Stoichiometry
Stoichiometry is simply the math behind chemistry. Given enough information, one can use stoichiometry to calculate masses, moles, and percents within a chemical equation.
In chemistry, we use symbols to represent the various chemicals. Success in chemistry depends upon developing a strong familiarity with these basic symbols. For example, the symbol "C"represents an atom of carbon, and "H" represents an atom of hydrogen. To represent a molecule of table salt, sodium chloride, we would use the notation "NaCl", where "Na" represents sodium and "Cl" represents chlorine. We call chlorine "chloride" in this case because of its connection to sodium.
You should have reviewed naming schemes, or nomenclature, in earlier readings.
A chemical equation is an expression of a chemical process. For example:
AgNO_{3}(aq) + NaCl(aq) > AgCl(s) + NaNO_{3}(aq)
In this equation, AgNO_{3} is mixed with NaCl. The equation shows that the reactants (AgNO_{3} and NaCl) react through some process (>) to form the products (AgCl and NaNO_{3}). Since they undergo a chemical process, they are changed fundamentally.
Often chemical equations are written showing the state that each substance is in. The (s) sign means that the compound is a solid. The (l) sign means the substance is a liquid. The (aq) sign stands for aqueous in water and means the compound is dissolved in water. Finally, the (g) sign means that the compound is a gas.
Coefficients are used in all chemical equations to show the relative amounts of each substance present. This amount can represent either the relative number of molecules, or the relative number of moles (described below). If no coefficient is shown, a one (1) is assumed.
On some occasions, a variety of information will be written above or below the arrows. This information, such as a value for temperature, shows what conditions need to be present for a reaction to occur. For example, in the graphic below, the notation above and below the arrows shows that we need a chemical Fe_{2}O_{3}, a temperature of 1000° C, and a pressure of 500 atmospheres for this reaction to occur.
The graphic below works to capture most of the concepts described above:
Given the equation above, we can tell the number of moles of reactants and products. A mole simply represents Avogadro's number (6.022 x 10^{23}) of molecules. A mole is similar to a term like a dozen. If you have a dozen carrots, you have twelve of them. Similarly, if you have a mole of carrots, you have 6.022 x 10^{23} carrots. In the equation above there are no numbers in front of the terms, so each coefficient is assumed to be one (1). Thus, you have the same number of moles of AgNO_{3}, NaCl, AgCl, NaNO_{3}.
Converting between moles and grams of a substance is often important.
This conversion can be easily done when the atomic and/or molecular
mass of the substance(s) are known. Given the atomic or molecular
mass of a substance, that mass in grams makes a mole of the substance.
For example, calcium has an atomic mass of 40 atomic mass units. So, 40
grams of calcium makes one mole, 80 grams makes two moles, etc.
Sometimes, however, we have to do some work before using the coefficients of the terms to represent the relative number of molecules of each compound. This is the case when the equations are not properly balanced. We will consider the following equation:
Al + Fe_{3}O_{4}> Al_{2}O_{3}
+ Fe
Since no coefficients are in front of any of the terms, it is easy to assume that one (1) mole of Al and one (1) mole of Fe_{3}O_{4} react to form one (1) mole of Al_{2}O_{3}. If this were the case, the reaction would be quite spectacular: an aluminum atom would appear out of nowhere, and two (2) iron atoms and one (1) oxygen atom would magically disappear. We know from the Law of Conservation of Mass (which states that matter can neither be created nor destroyed) that this simply cannot occur. We have to make sure that the number of atoms of each particular element in the reactants equals the number of atoms of that same element in the products. To
do this we have to figure out the relative number of molecules of each term expressed by the term's coefficient.
Balancing a simple chemical equation is essentially done by trial and error. There are many different ways and systems of doing this, but for all methods, it is important to know how to count the number of atoms in an equation. For example we will look at the following term.
2Fe_{3}O_{4}
This term expresses two (2) molecules of Fe_{3}O_{4}. In each molecule of this substance there are three (3) Fe atoms. Therefore in two (2) molecules of the substance there must be six (6) Fe atoms. Similarly there are four (4) oxygen atoms in one (1) molecule of the substance so there must be eight (8) oxygen atoms in two (2) molecules.
Now let's try balancing the equation mentioned earlier:
Al + Fe_{3}O_{4}> Al_{2}O_{3}+ Fe
Developing a strategy can be difficult, but here is one way of approaching a problem like this.
 Count the number of each atom on the reactant and on the product side.
 Determine a term to balance first. When looking at this problem, it appears that the oxygen will be the most difficult to balance so we'll try to balance the oxygen first. The simplest way to balance the oxygen terms is:
Al + 3 Fe_{3}O_{4}> 4 Al_{2}O_{3}+ Fe
Be sure to notice that the subscript times the coefficient will give the number of atoms of that element. On the reactant side, we have a coefficient of three (3) multiplied by a subscript of four (4), giving 12 oxygen atoms. On the product side, we have a coefficient of four (4) multiplied by a subscript of three (3), giving 12 oxygen atoms. Now, the oxygens are balanced.
 Choose another term to balance. We'll choose iron, Fe. Since there are nine (9) iron atoms in the term in which the oxygen is balanced we add a nine (9) coefficient in front of the Fe. We now have:
Al +3 Fe_{3}O_{4}> 4Al_{2}O_{3}+ 9Fe
 Balance the last term. In this case, since we had eight (8) aluminum atoms on the product side we need to have eight (8) on the reactant side so we add an eight (8) in front of the Al term on the reactant side.
Now, we're done, and the balanced equation is:
8Al + 3Fe_{3}O_{4} > 4Al_{2}O_{3} + 9 Fe
Sometimes when reactions occur between two or more substances, one
reactant runs out before the other. That is called the "limiting
reagent". Often,
it is necessary to identify the limiting reagent in a problem.
Example: A chemist only has 6.0 grams of C_{2}H_{2} and an unlimited supply of oxygen and he desires to produce as much CO_{2} as possible. If she uses the equation below, how much oxygen should she add to the reaction?
2C_{2}H_{2}(g) + 5O_{2}(g) > 4CO_{2}(g) + 2 H_{2}O(l)
To solve this problem, it is necessary to determine how much oxygen should
be added if all of the reactants were used up (this is the way to produce the maximum amount of CO_{2}).
First, we calculate the number of moles of C_{2}H_{2} in 6.0 g of C_{2}H_{2}. To be able to calculate the moles we need to look at a periodic table and see that 1 mole of C weighs 12.0 g and H weighs 1.0 g. Therefore we know that 1 mole of C_{2}H_{2} weighs 26 g (2 × 12 grams + 2 × 1 gram).
6.0 g C_{2}H_{2} x 
1 mol C_{2}H_{2}
(24.0 + 2.0)g C_{2}H_{2} 
= 0.25 mol C_{2}H_{2} 
Then, because there are five (5) molecules of oxygen to every two (2) molecules of C_{2}H_{2}, we need to multiply the result by 5/2 to get the total molecules of oxygen. Then we convert to grams to find the amount of oxygen that needs to be added:
0.25 mol C_{2}H_{2} x 
5 mol O_{2}
2 mol C_{2}H_{2} 
x 
32.0 g O_{2}
1 mol O_{2} 
= 20 g O_{2} 
It is possible to calculate the mole ratios (also called mole fractions) between terms in a chemical equation when given the percent by mass of products or reactants. percentage by mass = mass of part/ mass of whole
There are two types of percent composition problems problems in which you are given the formula (or the weight of each part) and asked to calculate the percentage of each element
and problems in which you are given the percentages and asked to calculate the formula.
In percent composition problems, there are many possible solutions. It is always possible to double the answer. For example, CH and C_{2}H_{2} have the same proportions, but they are different compounds. It is standard to give compounds in their simplest form, where the ratio between the elements is as reduced as it can be called the empirical formula. When calculating the empirical formula from percent composition, one can convert the percentages to grams. For example, it is usually the easiest to assume you have 100 g so 54.3% would become 54.3 g. Then we can convert the masses to moles; this gives us mole ratios. It is necessary to reduce to whole numbers. A good technique is to divide all the terms by the smallest number of moles. Then the ratio of the moles can be transferred to write the empirical formula.
Example: If a compound is 47.3% C (carbon), 10.6% H (hydrogen) and 42.0% S (sulfur), what is its empirical formula?
To do this problem we need to transfer all of our percents to masses. We assume that we have 100 g of this substance. Then we convert to moles:
Carbon: 
47.3 grams
1 
x 
1 mole
12.01 grams 
= 3.94 moles 
Hyrdrogen: 
10.6 grams
1 
x 
1 mole
1.008 grams 
= 10.52 moles 
Sulfur: 
42.0 grams
1 
x 
1 mole
32.07 grams 
= 1.310 moles 
Now we try to get an even ratio between the elements so we divide by the number of moles of sulfur, because it is the smallest number:
Carbon: 
3.94
1.310 
= 3 
Hydrogen: 
10.52
1.310 
= 8 
Sulfur: 
1.310
1.310 
= 1 
So we have: C_{3}H_{8 }S
Example: Figure out the percentage by mass of hydrogen sulfate, H_{2}SO_{4}.
In this problem we need to first calculate the total mass of the compound by looking at the periodic table. This gives us:
2(1.008) + 32.07 + 4(16.00) g/mol = 98.09 g/mol
Now, we need to take the weight fraction of each element over the total mass (which we just found) and multiply by 100 to get a percentage.
hydrogen: 
2(1.008)
98.09 
= 
2.016
98.09 
= 0.0206 ∗ 100 = 2.06% 
sulfur: 
32.07
98.09 
= 0.327 ∗ 100 = 32.7% 
oxygen: 
4(16.00)
98.09 
= 
64.00
98.09 
= 0.652 ∗ 100 = 65.2% 
Now, we can check that the percentages add up to 100%
65.2 + 2.06 + 32.7 = 99.96
This is essentially 100 so we know that everything has worked, and we probably have not made any careless errors. So the answer is that H_{2}SO_{4} is made up of 2.06% H, 32.7% S, and 65.2% O by mass.
While the empirical formula is the simplest form of a compound, the
molecular formula is the form of the term as it would appear in a chemical
equation. The empirical formula and the molecular formula can be the
same, or the molecular formula can be any positive integer multiple of the empirical
formula.
Examples of empirical formulas: AgBr, Na_{2}S, C_{6}H_{10}O_{5}. Examples of molecular formulas: P_{2}, C_{2}O_{4}, C_{6}H_{14}S_{2}, H_{2}, C_{3}H_{9}.
One can calculate the empirical formula from the masses or percentage composition of any compound. We have already discussed percent composition in the section above. If we only have mass, all we are doing is essentially eliminating the step of converting
from percentage to mass.
Example: Calculate the empirical formula for a compound that has 43.7 g P (phosphorus) and 56.3 grams of oxygen.
First we convert to moles:
43.7 grams P
1 
x 
1 mol
30.97 grams 
= 1.41 moles 
56.3 grams O
1 
x 
1 mol
16.00 grams 
= 3.52 moles 
Next we divide the moles to try to get an even ratio.
Phosphorus: 
1.41
1.41 
= 1.00 
Oxygen:
 3.52
1.41 
= 2.50 
When we divide, we did not get whole numbers so we must multiply by two (2). The answer = P_{2}O_{5}
Calculating the molecular formula once we have the empirical formula is easy.
If we know the empirical formula of a compound, all we need to do is divide
the molecular mass of the compound by the mass of the empirical formula.
It is also possible to do this with one of the elements in the formula;
simply divide the mass of that element in one mole of compound by the mass
of that element in the empirical formula. The result should always be a
natural number.
Example: if we know that the empirical formula of a compound is HCN and we are told that a 2.016 grams of hydrogen are necessary to make the compound, what is the molecular formula? In the empirical formula hydrogen weighs 1.008 grams. Dividing 2.016 by
1.008 we see that the amount of hydrogen needed is twice as much. Therefore the empirical formula needs to be increased by a factor of two (2). The answer is: H_{2}C_{2}N_{2}.
Density
refers to the mass per unit volume of a substance. It is a very common
term in chemistry.
The concentration of a solution is the "strength" of a solution. A solution typically refers to the dissolving of some solid substance in a liquid, such as dissolving salt in water.
It is also often necessary to figure out how much water to add to a solution to change it to a specific concentration.
The concentration of a solution is typically given in molarity.
Molarity is defined as the number of moles of solute (what is actually dissolved in the solution) divided by the volume in liters of solution (the total volume of what is dissolved and what it has been dissolved in).
Molarity = 
moles of solute
liters of solution 
Molarity is probably the most commonly used term because measuring a volume of liquid is a fairly easy thing to do.
Example: If 5.00 g of NaOH are dissolved in 5000 mL of water, what is the molarity of the solution?
One of our first steps is to convert the amount of NaOH given in grams into moles:
5.00g NaOH
1 
x 
1 mole
(22.9 + 16.00 + 1.008)g 
= 0.125 moles 
Now we simply use the definition of molarity: moles/liters to get the answer
Molarity = 
0.125 moles
5.00 L of soln 
= 0.025 mol/L 
So the molarity (M) of the solution is 0.025 mol/L.
Molality is another common measurement of concentration. Molality is defined as moles of solute divided by kilograms of solvent
(the substance in which it is dissolved, like water).
Molality = 
moles of solute
kg of solvent 
Molality is sometimes used in place of molarity at extreme temperatures because the volume can contract or expand.
Example: If the molality of a solution of C_{2}H_{5}OH dissolved in water is 1.5 and the mass of the water is 11.7 kg, figure out how much C_{2}H_{5}OH must have been added in grams to the solution.
Our first step is to substitute what we know into the equation. Then we try to solve for what we don't know: moles of solute. Once we know the moles of solute we can look at the periodic table and figure out the conversion from moles to grams.
Molality =
 moles solute
kg solvent 

Now we simply use the definition of molarity: moles/liters to get the answer
Molality = 
moles solute
kg solvent 

1.5 
mols
kg 
= 
moles solute
11.7 kg 

1.5 
moles
kg 
x 11.7 kg = 17.55 moles 

17.55 moles
1 
x 
(2 ∗ 12.01) + (6 ∗ 1.008) + 16
1 moles 
= 808.5 g C_{2}H_{5}OH 


It is possible to convert between molarity and molality. The only information needed is density.
Example: If the molarity of a solution is 0.30 M, calculate the molality
of the solution knowing that the density is 3.25 g/mL.
To do this problem we can assume one (1) liter of solution to make the numbers easier. We need to get from the molarity units of mol/L to the molality units of mol/kg. We work the problem as follows, remembering that there are 1000 mL in a Liter and 1000 grams in a kg. This conversion will only be accurate at small molarities and molalities.
0.3 mol
1 L 
x 
1 mL
3.25 g 
x 
1 L
1000 mL 
x 
1000 g
1 kg 
= 0.09 mols / kg 

It is also possible to calculate colligative properties, such as boiling point depression, using molality. The equation for temperature depression or expansion is
ΔT= K_{f} × m
Where:
ΔT is temperature depression (for freezing point) or temperature expansion (for boiling point) (°C)
K_{f} is the freezing point constant (kg °C/mol)
m is molality in mol/kg
Example: If the freezing point of the salt water put on roads is 5.2° C, what is the molality of the solution? (The K_{f} for water is 1.86 °C/m.)
This is a simple problem where we just plug in numbers into the equation. One piece of information we do have to know is that water usually freezes at 0° C.
ΔT = K_{f} * m
ΔT/K_{f} = m
m = 5.2/1.86
m = 2.8 mols/kg
Practice Problems
1. If only 0.25 molar NaOH and water are available, how much NaOH needs to be added to make 10 liters of 0.2 M solution of NaOH?
Check your work
2. If 2.0 moles of sucrose weighing 684 g is put in 1000 g of water and is then dissolved, what would be the molality of the solution?
Check your work.
3. If you have a 0.25 M solution of benzene with a density of 15 g/L, calculate the molality of the solution.
Check your work
4. If the density of mercury is 13.534 g/cm^{3} and you have 62.5 cm^{3} of mercury, how many grams, moles, and atoms of mercury do you have? (Mercury has a mass of 200.6 g/mol.)
Check your work
[Basic Index]
[Chemical Nomenclature]
[Atomic Structure]
[Periodic Table]
[Lewis Structure]
[Chemical Reactions]
[Stoichiometry]
[AcidBase Chemistry]
