Stoichiometry

Outline

Moles and Masses: Using Stoichiometric Coefficients
Practice with Concepts I: Stoichiometry Exercises
Practice with Concepts II: Limiting Reactants
Discussion Questions
Optional Website Reading

Moles and Masses: Using Stoichiometric Coefficients

Our next step in building this picture of how different kinds of matter interact requires that we start concentrating on the more practical problems of chemistry, the "how much do we need to do this" considerations that underlie modern industrial chemical processes as well as theoretical research. We need to be able to go from a chemical reaction equation and the number of atoms or molecules involved in a single instance of the reaction to the number of moles (and grams) involved in an actual situation.

Using Stoichiometric Coefficients and calculating masses

We've already used this kind of analysis to determine moles to grams or grams to moles for the components in a single compound. For example, water (H₂O) has 2 hydrogen atoms and 1 oxygen atom in each molecule, so it therefore has 2 moles of hydrogen and one of oxygen in each mole of water. We can use the atomic weights (grams/mole) to determine that at 1.0079 grams per mole for hydrogen and 15.9994 grams per mole for oxygen, a mole of water will mass (2 * 1.0079) + (1 * 15.9994) = 2.0158 + 15.9994 = 18.0152 grams.

Now we combine that information to determine molecular, molar, and mass information for different components in a chemical reaction when we know the stoichiometric coefficients of the reactants and products involved.

Suppose we need to know how many grams of one compound (A) we need for a certain reaction with a known quantity of another compound (B).

First, we balance the equation for the reaction. This gives us the molecular and molar ratios of the components.
Then, from the known mass of compound B, we determine the number of moles we have of B to start the reaction.
We can then determine the number of moles we need of compound A, and then calculate the grams we need of A for the reaction to proceed to completion.

For example, water reacts with aluminum carbide (a solid) to form aluminum hydroxide and methane gas. Suppose that we have 100 grams of aluminum carbide. How much water do we need to dissolve all the carbide?

Al₄C₃ (s) + H₂O (l) -> Al(OH)₃ (aq) + CH₄(g)

First, we balance the equation:

Al₄C₃ (s) + 12 H₂O (l) --> 4 Al(OH)₃ (aq) + 3 CH₄(g)

This tells us that we need 12 moles of water for every mole of aluminum carbide.
Now, we start with 100 grams of aluminum carbide. The molecular weight of aluminum carbide is 4 * 26.9815 + 3 * 12.011 = 107.9260 + 36.033 = 143.9590 ==> 143.96 (sig figs) grams/mole.

If we have 100 grams, then we have 100/143.96 moles, or .695 moles of aluminum carbide.

We need 12 times this many moles of water for the reaction, or 12 * .695 = 8.335 moles of water.

At 18.01 grams/mole, 8.335 moles of water weighs 150.12 grams, so we need 150 grams of water, more or less, to completely react with 100 grams of aluminum carbide, and we can now set up our methane production plant.

The important point to keep in mind here is that a chemical reaction always gives us the molar ratios. Mass ratios must be calculated from the atomic or molecular weights of the atoms and compounds involved in the reaction.

Practice with the Concepts I: Stoichiometry

Balance an equation where a reactant is in limited supply

Suppose we have drop 6.54 g of zinc metal reacting with a solution containing 5.47g of hydrochloric acid. How much hydrogen gas is produced?

Step 1: Write the reaction equation and balance it.

First we have to determine all the reactants and products. We knowthat the reactants are zinc metal, Zn, and hydrochloric acid, HCl. We know that one of the products must be hydrogen gas, H₂, so the remaining product or products contain the zinc and chlorine. This starts to look like a metal-halogen compound. If we look up zinc, we see that it has a charge of +2, so we need to Cl- ions to balance that. Zinc chloride thus has the formula ZnCl₂.

We know have enough information to write the unbalanced equation for the reaction: Zn (s) + HCl (aq) -> H₂(g) + ZnCl₂.

Balance the equation!

Step 2: Convert the mass data to moles

Zn has a molar mass of 65.39 grams/mole. How many moles of zinc are in 6.54 grams of the stuff?

HCl has a molar mass of 1.0079 + 35.4527 = 36.4606 = 36.46 grams/mole. How many moles of HCl do we have?

Step 3: Determine the limiting reactant

When we look at the reaction equation, we see that we have a ratio of 1 mole of Zn to 2 moles of HCl. But our masses give us a ratio of .100 moles to .150 moles, or 1:1.5. So which is the limiting reactant?

Step 4: Determine the amount of products formed.

The products are formed in a 1:1 ratio to the available zinc, and a 1:2 ratio to the available HCl. How many moles of H₂ will be formed, based on your determination of the limiting reactant?

Step 5: Moles of hydrodrogen back to mass.

Given your determination in step 4 of the moles of hydrogen produced, what is the mass of the hydrogen produced if hydrogen has a molar mass of 1.0079 grams?

Limiting Reagents

What happens to a reaction when not enough of one of the reactants is available? Some of the other reactants will be left over. The yield is limited by the reactant with the least amount available to react; this amount will be completely consumed in the reaction (under ideal conditions). In order to figure out which reactant is the limiting one, you must determine the amount of the products for the available amount of each reactant. The theoretical yield will be the lesser amount, limited by the reactant on which the calculation for that amount is based.

Take the reaction of silver with iodine: 2Ag (s) + I₂(s) --> 2AgI(s). This reaction has already been balanced. The periodic table gives 107.9 g/mole for the atomic weight of silver, and 126.9g/mole for the atomic weight of iodine, so the resulting molecule of AgI is 234.8 g/mole. To determine the theoretical yield, assume that you have only 1 gram of silver, and calculate the reaction amounts:

1.00g Ag * ( 234.7g AgI / 107.9 g Ag) = 2.18 g AgI

Now assume that you have only 1 gram of iodine:

1.00g I * (234.7g AgI / 126.9 g I) = 1.85 g AgI

Since the amount is less if only one gram of iodine is available, the limiting factor is the iodine. In the presence of any amount of iodine and an equal amount of silver, the whole of the iodine will be consumed, and there will be silver left over. If we have 1 g iodine and 1 g silver, we should get 1.85 g AgI.

However, most reactions do not run to their theoretical limits. Normally we would get less than the theoretical yield, say, 1.50g AgI was the actual result. This is a fraction of what we expected, and we can calculate it : 1.50/1.85 = .811 or 81.1%. This percent yield is a measure of the efficiency of the reaction. The close it is to 100%, the more efficient the process in causing all the reactants to, well, react and make the products.

Again, consider what happens if we only have 100 grams of aluminum carbide available, even if we have 200 grams of water? Obviously, not all of the water can react, and we will get an amount of methane limited by the amount of available aluminum carbide. So another practical consideration (one which occupies much of the time of industrial chemists) involves determining which reactant limits the amount of output if we have specific amounts of reactants.

Suppose we add 40.0 kg of magnesium to 85.2 kg titanium chloride in order to purify the titanium. Our reaction

TiCl₄ + Mg --> MgCl₂ + Ti

must first be balanced. We start with 4 chloride atoms, which must appear in the products, and since the chloride is combined with magnesium in the products, we are forced to balance magnesium in the reactants as well:

TiCl₄ + 2 Mg --> 2 MgCl₂+ Ti

So for every mole of titanium chloride, we need two moles of magnesium to support the reaction; we get two moles of magnesium chloride and a mole of pure titanium as a result of this reaction.

Magnesium's atomic weight is 24.3050 grams/mole, so 40000 gm/24.3050 gm/mole Mg = 1645.75 moles of magnesium. This amount of magnesium could react with twice as much titanium chloride, or 3291.5 moles of TiCl₄.

But how much do we actually have? TiCl₄has a molecular weight of (1 * 47.88) + (4 * 35.4527) = 189.6908 g/mole. We have 85.2 kg, or 85200/189.6908 = 449.130 moles of titanium chloride. This is much less than the 3291.5 moles of TiCl₄ with which the available magnesium can react, so the limiting reactant is the titanium chloride.

Another way to do this is to determine the molar ratios of the two reactants. If we set this up as magnesium to titanium chloride, we get 1645.75/449.130 = 3.66. The molar ratio of Mg to TiCl₄ according to the reaction is 2/1 = 2.0. Since the ratio of moles required is smaller than the ratio of moles available, the reagent in the DENOMINATOR is the limiting reagent.

Since the TiCl₄is the limiting reagent, and the reaction gives us 1:1 ratio between TiCl₄ and the resulting pure titanium, we will have as many moles of titanium as we had titanium chloride to start with, or 449.130 moles. If we want to be complete, we can calculate the mass of the resulting pure titanium:

449.130 moles * 47.88 g/mole Ti = 2.15 * 10⁴= 21.5 kg pure titanium.

The moles of magnesium involved number twice the moles of titanium, or 2 * 449.130 = 898.26 moles of magnesium used up, so we have 1645.75 - 898.26 = 747.49 moles of magnesium left--almost half of what we started with.

Percent Yield

In any reaction, the amount we should get from the available reactants and the amount we do get will differ, because no operation is 100% efficient. Both thermodynamic considerations and the process used to implement the reaction will take their toll on the energy available for the reaction.

Percent yield is a measure of this discrepancy, and is defined as the amount you get divided by the amount you should get in a perfect world:

Percent yield = actual yield/theoretical yield

Because percent yield is a pure number, it doesn't matter whether we measure actual and theoretical yield in moles or masses, as long as we use the same kind of units in both the denominator and numerator of our ratio.

Suppose we run the titanium chloride purification process. To our dismay, instead of the 21.5 kg we should have gotten, we end up with 10.4 kg of pure titanium. What is the percent yield (on which we must base our actual costs and profits as purveyors of pure titanium culled from TiCl₄ ores)?

actual/theoretical = 10.4 kg/21.5 kg = 48.4%

Our reaction process is less than 50% efficient. Maybe we better find a new process!

Practice with the Concepts II: Limiting Reactants

Discussion Questions

Why do we need to figure out limiting reactants in most practical situations?
Can you think of any situations where there is a (for all practical purposes) unlimited amount of one reactant?
In real chemical analysis situations, the products of the reaction are never 100% of the reactants. How does this reality complicate chemical analysis of substances?

Optional Readings

More reaction balancing.
ChemTeam has a *great* site with a number of stoichiometry tutorials.

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