Today is Tuesday,April 28th, 2015

Pre-Class:How does an airbag work?

Sherlock Holmes, in Sir Arthur Conan Doyle’s A Study in Scarlet

“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.”

Stuff You Need:Calculator

Periodic TableSmall Paper

Towel

In This Lesson:Stoichiometry(Lesson 4 of 4)

Today’s Agenda

• Mole Ratios• Stoichiometry• Percent Yield• Limiting Reagents and Excess Reagents

• Where is this in my book?– P. 359 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:

• 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

given moles [coeff] target moles[coeff] given moles

0.60 mol N2 2 mol NH3

1 mol N2

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:

• So we would need 0.60 * (3/1) = 1.8 mol H2.

0.60 mol N2 3 mol H2

1 mol N2

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) + 3H2 (g) 2NH3 (g)

• 16.81 g + 3.63 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:

• So we would need 4.99 * (1/2) = 2.495 mol N2.

4.99 mol NH3 1 mol N2

2 mol NH3

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

Atoms, Molecules, or Formula Units

Molar Mass

1 mol

1 mol

Molar Mass

22.4 L

1 mol

1 mol

22.4 L

6.02x1023

Rep. Part

1 mol

1 molxx xx x x

6.02x1023

Rep. Part

Volume (L)Mass (g)

The Mole Highway

Atoms, Molecules, or Formula Units

Molar Mass Molar Mass 22.4 L 22.4 L 6.02x1023

Rep. Part. ÷÷ ÷x x x 6.02x1023

Rep. Part.

Volume (L)Mass (g)

MOLE BRIDGE

Given Chemical Target Chemical

This is the Mole

Bridge.

The Mole Bridge is

guarded…

…by a Mole Troll.

You need to have units in

moles to enter the bridge.

You need to have units in moles to exit the bridge.

MOLES

MOLES

MOLES

MOLES

Grams Make it Moles

Liters

Rep. Part.

Make it

Moles

Make it Moles

Grams

Liters

Rep. Part.

Make it Grams

Make it

Liters

Make it Rep. Part.

MOLE RATIO

The Mole BridgeREQUIRES BALANCED EQUATIONS!

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 reacts 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.

• Step 4: Convert moles to grams.• 0.1205 mol Al2O3 = 12.3 g Al2O3

0.241 mol Al 2 mol Al2O3

4 mol Al= 0.1205 mol 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) 2 KCl (s) + 3O2 (g)

• Step 2: Convert grams to moles.• 50.0 grams KClO3 = 0.408 mol KClO3

2KClO3 (s) 2 KCl (s) + 3O2 (g)

• Step 3: Set up mole ratios and calculate moles of desired component.

• Step 4: Convert to liters.• 0.612 mol O2 = 13.7 L O2

0.408 mol KClO3 3 mol O2

2 mol KClO3

0.612 mol O2=

Now to practice…

• Benchmark Stoichiometry Problems– #2, 4, 6, 7

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:

• Percent Yield:

• NOTE: Percent yield should be calculated using grams or liters, not moles.

100)Value Accepted

|Value Accepted - alExperiment|( Error Percent

Percent Yield (Actual Yield

Theoretical Yield) 100

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?

Percent Yield (Actual Yield

Theoretical Yield) 100

(63

115) 100 54.78%

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 NoteThis 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

• 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.

Compare!

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 ExampleWRITE 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:

solution of liters

solute of molesM

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…

• 0.12 moles of HCl are in the beaker/your hand.

solution of liters

solute of molesM

L 0.020

solute of molesM 6

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