of the two solutions, by using
some simple mathematics, you can calculate the
concentration of the second solution. The type of
experiment used to determine the volumes required
is called “titration”.
Next Slide
Continued
Remember, if you know the concentration
of a solute, it may be expressed in different
ways.
Molarity is the concentration of the solute in
moles of solute per liter of solution.
Normality is the concentration of the solute
in equivalents per liter of solution.
Percent concentration can be the
percentage based on grams of solute in
grams of solution.
Next Slide
Continued
So, if the concentration of a solution is known
and the volume is known, you know how much
of the solute you have. You simply multiply the
concentration by the volume (making sure the
units are consistent).
One useful equation for this type of problem is
NA x VA = NB x VB
Where N refers to normality and V refers to
volume in liters; A and B represent two
reactants, like an acid and a base.
With the knowledge you have from previous
lessons, you now have the ability to do
stoichiometric calculations for solutions.
Calculations like these may be done for
neutralization reactions, precipitation reactions,
or reactions that produce gases.
Next Slide
Acid-Base Neutralization
What volume of 1.5M H3PO4 is necessary to
neutralize 20 ml of 2N KOH?
Solution:
2N KOH = 2 equivalents of KOH/L solution.
1.5M H3PO4 = 4.5 equivalents H3PO4/L solution
Since NA x VA = NB x VB
Solving for 𝑉𝐴 =
2𝑁𝑁𝑁𝑁𝑁𝑁
4.5𝑁
= 8.9𝑚𝑚
Next Slide
Precipitation Reaction
What is the minimum volume of 0.250M
AgNO3(aq)that will precipitate all the chromate
ion in 0.500L of 0.800M K2CrO4(aq)?
Solution:
The balanced equation is
K2CrO4(aq) + 2 AgNO3(aq) →
Ag2CrO4(s) + 2 KNO3(aq)
In 0.500L of 0.800M K2CrO4(aq),
𝑚𝑚𝑚𝑚
0.500𝐿 𝑥 0.800
= 0.400 𝑚𝑚𝑚𝑚𝑚 K2CrO4
𝐿
Next Slide
Precipitation cont’d
2𝑚𝑚𝑚 𝐴𝐴𝐴𝐴3
0.400𝑚𝑚𝑚 𝐾2𝐶𝐶𝐶4 𝑥
1𝑚𝑚𝑚 𝐾2𝐶𝐶𝐶4
= 0.800𝑚𝑚𝑚 𝐴𝐴𝐴𝐴3
The volume of 0.250 M AgNO3 solution
required is
1.00𝐿
𝑉 𝐴𝐴𝐴𝐴3 = 0.800𝑚𝑚𝑚 𝑥
= 3.20 𝐿
0.250𝑚𝑚𝑚
Next Slide
Gas Producing Reactions
How many moles of carbon dioxide gas will
form if 300. mL of 0.500 M HCl are added to
500. mL of 0.500 M NaHCO3?
Solution:
Balanced equation is:
𝑁𝑁𝑁𝑁𝑁3 𝑎𝑎 + 𝐻𝐻𝐻 (𝑎𝑎)
→ 𝑁𝑁𝑁𝑁 𝑎𝑎 + 𝐻2𝐶𝐶3 (𝑎𝑎)
→ 𝑁𝑁𝑁𝑁 𝑎𝑎 + 𝐻2𝑂 𝑙 + 𝐶𝐶2 (𝑔)
Looking at the data given and the 1:1 mole
ratio of the reactants, HCl is the limiting
reagent.
Next Slide
Gas Producing Reactions
cont’d
0.500𝑚𝑚𝑚𝑚𝑚𝑚 1𝑚𝑚𝑚𝑚𝑚2
𝑚𝑚𝑚 𝐶𝐶2 = 0.300𝐿 𝑥
𝐿
1𝑚𝑚𝑚𝑚𝑚𝑚
= 0.150𝑚𝑚𝑚 𝐶𝐶2
Note that at STP, we could use Avogadro’s
Law to determine the volume of CO2.
Review Lesson
Go to Quiz
Quiz Questions
1.
What volume of 0.250 M HCl is necessary
to precipitate the solute in 0.150 L of
0.100M Pb(NO3)2?
2.
How many mL of 0.500M sulfuric acid will
just neutralize 12.0g sodium hydroxide in
500 mL of water?
Review Lesson
Check Answers
Quiz Answers
Solution: 120 mL 0.100M HCl.
Must first write the balanced equation:
Pb(NO3)2(aq) + 2HCl(aq) →
PbCl2(s) + 2HNO3(aq)
Then calculate how many moles of
Pb(NO3)2 there are in 0.150L of solution. You
also know there is a 1:2 ratio of Pb(NO3)2 to
HCl so you multiply the moles of Pb(NO3)2
by 2 to obtain the final answer.
1.
Next Slide
Quiz Answers cont’d
Solution: 300mL 0.500M H2SO4.
Need balanced equation:
2NaOH(aq) + H2SO4(aq) →
Na2SO4(aq) + 2H2O(l)
There is a 2:1 ratio between the moles of
NaOH and moles of H2SO4. Have grams of
NaOH – convert this to moles by using molar
mass of NaOH (40.0g). Convert to moles of
1𝑚𝑚𝑚𝑚2𝑆𝑆4
H2SO4 by using ratio of
2.
2𝑚𝑚𝑚𝑚𝑚𝑚𝑚
Next Slide
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