Today is Wednesday, April 26th, 2017

Today is Wednesday, April 26th, 2017
In This Lesson: Stoichiometry (Lesson 4 of 4) Stuff You Need: Calculator Periodic Table Today is Wednesday, April 26th, 2017 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!

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 = g 0.60 mol H2 * 3 moles = 1.8 mol H2 = 3.63 g 0.60 mol NH3 * 2 moles = 1.2 mol NH3 = g N2 (g) H2 (g)  2NH3 (g) 16.81 g g = 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) = 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?

Atoms, Molecules, or Formula Units
The Mole Highway 6.02x1023 Rep. Part Molar Mass 1 mol 22.4 L 1 mol 1 mol x x x x x x 1 mol Molar Mass 1 mol 22.4 L 1 mol 6.02x1023 Rep. Part Atoms, Molecules, or Formula Units Mass (g) Volume (L)

Atoms, Molecules, or Formula Units
The Mole Highway 6.02x1023 Rep. Part. 6.02x1023 Rep. Part. x Molar Mass ÷ Molar Mass x 22.4 L ÷ 22.4 L x ÷ Atoms, Molecules, or Formula Units Mass (g) Volume (L)

The Mole Bridge REQUIRES BALANCED EQUATIONS!
This is the Mole Bridge. The Mole Bridge is guarded… Liters Liters Make it Moles Make it Liters MOLES MOLES Make it Moles MOLE RATIO MOLE BRIDGE Make it Grams Grams Grams …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 Make it Rep. Part. Make it Moles Rep. Part. Rep. Part. Given Chemical 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
Write and balance the equation. Convert mass or volume or particles to moles, if necessary. Set up mole ratios and calculate moles of desired component (product or reactant). 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 = moles Al

4Al + 3O2  2Al2O3 Step 3: Set up mole ratios and calculate moles of desired component. Step 4: Convert moles to grams. mol Al2O3 = 12.3 g Al2O3 0.241 mol Al 2 mol Al2O3 4 mol Al = 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 mol Al2O3 from 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 = mol KClO3

2KClO3 (s)  2KCl (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

The Mole Bridge REQUIRES BALANCED EQUATIONS!
This is the Mole Bridge. The Mole Bridge is guarded… Liters Liters Make it Moles Make it Liters MOLES MOLES Make it Moles MOLE RATIO MOLE BRIDGE Make it Grams Grams Grams …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 Make it Rep. Part. Make it Moles Rep. Part. Rep. Part. Given Chemical 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 g to 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  ?

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…

55 36 38 27.5 19 __C18H __O2  __CO2 + __ H2O C H O 2 18 36 18 1 18 36 76 38 2 38 76 110 55 2 3 37 55 91 110 Remix! Multiply everything by 2!

2 C18H O2  36 CO H2O Now to handle the stoichiometry part… The problem tells us that the candle (AKA wax) went from g to g after burning. In other words, the mass of wax went down g (38.89 g – g). The problem is trying to tell us, then, that we reacted g of wax. Now it’s just standard stoichiometry!

2 C18H O2  36 CO H2O Convert g wax into moles. 14.21 g C18H38 = mol C18H38 Now for the mole ratio according to the equation: mol C18H38 36 mol CO2 2 mol C18H38 = 1.005 mol CO2 1.061 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: Percent 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?

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
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 = mol NH3 = mol NO Step 3: Use the other reactant to determine yield. 4 g O2 = 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)
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 ALTERNATE METHOD
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 = mol NH3 So you need 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! 0.117 mol NH3 5 mol O2 = 0.146 mol O2 4 mol NH3

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 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 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:

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 L. The molarity is 6. So… 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 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

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