Moles, Atoms, Liters

Stoichiometry

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Stoichiometry is to chemistry what recipes are to cooking: it's all about ingredients and amounts.

The ingredients, in chemistry, are balanced chemical reactions. The 'amounts' part is what usually trips up chemistry students. Because chemical reactions occur between atoms and/or molecules, our amounts have to be in units that correspond to molecules or atoms. The unit we use is called the mole and it's a tricky little devil until you know what's what.

The mole

The mole is to chemistry what the dozen is to eggs: a unit specifying a quantity. One mole is equal to 6.02 x 10²³ atoms or molecules. This number is referred to as Avogadro's number and it is massively important. It allows us to 'count' molecules of different masses and establish the proper proportions on the basis of weight rather than counting molecules or atoms (which is something we cannot easily do).

Let's ponder the reaction above. It tells us that 1 molecule of propane gas will react with 5 molecules of oxygen to generate 3 molecules of carbon dioxide and 4 molecules of water. We can't measure out molecules for the 'recipe', so we have to rely on another method to get the proportions right. To do that, we use the molar mass.

Molar mass is the mass (in grams) of an Avogadro's number's worth of molecules or atoms, i.e., the mass of one mole.

So in our example, one mole's worth of propane molecules would react with five moles of oxygen gas molecules to make three moles carbon dioxide gas molecules and four moles of water molecules.

Molar mass is the sum of atomic masses for the atoms in a compound.

In our example, propane gas has a molar mass of 3x the atomic mass of carbon (12.011) plus 8x the atomic mass of hydrogen (1.008) = 36.033 + 8.064 = 44.097 grams/mole. We do exactly the same for all the other reactants and products.

• 1 C₃H₈ (g) = 1 x 44.09 grams/mole = 44.090 grams

• 5 O₂ (g) = 5 x 31.999 grams/mole = 159.995 grams

• 3 CO₂ (g) = 3 x 44.009 grams/mole = 132.027 grams

• 4 H₂O (g) = 4 x 18.015 grams/mole = 72.060 grams

So far, so good. But what happens if we don't have exactly one mole's worth of propane?

Ah, yes. This is where the fun begins. The balanced chemical reaction has given us the correct proportions of moles for the reactants and products, but we are almost never given the number of moles as the starting point of our reaction. Instead, we almost always know the amount of reactants in terms of mass (grams) or volume (liters). Liters is the easier of the two to convert. Since 22.4 liters of any gas is 1 mole's worth of gas (at standard temperature and pressure, i.e. STP), we can easily convert to moles by dividing the volume (in liters) by 22.4.

Let's convert 112 liters of propane to moles (at STP)

But what if some of the reagents or products are liquids or solids?

Having liquid water as a product of the reaction might seem to make it harder, but only a little. However, in order to look at liquids and solids we have to work with mass, which means working with molar mass. In this example we have 5 moles of propane gas. According to the reaction, we will generate 4x the number of moles of water, or 20 moles. Since the molar mass of water is 18.015 moles/gram, the total amount of water formed in the reaction would be 360.3 grams, or about 2/3 the volume of a standard half-liter soda bottle.

Real world example!

Let's say I'm cooking up some tasty burgers on the barbie and that it will take about 5 liters (.22 moles) of propane to get it done. That means we'll need 1.1 moles of O₂ (g). Now, one estimate of the amount of oxygen in the atmosphere is around 1.5 x 10²⁰ moles. So it looks like we can get these meat pucks done without worrying about the oxygen part of the reaction.

When one reagent is available in abundance and cannot limit the reaction, we say the reagent is present in excess. This is the usual case with combustion reactions.

Let's do a few practice problems to see how you're doing so far....

Oh...one thing. You'll have to balance your own reactions.....

Fill in the Blank

How many liters of CO₂ (g) would be created from the combustion of 336 liters of methane (CH₄ (g)) in open air? Report your answer to the nearest tenth (i.e., xx.x).

Remember...combustion reactions have oxygen (O₂) as a reactant and water as one of the products.

Multiple Choice

How many liters of carbon dioxide gas are formed by the combustion of 300.63 grams of liquid heptane (C₇H₁₆)?

• 470.4 liters

• 67.2 liters

• 250.9 liters

• 382.6 liters

Let's move on to solids. The only difference between working with solids, liquids or gases is the unit of measurement. We use mass units (usually grams) when working with solids and liquids. Otherwise, the principal is the same: figure out the number of moles for each reagent and product by using the molar mass and the proportions of each chemical species. Let's try a simple example.

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In this example, the reaction of 1 mole of ferric sulfate with 3 moles of calcium hydroxide will produce 2 moles of ferric hydroxide and 3 moles of calcium sulfate. If we know the number of moles of one chemical species, we can determine the number of moles of every other species. Since we are usually given the number of grams, solving the problem means first determining the number of moles by converting from grams using the molar mass.

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Suppose we just happen to have 74.093 grams of calcium hydroxide. By mere coincidence, the molar mass of calcium hydroxide is 74.093! Amazing! We have exactly 1 mole!

Now that we know one of the species, we can determine the number of moles of the other species by using the proportions as defined by the coefficients. We do this by setting the denominator of the proportion to equal the coefficient of the known species.

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So using this trick, the number of moles of ferric sulfate would be 1/3(1) = 0.333; the number of moles of ferric hydroxide would be 2/3(1) = 0.667, and the number of moles of calcium sulfate would be 3/3(1) = 1.000. Let's check this out with the law of conservation of matter.

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Using our proportions from the last slide, the mass of each species in the reaction should be:

• Fe₂(SO₄)₃ = 0.333*molar mass of Fe₂(SO₄)₃ = 0.333*399.88 = 133.293 grams

• Ca(OH)₂ = 74.093 grams (we already knew this)

• Fe(OH)₃ = 0.667 * 106.876 grams/mole = 71.245 grams

• CaSO₄ = 1 * 136.14 grams/mole = 136.14 grams

If we sum up the masses of the reactants together and compare it to the sum of the masses of the products...they match. As they should. As they must.

Let's do another practice problem.

Multiple Choice

How much ferric chloride is produced from 58.44 grams of NaCl by the following reaction?

$$Fe\left(OH\right)_3\ +\ 3NaCl\ \rightarrow\ FeCl_3\ +\ 3NaOH$$  

• 54.07 grams

• 162.22 grams

• 486.65 grams

• 27.04 grams

Multiple Choice

Using the same reaction, how much ferric hydroxide is needed to react with all of the 58.44 grams of NaCl?

$$Fe\left(OH\right)_3\ +\ 3NaCl\ \rightarrow\ FeCl_3\ +\ 3NaOH$$  

• 35.62 grams

• 106.87 grams

• 71.24 grams

• I missed the last one and I can't do this one so I'm going back a few slides to review....

So we're getting close to understanding all we need to know about stoichiometry.

There's just one thing left to grapple one: limiting reagents. They were called cautum est reagentia when I studied under Pliny the Elder back in 79 AD. Before he became the first scientist to learn that running towards an exploding volcano is a bad idea. Before time became the cautum est reagentia for Ol' Pliny...

In the modern era, limiting reagents are literally 'the thing that runs out first'. It's easy to figure out what the limiting reagent is. Again...it's all about the proportions.

Time to make some cheeseburgers. Our reactants are burger patties, buns, cheese slices, and tomato slabs according to the recipe:

1 patty + 1 cheese slice + 2 tomato slabs resting gently on one bun pair (see right).

In this example, it is clear that we're going to run out of burger patties first, and that the maximum number of burgers to be made is going to be the same as the number of burger patties. That makes burger patties the limiting reagent.

We treat chemical reactions in exactly the same way. Reaction can't proceed to infinity. SOMETHING is going to run out. The thing that runs out controls how much product can be made by the reaction. All we need to do is figure out what that SOMETHING is!

 $Fe_2\left(SO_4\right)_3\ +\ 3Ca\left(OH\right)_2\ \rightarrow\ 2Fe\left(OH\right)_3\ +3CaSO_4$  

Let's return to a reaction we've already worked with, and assume that we have 100 grams of each reactant. 100 grams of ferric sulfate is 100/399.88 = 0.25 moles. 100 grams of calcium hydroxide is 100/74.09 = 1.35 moles. Based on the reactant coefficients, if we use up all the ferric sulfate we'll need 0.25*3 = 0.75 moles of calcium hydroxide. If we look at it from the perspective of calcium hydroxide, we need 1.35/3 = 0.45 moles of ferric sulfate. One of these options works, and one of them doesn't.

 $Fe_2\left(SO_4\right)_3\ +\ 3Ca\left(OH\right)_2\ \rightarrow\ 2Fe\left(OH\right)_3\ +3CaSO_4$  

If you've already figured out that ferric sulfate is the limiting reagent, congratulations! Since we have more than enough calcium hydroxide to react with all the ferric sulfate, calcium hydroxide is present in excess.

The amount of product we can make is based on the limiting reagent. Since we have 0.25 moles of ferric sulfate, we can make 0.5 moles of ferric hydroxide and 0.75 moles of calcium sulfate.  Both of these are referred to as theoretical yields.

Theoretical vs actual yields

But sometimes stuff just doesn't turn out the way we want, for whatever reason. Let's go back to our cheeseburgers. We had three burger patties limiting the number of burgers we could assemble to three. But for some reason, we ended up with only two edible burgers. This means our actual yield was only 2/3, or 66.7%.

 $Fe_2\left(SO_4\right)_3\ +\ 3Ca\left(OH\right)_2\ \rightarrow\ 2Fe\left(OH\right)_3\ +3CaSO_4$  

 $Fe_2\left(SO_4\right)_3\ +\ 3Ca\left(OH\right)_2\ \rightarrow\ 2Fe\left(OH\right)_3\ +3CaSO_4$  

Stoichiometry problems can be daunting when first encountered, but if you apply some straightforward, basic principles and just a little bit of algebra, they are actually pretty easy. Organization is key. Just a little common sense...and an insane amount of luck...

There are some practice problems waiting for you. Enjoy!

Multiple Choice

Al(OH)₃ + 3HCl --> 3H₂O + AlCl₃

14 grams of Al(OH)₃ is completely reacted with excess hydrochloric acid. If you actually recovered 22.0 grams of aluminum chloride, what is the percent yield of the reaction?

• 100%

• 25%

• 92%

• 69%

Multiple Choice

Silver nitrate and sodium phosphate are reacted in equal amounts of 200 g each. How many grams of silver phosphate are produced? 
3AgNO₃  + Na₃(PO₄)  --->  Ag₃(PO₄)  + 3NaNO₃

• 164 g
• 64 g
• 146 g
• 164 moles

Multiple Choice

Theoretical yield = 73g
Actual yield = 62g
Calculate the percent yield.

• 1.16%
• 116%
• 85%
• 76%

Multiple Choice

P₄ + 3O₂ --> P₄O₆ 
What is the limiting reactant is 12 moles of P₄ react with 15 moles of O₂?

• P₄
• O₂
• P₄O₆ 
• none of the above

Multiple Choice

If 15 grams of copper (II) chloride react with 20 grams of sodium nitrate, how much sodium chloride can be formed? 

• 13.04 g
• 130.4 moles
• 130.4 g
• 13.04 moles

Multiple Choice

4 Al + 3 O₂ –> 2 Al₂O₃  How much aluminum would be needed to completely react with 45 grams of O₂?

• 1.05 moles
• 3.75 moles
• 1.875 grams
• 1.875 moles

Multiple Choice

B₂H₆ + 3O₂ -->2 HBO₂ + 2 H₂O
 What mass of O₂ will be needed to burn 36.1 g of B₂H₆?

• 13.8 g O₂
• 3.86 mol of O₂
• 124 g O₂

Multiple Choice

2 KClO₃ → 2 KCl + 3 O₂
How many moles of oxygen are produced when 6.7 moles of KClO₃ decompose completely?

• 6.7 mol
• 1.0 mol
• 10.1 mol
• 4.5 mol