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Topic 1: Stoichiometry
1.1 Mole Concept and Avogadro’s Constant
1.1.1. Describe the mole concept and apply it to substances.
The mole concept applies to all kinds of particles: atoms, molecules, ions, formula units
etc. The amount of substance is measured in units of moles. The approximate value of
Avogadro’s constant (L), 6.02 x 1023 mol-1, should be known.
A mole (often abbreviated mol) is the number equal to the number of carbon atoms in
exactly 12 grams of pure 12C. Techniques such as mass spectrometry, which count atoms
very precisely, have been used to determine this number as 6.02214 x 1023 (6.02 x 1023 is
good enough for IB). This number is called Avogadro’s number to honor his
contributions to chemistry. One mole of something consists of 6.02 x 1023 units of that
substance. Just as a dozen eggs is 12 eggs, a mole of eggs is 6.02 x 1023 eggs.
The magnitude of the number 6.02 x 1023 is difficult to imagine. To give you some idea,
1 mole of seconds represents a span of time 4 million times as long as the earth has
already existed, and 1 mole of marbles is enough to cover the earth to a depth of 50
miles! However, since atoms are so tiny, a mole of atoms or molecules is a perfectly
manageable quantity to use in a reaction.
The mole is also defined as such that a sample of a natural element with a mass equal to
the element’s atomic mass expressed in grams contains 1 mole of atoms. This definition
also fixes the relationship between the atomic mass unit and the gram. Since 6.02 x 1023
atoms of carbon (each with a mass of 12 amu) have a mass of 12 g, then
(6.02 x 1023 atoms)(12 amu/atom)= 12 g
6.02 x 1023 amu=1 g
1.1.2. Calculate the number of particles and the amount of substance (in moles).
Calculate between the amount of substance (in moles) and the number of atoms,
molecules or formula units.
In order to convert from x moles of anything to how many actual atoms, multiply x by
Avagadro’s constant to find how many actual particles you have.
For example, say you are given 2.3 moles of hydrogen. In order to find how many actual
particles of hydrogen you have, you do the following calculations…
2.3 mol H x (6.02 x 1023 particles H/1 mol H) = 1.38 x 1024
1.2 Formulas
1.2.1. Define the term molar mass (M) and calculate the mass of one mole of a species.
The molar mass of a substance is the mass in grams of one mole of the compound.
How can we calculate the mass of one mole of a substance? Let’s take methane for
example. Methane is CH4, which means in a molecule of methane there is one carbon
atom and four hydrogen atoms. So, the molar mass of CH4 would be, what is the mass in
grams of one mole of methane, or what is the mass of 6.02 x 1023 CH4 molecules? The
mass of 1 mole of methane can be found by summing the masses of carbon and
hydrogen.
Mass of 1 mol C: 12.01 g (This can be found on the periodic table)
Mass of 4 mol H: 4 x 1.008 g (If you don’t understand where the four came from, learn
what subscripts mean)
Add them.
Mass of 1 mol CH4: 16.04 g
Traditionally, the term molecular weight has been used for this quantity. The molar mass
of a known substance is obtained by summing the masses of the component atoms as we
did for methane.
1.2.2. Distinguish between atomic mass, molecular mass and formula mass.
The term molar mass (in g mol-1) can be used for all of these.
Molecular Mass: The mass in grams of one mole of molecules or formula units of a
substance; the same as molar mass.
Formula Mass: The mass of a formula, including ionic compounds. Technically,
molecular mass only applies to molecules (which are defined by covalent bonding).
Formula mass can also include ionic compounds, etc.
Atomic Mass: Atomic mass is somewhat difficult to define. According to Zumdahl,
atomic mass is technically the average atomic mass of an element, which is determined
by finding the different isotopes of an element present in nature, and how much percent
of the total amount of that element in the world they make up. For example, in nature
there are two primary isotopes of Carbon, 12C and 13C (there are other isotopes but they
are so rare they are insignificant at this level of precision. 98.89% of Carbon is 12C, but
1.11% is 13C. So to find the atomic mass of Carbon, we do the following…
(.9889)(12 amu (mass of 12C by definition)) + (.0111)(13.0034 amu (mass of 13C)) =
12.01 amu
This is the atomic mass of carbon.
1.2.3. Define the terms relative molecular mass (Mr) and relative atomic mass (Ar).
The terms have no units.
The relative atomic mass is the mass of one atom of an element compared to the mass of
Carbon 12. The relative atomic mass has no units because it is a ratio of masses and the
units cancel out. The relative molecular mass is the mass of the relative atomic masses of
all the atoms in a molecule added up. It’s basically the molecular mass without units as
far as I can tell.
1.2.4. State the relationship between the amount of substance (in moles) and mass, and
carry out calculations involving amount of substances, mass and molar mass.
The relationship between moles and mass is demonstrated in the mass formula.
Number of Moles (N)= mass (m)/Molecular Mass (M)…
N=m/M
So, say you are given 40 grams of Carbon Dioxide, how many moles do you have? The
molecular mass of Carbon Dioxide (CO2) is 44 g. So, the way to solve this would be to
divide 40 grams by 44 g.
Number of Moles: 40g/44= .91 mol
Note: I am not sure how the units work out for this. It would appear that the grams
cancel each other out leaving no value, but for some reason you are left with moles. This
may be something to ask your teacher about.
1.2.5. Define the terms empirical formula and molecular formula.
The molecular formula is a multiple of the empirical formula.
The empirical formula is best understood by knowing its usefulness. One way scientists
find the formulas for molecules is by determining the percent of the mass each element in
the molecule takes up. For example, let us take CH5N. A scientist is attempting to find
the formula but doesn’t know what it is. He breaks it up and finds that 38.67% Carbon,
16.22% Hydrogen, and 45.11% Nitrogen. By assuming the compound weighs 100
grams, each of these percentages now becomes gram units. Then by changing them to
moles and setting them up in ratio to each other, the scientist finds that the ratio of
Nitrogen to Hydrogen to Carbon is 1:5:1. So, he knows that the formula can be CH5N.
However, the formula could also be C2H10N2, or C3H15N3. That is because all he has is a
ratio. The empirical formula is CH5N. This is the simplest form. The exact formula is
called the molecular formula, and that will be (CH5N)n where n is an integer. Sometimes
the empirical and molecular formula will be the same, other times they will not be. If
you did not understand the whole percent composition and ratios, don’t worry, those will
be explained later. The important part is understanding the difference between empirical
and molecular formula and how they are related. The empirical formula is the formula
for the compound that has the lowest values possible for it to still work. The molecular
formula is the formula with the actual values for the compound, and it will always be the
empirical formula multiplied by some integer.
1.2.6. Determine the empirical formula and/or the molecular formula of a given
compound.
Determine the:
 Empirical formula from the percentage composition or from other suitable
experimental data.
 Percentage composition from the formula of a compound.
 Molecular formula when given both the empirical formula and the molar mass.
The steps taken in the previous objective will be examined in much more detail here. In
order to do these problems we will actually work the problems out, and taking the same
steps with any similar step should work out the same way.
1. Empirical formula from the percentage composition or from other suitable
experimental data.
When a new compound is prepared, one of the first items of interest is the formula of the
compound. This is most often determined by taking a weighed sample of the compound
and either decomposing it into its component elements or reacting it with oxygen to
produce substances such as CO2, H2O, and N2, which are then collected and weighed.
We will see how information of this type can be used to compute the formula of a
compound. Suppose a substance has been prepared that is composed of carbon,
hydrogen, and nitrogen. When 0.1156 gram of this compound is reacted with oxygen,
0.1638 gram of carbon dioxide (CO2) and 0.1676 gram of water (H2O) are collected.
Assuming that all the carbon in the compound is converted to CO2, we can determine the
mass of carbon originally present in the 0.1156-gram sample. To do this, we must use
the fraction (by mass) of carbon in CO2. The molar mass of CO2 is
C: 1 mol x 12.01g/mol=12.01 g
O: 2 mol x 16.00 g/mol=32.00 g
Added together, the molar mass of CO2 = 44.01 g/mol.
The fraction of carbon present by mass is
Mass of C/Total mass of CO2 = 12.01 g C/ 44.01 g CO2
This factor can now be used to determine the mass of carbon in 0.1638 gram of CO2:
0.1638 g CO2 x 12.01 g C/44.01 g CO2 = 0.04470 g C
Remember that this carbon originally came from the 0.1156-gram sample of unknown
compound. Thus the mass percent of carbon in this compound is
(0.04470 g C/0.1156 g compound) x 100= 38.67% C
The same procedure can be used to find the mass percent of hydrogen in the unknown
compound. We assume that all the hydrogen present in the original 0.1156 gram of
compound was converted to H2O. The molar mass of H2O is 18.02 grams, and the
fraction of hydrogen by mass of H2O is
Mass of H/Mass of H2O = 2.016 g H/18.02 g H2O
Therefore, the mass of hydrogen in 0.1676 gram of H2O is
0.1676 g H2O x (2.016 g H/18.02 g H2O)= 0.01875 g H
And the mass percent of hydrogen in the compound is
(0.01875 g H/0.1156 g) compound x 100 = 16.22%
The unknown compound contains only carbon, hydrogen, and nitrogen. So far we have
determined that is is 38.67% carbon and 16.22% hydrogen. The remainder must be
nitrogen.
100.00% - (38.67% + 16.22%) = 45.11% N
We have determined that the compound contains 38.67% carbon, 16.22% hydrogen, and
45.11% nitrogen. Next we use these data to obtain the formula. Since the formula of a
compound indicated the numbers of atoms in the compound, we must convert the masses
of the elements to numbers of atoms. The easiest way to do this is to work with 100.00
grams of the compound in the present case, 38.67% carbon by mass means 38.67 grams
of carbon per 100.00 grams of compound, etc. To determine the formula, we must
calculate the number of carbon atoms in 38.67 grams of carbon, the number of hydrogen
atoms in 16.22 grams of hydrogen, etc. You can do that using the mass formula (found in
1.2.4.). The answers are 3.220mol C, 16.09 mol H, 3.219 mol N. Thus 100.00 grams of
this compound contains 3.220 moles of carbon atoms, 16.09 moles of hydrogen atoms,
and 3.219 moles of nitrogen atoms. We can find the smallest whole number ratio
(empirical formula) of atoms in this compound by dividing each of the mole values above
by the smallest of the three:
C: 3.220/3.219 = 1.000 = 1
H: 16.09/3.219 = 4.998 = 5
N: 3.219/3.219 = 1.000 = 1
So, the empirical formula for the unknown compound is CH5N. Now, if you want to
know the molecular formula, you have to know the formula mass of the unknown
compound. If the formula mass of the unknown is the same as the formula mass of the
empirical formula, then the empirical formula is the molecular formula, but if the formula
mass of the unknown is say n times more then the formula mass of the empirical formula,
then the molecular formula is n times the empirical formula.
2. Percentage composition from the formula of a compound.
Say you are given the formula H2O, and you want to find out the percentage composition
of hydrogen. How you do this is you find the formula mass of water, and then find the
atomic mass of H2, and divide the latter by the former. Then multiply by 100 to get a
percent value. So, formula mass of H2O is 18, H2 is 2, so 2 g H2/ 18 g H2O=.11, x
100=11.1%.
3. Molecular formula given both the empirical formula and the molar mass.
Say you are given the empirical formula CH5N, and are told that the molar mass is 62.12
g, and are told to find the molecular formula. First you find the formula mass of CH5N,
which ends up being 31.06 g. You know that that multiplied by n equals the molecular
mass of the formula you want, and the n in this is case is 2 (you could find this by
dividing 62.12 g unknown/ 31.06 g empirical formula).
1.3 Chemical Equations
1.3.1. Balance chemical equations when all reactants and products are given.
Distinguish between coefficients and subscripts.
Say you are given the reaction C6H14 + O2 CO2 + H2O and are told to balance it. The
main idea behind balancing an equation is you have to have the same amount of each
element on both sides of the equation, and the only thing you can edit are the coefficients,
not the subscripts. So, for example, in this equation, in the hexane (reactant), you have 6
carbons, but in the products you only have one carbon. So, to balance the carbons you
have to place a 6 in front of the CO2. You now have 6 carbons on both sides. Next look
at the hydrogen (a good rule of thumb is whenever you have an element that is by itself,
such as the oxygen in the reactants, leave it for last). You have 14 hydrogen on one side
and only 2 on the other, so you have to multiply the water on the right by 7 (because the
hydrogen has a subscript of 2, which means there are two hydrogen in the water
compound.) You now have fourteen hydrogen on both sides. You now move onto the
oxygen. There is 2 oxygen in the reactants and 19 in the products. There is a rule when
balancing equations that you are not really supposed to use anything but integers
(although when you get into more advanced stuff you will occasionally use fractions to
make it easier on yourself, but you’re not supposed too). So, there is no way to balance it
as is. So, you’ll have to go back and change your original coefficients. If you put a 2 in
front of the hexane and a 12 in front of the carbon dioxide you’ll have balanced the
carbons. Then to balance the hydrogen put a 14 in front of the water. Now, you have 28
hydrogen on both sides. You now have two oxygen in the reactants and 26 in the
products. So, by multiplying the oxygen in the reactants by 13, you have 26 oxygen on
both sides and you have balanced the equation.
2C6H14 + 13O2 12CO2 + 14H2O
In reality, practice makes perfect when it comes to balancing equations. The only way to
learn is to keep trying, and after a few you’ll get pretty good and fast at it. Some other
rules are that if all of the coefficients have a common denominator, you can divide
everything by that common denominator (so if the reactants had coefficients of 2 and 4
and the products had coefficients of 6 and 10, you could divide all of them by 2). And
remember that you cannot change a subscript, and when you multiply a compound by a
coefficient the whole compound is multiplied, not just the front element.
1.3.2. Identify the mole ratios of any two species in a balanced chemical equation.
Use balanced chemical equations to obtain information about the amounts of reactants
and products.
Using the reaction from above, 2C6H14 + 13O2 12CO2 + 14H2O.
The mole ration from hexane to Oxygen is 2 mol hexane/13 mol O2. Assuming
stoichiometric equilibrium (there are no limiting or excess reagents) it takes 2 moles of
hexane to react with every 13 moles of oxygen. This can also be done from reactants to
products. Another mole ratio is 13 mol O2/12 mol CO2. Once again assuming
stoichiometric equilibrium, for every 13 mol O2 you get 12 mol CO2. This can also be
done for products and products (it can be done for any two species in a reaction.)
1.3.3. Apply the state symbols of (s), (l), (g), and (aq).
Encourage the use of state symbols in chemical equations.
The state symbols are placed after each species to specify what state each each species in
a reaction is in. The symbol (s) stands for the solid state, (l) stands for liquid, (g) stands
for gaseous, and (aq) stands for in aqueous solution (dissolved in water). Often times you
don’t have to write the state symbols and they won’t have much importance, but when
they are needed it is really bad to miss them. For that reason I HIGHLY recommend you
get used to using them every time you know what they are, it’ll save you later on from
making the stupid mistake of forgetting them when they are needed.
1.4 Mass and Gaseous Volume Relationships in Chemical Reactions
1.4.1. Calculate stoichiometric quantities and use these to determine experimental and
theoretical yields.
Mass is conserved in all chemical reactions. Given a chemical equation and the mass or
amount (in moles) of one species, calculate the mass or amount of another species.
The first rule to remember is mass is always conserved (at least at this point in the
syllabus). So the total mass of one side should always add up on another side. Now,
once again we will use the chemical equation from above.
2C6H14 (g) + 13O2 (g) 12CO2 (g) + 14H2O (l)
Say you are told that you have 40 grams of hexane, how many grams of oxygen do you
need to completely react with all the hexane? Well, in order to solve this you use mole
rations.
2 moles hexane/13 moles O2 = 40 g hexane/x moles O2. You then solve this like a
normal ratio problem (multiply diagonally then solve for x). You answer is 260 grams of
oxygen is needed to react with 40 g of hexane. Note: This is not the best example because
you normally don’t use the measurement grams for gases, but it would be the same way
for liters.
1.4.2. Determine the limiting reactant and the reactant in excess when quantities of
reacting substances are given.
Given a chemical equation and the initial amounts of two or more reactants:
 Identify the limiting reactant
 Calculate the theoretical yield of a product
 Calculate the amount(s) of the reactant(s) in excess remaining after the reaction
is complete.
1. Identify the limiting reactant.
There are many ways to solve for the limiting reactant, and everyone tends to like their
own way. I will show mine below but if you don’t like it feel free to ask your teacher or
someone else for another way to do it. Say you have the equation
4FeCr2O4(s) + 8K2CO3(s) + 7O2(g)  8K2CrO4(s) + 2Fe2O3(s) + 8CO2(g)
The masses are 169 kg, 298 kg, and 75 kg respectively for the reactants, and you want to
find out which one is the limiting reagent. In order to do this, you first have to find out
how many moles of each of the reactants you have (using the mass formula), we find
their to be 7.55 x 102 mol chromite ore, 2.16 x 103 mol potassium hydroxide, and 2.34 x
103 mol O2.
You then set up the mole ratios you need to react correctly and the mole ratios you have.
So, to react correctly you should 4 moles of chromite ore for every 8 moles of potassium
carbonate. That is, 4 mol FeCr2O4/8 mol K2CO3(s), which equals .5. You then take the
values that you actually have, which are 7.55 x 102 mol chromite ore/2.16 x 103 mol
potassium hydroxide, which equals .35. This is smaller then .5, and that means that the
numerator is too small (if the answer was bigger then .5, the denominator would be too
small). Whichever value is too small, that number is the limiting reagent FOR THAT
PAIR. You then have to take that reagent and compare it with the other reactant, or the
rest of the reactants if there are more then three. So, taking chromite ore we do the same
thing comparing it with oxygen. We have 7.55 x 102 mol chromite ore/ 2.34 x 103 mol
O2 which then equals .32. The ratio we want is 4/7, or .57, which is bigger then the
actual value we got, which means O2 is the limiting reagent of the entire reaction (you
don’t have to then compare it to the first reactant tested). So, oxygen is the limiting
reagent of the formula.
2 and 3. Calculate the theoretical yield of a product and amount of excess reagent
remaining after reaction is complete.
Theoretical yield of a product is the amount of a product formed when the limiting
reactant is completely consumed. In order to obtain it, you go through several steps, that
we will go through with the reaction 2C2H6 + 7O2  4CO2 + 6H20 starting with 20 g of
ethane and 40 grams of oxygen. So, the first step is to identify the limiting reagent.
2 mol ethane/7 mol oxygen=.29 .66 mol ethane/1.25 mol oxygen=.528. Since the
number is too big, that means the denominator is the limiting reagent, or oxygen is the
limiting reagent. We then have to find how much ethane is used up before the oxygen is
consumed. We do this by multiplying the actual amount of oxygen by the mole ratio of
the two reactants.
1.25 mol oxygen x (2 mol ethane/ 7 mol oxygen)= .36 mol ethane.
This is how much ethane is used up in the reaction, so now we know how much ethane
was used and we know how much ethane we started with, so to find how much ethane is
left over we do a simple subtraction.
.66 mol ethane started with - .36 mol ethane used = .3 mol ethane left over.
So, we have .3 mol ethane left over after the reaction. Then, to find theoretical yield, you
have two options. You can use how much oxygen use, or you can use how much ethane
you use in the reaction, either should gain you the right answer. I would recommend
using the value given to you by the problem, so that way if you made a mistake in finding
the excess reagent left over you won’t mess up your second answer. You know you used
1.25 mol oxygen, so you do another mole ratio.
1.25 mol oxygen x (4 mol carbon dioxide/7 mol oxygen)=.714 mol carbon dioxide.
So, .714 mol carbon dioxide is your theoretical yield for that product.
Note: Rarely do you ever calculate the theoretical yield, because side reactions normally
occur that decrease from the theoretical yield. The actual yield is sometimes compared
to the actual yield by doing the following calculations: actual yield/theoretical yield x
100= Percent Yield.
1.4.3. Apply Avagadro’s law to calculate reacting volumes of gases.
Avagadro’s law is V=an, where V stands for volume, a stands for proportionality
constant, and n is the number of moles of gas particles. Basically, what this equation
states is that for constant temperature and pressure, the volume is directly proportional to
the number of moles of gas. This relationship is obeyed closely by gases at low
pressures. An example of this can be exemplified by the following reaction 3O2(g) 
2O3(g) . The temperature and pressure of this reaction are constant (25 degrees C and 1
atm). Say we start with a 12.2-L sample containing 0.50 mol oxygen gas. What would
be the volume of the ozone?
0.50 mol oxygen x 2 mol ozone/3 mol oxygen= 0.33 mol ozone.
Avagadro’s law can be rearranged to show V/n=a. Since as is a constant, an alternative
representation is V1/n1= a =V2/n2
So, for this reaction n1 equals 0.50 mol, V1 equals 12.2 L, n2=0.33 mol, and V2=? By
setting up the above rearrangement of avagadro’s law, 12.2/0.5=?/0.33, so the volume is
8.052 L.
1.5 Solutions
1.5.1. Define the terms solute, solvent, solution and concentration (g dm-3 and mol dm3).
Concentration in mol dm-3 is often represented by square brackets around the substance
under consideration, eg. [CH3COOH].
Solute: A substance dissolved in a liquid to form a solution.
Solvent: The dissolving medium in a solution.
Molarity: moles of solute per volume of solution in liters. This often times written by IB
with units mol dm-3, which means moles/dm3, and a dm3 equals a liter. Same meaning,
just different ways of writing it. I recommend writing it mol dm-3 whenever your in IB
class because it gets you well practiced, but if you are working with people outside of the
IB program write it per liters, because I have found they have a hard time understanding
IB notation. This is also often times written short hand in square brackets such as
[CH3COOH]. That means the molarity of that compound. This could also be written as g
dm-3,(molality) however I have never seen it written that way, even by IB. If it is ever
written that way, I would recommend converting it to moles (by just using Avagadro’s
number), however I could see how having it in grams may help calculate it at some point.
So, if you dissolve 5 moles of HCl in 3 Liters of water, your molarity [HCl] is 5 mol/3 L,
or 1.66 mol L-1, or 1.66 mol dm-3.
1.5.2. Carry out calculations involving concentration, amount of solute and volume of
solution.
These are fairly easy. Just remember that concentration equals amount of solute/volume
of solution. So say you know the concentration of a solution is 4 mol dm-3, and you
know the volume is 5 L. That means x mol/ 5 L=5 mol/L. That means x must equal 20
mol.
Note: A good rule to remember is the equation m1L1=m2L2. m stands for the molarity,
and L stands for volume. So, if you had a 2 L concentration of 5 M HCl solution (M is
often used to stand for molarity) and you wanted to dilute it to a 2 M HCl, how much
water would you need to add? Well, m1=5, L1=2, m2=2, and L2=?. So, (5 mol/L)(2
L)=(2 mol/L)(? L). By solving for the unknown, you find your final solution will be a
volume of 5 liters, and since you started with 2 liters, you need to add 3 liters to get to 5
liters.
1.5.3. Solve solution stoichiometry problems.
Given the quantity of one species in a chemical reaction in solution (in grams, moles or
in terms of concentration), determine the quantity of another species.
When it is given to you in grams or moles, it’s just like a regular stoichiometric problem.
However, if it is given to you in terms of concentration, work your way around it to find
moles or grams (so, if you have a concentration of 5 M, then you have 5 moles per liter,
and if you have 3 liters, that means you have 15 moles.)