# Chemical Reactions

### Introduction

In previous lecture notes, we discussed how to obtain chemical formulas (Chemical Nomenclature) and how to write and balance chemical equations (see Counting Atoms in Chemical Formulas and Balancing Chemical Equations).  We learned that the formula H2O tells us that a water molecule contains 2 atoms of hydrogen and 1 atom of oxygen, and that the chemical equation

2H2(g)   +   O2(g)   ---------->   2H2O(g)

tells us that it takes 2 molecles of H2 to react with 1 molecule of O2.

But we normally look at chemistry at the macroscopic level.  We can't see individual atoms, molecules, and formula units, so we certainly can't count them.  Yet the chemical equation above tells us that we must have twice the number of H2 molecules as O2 molecules, if we don't want to have either reactant left over at the end of the reaction.  How can we arrange the proper number of molecules for the reactants if we can't count them?  That's where stoichiometry comes in.

Stoichiometry is the study of mass relationships in chemical formulas and chemical equations.  Macroscopically, we measure out our chemical substances on a mass basis, even though chemicals react on an atomic or molecular basis.  Stoichiometry is the bridge that lets us connect the macroscopic world in which we make our observations, to the submicroscopic world where the action in chemistry really takes place.  Stoichiometry lets us answer questions such as

• What percent of the mass of water is due to the presence of hydrogen atoms in the water?

•
• What mass of water will be formed if we burn 5.000 g of hydrogen, as described by the reaction shown above in these notes?

•
• If a compound of nitrogen and oxygen is 63.65% nitrogen by mass, what is its empirical (lowest terms) formula?
Notice that all of these questions have something to do with mass.  As was noted above, stoichiometry deals with mass relationships.  But we can not work out mass relationships in chemistry unless we know the masses of the atoms involved in the chemical reaction.

### The Masses of Atoms

I'm sure I don't have to tell you that atoms are extremely tiny.  We all know that.  What is worth discussing, however, is how we can determine the mass of something so small.  Atomic masses were originally determined by placing them on a relative scale.  By this I mean that one particular element was chosen as a "reference" element, and the masses of atoms of other elements were assigned a mass on the basis of how they compared to the reference element.  This relative scale was necessary because atoms are much too small to conveniently measure in any of the units we normally use.  Even a mass as small as a microgram, is HUGE when compared to the mass of an atom.  In the lecture notes on Unit Conversions, I pointed out that it is desirable to measure things using units that are comparable in size to what is being measured.  Unfortunately, an atom is so small that the only thing comparable in mass to an atom is another atom.  In the notes on Classification of Matter, we saw that chemical compounds have a fixed elemental compostion by mass.  Chemists took advantage of these fixed combinig ratios to work out a relative atomic mass scale.  The following example shows how it works.

The elements hydrogen and fluorine react with each other to form a compound called hydrogen fluoride.

hydrogen   +   fluorine   ---------->   hydrogen fluoride

Experimentally, it can be shown that any given mass of hydrogen requires 19 times as much mass of flourine to react with it.  If you have 1 gram of hydrogen and you want to convert it all to hydrogen fluoride -- having no hydrogen or fluorine left over -- you will need 19 grams of fluorine.  Any less fluorine and there will be hydrogen left over.  Any more fluorine and there will be fluorine left over.  Of course, if you have 2 grams of hydrogen, then you will need 38 grams of fluorine, because the combining ratio is always 1 to 19.  (A ratio of 2 to 38 is equivalent to 1 to 19 when reduced to lowest terms -- divide by 2).

If we assume that hydrogen and fluorine atoms react in a one to one ratio to make hydrogen fluoride, then the only way the macroscopic combining ratio could be 1 to 19 is if the masses of the individual atoms are also in a 1 to 19 ratio.  Therefore, if we arbitrarily assign the hydrogen atom 1 unit of mass, the fluorine atom must be assigned 19 units of mass.  We see this in Figure 1 (we are temporarily ignoring the fact that these elements are diatomic).  In the top pannel, a single hydrogen atom reacts with a single fluorine atom.  In terms of masses, 1 mass unit of hydrogen reacts with 19 mass units of fluorine to form 20 mass units of hydrogen fluoride.  In the second pannel, 2 atoms of hydrogen react with 2 atoms of fluorine to form 2 molecules of hydrogen fluoride.  In terms of masses, 2 mass units of hydrogen react with 38 mass units of fluorine to form 40 mass units of hydrogen fluoride.  The ratio 2 to 38 to 40 is the algebraically the same as the ratio 1 to 19 to 20.  You just divide all the numbers in the "larger" ratio by 2 and it is reduced to lowest terms. We could continue in this way indefinitely.  That is, we could go on to take 3 atoms of each element, 4 atoms of each element, and so on.  We would find that the mass ratios are always the same -- that is, when reduced to lowest terms, we always recover our original ratio of 1 to 19 to 20.

Now suppose we take such a LARGE number of hydrogen atoms that the collection has a mass of 1 gram.  If we then take the same number of fluorine atoms, the collection must have a mass of 19 grams.  This is because the mass ratio of the collections of atoms is the same as the mass ratio of one atom to another, as long as the number of atoms in each collection is the same.  This is shown in Figure 2, where we imagine that the number of atoms of each element present is large enough to make the collection weigh in grams what just one atom weighs on the atomic mass scale.  The large number that accomplishes this is called a mole.  In the figure, I have only drawn 21 of each atom.  We have to imagine this to be a large number, because it is clearly not possible to draw anywhere near the number of atoms that it actually takes to accomplish this. We shall look at a precise definition of the mole very soon.  As you can see by the reasoning we have done here, the mass relationships among the individual atoms determine the mass relationships that will exist when substances react in bulk. Now that we have relative masses for hydrogen and fluorine, we can find other elements that will react with either of those two and again measure the bulk combining ratios.  We can then determine the relative masses for the atoms of those elements.  Proceeding in this way, we can place more and more elements on the atomic mass scale.

The preceeding discussion has been somewhat oversimplified, but hopefully, it will at least give you an idea of how the masses of the elements can be established.  One problem that was glossed over as the above method was presented is that you must know the formula of the compound formed from the elements in order to get the correct masses.  For example, consider the reaction

hydrogen   +   oxygen   ---------->   water

Experimentally, 1 gram of hydrogen requires 8 grams of oxygen to react with it.  If we assumed the combining ratio was 1 to 1, we might represent the reaction (temporarily ignoring that hydrogen and oxygen are diatomic) like this:

H   +   O   ---------->   HO

Following the same line of reasoning we used with the hydrogen fluoride example, we would conclude that the relative mass of an oxygen atom is 8 units -- that is, that an oxygen atom is 8 times heavier than a hydrogen atom.  Once we realize that the combining ratio is 2 to 1 (and for the moment, again ignoring the diatomic nature of these elements),we might represent the reaction as

2H   +   O   ---------->   H2O

Since we need 8 times as much mass of oxygen as we have hydrogen, the relative mass of an oxygen atom must be 16, because in taking TWO hydrogen atoms, we already have 2 mass units of hydrogen, and 8 times that amount is 16 mass units.  In the early days of chemistry, attempts to determine relative masses of the atoms were plagued by a lack of knowledge of the chemical formulas of the compounds being formed.  The development of a reliable atomic mass scale had to wait until chemists could accurately determine chemical formulas.

Another problem with this method is that for most elements, not all atoms of the element have the same mass.  You saw this back in the lecture notes Chemical Nomenclature, where the concept of isotopes was presented.  Two atoms of the same element can differ in the number of neutrons they have, and therefore have different masses.  Thus, if you really want to be precise, you must not only pick a reference element for the atomic mass scale -- you must also specify which particular nuclide of that element you are going to use.

The modern atomic mass scale is based on the carbon atom that has 6 protons (as all carbon atoms do) and 6 neutrons for a total of 12 nuclear particles.  By definition (not measurement) this atom has a mass of EXACTLY 12 atomic mass units.  We use the letter u to abbreviate atomic mass units, so a single atom of carbon-12 has a mass of 12 u.

Now consider an atom that has a mass that is EXACTLY twice that of a carbon-12 atom.  Its mass would be EXACTLY 24 u.  If we divide the mass of one of these atoms by the mass of a carbon-12 atom, we get a ratio of 2.

24 u    =     2
12 u

Suppose we take two atoms of each element instead of only one.  Then for our hypothetical atom that weighs twice as much as a carbon-12 atom, we have 48 units of mass (total for 2 atoms) and for carbon-12, we have 24 units of mass (total for 2 atoms).  If we again compute the mass ratio, we find it has not changed.

48 u     =     2
24 u

The ratio will remain constant as long as we compare the same number of atoms of each element.  A very important concept that we make use of over and over in chemistry is that RATIOS DO NOT DEPEND ON SAMPLE SIZE.

Now let us take enough carbon-12 atoms to make the collection weigh 12 grams.  We want to do this because we want to retain the numerical value of 12 that has been assigned to carbon, but we are more comfortable measuring in grams than we are in atomic mass units.  So instead of talking about the mass of just 1 carbon-12 atom (12 u), we take enough carbon-12 atoms to make the collection weigh 12 grams.  The number of carbon-12 atoms it takes to accomplish this is called a mole.  That is, the formal definition of a mole is the number of carbon-12 atoms in exactly 12 grams of carbon-12.  A mole, then, is a definite number of something, similar to a dozen.

A mole does not always have the same mass, just as a dozen does not always have the same mass.  If I were to ask you, "How much does a dozen weigh?" you would have to answer my question with a question: "A dozen what?"  Certainly, a dozen peanuts does not weigh the same as a dozen watermellons!  But if I ask you "How many in a dozen?", you can immediately answer 12, without knowing what I have a dozen of.  We can think of the mole as "the chemist's dozen".  Of course, where something as small as atoms and molecules are concerned, 12 is too small a number to deal with.  Clearly, just 12 carbon-12 atoms can't weigh 12 grams.  The "chemists dozen" or mole, is 6.0221367 x 1023 of something.  In principle, we could have a mole of anything.  For example, 6.0221367 x 1023 pennies would be a mole of pennies.  Because of its VERY large size, however, we really can only have a mole of something extremely small, like atoms, ions, molecules, electrons, and so on.

Now lets return to the mass ratios we were talking about earlier.  The formal definition of the mole only gives us the mass of a mole of carbon-12.  So if we have a mole of that hypothetical atom that is twice as heavy as a carbon-12 atom, the mole definition alone won't give us the mass.  The definition only addresses carbon-12.  However, we can use the fact that the ratio of the atomic masses does not change, as long as I compare collections of the two atoms with the same number of atoms in each collection.  So if I compare one mole of each atom

x         =     2
12 g

I have written x for the mass of the mole of the hypothetical atom.  Since our formal mole definition does not address anything but carbon-12, in principle, we don't know what that mass is.  For the mass of the mole of carbon-12, I can write 12 g, because that is given by the definition of a mole.  However, the only way the ratio can remain equal to 2 is if x is 24 g.

Think about what that means.  A single one of my hypothetical atoms weighs 24 u and one mole of them weighs 24 g.  The situation runs parallel to that for carbon-12.  A single carbon-12 atom weighs 12 u and one mole of them weighs 12 g.  I happened to let the hypothetical atom in this example have a mass of 24 u, but there is nothing special about that number.  The same pattern would hold true no matter what I let the mass of the hypothetical atom be.  This pattern is worth making special note of.

One mole of ANYTHING will weigh, in grams, what just one of them weighs on the atomic mass scale.

This is not the formal definition of the mole, but it is a principle that is perhaps even more important than the formal definiton.  It is a principle we use every time we do stoichiometric calculations.

As was pointed out earlier, not all atoms of the same element have the same mass, due to the existence of isotopes.  When you take a mole of a naturally occurring element, the mass in grams will be equal to the average mass (in u) of the atoms that make up that element.  This will normally not be a problem.  For most elements, the isotopic distribution is constant for all samples of the element, and any macroscopic sample of an element contains such an enormous number of atoms that we are virtually guaranteed to get the correct statistical distribution of atoms.  Thus, in chemical reactions, even though all atoms of the same element don't have the same mass, they behave as if they do.

Thus, when we look in the periodic table and see that the atomic weight of carbon is 12.011 (units can be either u/atom or g/mol) we know that this is the average mass of a carbon atom, but in our calculations we treat it as if all carbon atoms had this mass.

Another important principle is that the atomic ratio and the mole ratio are the same.  For example, in a water molecule, H2O, we can say there is a ratio of 2 hydrogen atoms to 1 oxygen atom.  Now suppose we have 12 water molecules.

H2O     H2O     H2O     H2O     H2O     H2O

H2O     H2O     H2O     H2O     H2O     H2O

Here, we have a ratio of 24 H atoms (12 x 2) to 12 O atoms (12 x 1).  Notice that when reduced to lowest terms, this is still a 2 to 1 ratio.  Remember, ratios don't depend on sample size.  We can get our original 2 to 1 ratio back by defining 12 as a dozen.  Then 24 is 2 dozen, and 12 is 1 dozen.

Now suppose we have 6.0221367 x 1023 water molecules.

H2O     H2O     H2O     H2O     H2O     H2O     H2O     H2O     H2O     H2O

H2O     H2O     H2O     H2O     H2O     H2O     H2O     H2O     H2O     H2O

H2O     H2O     H2O     H2O     H2O     H2O     H2O     H2O     H2O     H2O

H2O     H2O     H2O     H2O     H2O     H2O     H2O     H2O     H2O     H2O

(you get the idea -- you didn't really think I was going list ALL of them did you?!)

In this collection, there are  1.20442734 x 1024 (2 x 6.0221367 x 1023) H atoms and 6.0221367 x 1023 (1 x 6.0221367 x 1023) O atoms.  This is still a ratio of 2 to 1.  We can get our original 2 to 1 ratio back by defining 6.0221367 x 1023 to be a mole.  Then 1.20442734 x 1024 will be 2 moles and 6.0221367 x 1023 will be 1 mole.  Thus there are 2 moles of H atoms for every 1 mole of O atoms, just as there are 2 H atoms for every 1 O atom.  The atomic and molar ratios are the same.

This principle not only applies to individual chemical formulas, it also extends to entire chemical equations.  For example, the equation

2H2(g)   +   O2(g)   ---------->   2H2O(g)

was presented at the beginning of these lecture notes.  Back then, we had interpreted it to mean that 2 molecules of H2 was reacting with 1 molecule of O2 to form 2 molecules of H2O.  If we put the "molecular interpretation" on the formulas in this equation, the numbers will get very large when we work with macroscopic samples.  With our present interpretation, the equation

24H2(g)   +  12O2(g)   ---------->   24H2O(g)

represents the reaction of 24 H2 molecules with 12 O2 molecules to make 24 H2O molecules.  Notice this equation is still balanced, just not in lowest terms.  Suppose we now agree to let the chemical formula of a substance represent one dozen units of the substance (units are atoms, molecules, or formula units as the case may be).  With the "dozen interpretation", the equation we just wrote can be represented by

2H2(g)   +   O2(g)   ---------->   2H2O(g)

This equation and the one just before it represent the same amounts of chemicals, because we have shifted our interpretation from one of individual molecules to dozens of molecules.

Now lets go back to the "molecular interpretation" and write the equation

1.20442734 x 1024 H2   +   6.0221367 x 1023O2(g)   ---------->1.20442734 x 1024 H2O

Now suppose we interpret the chemical formula to stand for 1 mole of each substance, rather than 1 molecule.  We can then represent the same quantity of reaction by writing

2H2(g)   +   O2(g)   ---------->   2H2O(g)

This equation and the one just before it represent the same amount of chemical, because we have shifted our interpretation from one of individual molecules to moles of molecules.

The above analysis shows that whatever ratio the substances react in at the atomic or molecular level, they will also react in that ratio on a molar basis.  The significance of this is that we can easily calculate the mass of a mole, so now we can work out mass relationships in chemical formulas and chemical equations.

### Sample Calculations

Problem 1

Calculate the molecular weight of water, H2O.

Solution

The weight of any formula will be the sum of the weights of the atoms it contains.  We treat all atoms of a given element as if they have the same mass -- that given in tables of atomic weights and in periodic tables.  In the calculation below, MWT stands for "molecular weight".

For H2O:

MWT = 2  x  1.00794   +   1  x  15.9994   =   18.01528

The units in the above calculation can be taken to be either u / molecule or g / mol.  It is normally more convenient to think in terms of g / mol.  When we do think in terms of u / molecule, we must remember that not all atoms of the same element have the same mass, due to the existence of isotopes.  Therefore, if the say the molecular weight of water is 18.01528 u / molecule, we mean this is the average mass of a water molecule, not that this is the mass of any particular water molecule.  We don't have this problem when thinking in terms of g / mol.  Every mole of water will weigh 18.01528 g.  This is because a mole is such a large number of molecules that we are virtually guaranteed to get the average masses that are reported in tables.  With a large number of atoms, the mass of the collection is the same as it would be if every atom in the collection had the average mass.  And since one mole of anything will weigh in grams, what just one of them weighs in atomic mass units, the mass of one mole will weigh in grams, what the average mass is in atomic mass units.

Problem 2

Calculate the formula weight of ammonium nitrate, NH4NO3.

Solution

In the calculation below, FWT stands for "formula weight".  It would not be correct to use the term molecular weight here, because ammonium nitrate is an ionic compound and does not exist as molecules.

FWT = 2  x  14.00674  +   4  x  1.00794  +   3  x  15.9994   =   80.04344 g / mol

The numbers 14.00674, 1.00794, and 15.9994 also carry units of g / mol, but they have been omitted to conserve space, and hopefully, make the calculation fit on one line of your screen.

If you understand the above two examples, you should now be able to calculate the weight of any formula you are given.  I shall not present any more example of simple formula weight calculations here, because we have much more to cover, and because calculation of formula weights will be part of other problems we will be working.  If the above two examples are giving you difficulty, you should contact Mr. Robinson for assistance.  Our next examples will involve converting between grams and moles.

Problem 3

How many moles of CO2 are in 54.835 g of CO2?

Solution

We will need the molecular weight of CO2, because this gives us a conversion factor to relate grams and moles.

MWT = 1  x  12.011   +   2  x  15.9994   =   44.0098 g / mol

That is, each mole of CO2 will weigh 44.0098 g.  We can write this in equation form as

1 mol CO2   =   44.0098 g CO2

If you will think back to the notes on Unit Conversions (review them again if you need to) you will remember that equations like the one above can be used to construct conversion factors.  From the above equation we can write

1 mol CO2               =      1      and    44.0098 g CO2          =       1
44.0098 g CO2                                         1 mol CO2

Since we start out knowing the number of grams of CO2 we need the first conversion factor -- the one with g CO2 on the bottom.  By multiplying by this factor, we can cancel out g CO2 and get mol CO2 in its place.  The calculation looks like this:

54.835 g CO2      x    1 mol CO2               =      1.2460 mol CO2
44.0098 g CO2

Problem 4

What is the mass of 2.831 moles of SO3?

Solution

We always need the formula weight when we want to covert between grams and moles for a substance.  For SO3 we have

1  x  32.066   +   3  x  15.9994   =   80.0642 g / mol

The equation relating grams and moles for this substance is

1 mol SO3      =      80.0642 g SO3

The conversion factor we will need for this calculation must have moles in the denominator so they can cancel out and give us grams in the answer.  That is, the factor needed is

80.0642 g SO3           =      1
1 mol SO3

Remember that since conversion factors are equal to 1, we can multiply by them without changing the actual value of the quantity we start with, even though we change the units.  The calculation looks like this:

2.831 mol SO3      x    80.0642 g SO3           =      226.7 g SO3
1 mol SO3

Problem 5

How many moles are in 31.17 g of carbon disulfide, CS2?

Solution

First, we need the molecular weight for the conversion:

MWT = 1  x  12.011   +   2  x  32.066   =   76.143 g / mol

The equation relating grams and moles for this substance is

1 mol CS2      =      76.143 g / mol

and the conversion factor needed is

1 mol CS2             =      1
76.143 g CS2

The calculation is

31.17 g CS2      x    1 mol CS2             =      0.4094 mol CS2
76.143 g CS2

Next, we consider how to calculate the precent by mass composition from a chemical formula.

Problem 6

What is the percent by mass of sulfur in sulfur dioxide, SO2?

Solution

Percentages are ratios (how many parts for each 100 parts) and ratios don't depend on sample size, so we can assume any amount of SO2 that is convenient.  Let's assume we have one mole of SO2.  As the chemical formula shows, in one mole of SO2 molecules, there will be one mole of S atoms and two moles of O atoms.  Remember that the mole ratio is the same as the atomic ratio, and the formula shows that a molecule of SO2 consists of one S atom and two O atoms.  We can easily calculate the mass contribution of each element, and the total mass of the mole.

FWT   =   mass of sulfur    +   mass of oxygen

=    1  x  32.066     +     2  x  15.9994

=       32.066          +         31.9988        =        64.0648 g / mol

Notice that sulfur contributes 32.066 g and oxygen contributes 31.9988 g of mass to the total mass of 64.0648 g for the mole.  The formula for percentage is

percent   =   100%   x   part
whole

The problem is asking for the percent of sulfur, so the "part" we are interested in is the mass of sulfur, and the "whole" is the mass of the mole.  Using the numbers from the calculation above, we have

percent   =   100%   x   32.066 g         =     50.052%
64.0648 g

Problem 7

What percent of the mass of ammonia, NH3 is due to hydrogen?

Solution

Assume one mole of NH3 and calculate the mass of each element in that mole, and the total mass of the mole.

MWT   =   1  x  14.00674   +   3  x  1.00794   =

=        14.00674      +         3.02382     =      17.03056 g / mol

In one mole of ammonia, there is a total mass of 17.03056 g, and 3.02382 g of this is hydrogen.

percent   =   100%   x   part
whole

percent   =   100%   x   3.02382 g         =      17.7553%
17.03056 g

Now that we have seen how to go from a chemical formula to a mass percentage, the next logical question is, can we go the other way?  That is, given the mass percentages of the elements in a compound, can we work out its chemical formula?

We can, but what we will always get is the formula in lowest terms.  Such a formula is called an empirical formula.  It may not be the true molecular formula.

Problem 8

A compound known to contain only the elements carbon and hydrogen was analyzed and found to contain 74.87% carbon by mass.  What is the empirical formula of this compound?

Solution

Since the atomic ratio and the mole ratio are identical, we can solve this problem by calculating the number of moles of each element in any convenient size sample of this compound.  Since percentages are based on 100, the most convenient thing to do is assume 100.00 g of the compound.  Then the 74.87% carbon becomes 74.87 g of carbon.  The remainder of this mass, 25.13 g, is hydrogen.  Now we can calculate the number of moles in each of these masses of element.

74.87 g C      x     1 mol C             =      6.233 mol C
12.011 g C

25.13 g H      x     1 mol H            =      24.93 mol H
1.00794 g H

Notice that we get decimal numbers rather than integers.  Where moles are involved, there is nothing wrong with having decimal numbers.  We can have a fraction of a mole -- they don't have to come in integer amounts.  However, we must be able to interpret our formula both on a molar basis and on an atomic basis.  We can't write the formula as C6.233H24.93 because we can't have fractions of an atom.  We must find the equivalent integer ratio.  We can express a ratio using different numbers (but without actually changing the ratio mathematically) if we multiply or divide both numbers in the ratio by the same number.  We can often get both numbers to be integers by dividing both numbers by the smaller of the two.  This of course, forces one of the numbers to become 1, so this technique works for formulas in which one of the subscripts is 1.  We will see in the next example what to do if this is not the case.  For the present problem, we proceed as follows:

For C      6.233      =      1
6.233

For H      24.93      =      4
6.233

The above calculation shows that the mole ratio is 1 to 4.  Therefore, the empirical formula is CH4.  The subscript of 1 on C is understood and not written.

Problem 9

A compound known to contain only the elements iron and oxygen was analyzed and found to contain 69.94% iron by mass.  What is the empirical formula of this compound?

Solution

Assume 100.00 g of the compound.  Then there will be 69.94 g of iron and 30.06 g of oxygen.  Now convert these masses to moles.

69.94 g Fe      x     1 mol Fe            =      1.252 mol Fe
55.847 g Fe

30.06 g O       x    1 mol O             =      1.879 mol O
15.9994 g O

We can't write the formula as Fe1.252O1.879 because we must be able to interpret the formula on an atomic basis, as well as a molar basis.  We must find the equivalent integer ratio.  If we try what we did in the last problem we get

For Fe      1.252      =      1
1.252

For O       1.879      =      1.501
1.252

The second number, 1.501, is too far from an integer to round it to either 1 or 2.  Our data has 4 significant figures, so the integers 1 and 2 are both outside the window of uncertainty around 1.501.  What we do in this case is try to recognize a fraction in the decimal part of the number and multiply both numbers by the denominator to clear the fraction.  Since the decimal equivalent of 1/2 (one half) is 0.5, we assume that .501 is about 1/2.  Thus the number for O is 1 1/2 or 3/2.  The number 1 for Fe can be written as 2/2.  With the ratio 3/2 to 2/2, we can multiply by 2 and get a ratio of 3 to 2.

2      x      2      =      2      (For Fe)
2

3      x      2      =      3      (For O)
2

We can also show this with the original decimal (non-fractional) numbers:

For Fe      1   x   2   =   2

For O      1.501   x   2   =   3.002

The number 3.002 is acceptably close to the integer 3 that it can be rounded off to 3 within reasonable roundoff error.

Therefore, the empirical formula is Fe2O3.

Sometimes, instead of being given percentage data, we are given a particular mass of the sample and the mass of one of the elements it contains.  The following problem illustrates this.

Problem 10

A 5.000 g sample of a compound known to contain only the elements carbon and oxygen was analyzed and found to contain 1.365 g of carbon.  What is the empirical formula of this compound?

Solution

If you feel you must work with percentages like we've been doing so far, we could always turn the masses given into percentages.  There is 1.365 g of carbon and the remainder of the mass, 3.635 g, must be oxygen.

For C      100%      x      1.365 g      =      27.30%
5.000 g

For O      100%      x      3.635 g      =      72.70%
5.000 g

We could then assume a 100.00 g sample (even though we were given a 5.000 g sample) and proceed as we have before.  The reason we can change sample mass like this is because the information conveyed by a chemical formula is a ratio of one element to another, and ratios do not depend on the sample size.  Of course, once we choose a sample size, we must stick with it throughout the calculation.  You will have problems if you change sample size in mid-problem.  Since we have already used percentages in the previous two problems, lets use the actual masses this time.  As usual, we convert the masses to moles.

1.365 g C      x     1 mol C           =      0.1136 mol C
12.011 g C

3.635 g O      x     1 mol O           =      0.2272 mol O
15.9994 g O

For C      0.1136      =      1
0.1136

For O      0.2272      =      2
0.1136

Therefore, the empirical formula is CO2

In the previous calculations, we obtained empirical formulas.  The ones presented here (CH4, Fe2O3, and CO2) are probably the true formulas, because these are all known compounds.  For some compounds such as methane (CH4), and carbon dioxide (CO2) the the empirical formula and the molecular formula are the same.  For methane, CH4 is an empirical formula because a ratio of 1 to 4 can not be expressed in any lower terms.  It is also a molecular formula, because a methane molecule truely does  have 1 carbon atom and 4 hydrogen atoms.  Likewise, for carbon dioxide, CO2 is an empirical formula because a ratio of 1 to 2 can not be expressed in any lower terms.  It is also a molecular formula because a carbon dioxide molecule really does have 1 carbon atom and 2 oxygen atoms.

But for other substances the empirical formulas and molecular formulas are different.  Where possible, we prefer to have the molecular formula because it conveys more information.  For example, the gas acetylene (molecular formula C2H2) and the liquid benzene (molecular formula C6H6) both have the empirical formula CH.  That is, a ratio of 2 to 2 and a ratio of 6 to 6 both reduce to a ratio of 1 to 1.  If I tell you I am working with a compound that has an empirical formula of CH, you don't know whether I'm talking about acetylene or benzene (or maybe even some other compound that also has this empirical formula).  But if I tell you I am working with C2H2 you know I am talking about acetylene.  I mentioned earlier in these notes that starting from a percentage, you can only calculate the empirical formula, not the true molecular formula (unless the molecular formula happens to be the same as the empirical formula).  To see that a percentage can not tell us uniquely which molecule we have, let's calculate the mass percentages for acetylene (C2H2) and benzene (C6H6).

For C2H2

2  x  12.011   +   2  x  1.00794   =

24.022       +        2.01588     =     26.03788

% C   =   24.022          =      92.26 %
26.03788

% H   =   2.01588        =        7.74 %
26.03788

For C6H6

6  x  12.011   +   6  x  1.00794   =

72.066       +         6.04764    =      78.11364

% C   =   72.066          =      92.26 %
78.11364

% H   =   6.04764        =        7.74 %
78.11364

Notice that C2H2 and C6H6 have the same percentage composition.  Thus, if we are told only that we have a compound that is 92.26% carbon and 7.74% hydrogen we can not determine whether it is C2H2 or C6H6 since both of these formulas are consistent with those percentages.  What these formulas have in common is their empirical formula, which is CH.

For CH

1  x  12.011   +   1  x  1.00794   =

12.011      +        1.00794     =      13.01894

% C   =   12.011           =      92.26 %
13.01894

% H   =   1.00794          =        7.74 %
13.01894

As these calculations show, all formulas that have the same empirical formula will have the same percentage composition.  Thus, given only the percentage composition, we can only calculate the empirical formula.  If we are given the molecular weight of the compound, we can also calculate the true molecular formula.  The following problem illustrates how this is done.

Problem 11

A compound known to contain only the elements carbon and hydrogen was analyzed and found to contain 82.66% carbon by mass.  In a separate experiment, the molecular weight of this compound was determined to be 58.12 g / mol.  What is the molecular formula of this compound?

Solution

Most of the work in solving this problem is the same as we have done before.  We must first calculate the empirical formula before we can calculate the molecular formula.  We just add a couple of steps to the procedure we have already been following to get the molecular formula.

Let's assume 100.00 g of the compound so that we have 82.66 g of carbon and 17.34 g of hydrogen.  Calculate the moles of each element:

82.66 g C      x     1 mol C          =       6.882 mol C
12.011 g C

17.34 g H      x     1 mol H          =       17.20 mol H
1.00794 g H

Now get the equivalent integer ratio:

For C      6.882      =      1
6.882

For H      17.20      =      2.499
6.882

The number 2.499 is too far away from 2 to round it off and get a ratio of 1 to 2.  Note that the decimal part of the number, .499, is close to .5 which is 1/2.  Therefore, try multiplying both numbers in the ratio by 2.

For C      1  x  2   =   2

For H      2.499  x  2   =   4.998

The number 4.998 is acceptably close to 5 that it can be rounded off to the integer 5.  Therefore the empirical formula is C2H5.  Could this be the true molecular formula also?  We can see very quickly that it is not.  Let's calculate the formula weight for this empirical formula.

For C2H5

FWT   =   2  x  12.011   +   5  x  1.00794   =   29.0617 g / mol

This answer is not consistent with the information given in the problem.  We were told that the molecular weight was 58.12, and the formula C2H5 does not have this weight, so it can't be the true molecular formula.  But since C2H5 is the empirical formula, we know that whatever the molecular formula is, it must reduce to this formula.  Possibilities are C4H10, C6H15, C8H20, and so on.  An inefficient way to solve problems like this would be to list a fairly large number of possibilities for what the molecular formula could be, then check the molecular weight of each one until we obtained a molecular weight that matched what was given in the problem.

There is a much better way, however.  If we divide the molecular weight given in the problem by the empirical formula weight, we can see immediately how many times larger the molecular formula is than the empirical formula.  Every subscript in the empirical formula must be multiplied by this factor to get the subscript in the molecular formula.

58.12           =      2
29.0617

In this problem, the true molecular formula is twice as large as the empirical formula.  So the molecular formula is C2x2H2x5 which is C4H10.

Problem 12

A 5.000 g sample of a compound known to contain only the elements carbon and hydrogen was analyzed and found to contain 4.281 g of carbon.  In a separate experiment, the molecular weight of this compound was determined to be 84.16 g / mol.  What is the molecular formula of this compound?

Solution

We must first obtain the empirical formula.  In this compound, there are 4.281 g of carbon and the remaining mass, 0.719 g, is hydrogen.  Calculate the moles of each element:

4.281 g C      x     1 mol C           =      0.3564 mol C
12.011 g C

0.719 g H      x     1 mol H           =      0.713 mol H
1.00794 g H

For C      0.3564      =      1
0.3564

For H      0.713        =      2
0.3564

Therefore, the empirical formula is CH2.  Could this be the true molecular formula also?  Let's calculate the weight and find out.

For CH2

1  x  12.011   +   2  x  1.00794   =

12.011      +       2.01588      =      14.02688 g / mol

This does not match the molecular weight of 84.16 g / mol that was given in the problem.  Therefore, CH2 is not the molecular formula.  How many times larger is the molecular formula than the empirical formula?

84.16              =      6
14.02688

Therefore, for each element, there are 6 times as many atoms in the molecular formula as are indicated in the empirical formula.  The molecular formula is C6x1H6x2 which is C6H12.

Having done quite a bit of work with individual chemical formulas, it is now time to turn our attention to whole chemical equations.  In the next problem, we address a question raised in the beginning of these notes.

Problem 13

The balanced chemical equation for the combustion of hydrogen in oxygen to form water is

2H2(g)   +   O2(g)   ---------->   2H2O(g)

If 5.000 g of hydrogen is burned as described by the above chemical equation, what mass of water will be formed?

Solution

We once interpreted the chemical equation in this problem to refer to the number of molecules.  But now we know we can also interpret it on a molar basis, and this will be much more convenient.  It says 2 moles of H2 will react with 1 mole of O2 to form 2 moles of H2O.  Let's first solve this problem in steps, to be sure we understand how it works, then we'll go back and do it all in one step, which is the way we will usually do it.

Since the equation relates the substances in moles, NOT grams, we must first convert our mass of hydrogen to moles.  It is important to note that we must calculate the molecular weights for the substances as they exist in the chemical equation.  Hydrogen is given as H2 not H, so we must use 2.01588 g / mol, not 1.00794 g / mol, for the moleclar weight.

You may be wondering, then, why I just used the atomic weights in the earlier calculations with chemical formulas.  The reason is, when we work with just the chemical formulas, we count the individual atoms in those formulas, so we have to use the individual atomic weights.  But when we deal with balanced chemical equations, we represent the reactants and products by the formulas that represent them as they truly exist.  The elements hydrogen and oxygen exist as diatomic molecules, so that's the way these elements will be represented in chemical equations.  When these elements react to form compounds, however, it's clear that the paired atoms can become separated.  If that was not the case, there could be no molecules with an odd number of either H or O atoms.  But the molecular formula of water is H2O, having only a single O atom.

5.000 g H2      x    1 mol H2              =      2.480 mol H2
2.01588 g H2

Now that we know how many moles of H2 we have, we can use the balanced chemical equation to determine how many moles of H2O we can make from the H2.  From the chemical equation, we see that 2 moles of H2 will produce 2 moles of H2O.  In other words, the number of moles of H2O formed is equal to the number of moles of H2 we started with.  So from 2.480 moles of H2 we can make 2.480 moles of H2O.  Rather than having to go through this reasoning each time we work with a chemical equation, we will normally just use the coefficients in the balanced equation to construct a conversion factor.  For the present equation, we can write the equivalence

2 mol H2      =      2 mol H2O

The equality is used somewhat loosely here.  Clearly H2 can't be equal to H2O in the sense that equal means identical.  H2 is NOT identical to H2O -- they are two completely different substances.  However, they correspond to each other, because having 2 moles of H2 means you will be able to make 2 moles of H2O.  In the context of this reaction, then, we can use the above "equality" as a conversion factor.

2.480 mol H2      x    2 mol H2O           =      2.480 mol H2O
2 mol H2

Of course, we can cancel out the 2's in the above conversion factor.  I left them in because I am teaching a "recipe" that we can always use in problems like this.  We just use the coefficients in the balanced equation to form the conversion factor relating moles of one substance to moles of another substance.  We can always do cancellations later if they are possible.

We must remember that conversion factors like that above, relating moles of one substance to moles of another substance, are only defined in terms of a balanced chemical equation.  There is no inherent relationship between moles of H2 and moles of H2O.  In this equation, they appear in equal amounts, but in other equations they might not.

In solving problems like this, we encounter inherent relatiohsips and relationships that are only defined within the context of the problem.  For example, when I use the fact that one mole of H2 has a mass of 2.01588 g, this is an inherent relationship.  One mole of H2 always has this mass, regardless of the reaction in which the H2 appears.  On the other hand, the relationship between the moles of H2 and the moles of H2O is not inherent -- it depends on the reaction.  Perhaps a simpler example that would make this point clear is that the relationship between feet and inches is inherent and the relationship between hours and dollars in not.  A foot is always 12 inches, regardless of what is being measured, but the monetary equivalent of a dollar depends on what you get paid per hour.

Returning to our calculation, we have one more step.  We now have the number of moles of H2O we can make, and we only need to convert this to grams to arrive at the final answer:

2.480 mol H2O      x    18.01528 g H2O      =      44.68 g H2O
1 mol H2O

Rather than having to do this step by step, we can simply chain all the conversion factors together.  In this way, it will not be necessary to work out the intermediate answers.  The chained calculation looks like this:

5.000 g H2  x  1 mol H2          x  2 mol H2O     x 18.01528 g H2O   =   44.68 g H2O
2.01588 g H2      2 mol H2                 1 mol H2O

Sometimes the answer to the chained calculation may differ slightly from the one obtained when the problem is worked step by step.  The answer to the chain calculation is the better answer, because it does not suffer from the cumulative effects of rounding off each intermediate answer in the step by step method.

In the problem just worked, we did not worry about the mass of O2.  When the mass of only one reactant is given, and you are asked for the mass of a product that can form, you assume that the reactant not mentioned is present in excess.  Clearly, it will require a certain amount of O2 to react with the 5.000 g of H2 and if we don't have enough O2 available, not all 5.000 g of H2 can be burned.  If we have more O2 available than what we need, then the amount in excess of what we need for the reaction will simply not be used.  The reactant that runs out first is called the limiting reactant, because its depletion stops the reaction and thus limits the amount of product(s) we can make.

Problem 14

What mass of oxygen (O2) was used in the reaction in problem 13?

Solution

Stoichiometry will allows to relate the reactive masses of ANY two substances in a chemical equation.  They might be on opposite sides, like they were in problem 13, or they could be on the same side, like they are in this problem.  Since I demonstrated the calculation both step by step and in chain form in the last problem, I will work this one in chain form only.  This is the way you should become accustomed to solving these types of problems.  To help us obtain the correct conversion factors and chain them together properly, we can write a memory device like that used in the notes on Unit Conversions.  For this problem, we start with the mass of H2 and seek the mass of O2 that would be required to react with it.

g H2   ----->   mol H2   ----->   mol O2   ----->   g O2

The above "roadmap" tells us that we must start with the mass of H2 and convert it to moles.  We must do this because the chemical equation relates the substances in moles, not grams.  The roadmap then tells us to convert from the moles of H2 we have to the moles of O2 required to react with it.  Finally, it tells us that we must convert our moles of O2 to grams of O2.  Recall from the notes on Unit Conversions that each arrow is replaced by a conversion factor, and whatever units appear at the tail of the arrow are in the denominator of the corresponding conversion factor, and whatever units appear at the head of the arrow are in the numerator of the corresponding conversion factor.  Looking at the chain indicated by the succession of 3 arrows above, we write

5.000 g H2   x   1 mol H2        x  1 mol O2      x  31.9988 g O2  =  39.68 g O2
2.01588 g H2     2 mol H2            1 mol O2

The first and last of the coversion factors above are inherent relationships between grams and moles.  The middle factor is determined from the balanced chemical equation.  Notice that the numbers 1 and 2 in the middle factor are the coefficients in the balanced chemical equation.

We can verify that the answer we just obtained is correct on the basis of the law of conservation of mass.  Since H2O is the sole product of the reaction, its mass must be equal to the masses of H2 and O2 that reacted.

5.000 g   +   39.68 g   =   44.68 g

Recall that 44.68 g is the mass of water that we calculated should form in problem 13.

The next problem is more complicated because we don't know in advance which reactant is the limiting reactnat.

Problem 15

Ammonia (NH3) can be formed from the elements nitrogen (N2) and hydrogen (H2).  The balanced chemical equation is

N2(g)   +   3H2(g)   ---------->   2NH3(g)

If a closed reaction vessel contain 5.992 g of N2 and 3.185 g of H2 what is the maximum mass (theoretical yield) of NH3 that can be formed?

Solution

We don't know in advance with reactant will be the limiting reactant.  It is not necessarily the H2, even though we seem to have less of it (3.185 g versus 5.992 g).  We must remember that chemicals react on a molar basis, not a mass basis.  A more appropriate question would be which substance we have fewer moles of, but even that's not necessarily the limiting reactant.  For example, suppose we had 1 mole of N2 and 2 moles of H2 available for the reaction.  From the balanced chemical equation, we see that 1 mole of N2 requires 3 moles of H2 so if we only have 2 moles of H2 we are going to run out of it before all the N2 is used up.  In this example, the H2 will run out first, making it the limiting reactant, even though we had more moles of it (2 versus 1).  Only if there is a one to one mole ratio between the reactants can you assume that the substance present in fewer moles is the limiting reactant.

One way to solve this problem is to calculate the amount of NH3 twice -- calculate how much NH3 can be produced from the N2 and then do a second calculation to see how much NH3 can be produced from the H2.  The reactant that restricts us to making the smaller mass of NH3 is the limiting reactant.

g N2   ----->   mol N2   ----->   mol NH3   ----->   g NH3

5.996 g N2  x  1 mol N2      x  2 mol NH3    17.03 g NH3     =  7.290 g NH3
28.01 g N2        1 mol N2            1 mol NH3

g H2   ----->   mol H2   ----->   mol NH3   ----->   g NH3

3.185 g H2  x  1 mol H2    x  2 mol NH3   x  17.03 g NH3   =  17.94 g NH3
2.02 g H2      3 mol H2            1 mol NH3

In the above calculations, the molecular weights displayed have been rounded to help make them fit on one line of the screen, but the full unrounded values were used to calculate the answers.

Since N2 restricts us to the smaller mass of NH3, it is the limiting reactant, and the maximum mass of NH3 that can be formed in this system is 7.290 g.  Notice that 17.94 g is clearly impossible.  We started with 5.996 g of N2 and 3.185 g of H2 for a total system mass of only 9.181 g.  Certainly, the mass of NH3 formed could not exceed 9.181 g, and it could reach this mass only if all the N2 and H2 were used in the reaction.  But part of the H2 is not used because we run out of N2 before all the H2 is used.  Usually, one reactant will run out before the other.  If both reactants run out at the same time, we say we have a stoichiometric mixture of the reactants -- that is, the amounts present are in the same ratio as the ratio in which they react in the balanced equation.

Problem 16

What mass of H2 is left over (unreacted) after the reaction in problem 15 is complete?

Solution

Stoichiometry can only calculate REACTIVE masses, not the mass of something left over.  We can solve this problem by using stoichiometry to calculate the mass of H2 that reacts.  We then subtract this mass from the mass of H2 initially present to get the mass remaining.

Since N2 is the limiting reactant, we know that all of it (5.996 g) was used in the reaction.  We can calculate the mass of H2 that reacts with N2 as follows:

g N2   ----->   mol N2   ----->   mol H2   ----->   g H2

5.996 g N2   x  1 mol N2    x  3 mol H2   x  2.02 g H2   =  1.294 g H2
28.01 g N2     1 mol N2       1 mol H2

Now we subtract this mass from the mass initially present to get the mass left over:

H2 available     -     H2 used     =     H2 remaining

3.185 g           -       1.294 g    =          1.891 g

In the stoichiometric calculation above, the displayed molecular weights have been rounded off to help make the calculation fit on one line, but the full unrounded values have been used to obtain the answer.

We can check the consistency of our calcuations in problems 15 and 16 by filling out a mass table for the reaction.  Such a table shows the initial masses, change in masses, and final masses for each substance in the reaction, as well as the total mass.  The mass table for the reaction we have been considering is shown below:

 TIME N2 H2 NH3 TOTAL INITIAL 5.996 g 3.185 g 0.000 g 9.181 g CHANGE -5.996 g -1.294 g +7.290 g 0.000 g ENDING 0.000 g 1.891 g 7.290 g 9.181 g

Just looking at the completed table, you can't see the order in which I filled in the entries.  So in case you find it helpful, here is how I did it.

First, I filled in the INITIAL row, because the initial values are given.  We are told we started with 5.996 g of N2 and 3.185 g of H2.  Presumably, no NH3 was initially present, so that our sole source of NH3 would be that formed in the chemical reaction.  Therefore, I entered 0.000 g for initial NH3.  It was a simple matter to add the starting masses for the reactants and determine that the total mass of the chemical system is 9.181 g.  This completes the entries for the entire INITIAL row.

Next, I made use of the law of conservation of mass.  This law tells us that the total mass of a chemical system does not change during a chemical reaction, so I entered 0.000 g for the CHANGE in the TOTAL column.  Since the total mass can not change, the system must have a total mass of 9.181 g at the end of the reaction, just like it did at the beginning of the reaction.  Therefore, I entered 9.181 g for the ENDING mass in the TOTAL column.  This completes the entries for the entire TOTAL column.

Next, I worked on the N2 column.  Since N2 is the limiting reactant, its ENDING mass must be 0.000 g.  The limiting reactant is entirely used up.  This means the CHANGE in mass must be -5.996 g, because the ENDING  mass should be the sum of the INITIAL mass and the CHANGE in mass.

ENDING = INITIAL + CHANGE

This completes the entire N2 column.

Then I worked on the NH3 column.  The theoretical yield of NH3 was calculated in problem 15 to be 7.290 g.  Therefore, I entered 7.290 g as the ENDING mass of NH3.  This means the CHANGE in NH3 must be +7.290 g.  This completes the entire NH3 column.

Only the CHANGE and ENDING entries in the H2 column remain to be established.  Both of these can be established by noting that in all cases the TOTAL mass must be the sum of the masses of N2 , H2 , and NH3.  This is true of the changes in mass as well as the absolute masses.  For all horizontal rows of the table,

TOTAL = N2 + H2 + NH3

I can solve this for H2 to get

H2 = TOTAL - N2 - NH3

substituting the values in the ENDING row of the table,

H2 = 9.181 g - 0.000 g - 7.290 g = 1.891 g

Finally, I can get the CHANGE in H2 in either of two ways:

I can solve for it using the vertically based equation

ENDING = INITIAL + CHANGE

or using the horizontally based equation

TOTAL = N2 + H2 + NH3

Soving the first of these and substituting values, I get

CHANGE = ENDING - INITIAL

=    1.891 g  -    3.185 g

=     -1.294 g

Solving the second of these and substituting values, I get

H2 = TOTAL - N2 - NH3

=    0.000 g - (-5.996 g) - 7.290 g

=     -1.294 g

Notice I get the same answer either way.

It is ok to have negative CHANGES in mass, but of course, the absolute masses (INITIAL and ENDING) must be positive.  A negative change means that the mass is decreasing, and a positive change means the mass is increasing.  Thus we expect negative changes for the reactants, and positive changes for the products, which is what we have found.

A mass table can be a useful device for checking that all your calculated masses are correct, and gives a nice at-a-glance summary of the chemical system.  I like to call these constructions ICE tables, where the letters in the abbreviation come from the names of the table rows -- Initial, Change and Ending.

There is one final calculation that must be discussed before these notes can be considered complete.  In problem 15, we calculated the theoretical yield of NH3.  This is the maximum mass of NH3 that we could possibly obtain from the chemical system.  Usually, however, we obtain less that what we ideally could get.  You can think of the theoretical yield as a perfect exam score.  Theoretically, you can get 100 points on my exams, but for most students, the reality is they will get less than this number of points.  Obtaining the theoretical yield from a reaction is like getting a perfect score on an exam -- it is difficult.  There are several reasons why you probably won't actually recover the theoretial yield of product from a chemical reaction.

• The reactants may not be pure, so the actual masses of reactants are not really as large as what the chemist assumes to be present.

•
• The chemical reaction may be reversible, so that once the product(s) are formed, there can be decomposition to re-form the original reactants.  This results in less than 100% conversion of limiting reactant to product(s).

•
• There may be a loss of product either during the reaction or when the chemist attempts to recover it from the reaction system.
Most instructors (Mr. Robinson included) grade on a 100 point scale.  In this way, what we do is express your grade as a percent yield.  For example, if you get a grade of 91 on one of my exams, it means you earned 91% of all the points that were available.  I guess you could say chemists "grade" chemical reactions when they express the percent yield of the reaction.  For a chemical reaction to have a 100% yield, the actual yield must be equal to the theoretical yield.  That is, we must recover form the reaction, every bit of product it is capable of making.  The formula relating all these yields is

percent yield =   100%    x   actual yield
theoretical yield

Now consider the following problem:

Problem 17

If a chemist recovered 5.192 g of NH3 from the reaction in problem 15, what was the percent yield of the reaction?

Solution

percent yield  =  100%   x   5.192 g     =     71.22%
7.290 g

We obtained only 71.22% of the NH3 that theoretically could have been produced by the reaction.