Download mole ratio

In This Lesson:
Stoichiometry
(Lesson 4 of 4)
Today is Wednesday,
th
April 26 , 2017
Stuff You Need:
Calculator
Periodic Table
Pre-Class:
How does an airbag work?
“In solving a problem of this sort, the grand
thing is to be able to reason backward. This
is a very useful accomplishment, and a very
easy one, but people do not practice it
much.”
Sherlock Holmes, in Sir Arthur Conan Doyle’s A Study in Scarlet
Today’s Agenda
•
•
•
•
Mole Ratios
Stoichiometry
Percent Yield
Limiting Reagents and Excess Reagents
• Where is this in my book?
– P. 390 and following…
By the end of this lesson…
• You should be able to determine the expected
and experimental yields of a chemical
reaction.
• You should be able to determine the correct
quantities of reactants for a desired amount of
products.
• You should be able to determine the limiting
and excess reagents of a reaction.
Airbags
• Sure, airbags fill with “air.” Where do they
get the air?
• It’s from three very rapid chemical
reactions:
• NaN3 (s) (sodium azide)  Na (s) + N2 (g)
• Na (s) + KNO3 (s)  K2O (s) + Na2O (s) + N2 (g)
• This one inflates the bag with nitrogen gas.
• K2O (s) + Na2O (s) + SiO2 (s)  glass
• This one melts the reactants into glass for safety.
• Why do I mention this? It’s stoichiometry!
• Air Bag Slow Motion video!
http://blog.electricbricks.com/wp-content/uploads/airbag.jpg
Les Big Ideas
• Stoichiometry is calculating the mass of
products or the mass of reactants given one of
the two.
– “If I have X grams of this, how much of that can I
make?”
Les Big Ideas
• When we balanced equations, we used
coefficients.
• For stoichiometry, we can consider these
coefficients equivalent to moles.
• Example: 2H2 + O2  2H2O
– Two moles of hydrogen gas react with one mole of
oxygen gas to form two moles of water.
• NOTE: You don’t need two moles of hydrogen
for a reaction to occur; they just react in that
proportion.
Ratios
• How many wheels are on the average car?
• 4
• So the ratio of wheels to cars is 4:1; every one car
has four wheels.
• How many spikes are on Bart Simpson’s head?
• 9
• So the ratio of spikes to heads is 9:1; every one
head has nine spikes.
Ratios
• What is the ratio of fingers to hands in a standard
human being?
• 10:2
• There are 10 fingers (technically 8) for every two
hands in a human.
• Alternatively, you could say 5:1.
• What is the ratio of carbon atoms to oxygen
atoms in carbon dioxide?
• 1:2
• There is one carbon atom for every two oxygen atoms.
Mole Ratios
• Another idea we need to explore further is the
mole ratio.
• A mole ratio is a relationship between the
amounts of two different quantities in a given
equation. Here’s a non-chemistry example:
– Suppose a cake recipe calls for two eggs and one
box of cake mix.
• 2 eggs + 1 box of mix  1 cake
Mole Ratios
• 2 eggs + 1 box of mix  1 cake
• From this equation, we can say things like:
– There are two eggs needed for every one box of
cake mix.
• 2 eggs / 1 box of mix
– There is one cake produced from every one box of
mix.
• 1 cake / 1 box of mix
– There is one box of mix needed to make one cake.
• 1 box of mix / 1 cake
Mole Ratios
• Here’s a chemistry example:
– Fertilizer companies often react nitrogen gas with
hydrogen gas to create ammonia gas for use in
their products (known as the Haber process).
– N2 (g) + H2 (g)  NH3 (g)
Mole Ratios
• Balanced: N2 (g) + 3H2 (g)  2NH3 (g)
• From this equation, we can say things like:
– There is one mole of nitrogen gas needed for every
three moles of hydrogen gas.
• 1 mol N2 / 3 mol H2
– There are two moles of ammonia gas produced from
every one mole of nitrogen gas.
• 2 mol NH3 / 1 mol N2
– There are three moles of hydrogen gas needed to
make two moles of ammonia gas.
• 3 mol H2 / 2 mol NH3
N2 (g) + 3H2 (g)  2NH3 (g)
• Example:
– If I have 1 mole of N2, how many moles of NH3 can
I make?
• 2 mol NH3
– If I have 2 moles of N2, how many moles of NH3
can I make?
• 4 mol NH3
– If I have 2.5 moles of N2, how many moles of NH3
can I make?
• 5 mol NH3
Mole Ratios
• To summarize this in general form, we can
make a mole ratio equation:
given moles
[coeff] target moles
[coeff] given moles
Moles to Moles
• N2 (g) + 3H2 (g)  2NH3 (g)
• How many moles of ammonia are produced when
0.60 moles of nitrogen react with an excess of
hydrogen?
• First, let’s identify the key points:
– We’re going to need the mole ratio from the previous
page.
– Excess means just that – we’ve got lots and lots of
hydrogen so that quantity basically doesn’t matter for
this problem.
• NOTE: The “opposite” of excess is a limiting reagent. More
on that later.
N2 (g) + 3H2 (g)  2NH3 (g)
• Let’s use our previous formula:
given moles
[coeff] target moles
[coeff] given moles
0.60 mol N2
2 mol NH3
1 mol N2
• We are given 0.60 moles, and need to multiply it
by (the coefficient of the wanted compound, 2,
divided by the coefficient of the given compound,
1).
• 0.60 x (2/1) = 0.60 x 2 = 1.2 mol NH3
N2 (g) + 3H2 (g)  2NH3 (g)
• Last step: Check your work!
• According to the formula, we should have 2
moles of NH3 for every one mole of N2.
– In other words, we have twice as much NH3.
• Sure enough, we got 1.2 moles of NH3 – twice
as much as our moles of N2 (0.60).
– Wooo! (kinda)
N2 (g) + 3H2 (g)  2NH3 (g)
• [Here’s another way questions might look]
• Given the above equation, how many moles of
hydrogen gas do you need to react completely
with 0.60 moles of nitrogen gas?
• Solve this problem the same way but with a
new mole ratio:
0.60 mol N2
3 mol H2
1 mol N2
• So we would need 0.60 * (3/1) = 1.8 mol H2.
N2 (g) + 3H2 (g)  2NH3 (g)
• Wait a second…what about Conservation of
Mass?
• 0.60 mol N2 = 16.81 g
• 0.60 mol H2 * 3 moles = 1.8 mol H2 = 3.63 g
• 0.60 mol NH3 * 2 moles = 1.2 mol NH3 = 20.44 g
• N2 (g)
+
• 16.81 g +
3H2 (g)
3.63 g
 2NH3 (g)
= 20.44 g
– It works! (but notice that moles are NOT conserved).
N2 (g) + 3H2 (g)  2NH3 (g)
• [Here’s yet another way questions might look]
• Given the above equation, how many moles of
nitrogen gas are needed to produce 4.99
moles of ammonia?
• Solve this problem the same way but with a
new mole ratio:
4.99 mol NH3
1 mol N2
2 mol NH3
• So we would need 4.99 * (1/2) = 2.495 mol N2.
Summary
• We solved a “given reactants, how much
product” problem.
• We solved a “given reactants, how much other
reactant” problem.
• We solved a “given products, how much
reactant” problem.
• USE THE SAME PROBLEM-SOLVING
TECHNIQUE REGARDLESS OF THE “NATURE”
OF THE QUANTITIES.
Now to practice…
• Benchmark Stoichiometry Problems
– #1
“But wait,” said the class…
• What if your problem gives you grams and not
moles?
– Just convert to moles first. This is why it was so
important to learn mole conversions when we did.
• What if your problem gives you representative
particles and not moles?
– Just convert to moles first. This is why it was so
important to learn mole conversions when we did.
The Mole Highway
Molar Mass 1 mol
x
x
1 mol
Molar Mass
Mass (g)
x
22.4 L
1 mol
x
1 mol
22.4 L
Volume (L)
x
6.02x1023
Rep. Part
1 mol
1 mol
x
6.02x1023
Rep. Part
Atoms, Molecules,
or Formula Units
The Mole Highway
x Molar Mass ÷ Molar Mass x 22.4 L
Mass (g)
÷ 22.4 L
Volume (L)
6.02x1023
x
Rep. Part.
6.02x1023
÷
Rep. Part.
Atoms, Molecules,
or Formula Units
The Mole Bridge
REQUIRES BALANCED EQUATIONS!
The Mole
Bridge isLiters
guarded…
This is the
Liters Mole
Bridge.
MOLES
MOLES
Grams
Make it
Moles
You need toMOLES
have units in
Rep.
moles to enter
Part.
the bridge.
Given Chemical
MOLE
RATIO
BRIDGE
Make it
Grams
MOLES
Grams
…by
You
needato
have units
in
Mole
Troll.
Rep.
moles to exit
Part.
the bridge.
Target Chemical
Stoichiometry Road Map
Important Notes
• Before we get even further into stoichiometry,
you need to know these key points:
– Any kind of stoichiometry problem requires you to
use mole ratios. You can’t make a gram ratio or
anything else.
– Having a coefficient in front of a term in a
chemical reaction does not change its molar mass.
How to Work Stoichiometry Problems
1. Write and balance the equation.
2. Convert mass or volume or particles to
moles, if necessary.
3. Set up mole ratios and calculate moles of
desired component (product or reactant).
4. Convert moles to mass or volume or
particles, if necessary.
Stoichiometry Example
• 6.50 grams of aluminum react with an excess of
oxygen. How many grams of aluminum oxide are
formed?
• Step 1: Write and balance the equation.
• 4Al + 3O2  2Al2O3
• Step 2: Convert mass to moles.
• 6.50 grams Al = 0.241 moles Al
4Al + 3O2  2Al2O3
• Step 3: Set up mole ratios and calculate moles
of desired component.
0.241 mol Al
2 mol Al2O3
4 mol Al
= 0.1205 mol Al2O3
• Step 4: Convert moles to grams.
• 0.1205 mol Al2O3 = 12.3 g Al2O3
4Al + 3O2  2Al2O3
• Check Your Work!
• For every 4 moles of aluminum, we produce 2
moles of aluminum oxide.
• In other words, we produce half as much
aluminum oxide as aluminum with which we
start.
• We produced 0.1205 mol Al2O3 from 0.241
mol Al, so it works!
Now to practice…
• Benchmark Stoichiometry Problems
– #3, 5, 8
Gas Stoichiometry
• You can also do stoichiometry with gases.
• As you can imagine, you’ll probably need the
22.4 L = 1 mol conversion.
• The good news is, if reactants and products
are at the same temperature and pressure,
you can use volume ratios in place of mole
ratios.
– You can skip Steps 2 and 4.
Gas Stoichiometry
• How many liters of ammonia can be produced
when 12 liters of hydrogen react with an
excess of nitrogen?
• Step 1: Write and balance the equation.
• N2 + 3H2  2NH3
• Step 3: Set up volume ratios and calculate
volume of desired component.
12 L H2
2 L NH3
3 L H2
=
8 L NH3
Gas Stoichiometry?
• How many liters of oxygen gas can be collected from
the complete decomposition of 50.0 grams of
potassium chlorate?
• NOTE: There is a solid here, so you must use all four
steps.
• Step 1: Write and balance the equation.
• 2KClO3 (s)  2KCl (s) + 3O2 (g)
• Step 2: Convert grams to moles.
• 50.0 grams KClO3 = 0.408 mol KClO3
2KClO3 (s)  2KCl (s) + 3O2 (g)
• Step 3: Set up mole ratios and calculate moles
of desired component.
0.408 mol KClO3
3 mol O2
2 mol KClO3
• Step 4: Convert to liters.
• 0.612 mol O2 = 13.7 L O2
=
0.612 mol O2
The Mole Bridge
REQUIRES BALANCED EQUATIONS!
The Mole
Bridge isLiters
guarded…
This is the
Liters Mole
Bridge.
MOLES
MOLES
Grams
Make it
Moles
You need toMOLES
have units in
Rep.
moles to enter
Part.
the bridge.
Given Chemical
MOLE
RATIO
BRIDGE
Make it
Grams
MOLES
Grams
…by
You
needato
have units
in
Mole
Troll.
Rep.
moles to exit
Part.
the bridge.
Target Chemical
Now to practice…
• Benchmark Stoichiometry Problems
– #2, 4, 6, 7
Testing your knowledge…
A Tough Sample Problem – Write It Down
• The formula for wax is C18H38. A candle left to
burn decreases in mass from 38.89 g to 24.68 g.
How much carbon dioxide, in moles and liters,
was produced?
• First step: Whoa. Let’s get our bearings.
• The problem doesn’t provide us an equation so
we’ll start there.
• Since it says the wax burns:
– C13H33 + O2  ?
Testing your knowledge…
A Tough Sample Problem – Write It Down
• C18H38 + O2  ?
• What type of reaction is this?
– Combustion!
– Since the wax formula (C13H33) is a hydrocarbon, so
the products of this combustion reaction are CO2
and H2O.
• C18H38 + O2  CO2 + H2O
• Now we need to balance…
Testing your knowledge…
A Tough Sample Problem – Write It Down
55
2 18H38 + __O2
• __C
36
76
110 55
36
38
19 H2O
18 2 + __
 __CO
18 C 1 18 36
38 H 2 38 76
2 O 3 37 55 91 110
Remix! Multiply
everything by 2!
Testing your knowledge…
A Tough Sample Problem – Write It Down
• 2 C18H38 + 55 O2  36 CO2 + 38 H2O
• Now to handle the stoichiometry part…
• The problem tells us that the candle (AKA wax)
went from 38.89 g to 24.68 g after burning.
– In other words, the mass of wax went down 14.21 g
(38.89 g – 24.68 g).
– The problem is trying to tell us, then, that we reacted
14.21 g of wax.
• Now it’s just standard stoichiometry!
Testing your knowledge…
A Tough Sample Problem – Write It Down
• 2 C18H38 + 55 O2  36 CO2 + 38 H2O
• Convert 14.21 g wax into moles.
– 14.21 g C18H38 = 0.05583 mol C18H38
• Now for the mole ratio according to the
equation:
0.05583 mol C18H38
1.061 mol CO2
36 mol CO2
2 mol C18H38
=
1.005 mol CO2
22.4 L CO2
1 mol CO2
=
23.77 L CO2
Cooking
• You know how when you’re cooking, stuff
usually sticks to the spoon or the bowl and
you can never quite use all of your
ingredients?
• So when a recipe says you can make three
dozen cookies, you might end up only making
30?
• And you know how 30/36 is like 83%?
Percent Yield
• Stoichiometry, like a recipe, is great for
figuring out how much product a reaction will
produce…theoretically.
– As we’ve learned, that probably won’t happen.
• Enter percent yield.
• Percent yield, like percent error, is a measure
of how much you should have gotten
compared to how much you got.
– Like percent error, but for “production” of stuff.
– Remember this from the MgO lab?
Percent Yield
• Note the similarity:
• Percent Error:
| Experiment al - Accepted Value |
Percent Error  (
) 100
Accepted Value
• Percent Yield:
Actual Yield
Percent Yield  (
)  100
Theoretical Yield
• NOTE: Percent yield should be calculated
using grams or liters, not moles.
Percent Yield Example
• Imagine you perform a reaction designed to
produce 115 grams of product but you are
only able to produce 63.0 grams. What is your
percent yield?
Actual Yield
Percent Yield  (
)  100
Theoretical Yield
63
(
)  100  54.78%
115
Percent Yield
• One other thing: Here’s a friendly translation of
the text many problems will give you:
– “What is the theoretical yield of…?” really means:
• Solve the stoichiometry problem like we’ve been doing, and
answer in grams or liters (not moles).
– “Theoretical yield” or “Expected yield”
• What you would “expect” to produce “in theoretical” perfect
conditions.
– “Actual yield” or “Experimental yield”
• What you “actually” produced in the “experiment” under
real world imperfect conditions.
The Big Important Note
• When you solve a stoichiometry problem like
we just did…and get the answer in
grams/liters…that answer is the theoretical
yield!
– If you really did the experiment, you probably
won’t make that much.
– That amount is the actual yield – the amount you
actually make in reality (not stoichiometry).
Now to practice…
• Stoichiometry Practice Problems
– #4-9; 9 is interesting.
The Other Big Important Note
This comes into play a lot for labs. Write it in an obvious location!
• Occasionally you’ll run into a question about “mole
ratios.”
• As you’ve seen, theoretical mole ratios are found
using coefficients in the balanced equation (like 3:2,
1:1, 2:1, et cetera).
• Experimental mole ratios are found by dividing one
actual mole quantity by another actual mole quantity
to get a number.
– The bigger one should be the numerator.
– The number tells you how many more moles of one
substance you have than another substance.
• Neither ratio has units.
Limiting Reagents
• Earlier I mentioned something about limiting
reagents (and things being in “excess”).
• Here’s a more concrete definition:
– Excess (or the excess reagent/reactant) is when
there’s more than enough of something.
• Example: When you light a Bunsen burner, you’re
reacting methane with oxygen. We’ve got an excess of
oxygen ‘cause we’re probably never going to run out.
– Limiting reagents/reactants are things that will
run out during the reaction, determining when
the reaction stops.
Limiting Reagents
• Suppose you’re baking a cake and the recipe calls
for one box of cake mix and 2 eggs per cake.
– You have 50 boxes of cake mix on hand.
– You have 4 eggs on hand.
• How many cakes can you make?
– 2.
• What is this “reaction’s” limiting reagent?
– The eggs. You have an excess of cake mix.
• How much excess reactant is left over?
– 48 boxes of cake mix.
Identifying Limiting Reagents
• To identify a limiting reagent in a problem, you
will have to compare theoretical yields of each
reactant:
– Step 1: Pick a product. It can be any one, but pick the
one the problem mentions if possible. Stick with it.
– Step 2: Use one reactant to determine how much of
that product you’d make.
– Step 3: Use the other reactant(s) to determine how
much of that same product you’d make.
– Step 4: Compare. Whichever one led to fewer
product moles is the limiting reagent.
– Step 5: If necessary, use the limiting reagent to
calculate percent yield.
Identifying Limiting Reagents
• How to solve one of these problems in plain
English?
• Do two stoich problems in one.
– One is for one reactant, the other is for the other
reactant.
– The reactant that makes the lesser amount of
product is the limiter.
– The other is the excess reagent.
Limiting Reagent Example
• A 2.00 g sample of ammonia (NH3) is mixed
with 4.00 g of oxygen. Which is the limiting
reactant?
• 4NH3 (g) + 5O2 (g)  4NO (g) + 6H2O (g)
• Step 1: Pick a product.
– The problem doesn’t specify either, so let’s go
with NO (nitrogen monoxide).
4NH3 (g) + 5O2 (g)  4NO (g) + 6H2O (g)
• Step 2: Use one reactant to determine yield.
– This is like the stoichiometry stuff we practiced.
• 2 g NH3 = 0.117 mol NH3 = 0.117 mol NO
• Step 3: Use the other reactant to determine yield.
• 4 g O2 = 0.125 mol O2 = 0.1 mol NO
• Step 4: Compare. Less product = limiting reagent.
– Since the 4 g of O2 produced less product, oxygen is the
limiting reagent.
4NH3 (g) + 5O2 (g)  4NO (g) + 6H2O (g)
• So, if oxygen is the limiting reagent and leads
to 0.1 mol NO, what is the theoretical yield of
NO (in grams) we would produce?
• 0.1 mol NO = 3.01 g NO
• And if we performed this experiment and
formed 2.56 g NO, what is our percent yield?
• (2.56/3.01) * 100 = 85.05% yield
Summary
Compare!
• 4NH3 (g) + 5O2 (g)  4NO (g) + 6H2O (g)
• Notice that we went from NH3 to NO.
• Then we went from O2 to NO.
• Then we compared.
– Smaller amount of NO comes from the limiting
reagent.
Alternate Method
• METHOD WE LEARNED
– Reactant 1 to Product 1
– Reactant 2 to Product 1
• ALTERNATE METHOD
– Reactant 1 to Reactant 2
– Compare
• Let’s re-solve the last problem as an example.
Alternate Method
• A 2.00 g sample of ammonia (NH3) is mixed
with 4.00 g of oxygen. Which is the limiting
reactant?
• 4NH3 (g) + 5O2 (g)  4NO (g) + 6H2O (g)
• 2 g NH3 = 0.117 mol NH3
0.117 mol NH3
5 mol O2
4 mol NH3
= 0.146 mol O2
• So you need 0.146 mol O2, or 4.67 grams.
• Since you don’t have enough O2 for NH3, O2 is the
limiting reagent.
• If you like this method, try it!
Now to practice…
• Stoichiometry Practice Problems
– #10-11 (do 11 first); try #2-3 if you finish early.
• Limiting Reagent and Percent Yield Practice
Problems
– #1-3; try others if you finish early – your choice
which ones.
One last thing…
• On some occasions at the end of a limiting
reagent problem, you’ll be asked how much of
the excess reactant is leftover when the reaction
completes. To do this:
– Step 1: Use the moles of limiting reagent.
– Step 2: Calculate how many moles of excess reagent
are needed to react with the limiting reagent.
– Step 3: Convert to grams.
– Step 4: Subtract this number from the quantity in the
problem.
Excess Reagent Concept
• To do a limiting reagent problem, you do two
stoich problems in one and compare results.
– Compare the amount of a product made by each
of two reactants.
• To do an excess reagent problem, do a third
stoich problem.
– Find how much excess reactant was used up by
the limiting reagent.
Excess Reagent Example
WRITE THIS DOWN!
• 4NH3 (g) + 5O2 (g)  4NO (g) + 6H2O (g)
• In our last problem, how much of the excess
reagent is left when the reaction ends?
• Step 1: Use the moles of the limiting reagent.
– We had 0.125 moles of limiting reactant – O2.
• Step 2: Calculate how many moles of the excess
reactant must react with it.
• 0.125 mol O2 * (4 mol NH3 / 5 mol O2) = 0.1 mol NH3
4NH3 (g) + 5O2 (g)  4NO (g) + 6H2O (g)
• Step 3: Convert to grams.
• 0.1 mol NH3 = 1.70 g NH3 used.
• Step 4: Subtract from starting quantity.
– We started with 2.00 g NH3, so there are 2.00 g 1.70 g = 0.30 g NH3 remaining.
Summary
• 4NH3 (g) + 5O2 (g)  4NO (g) + 6H2O (g)
• Our limiter is O2.
• Notice that we went from O2 to NH3.
– That gives us NH3 used.
• Then we subtracted NH3 from the starting
amount of NH3.
Now to practice…
• Stoichiometry Practice Problems
– #12
Last thing…I promise…
• Before we get to the end of this unit, I do
need to give you one last little formula.
• It’s for molarity (M), which is a measure of
concentration:
moles of solute
M
liters of solution
Molarity Example
• Suppose I hand you 20 mL of 6 M HCl.
– In a beaker.
• How many moles of HCl are in there?
• 20 mL is 0.020 L.
• The molarity is 6. So…
moles of solute
moles of solute
M
6M 
liters of solution
0.020 L
• 0.12 moles of HCl are in the beaker/your hand.
Molarity
• Other than that, solution stoichiometry, as
this is called, is just like regular stoichiometry.
– Once you know moles, it doesn’t matter that it’s in
a solution. Just calculate as usual…
• You’ll need this for the occasional
stoichiometry problem, like in the Lead (II)
Nitrate and Potassium Iodide Lab we’re gonna
do.
• Actually, let’s do that lab now.
Closure
• Remember that this is the skeleton equation for the
first reaction that occurs in an airbag:
– NaN3 (s)  Na (s) + N2 (g)
• A typical airbag contains 132.0 g of sodium azide.
Use stochiometry to determine the number of liters
of nitrogen gas and the number of moles of solid Na
produced in the 1st decomposition reaction of NaN3.
– 2.030 mol Na
– 68.21 L N2