Section Two

Chapter Four, Section Two


OBJECTIVES

  1. Learn how to calculate the amounts of compounds formed and used up in chemical reactions.
  2. Learn how to calculate percent yields in chemical reactions.
  3. Understand what happens to ionic compounds in aquous solutions.

I. Mass Relationships in Chemical Equations

Now we want to ask (and be able to answer) some very important questions. For example, suppose that I am burning a specific amount of Mg in the presence of oxygen and I want to know how much product I will have. The chemical equation is

Mg(s) + O2(g) arrow MgO(s)

Before I can answer the question, I must balance the equation. Just as numbers without units are meaningless, an unbalanced equation is insufficient. So we balance the equation:

2Mg(s) + O2(g) arrow 2MgO(s)

Now if I have 0.145 g of Mg, how much MgO will this produce?

The key to answering this question is to understand that reactions occur on the same mole basis as the balanced stoichiometric coefficients. This equation says that 2 moles of Mg produce 2 moles of MgO. So how many moles of Mg did I start with?

.145 g Mg/(24.31 g/mole) = .00596 moles of Mg

So how many moles of MgO will this reaction produce?

.00596 moles Mg * (2 moles MgO/2moles Mg) = .00596 moles of MgO

which is .00596 moles of MgO *(40.31 g MgO/mole MgO) = .240 gms of MgO

Similarly we can calculate the number of grams of oxygen needed to complete the reaction since 2 moles of Mg react with 1 mole of oxygen:

.00596 moles of Mg * (1 mole of O2/2 mole Mg) = .00298 moles O2

and the grams of oxygen would be

.00298 moles O2 * (32.00 gms O2/mole O2 = .0954 g O2

Note that we had to use 32.00 for the molar mass of oxygen as a diatomic.

Also note above that I said "needed to complete the reaction". In these calculations we are assuming that the reaction is going to completion (at least one of the reactants is completely used up). How would we know if the reaction were not completed? We could measure the amount of MgO that we get and see if it is 0.240 g in this case (the theoretical yield). However don't forget that you may not be perfect in carrying out the reaction and in recovering your product so that the reaction could go to completion and you still not measure 0.240 g of MgO because you lost some of it. But let's use this procedure to introduce a new concept: percent yield.

Let's assume that we use 0.145 g of Mg and then we measure the amount of MgO that we get to be 0.198 g (the actual yield). What is the percent yield?

Percent yield = (actual yield)/(theoretical yield) * 100

So in this case we would have a percent yield of

(0.198/0.240) * 100 = 82.5% ......... actually considered quite good for experimental error.

Let's suppose that we are making aspirin:

2 C7H6O3(s) + C4H6O3(l) arrow 2 C9H8O4(s) + H2O(l)

where C7H6O3 is salicylic acid, C4H6O3 is acetic anhydride, and C9H8O4 is aspirin. Note that I have already balanced the equation. Now suppose that we start with 14.43 g salicylic acid and that we obtain 6.26 grams of aspirin. What is the percent yield?

First we have to calculate the theoretical yield which is the amount that we would get if the reaction went to completion (just like we did above for MgO). So first we calculate the moles of salicylic acid that we started with:

14.43g/(138 g/mol) = .105 moles of salicylic acid

Now we multiply this by the mole ratio of 2 moles of aspirin formed for 2 moles of salicylic acid to get the number of moles of aspirin:

0.105 moles salicylic acid * (2 moles of aspirin/(2 moles of salicylic acid)) = 0.105 moles aspirin

Now the theoretical yield (in grams) of aspirin would be

0.105moles aspirin * (180 g/mol) = 18.9 grams aspirin

Then the % yield that we have obtained is

6.26 g/18.9 * 100 = 33.1 %

This also is really not a bad yield for such a reaction.

Now it is time to complicate this a bit! Sometimes you don't have enough of one of the reagents to react in the mole ratio indicated by the stoichiometric coefficients. If so then the reagent in short supply is called the limiting reagent.

Before we begin a chemical equation example of the limiting reagent concept, let's look at a limiting ingredient problem. Consider the following recipe written in a different form than usual:

1 cup water + 2 cups flour + 4 squares of chocolate + 2 cups of sugar + 4 oz butter + 2 eggs --> 1 cake

Now suppose that you have the following amount of each reactant: unlimited water, 12 squares of chocolate, 4 cups of sugar, 4 cups of flour, 8 oz of butter, and 3 eggs. Answer the following questions:

Now apply this reasoning that you used to answer the above questions to a "standard" chemical reaction. How do you determine if you have a reagent in short supply? You proceed exactly as you did in the cake example, except now we must work on a mole basis. You have to calculate the moles of each reagent and then compare with the mole ratios of the stoichiometric coefficients (the recipe). Let's work an example:

3 CCl4(l) + 2 SbF3(l) arrow 3 CCl2F2(l) + 2 Sb Cl3(s)

Note that I have balanced the equation. This is a reaction for producing freon-12 (CCl2F2). Let's suppose that we start with 150 g of CCl4 and 100 g of SbF3 and see if we have a limiting reagent problem.

The correct mole ratios would be 3 moles of carbon tetrachloride to 2 moles of SbF3. So let's see what we have.

150 g CCl4/(154 g/mol) = 0.974 moles of CCl4

100 g SbF3/(179 g/mol) = 0.559 moles of SbF3

These moles would give a SbF3/CCl4 ratio of 0.559/0.974 = 0.574 whereas what we should have (from the stoichiometric ratios) is 2/3 = 0.667

So we have a problem. Which is the limiting reagent? The number 0.574 is less than 0.667 so either the numerator is too small or the denominator is too large which says that either the amount of SbF3 is either too small or the amount of CCl4 is too large. Either way of stating this says that we don't have enough SbF3 so this is the limiting reagent. This means that SbF3 will be used up before the CCl4 is used up so in calculating the amount of freon-12 formed we will have to use the SbF3 data.

So now let's calculate the amount of freon formed and the amount of CCl4 left over after the reaction is complete.

100 g SbF3/(179 g/mol) * (3 moles freon-12 produced/2 moles of SbF3) = 0.838 moles of freon-12 produced and if we multiply this by the molar mass of freon-12 which is 121 g/mol, we will get 101 grams of freon-12 formed in the reaction.

How much CCl4 will be left over after the reaction is complete?

0.559 moles of SbF3 * (3 moles CCl4/2 moles of SbF3) = 0.839 moles of CCl4 which will be used up in the reaction. We started with 0.974 moles of CCl4 so we have 0.135 moles of CCl4 left over which is

0.135 moles CCl4 * (154 g/mol) = 20.8 grams of CCl4 left over.

II. Reactions Between Ions in Aquous Solutions

Some chemical reactions produce an insoluble product -- called a precipitate. These types of reactions are usually carried out in a water solution where the reactants dissociate in water to give the appropriate cations and anions. Then if a cation and an anion combine to form a compound which is insoluble in water, then a precipitation reaction occurs.

An example is the reaction

AgNO3(aq) + KCl(aq) arrow AgCl(s) + KNO3(aq)

where we have introduced another nomenclature using the symbol "aq" to mean that the compound is water soluble.

If we write this in terms of the ions formed in solution we see that there is an intermediate step in the reaction:

AgNO3(aq) + KCl(aq) arrow Ag+(aq) + NO3-(aq) + K+(aq) + Cl-(aq)

Then the silver ion and the nitrate ion can combine to give the following:

Ag+(aq) + NO3-(aq) + K+(aq) + Cl-(aq) arrow AgCl(s) + K+(aq) + NO3-(aq)

So how do I know that AgCl is not water soluble and that KNO3 is water soluble so that the potassium and nitrate ions remain in solution? You probable were able to guess the answer to that question: Memorize a solubility table. I like the following one (you don't have to memorize this table, just be able to use it): -----------------------------------------------------------------------------------------

Soluble Compounds                             Exceptions

Almost all salts of Na+, ____________________________ K+, and NH4+ ----------- All salts of Cl-, Halides of Ag+, Pb2+, Br-, and I- Hg22+ ----------- Compounds containing F- Fluorides of II A except Be++ ----------- Salts of nitrate, chlorate, ____________________________ perchlorate, acetate ----------- Salts of sulfate Sulfates of Sr2+, Ba2+, Pb2+

Insoluble Compounds Exceptions

All salts of carbonate, phosphate, Salts of NH4+ and oxalate, chromate, sulfide and most the metal cations metal hydroxides and oxides

-----------------------------------------------------------------------------------------

Net Ionic Equations

Let's consider again the reaction

AgNO3(aq) + KCl(aq) arrow AgCl(s) + KNO3(aq)

If we write this as we did above which is called the complete ionic equation:

Ag+(aq) + NO3-(aq) + K+(aq) + Cl-(aq) arrow AgCl(s) + K+(aq) + NO3-(aq)

Writing it this way we see that there are ions of the same type on both sides of the equation:

NO3- and K+(s). These ions do not directly participate in the reaction and are called spectator ions. If we remove the spectator ions we are left with what is called the net ionic reaction:

Ag+(aq) + Cl-(aq) arrow AgCl(s)

Of course the spectator ions are important -- without them we never would have had the silver and chlorine ions in solution. However the anion to go with silver initially could have been any ion as long as it forms water soluble products with silver and potassium. Similarly the cation initially with chlorine could have been any cation as long as it forms water soluble products with potassium and silver.

Practice, practice, practice.

ASSIGNMENTS

QUIZ TWO

After you have studied this material and practiced some problems, take quiz two. If you score at least 80 on the test then you are ready to continue to the next section.


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Web Author: Dr. Leon L. Combs
Copyright 2001 by Dr. Leon L. Combs - ALL RIGHTS RESERVED