## Lecture Notes

The topic of stoichiometry (from the Greek, meaning to measure the elements) is central to an understanding of chemical reactions and equilibrium. It is based on two ideas: the chemical equation and the concept of the mole. This module begins by exploring both of these fundamental ideas.

The Chemical Equation

When sodium chloride is mixed with sulfuric acid, there is a chemical reaction and hydrogen chloride and sodium hydrogen sulfate are formed. You probably remember the formulas for each of these compounds, but just in case you have forgotten, sodium chloride = NaCl, sulfuric acid = H2SO4, hydrogen chloride = HCl, and sodium hydrogen sulfate = NaHSO4.

Problem One

Write an expression of the type:
compound 1 + compound 2 produce compound 3 + compound 4
that represents what happens during the reaction.

NaCl + H2SO4 produce HCl + NaHSO4

Problem Two
The above expression is a chemical equation. Take a look at it again and see if you can figure out which of the following is happening in the reaction at the atomic level.

Incorrect
If this were true there would be sodium metal (Na) on the product (right-hand) side of the equation.

Correct
This interpretation requires that you know that NaCl consists of ions and that sulfuric acid easily releases a hydrogen ion. See also answer (c).

Correct
It can also be called a double displacement reaction, all of which have the general form:
AX + BY produce AY + BX

Now let's substitute an arrow for the word produce and also check to see if this is really an equation.

NaCl + H2SO4 → HCl + NaHSO4

Insertion of the arrow was simple enough, but what do we mean by an equation? We know that something like:

2 + 2 = 4

is an equation because what's on the left side numerically equals what's on the right side of the equal sign. Now if we think of the arrow as an equal sign, we must ask ourselves, how can we equate NaCl and H2SO4 with HCl and NaHSO4? Certainly, the compounds on both sides are not the same; if they were there would be little sense in writing the expression.

Perhaps a reminder of Dalton's atomic theory and the law of conservation of matter will provide some help. One of Dalton's hypotheses was that compounds are formed by combinations of atoms, and the conservation law tells us that in any chemical or physical process matter can not be destroyed (actually, this is not true in nuclear reactions, where matter is converted to energy). So how does this apply to our reaction?

Let's suppose that we have exactly five chloride ions (and, of course, five sodium ions) and during the reaction all five of the ions react with sulfuric acid molecules. That is, when the reaction is over there are no chloride ions left.

Problem Three

What happened to the five chloride ions?

They were converted to hydrogen chloride molecules.

If we must have the same amount of matter before the reaction as we do when the reaction is complete, then the conservation law must mean that the five chloride ions were converted to five hydrogen chloride molecules. There are no other compounds on the right side of the equation that contain chlorine atoms and, of course we can not convert chlorine atoms to, say, sulfur atoms (this is the kind of thing that the Alchemists tried to do), so the chlorine ions on the left must end up in HCl molecules on the right.

In general, we can say that both sides of the equation must contain the same numbers of atoms of each element.

Problem Four

Count the atoms of each element in our expression:
NaCl + H2SO4 → HCl + NaHSO4
and determine whether this really is an equation.

left side right side
1 Na   1 Na
1 Cl   1 Cl
2 H   2 H
1 S   1 S
4 O   4 O

Thus, we have the same number of atoms of each element on both sides of the arrow and this therefore is an equation.

In fact, it is frequently called a "balanced" chemical equation, although the use of both words - balanced and equation - is redundant.

Let's write an expression for a preparation of hydrogen chloride from magnesium chloride and see if this expression is also an equation. The expression is very similar to the one above, as you might expect because we are just substituting one ionic chloride for another.

MgCl2 + H2SO4 → HCl + Mg(HSO4)2

Problem Five

What does your count of atoms show for the above equation?

left side right side
1 Mg   1 Mg
2 Cl   2 Cl
2 H   2 H
1 S   1 S
4 O   4 O

This expression is not balanced because the numbers of chlorine, hydrogen, sulfur, and oxygen atoms are not the same on both sides. Notice that in making our count we must be sure to take into account the fact that there are two hydrogen sulfate ions associated with one magnesium ion. That means, for example, that there are a total of 2 x 4 = 8 atoms of oxygen on the right side.

Our challenge now is to balance this expression. Can we simply balance the equation by changing HCl into HCl2 and Mg(HSO4)2 into MgHSO4, thereby producing the equation:

MgCl2 + H2SO4 → HCl2 + MgHSO4

Absolutely not! This would balance the atoms, but it would produce a meaningless expression. The compounds HCl2 and MgHSO4 do not exist. We can not change the formulas of the compounds (assuming that they are correct to begin with) in the expression.

If we think about it for a minute, isn't it true that in this case both chloride ions on the left side will be converted into HCl molecules; that is, they will be snatched up by hydrogen ions provided by the sulfuric acid? And isn't it also true that according to our expression one molecule of sulfuric acid is required to snatch one chloride ion? Yes, of course, both of these statements are true, but how do we "say" that in our expression? We produce and equation by putting a number in front of the H2SO4 and HCl to indicate how many of these molecules are used or, in the case of the HCl, produced. Thus, our equation is:

MgCl2 + 2 H2SO4 → 2 HCl + Mg(HSO4)2

Notice that when you count up the atoms you must multiply by the number (we will frequently call it the coefficient) in front of the compound. For example, in counting the hydrogens on the left side, we must multiply the two hydrogens that are present in each molecule of sulfuric acid by 2.

If you've really been paying attention, you have probably been wondering why both hydrogens on one sulfuric acid molecule do not react with chloride ions. In fact, under certain conditions they do, and in that case, we could write the expression as:

MgCl2 + H2SO4 → HCl + MgSO4

Problem Six

What is the correct equation for this reaction?

MgCl2 + H2SO4 → 2 HCl + MgSO4

We can verbalize the correct equation as "one magnesium ion and two chloride ions react with one sulfuric acid molecule to produce two molecules of hydrogen chloride, one magnesium ion, and one sulfate ion." Of course, to say this we have to know that MgCl2 and MgSO4 are ionic and that, therefore, we can not call MgCl2 a molecule. On the other hand, H2SO4 and HCl are covalent molecules. If you have not been exposed to a discussion of bonding, it may be useful to remember the following generalization: a compound containing a metallic element is likely to be ionic. Hydrogen chloride and sulfuric acid contain only nonmetallic elements and therefore are molecular (but not all covalent compounds are molecular).

Let's try one more equation before we introduce moles. Methane (CH4) reacts with elemental oxygen to form CO2 and water.

Problem Seven

Write the correct equation for this reaction.

CH4 + 2 O2 → CO2 + 2 H2O

In order to produce the correct equation we had to know that the elemental form of oxygen is O2; that is, that the common elemental form of oxygen consists of diatomic molecules. [Remember that H2, N2, O2, F2, Cl2, Br2, and I2 are the diatomic elemental forms of these elements.]

All of these substances are molecular and we would therefore say "one molecule of methane reacts with two molecules of O2 to form one molecule of CO2 and two molecules of water."

Problem Eight

Suppose that we have five molecules of methane. How many molecules of O2 will we need to combine with these five?

10 molecules of O2
If two molecules of O2 combine with one of CH4, then the combining ratio will always be two to one. If we are devotees of dimensional analysis we would write:
5 molecules CH4 x (2 molecules O2/1 molecule CH4) = 10 molecules O2

Problem Nine

If we have 100 molecules of methane combining with 200 molecules of O2, how many molecules of CO2 will be formed if the conversion is complete?

100 molecules

You probably looked at the equation and said, "Well, if one molecule of methane produces one molecule of CO2, then 100 molecules of methane must form 100 of CO2." Or:

100 molecules CH4 x (1 molecule CO2/1molecule CH4 ) = 100 molecules CO2

Notice that we could just as easily have phrased our solution in terms of O2: "If 2 molecules of O2 produce 1 molecule of CO2, then 200 molecules of O2 must produce 100 of CO2." Or:

200 molecules O2 x (2 molecule CO2/1 molecule O2 ) = 100 molecules CO2

There will be some problems that will require that we look carefully at how much of each reactant is present. In this problem, however, we have exactly enough O2 to allow all of the CH4 to be completely consumed without having any of the O2 left over at the end of the reaction.

Problem Ten

If the reaction between methane and O2 produces 150 molecules of CO2, how many molecules of O2 are consumed?

300 molecules
150 molecules of CO2 x (2 molecules of O2/ 1 molecules CO2) = 300 molecules O2

Mole-Mole Relationships

Now for the moles. Chemists usually work with gram amounts of substances. It was partially a consequence of this that the number of atoms in the atomic weight of an element expressed as grams was determined. For example, we know that in 12.01 grams of carbon there are 6.02 x 1023 atoms of carbon. This number--6.02 x 1023--is called Avogadro's number and is also called a mole. The term mole is analogous to terms such as dozen (12), a gross (144), or a ream (500).

To make sure that you understand the mole concept, try the following drill questions.

Problem Eleven

How much does 2.5 moles of fluorine atoms weigh?

47.5 g
One mole of fluorine is the atomic weight of fluorine in grams (sometimes called the gram atomic weight). Thus, one mole of fluorine is 19 g. In that 19 grams there are 6.02 x 1023 atoms of fluorine.

Problem Twelve

How many moles of oxygen atoms are there in 32 g of O2?

2 moles
One mole of O2 molecules has a mass of 32 grams. This is the sum of the atomic weight of two moles of oxygen atoms and is called the molecular weight or molar mass. Because each molecule of O2 contains two atoms, each mole of O2 must contain two moles of oxygen atoms.

Problem Thirteen

How many atoms are there in 6.0 grams of carbon?

3 x 1023 atoms
One mole of carbon has a mass of 12 grams. The number of moles of carbon in 6 grams is obtained by dividing this mass by the atomic weight--
6 g/ 12 g/mole = 0.5 mole
Thus, 6 g of carbon is 0.5 mole of carbon. One mole of carbon is 6.02 x 1023 atoms and 0.5 moles is:
0.5 x 6 x 1023 atoms = 3 x 1023 atoms

We can also apply the mole concept to equations. Consider our old friend:

CH4 + 2 O2 → CO2 + 2 H2O

Rather than talk about 1 molecule of methane or 100 molecules of methane, let's use 6 x 1023 molecules.

Problem Fourteen

What do we call this number of methane molecules?

One mole, of course
6 x 1023 of anything is one mole. My wife has a mole of shoes in our closet --- 6 x 1023 shoes (just kidding, dear!).

We would verbalize our methane equation using moles as, "One mole of methane reacts with two moles of O2 to form one mole of CO2 and two moles of water."

Okay, now we're cookin'!

Problem Fifteen

If we mix two moles of methane with four moles of O2, how many moles of CO2 and water will we get if the conversion is complete?

Two moles of CO2 and four moles of H2O. We could use dimensional analysis as follows:
2 moles CH4 x (1 mole CO2/ 1 mole CH4) = 2 moles CO2
2 moles CH4 x (2 mole H2O/ 1 mole CH4 ) = 4 moles H2O

Problem Sixteen

How many moles of water could we get from the reaction of 2.3 moles of methane?

4.6 moles
2.3 moles CH4 x 2 moles H2O/ 1 mole CH4 = 4.6 moles CH4

Problem Seventeen

How many moles of water could we get from the reaction of 8 g of methane?

1.0 mole
First we have to figure out how many moles of methane we have. The molecular weight of methane is 12 + 4 x 1 = 16. Hence, one mole of methane has a mass of 16 g. We only have 8 g of methane, which is:
8 g CH4/ 16 g/mol = 0.5 mole CH4
Next, we apply the coefficients in the equation:
0.5 mole CH4 x (2 moles H2O/1 mole CH4 ) = 1 mole H2O

All right, let's try something just a tad harder. Ethane (C2H6) also reacts with diatomic oxygen to form carbon dioxide and water.

Problem Eighteen

Write an equation for this reaction and then determine how many moles of ethane are required to react with one mole of oxygen.

The equation for the reaction is:
2 C2H6 + 7 O2 → 4 CO2 + 6 H2O
Thus, 1 mole O2 x (2 mole C2H6/ 7 mole O2) = 2/7 mole C2H6

If you had trouble balancing the equation in the problem above, start it again and just balance the carbon and hydrogen, like:

C2H6 + x O2 → 2 CO2 + 3 H2O

Now, count the number of oxygen atoms on the right. There are seven. But, how do we get seven oxygens on the left? If we use a 7/2 in front of O2, we will have 7 oxygen atoms (only 7/2 O2 molecules, but each molecule contains 2 atoms, so we have a total of 7 oxygen atoms). That makes the equation:

C2H6 + 7/2 O2 → 2 CO2 + 3 H2O

It is perfectly acceptable to have a fractional coefficient in an equation, but usually chemists like to see equations that have whole number coefficients. This can easily be accomplished by multiplying each coefficient by two to give:

2 C2H6 + 7 O2 → 4 CO2 + 6 H2O

This process raises an important point. Like any equation, you can multiply each side by the same number, or divide each side by the same number, or add or subract a number for each side of the equation.

Problem Nineteen

Suppose that we have 0.3 mole of ethane and we want to mix it with exactly enough O2 to allow all of the ethane to be converted to products. How many moles of O2 will we need?

1.05 moles
The equation tells us that 7 moles of O2 are required to react with 2 moles of C2H6. Thus:
0.3 mole C2H6 x (7 mole O2/ 2 mole C2H6 ) = 1.05 mole O2

It is sometimes useful to divide all of the coefficients in an equation by the coefficient in front of the substance that you start with. For example, in the problem above, divide all of the coefficients by two to get:

C2H6 + 7/2 O2 → 2 CO2 + 3 H2O

which makes it easy to see that 7/2 mole of O2 is required to react with one mole of C2H6. If we have not one, but 0.3 mole C2H6, we know that we will need 7/2 times that amount of O2:

0.3 x 7/2 = 1.05 mole O2

Let's try this process again.

Problem Twenty

Determine how many moles of CO2 will be formed from 1.5 moles of O2 (assuming we have enough ethane).

0.86 mole
In this case it is convenient to divide through by the coefficient in front of O2 (remember we start the reaction with O2), giving
2/7 C2H6 + 1 O2 → 4/7 CO2 + 6/7 H2O
We can read this as 2/7 mole will react with one mole of O2 to give 4/7 mole CO2 and 6/7 mole H2O. If we start with 1.5 moles O2 we will get only four seventh of that amount:
1.5 moles O2 x (4 moles CO2/ 7 moles O2) = 0.86 moles CO2
Notice that the units cancel appropriately to give moles CO2 when we use dimensional analysis.

Just to make sure that you've got a good start on this business of moles, the heart of stoichiometry, do at least some of the following drill problems.

Problem Twenty One

In the combustion of ethane, how much O2 is required to produce 0.5 moles of water?

0.58 mole
Notice that dividing by the coefficient in front of water gives the equation
2/6 C2H6 + 7/6 O2 → 4/6 CO2 + 1 H2O
We can see immediately that we will need more moles of O2 than the moles of water that we want, and that we need to multiply 0.5 by 7/6.

Problem Twenty Two

How many moles of N2O5 are produced from the complete reaction of 0.3 moles of AgNO3 in the reaction:
4 AgNO3 + 2 Cl2 → 4 AgCl + 2 N2O5 + O2

0.15 mole

Problem Twenty Three

How many moles of aluminum are required to prepare 0.15 moles of H2?
2 Al + 3 H2SO4 → Al2(SO4)3 + 3 H2

0.10 mole

Mass-Mass Conversions

Up to this point we have dealt only with the number of moles, rather than the mass of reactants or products. If you are comfortable working with moles, it should now be a simple matter to convert to mass.

Problem Twenty Four
To review, give the mass of 0.5 mole of O2.

Incorrect
This is the mass of one mole of O2, which has a molar mass of 32 g/mole.

Correct
The molar mass of O2 is 32 g/mole, and 0.5 mole has a mass of 0.5 mole x 32 g/mole = 16 g.

Incorrect
This is one-half the molar mass of oxygen atoms. O2, the diatomic molecular version of oxygen, has a molar mass of 32 g/mole.

Okay, so we're ready to do some mass-mass conversions. Let's try the Haber process for preparing ammonia from N2 and H2 (this is a very important industrial and biological process). Determine the mass of ammonia that can be obtained from 1.0 gram of N2.

The first thing we must do is write the equation for the reaction.

N2 + 3 H2 → 2 NH3

1.0 g/ 28 g/mole N2 = 0.036 mole N2

The heart of the problem is the mole-mole conversion that we have discussed earlier. The equation tells us that 2 moles of ammonia are obtained from N2, so we need only multiply 0.036 mole N2 by 2.

0.036 moles N2 x (2 moles NH3/1 mole N2) = 0.072 moles NH3

Finally, we convert 0.072 moles of ammonia to mass of ammonia by multiplying 0.072 moles by the molar mass of ammonia (17 g/mol):

0.072 moles ammonia x 17 g/mol = 1.2 grams NH3

Problem Twenty Five
Determine the mass of hydrogen necessary to obtain 1.7 g ammonia in the Haber process.

Correct
1.7 g ammonia is:
(1.7 g/17 g/mol) = 0.10 mol NH3
The equation tells us that three-halves mole of H2 is required to prepare one mole of ammonia:
1/2 N2 + 3/2 H2 → NH3
Thus, we need only multiply 0.10 mole NH3 by 3/2 - and obtain 0.15 mole H2. The mass of H2 is 0.15 mole x 2.0 g/mol = 0.30 g H2.

Incorrect
You have probably used a molar mass of 1.0 g/mole for H2.

Incorrect
You have probably used a mole to mole conversion of 2/3 H2 for each NH3.

Problem Twenty Six

The following is an important industrial reaction used to make methanol (CH3OH):
CO + 2 H2 → CH3OH
Determine the number of grams of H2 that will be used up during the preparation of 3.2 g of CH3OH.

0.40 g
The molar mass of methanol is 32 g/mole. Thus, we have:
3.2 g / 32 g/mole = 0.10 mole methanol
This must be multiplied by the 2 to 1 ratio between moles of H2 and moles of methanol:
0.10 mole CH3OH x (2 mole H2/1 mole CH3OH) = 0.20 mole H2
This 0.20 mole H2 must now be converted to mass by recognizing that one mole of H2 weighs 2.0 g.
0.20 mole H2 x 2.0 g/mole = 0.40 g H2

Problem Twenty Seven

Potassium chlorate (KClO3) decomposes to KCl and O2. Determine the mass of O2 obtained from the decomposition of 12 g of KClO3.

We first write the equation:
2 KClO3 → 2 KCl + 3 O2
The molar mass of KClO3 is 123 g/mole and therefore 12 g is:
12 g/ 123 g/mole = 0.10 mole
Now,
0.10 mole KClO3 x (3 mole O2/ 2 mole KClO3 ) = 0.15 mole O2
Finally, we convert moles of O2 to mass of O2 by multiplying by the molar mass:
0.15 mole O2 x 32 g/mole = 4.8 g O2

Mass/Mole - Volume Relationships for Gases

When a gas is formed or consumed during a reaction it is often convenient to measure the quantity of it in terms of its volume rather than its mass. To accomplish this conversion we can make use of the fact that one mole of a gas occupies 22.4 liters at standard temperature and pressure (0° C and 1 atmosphere pressure). For example, 0.15 mole of H2 occupies:

0.15 mole x 22.4 L/mole = 3.4 L (at 0°C and 1 atmosphere pressure)

If the temperature of the gas is increased, the volume will increase, and, if the pressure of the gas is increased, the volume will decrease. Thus, the actual volume of the gas depends upon the temperature and pressure of the container in which it is contained. For the moment the simplest way to deal with these non-standard conditions is to use the ideal gas equation:

PV = nRT

where P is pressure, V is volume, n is number of moles, R is the ideal gas constant (0.0821 L-atm/K-mol), and T is absolute temperature. Notice that the units of the gas constant tell you what units should be used for all of the variables; that is, pressure must be expressed in atmospheres, volume in liters, and temperature in degrees Kelvin. Suppose that we have been asked for the volume of 0.15 moles of H2 at 50 °C and 300 torr? First, we solve the ideal gas equation for volume:

V = nRT / P

Second, we convert 50° C to Kelvin by adding 273° (50° C = 323 K) and 300 torr to atmospheres. Because 760 torr (or 760 mm Hg) is one atmosphere, 300 torr is:

300 torr / 760 torr/atm = 0.395 atm

Now, we can enter values for all the variables -

V = nRT / P = (0.15 mole x 0.0821 L-atm/K-mol x 323 K / 0.395 atm) = 10 L

Notice that the increased temperature and lower pressure both lead to a larger volume (remember the volume at STP was 3.4 L).

Problem Twenty Eight

Determine the volume of 0.50 moles of N2 at 150 °C and 800 torr?

V = nRT / P = (0.50 mole x 0.0821 L-atm/K-mol x 423 K / 1.05 atm) = 17 L

A common example of a decomposition reaction (and also one of the first ways of preparing oxygen) is:

2 HgO → 2 Hg + O2

Problem Twenty Nine

If 43 g of mercuric oxide decomposes completely and the oxygen is collected in a balloon at 25 °C and 730 torr, what is the volume of oxygen collected?

First, we determine the number of moles of HgO by dividing 43 by its molar mass of 216.6 g/mole:
43 g / 216.6 g/mole = 0.20 mole
Next, we apply the stoichiometry of the equation to get the number of moles of O2:
0.20 mole HgO x (1 mole O2/ 2 mole HgO) = 0.10 mole O2
Finally, we use the ideal gas equation to get the volume:
V = nRT / P = 0.10 mole x 0.0821 L-atm/K-mol x 298 K / 0.96 atm = 2.5 L

Problem Thirty

In some cases, the number of moles of the gas is required. Consider the reaction of zinc with oxygen:
2 Zn + O2 → 2 ZnO
How much ZnO is formed in the complete reaction of zinc with 4.5 L of O2 at 1.3 atm and 200 K?

We first determine the number of moles of O2 by using the ideal gas equation:
to solve for n:
n = PV / RT = (1.3 atm x 4.5 L) / (0.0821 L-atm/K-mol x 200 K)
= 0.36 mole
Then, we can use the 1 to 2 mole ratio to determine that 2 x 0.36 = 0.72 moles of ZnO will be formed. Finally, we convert 0.72 moles of ZnO to mass:
0.72 moles x 81.4 g/ mole = 59 g ZnO

Problem Thirty One

How much sodium oxide will be formed in the complete reaction of sodium metal with 3.5 L O2 contained at 650 torr and 35 °C?
4 Na + O2 → 2 Na2O

4 Na + O2 → 2 Na2O

Problem Thirty Two

What volume of CO2 at STP is produced in the complete decomposition of 0.10 mole of calcium carbonate (limestone) according to the equation:
CaCO3 → CaO + CO2

2.2 L
If you had trouble with this one, you should go back to mole-mole relationships and do some serious reviewing.

Mole Relationships and Solutions

Most reactions are carried out in solution. The reagents mix much more rapidly in solution than in the solid state, and they, therefore, react more rapidly. When reactions are carried out in solution it is important to know the concentration of the solution. Suppose that we prepare a solution by mixing 1.0 g of NaBr with enough water to give us a total of 100 mL of solution. The NaBr is called the solute, the water is the solvent, and the mixture of the two is the solution. The solute is the substance present in the smallest amount. There are two common ways of expressing the amount of solute in the solution: weight percent and molarity.

Let's deal with weight percent first. The word percent means parts per hundred. Weight percent is mass per 100 g, per 100 pounds, per hundred whatever the unit. A 1% NaBr solution contains 1 g of NaBr per 100 g of solution, or 1 pound of NaBr per 100 pounds of solution, etc. If you have a certain weight of solution and know the weight percent you can always determine the weight of solute by multiplying the weight of solution by the decimal equivalent of the percent. Thus, in 12 g of a 1% solution there is:

12 g x 0.01 = 0.12 g NaBr

In our solution of 1.0 g of NaBr in 100 mL of solution we do not know the weight of the solution, we only know its volume (100 mL). Of course, we could weigh the solution, but usually we will know (because someone else has determined it) the density of the solution. In fact, our solution has a density of 1.0 g/mL. If one mL weighs 1.0 g, then 100 mL must weigh:

(1.0 g/ mL) x 100 mL = 100 g

Thus, our solution contains 1.0 g of NaBr in 100 g of solution.

Problem Thirty Three

What is the weight percent of the NaBr in this solution?

(1.0 g/100 g) x 100 = 1.0%

Problem Thirty Four

Sea water contains about 0.0070% by weight NaBr. How many grams of the solution will contain 1 g of NaBr?

(1.0 g/ 0.000070) = 1.4 x 104 g. You might be able to see this more easily by using proportions:
0.0070 g/ 100 g = 1 g/ x g
x = 100 g/ 0.0070 g = 1.4 x 104 g solution
To make sure that our answer is correct, we can multiply 1.4 x 104 g solution by 0.000070 to see if we get 1 gram -
1.4 x 104 g solution x 7 x 10-5 = 1 g NaBr

Problem Thirty Five

How many grams of a 10% Na2CO3 solution must be used to provide 0.50 moles of carbonate ion?

One mole of Na2CO3 contains one mole of carbonate ion. One mole of Na2CO3 has a mass of 106 g. Thus, 0.50 mole is 53 g (0.50 x 106 g/mole = 53 g). The weight of the solution that contains 53 g is:
0.0070 g/ 100 g = 1 g/ x g
10 g/100g = 53 g/ x g
x = (53 x 100)/10 = 530 g
Notice that 10% of 530 g is 53 g. Notice also that we could have solved the problem by dividing 53 by 0.1; that is, 10%.

Problem Thirty Six

How much Na2CO3 must be added to 50 g of water in order to prepare a 20% solution?

x/(50 + x) = 0.20 x = 12.5 g

The other measure of concentration that is very often used is molarity. Molarity is the number of moles of solute per liter of solution. For example, a 1 molar solution of sodium carbonate contains 106 g of Na2CO3 per liter of solution. It would be prepared by dissolving 106 g of Na2CO3 in perhaps 200 mL of water and then adding, with constant stirring to made the solution homogeneous, enough water to produce a total of 1.0 L of solution.

It is frequently necessary to determine the number of moles of the solute in a given volume of solution. This is easily accomplished by multiplying the molarity by the number of liters. For example, the number of moles of sodium carbonate in a 2 liters of a 0.5 M solution is:

2 L x 0.5 moles/L = 1 mole

Problem Thirty Seven

What volume of a 0.1 M HCl solution must be used to obtain 1 millimole (1 x 10-3 mole) of HCl?

0.001 mole / 0.1 mole/L = 0.01 L, or 10 mL

Problem Thirty Eight

How many moles of HCl are there in 50 mL of 0.20 M HCl?

0.050 L x 0.20 mole/L = 0.010 mole HCl

Now we can solve some stoichiometry problems involving solutions. The elementary substance bromine can be prepared from an aqueous solution of NaBr by reaction with Cl2:

2 NaBr(aq) + Cl2 → 2 NaCl(aq) + Br2

Problem Thirty Nine

How many grams of a 10% NaBr solution is required to react with 0.10 mole of Cl2?

0.10 mole Cl2 requires:
0.10 mole Cl2 x (2 mole NaBr/1 mole Cl2) = 0.20 mole NaBr
0.20 mole NaBr has a mass of:
0.20 mole x 103 g/mole = 21 g NaBr
21 g NaBr is contained in:
21 g/ 0.10 = 210 g of 10% NaBr solution

Problem Forty

How many liters of a 0.15 M HCl solution are required to react with 11 g of Na2CO3 according to the equation:
Na2CO3 + 2 HCl → CO2 + 2 NaCl + H2O

0.050 L x 0.20 mole/L = 0.010 mole HCl

Problem Forty One

If 35.2 mL of HCl solution is required to react with 2.12 g Na2CO3, what is the molarity of the HCl solution?

2.12 g/ 106 g/mol Na2CO3 = 0.0200 moles of Na2CO3
0.0200 moles of Na2CO3 x (2 mole HCl / 1 mole Na2CO3 ) = 0.0400 mole HCl
0.0400 mole HCl / 0.0352 L = 1.14 M HCl

Reactions With an Excess of One Reagent (Limiting Reagent)

Up to this point, we have assumed that each of the reactants in a reaction is completely consumed in a reaction. For example, when we ask how many moles of ammonia are formed from 0.1 mole of N2, we assume that at least 0.3 mole of H2 are present because the equation:

N2 + 3 H2 → 2 NH3

tells us that each mole of N2 will react with 3 moles of H2.

Problem Forty Two

What would happen if more than 0.3 mole of H2 were present? Will all of the hydrogen react? Will all of the nitrogen react?

Not all of the hydrogen will react. The 0.1 mole of N2 could be completely consumed, but only 0.3 mole of H2 would be consumed. Thus, some H2 would be left when the reaction is complete.

At the molecular level we can visualize the beginning of the reaction when there are stoichiometric amounts of N2 and H2 (that is, a 1 to 3 ratio as dictated by the equation) as follows: and when the reaction is over, all of the N2 and H2 has been converted to NH3 molecules. If the reaction is started with one molecule of N2 and five molecules of H2, what will the reaction mixture look like when the reaction is complete? Thus, two molecules of H2 are not needed for the reaction and simply remain in the reaction mixture at the completion of the reaction.

Produce a similar diagram for a mixture of two N2 molecules and four H2 molecules.

In this case there is not enough H2 to react with both N2 molecules because three H2 molecules are required to convert each N2 to ammonia. When the reaction is complete there will be two ammonia molecules, one N2 and one H2 molecule. Notice that we can not break up an N2 and allow one nitrogen atom to combine with two hydrogens to give NH2 because NH2 is unstable and is not part of our chemical equation.

In the first example, where we mixed one molecule of N2 with five molecules of H2, the H2 is said to be present in excess and the N2 is called the limiting reagent. In other words, the amount of N2 controls or limits the amount of product obtained.

In the second example, where we mixed two N2 molecules with four H2 molecules, there are not enough H2 molecules to convert both N2 molecules to ammonia and the H2 is therefore the limiting reagent.

Take a look at this last example on the macroscopic level. If we mix two moles of N2 with four moles of H2 and the reaction goes to completion, how much of each reactant will be left? According to the equation:

N2 + 3 H2 → 2 NH3

six moles of H2 would be required to react with two moles of N2, but only four moles of H2 are present. That means that the N2 is in excess and the H2 is the limiting reagent. Because there are four moles of H2:

4 moles H2 x (1 mole N2 / 3 mole H2 ) = 4/3 mole N2

4/3 mole of N2 will be consumed. We started with two moles of N2 and therefore 2 - 4/3 = 2/3 mole of N2 will remain. Of course, all of the H2 will be converted to ammonia.

Let's continue with our discussion of limiting reagent by considering another reaction. Most metals will react with acids to produce H2 and the metal cation. This is an oxidation-reduction reaction in which the metal loses electrons to the hydrogen ion that results from the acid. For the reaction of zinc with HCl, the equation is:

Zn + 2 HCl → ZnCl2 + H2

This reaction is usually carried out in water. [We call solutions in water aqueous solutions.] The hydrogen chloride is commercially available as an aqueous solution. Concentrated HCl as obtained from most chemical supply companies contains 12 moles of HCl per liter of solution. As you know, one of the most common ways to express solution concentrations is molarity - moles of solute per liter of solution.

Just to make sure you understand molarity, answer the following questions:

Problem Forty Three

How many moles of HCl are present in 250 mL of concentrated HCl solution?

3 moles
If one liter contains 12 moles, then 0.250 L must contain 0.250 x 12 = 3 moles.

Problem Forty Four

What volume of concentrated HCl is necessary to make 100 mL of a 0.1 M HCl solution?

0.8 mL
100 mL of a 0.1 M solution contains 0.1 L x 0.1 mole per liter = 0.01 mole. In order to get 0.01 mole of HCl from a 12 M solution, we must use:
0.01 mole/(12 mole/L) = 0.0008 L or 0.8 mL

If your answers to these questions were correct, we can tackle the following problem: How many grams of H2 would be obtained when 6.5 g of zinc are mixed with 100 mL of 1 M HCl?

0.1 g H2

We first determine the number of moles of each of the reactants:

6.5 g Zn / 65 g/mol = 0.10 mol Zn
0.1 L x 1 mole/L = 0.10 mol HCl

Next, we examine the equation:

Zn + 2 HCl → ZnCl2 + H2

to determine if we have enough HCl to allow the 0.10 mole of Zn to be completely consumed. Because the equation tells us that 2 moles of HCl are required for each mole of Zn, we realize that 0.20 moles of HCl would be required for 0.10 mole of Zn. We only have 0.10 mole of HCl and this therefore is the limiting reagent.

Now we look again at the equation to see that 2 moles of HCl produce 1 mole of H2. Therefore, we will have:

0.10 mole HCl x (1 mole H2 / 2 mole HCl) = 0.05 mole H2

To convert this to mass we multiply by the molecular weight of 2 g/mol.

0.05 mol x 2 g/mol = 0.1 g

Problem Forty Five

If 10 mL of 2.0 M HCl is mixed with 1.0 g of BaCO3, how many moles of CO2 will be produced in the following reaction:
BaCO3 + 2 HCl → CO2 + BaCl2 + H2O

5.1 x 10-3 mole CO2
1.0 g / 197 g/mole = 5.1 x 10-3 mole BaCO3
0.010 L x 2.0 mole/L = 0.020 mole HCl
5.1 x 10-3 mole BaCO3 x (2 mole HCl / 1 mole BaCO3) = 0.010 mole HCl
Thus, 0.010 mole HCl is required to react completely with the BaCO3. This is more than provided for by the 0.02 mole of HCl in the 10 mL of 2.0 M HCl. Therefore, the BaCO3 is the limiting reagent.
5.1 x 10-3 mole BaCO3 x (1 mole CO2 / 1 mole BaCO3) = 5.1 x 10-3 mole CO2

Reactions That Do Not Go To Completion

Up to this point we have assumed that all reactions go to completion. In other words that all of the limiting reagent is completely used up in the reaction. In the example above, we assume that all of the HCl has been consumed at the end of the reaction. In reality, many reactions do not go to completion. In fact, one of the major challenges of the synthetic chemist in the pharmaceutical industry, for example, is to figure out how to force each reaction in a series of reactions that ultimately yield some drug to go as far to completion as possible. You may be thinking, "Why don't some reactions go to completion?" The answer to this is fairly complicated, but for now let us just say that each reaction is somewhat like a scale: When the reactants and products are balanced in terms of something called their free energy, then the reaction will go only part way to completion. The reaction comes to an equilibrium state before all of the reactants are converted to products. If the reaction is unbalanced like this: (the free energy of the products is lower than that of the reactants), then the products are favored and the reaction goes almost to completion. Obviously, if the free energy balance looks like this: the reactants will have the lower free energy and the reaction will come to equilibrium when only a small amount of the products have been formed.

What would the balance of free energy look like for a reaction that essentially goes to completion? Suppose that our reaction between NaCl and sulfuric acid has a 30% extent at 25 °C (the balance of free energy depends upon the temperature). This means that only 30% of the product that we would expect if the reaction went to completion is actually produced. If we start with 1.0 mole of NaCl and 1.0 mole of H2SO4 the reaction would produce 1.0 mole of HCl if the reaction goes 100% to completion according to the equation:

NaCl + H2SO4 → HCl + NaHSO4

Problem Forty Six

If the reaction only goes 30% to completion, how many moles of HCl will be present when the reaction is over?

0.3 mole
This is an easy one because we just multiply the 100% completion amount by the extent--
1 mole x 0.30 = 0.3 mole

Synthetic chemists usually call the 100% amount the theoretical yield. The actual amount obtained in a given experiment is called the actual yield. For example, suppose that an organic chemist is making aspirin by mixing 0.30 moles of salicylic acid with 0.30 moles of acetic anhydride. After "working up" the reaction, the chemist manages to isolate 10 grams of the product. This 10 grams of acetyl salicylic acid, carefully secured in a closed vial, is the actual yield. What is the percent yield or extent of the reaction? (The reaction occurs in a one to one mole ratio and the molecular weight of aspirin is 180)

18.5%

The theoretical yield would be 0.30 mole of aspirin or 0.30 mole x 180 g/mole = 54 g. The percent yield is the actual yield (in this case 10 grams) divided by the theoretical yield - 10 g/54 g x 100 = 18.5%

Sometimes the synthetic chemist knows what the percent yield is for a reaction done under a particular set of conditions and needs to know how much of the starting materials (reactants) to use in order to obtain a particular amount of product. In the reaction of elemental boron with elemental chlorine to produce B2Cl4, the percent yield might be 15% under certain conditions.

Problem Forty Seven

If we start with 0.1 mole of boron and 0.2 mole Cl2, how much B2Cl4 would we expect to obtain if the reaction goes to completion?

0.05 mole
We must first write the equation for the reaction:
2 B + 2 Cl2 → B2Cl4
The equation tells us that we need equimolar amounts (a one to one molar ratio) of boron and Cl2. Thus, we have excess Cl2 and the amount of boron controls the amount of product. We would obtain:
0.1 mole B x (1 mole B2Cl4 / 2 mole B) = 0.05 mole B2Cl4
if the reaction goes to completion.

Problem Forty Eight

Now suppose that we want to get this same amount (0.05 mole B2Cl4) in the actual reaction, which, as we know, has a percent yield of 15%. How much boron must we start with, assuming that we will have excess Cl2?

First, notice that we must start with more than 0.1 mole boron. If we start with 0.1 mole B we will get only 0.05 x 0.15 = 0.0075 mole B2Cl4. But we want 0.05 moles, so must divide the amount of boron that gives us 0.05 mole of B2Cl4 by 0.15. Or, to look at it differently, we ask "How much boron when multiplied by 0.15 will produce 0.05 mole B2Cl4?" That is:
X moles B x 0.15 = 0.05 mole B2Cl4
X = 0.05/0.15 = 0.33 mole

Problem Forty Nine

At some other temperature we perform the same reaction starting with 0.2 moles boron and 0.1 mole Cl2 and obtain 4.5 g B2Cl4. What is the percent yield?

55%
Boron is in excess and the 0.1 mole Cl2 would produce 0.05 mole B2Cl4 if the reaction went to completion. The theoretical yield therefore is:
0.05 mole x 163 g/mol = 8.2 g B2Cl4
The actual yield is 4.5 g and the percent yield is 4.5 g/8.2 g x 100 = 55%

Problem Fifty

Copper (II) phosphide (Cu3P2) may be prepared by reaction of phosphine with an aqueous solution of copper sulfate, according to the following equation:
3 CuSO4 + 2 PH3 → Cu3P2 + 3 H2SO4
When 10.0 g of PH3 were added to a solution containing 50.0 g of CuSO4, 20.0 g of Cu3P2 were obtained. Calculate (a) the theoretical yield and (b) the percentage yield.

(a) 26.4 g Cu3P2
(b) 75.8%

Problem Fifty One

Copper reacts with elemental iodine (I2) to form CuI. When 6.3 g of copper are heated with 25 g I2, 12 g of CuI were obtained. Calculate the theoretical and percentage yield.

19 g CuI, 63%

Gravimetric and Volumetric Analyses

Many quantitative analyses of materials are obtained through chemical reactions. Although there are many more sophisticated methods for analyzing substances and mixtures today, particularly spectroscopic methods, many laboratories still employ these classical techniques because they are inexpensive, reliable, and fairly rapid.

Gravimetric analyses always involve weighing the material to be analyzed, carrying out a reaction that converts some part of the material to something else, followed by weighing the new material. Clearly, the technique gets its name from the Latin root for gravity (gravis = heavy), upon which the technique of weighing depends.

Let's look at a simple example of a gravimetric analysis. Suppose that we have just prepared a sample of aluminum in the laboratory by reacting aluminum metal with potassium hydroxide, followed by the addition of sulfuric acid. The product is a double salt - a one to one "chemical mixture" of potassium sulfate and aluminum sulfate. When it forms in aqueous solution, it also incorporates a certain number of water molecules within the crystal lattice. Because the compound must be electrically neutral we know that the empirical formula must contain one potassium ion, one aluminum ion, and two sulfate ions. Thus, a part of the formula is KAl(SO4)2. The water of crystallization or hydration is also part of the formula and is usually written as a suffix with a dot between the ions and the waters. For example cupric sulfate pentahydrate is written as CuSO4•5H2O. This could also be written as CuSO4(H2O)5, but tradition decrees that we use the nH2O designation, and, in fact, this tradition is a valuable one because these waters of hydration are fairly easily removed by gentle heating. So, in fact, there is a chemical and physical difference between the ions and the water molecules.

Back to our project. We know that our alum has a formula KAl(SO4)2•nH4O, but we don't know how many moles of water are present per mole of compound. We now weigh out 0.5487 g of the compound into a porcelain crucible and then heat it gently until all of the water has been driven from the compound. The weight of the residue left in the crucible is 0.2806 g.

Problem Fifty Two

What compound remains in the crucible after the heating and what is the formula of the compound?

KAl(SO4)2 remains in the crucible
The formula is KAl(SO4)2•nH2O.
The amount of water is obtained by subtracting the weight of the residue from the weight of the original sample: 0.5487 - 0.2806 = 0.2681 g. This weight represents the weight of the water lost. In order to obtain the the value of n, we must convert the mass of water and the mass of KAl(SO4)2 to moles and then simply find the ratio of the moles of water to moles of KAl(SO4)2. The value of n is 12 and the formula is:
KAl(SO4)2•12H2O

If we were interested in determining the percentage of sulfate in our sample, we could add a reagent to an aqueous solution of the alum that would quantitatively precipitate out the sulfate ions. A suitable reagent is barium nitrate (or any souble barium salt). Barium sulfate is quite insoluble and therefore the amount of sulfate ions left in solution after the addition of excess barium ions will be tolerably small (meaning that it will not affect the accuracy of our analysis to within the number of significant figures obtainable on most analytical balances.

Imagine, then, that we dissolve a 1.2053 g sample of our alum in water, and then add a solution of barium nitrate until we are convinced that no more barium sulfate precipitates (not an easy matter because the solution becomes very cloudy and one must wait for the precipitate to settle before adding more barium solution). We can then filter our preciptate of barium sulfate. Barium sulfate forms very small crystals and, consequently, it is important to give the precipitate time for "digestion" while the crystals grow larger. Moreover, very fine filter paper or a glass-frit crucible of very small pore size must be used. After drying the barium sulfate we find that it weighs 1.2724 g.

Problem Fifty Three

Calculate the percentage of sulfate in the alum sample.

43.44 %
The number of moles of BaSO4 is:
1.2724 g / (233.3 g/mol) = 0.005454 moles.
Because all of the sulfate ions were converted to BaSO4, this must also be the number of moles of sulfate ions in the sample. Notice that we can make this assertion without knowing the formula of the alum sample. We need only know that all of the sample winds up as BaSO4. Finally, the percentage of sulfate is obtained by converting moles of sulfate to mass of sufate:
0.005454 moles x 96.00 g/mole SO42- = 0.5236 g SO42-
The percentage is then obtained by dividing this mass by the mass of the original sample:
(0.5236 g /1.2053 g) x 100 = 43.44 %

Problem Fifty Four

Hardware brass is dissolved in nitric acid, and then H2SO4 is added to the solution to precipitate PbSO4. A 1.6841-g sample of brass yielded 0.4671 g of PbSO4. Calculate the percentage of lead in hardware brass.

18.95 % Pb

Problem Fifty Five

A sample of aluminum ore weighing 0.6036 g was dissolved and the aluminum precipitated as Al(OH)3. Heating converted Al(OH)3 to Al2O3 and gave 0.2105 g of Al2O3. What was the percentage of Al in the ore?

18.46% Al

In volumetric analyses, the sample to be analyzed is dissolved and then a reagent is added to the sample from a buret. A typical setup is shown below: In a gravimetric analysis, the reagent is generally added in excess to insure complete precipitation of the ion being analyzed. The excess reagent is then simply filtered off. In a volumetric analysis, the addition of the reagent from the buret must be stopped at exactly the point where the reaction is complete.

Let's consider a common acid-base titration, the titration of acetic acid with sodium hydroxide. We'll suppose that the acetic acid, perhaps in a vinegar sample, is being analyzed, and the sodium hydroxide is in the buret. The reaction between the two:

CH3COOH + NaOH → CH3CO2Na + H2O

occurs in a one to one stoichiometric ratio; that is, one mole of acetic acid reacts with one mole of sodium hydroxide.

Problem Fifty Six

If 0.0050 mole of acetic acid is present in the flask, how much of the titrant (0.10 M NaOH) will be required to convert all of the acetic acid to sodium acetate?

50 mL
0.0050 mole / 0.10 moles/L = 0.050 L

You are probably wondering how one determines when exactly enough sodium hydroxide has been added to convert all of the acetic acid to sodium acetate. Surely, no visible change will occur in this reaction because all of the reactants and products are colorless and soluble.

Problem Fifty Seven

Even though we have not yet discussed acid-base chemistry, can you think of some property of the solution that does change as acetic acid is converted to sodium acetate?

The acidity, which is usually measured by the pH of the solution.

We'll find later that the pH = -log [H3O+] and, therefore, is a measure of the concentration of hydronium ions in solution. In a solution of acetic acid about 1 percent of the acetic acid molecules react with water to form hydronium ions:

CH3COOH + H2O → CH3CO2- + H3O+

Thus, at the beginning of the titration the solution is acidic and has a pH of about 3 (acidic solutions have a pH below 7.0). As the acetic acid is converted to sodium acetate, there are fewer and fewer acetic acid molecules left to react with H2O and, therefore, the hydronium ion concentration decreases. In fact, when all of the acetic acid has been converted to sodium acetate, the sodium acetate reacts with the water to form hydroxide ions and the solution becomes basic, and the pH goes above seven (at a pH of 7 the hydronium ion concentration equals the hydroxide concentration and the solution is neutral).

Thus, the acidity (pH) changes drastically when the acetic acid has been totally converted and we can find out when this point has been reached by adding a substance that changes color when the pH changes rapidly. There are many such substance, called indicators, that change color at a variety of different pH values. We'll discuss how one chooses an indicator in our module on acid-base chemistry, but for now we assume that the correct indicator has been chosen and we simply slowly add titrant until the color of the solution changes.

Problem Fifty Eight

A 10.00 mL sample of vinegar is titrated with 0.1045 M NaOH. At the endpoint (the point at which the indicator changes color), 39.71 mL of the titrant have been added. What is the concentration of the acetic acid in the vinegar sample?

0.4150 M
The 39.71 mL of 0.1045 M NaOH provides:
0.03971 L x 0.1045 mole/L = 0.004150 moles of NaOH
The one to one stoichiometry requires that the same number of moles of acetic acid are present in the sample. Thus, the concentration is:
0.004150 mole / 0.01000 L = 0.4150 M

The equivalence point in a titration is the point at which exactly the correct amount of the titrant has been added. This is called the stoichiometric amount of titrant.

Problem Fifty Nine

How many moles of HCl are required to reach the equivalence point in the titration of 0.20 moles of Na2CO3 with HCl?
Na2CO3 + 2 HCl → H2CO3 + 2 NaCl

0.40 moles HCl

Is the equivalence point the same as the endpoint? If the indicator is correctly chosen the endpoint - that is, the point at which the change in color "tells" you that the reaction is complete - will be very similar to the equivalence point. If the indicator changes color too early or too late, the equivalence point will not be reached and an error will be made in the titration.

Problem Sixty

A sample of pure NaCl weighing 0.2500 g was dissolved in water, and the resulting solution was titrated with a AgNO3 solution. If 32.85 mL of the AgNO3 solution were required, what was the molarity of that solution?
NaCl + AgNO3 AgCl + NaNO3

Problem Sixty One

A sample of impure Ca(OH)2 (slaked lime) weighing 0.2194 g required 42.42 mL of 0.1200 M HCl solution when titrated, using the reaction:
Ca(OH)2 + 2 HCl CaCl2 + 2 H2O
What was the percentage of Ca(OH)2 in the sample?