MOLS, PERCENTS, and STOICHIOMETRY
Why do we need mols?
Percents by weight.
Density times volume of a pure material.
Atoms or molecules to mols.
Concentration times volume of a
Mol and percent worksheet.
Stoichiometry problems with mass and gas at STP.
Stoichiometry problems on concentration and density.
Stoichiometry problems using complete roadmap.
WHY DO WE NEED MOLS?
Every chemist has dreamed that atoms were large enough to see
and manipulate one at a time. The same chemist realizes
after considering it, that if individual molecules were available
for manipulation, it would take far too long to get anything
done. The view from the atom is very different from the view of
trillions and trillions of atoms. The mass action of the atoms
that we see on our "macro" view of the world is the
result of the action of an incredibly large number of atoms
averaged in their actions. The most usual way we count the atoms is by
weighing them. The mass of material as weighed on a balance
and the atomic weight of the material being weighed is the way we
have of knowing how many atoms or molecules we are
working with. Instead of counting eggs, we can count cartons of
eggs, each carton of which has a given number, a dozen.
Instead of counting B-B's,
we can count liters of B-B's and find out how many B-B's are in a liter. Instead of
counting rice grains, we buy kilograms or pounds of rice and have an idea of how many
rice grains are in the container.
There are less than one hundred naturally occurring elements.
Each element has a characteristic atomic weight. Most
Periodic Charts include the atomic weight of an element in the
box with the element. The atomic weight is usually not an
integer because it is close to being the number of protons plus
the average number of neutrons of an element. Let's use the
atomic weight as a number of grams. This will give us the same
number of any atom we choose. If we weigh out 1.008
grams of hydrogen and 35.45 grams of chlorine and 24.3 grams of
magnesium, we will have the same number of atoms of
each one of these elements. The neat trick with this system is
that we can weigh the atoms on a grand scale of number of
atoms and get a count of them. This number of atoms that is the
atomic weight expressed in grams is Avogadro's number,
6.022 E 23. The
name for Avogadro's number of ANYTHING is a
mole or mol. A mol of aluminum is 27.0 grams of
aluminum atoms. Aluminum is a metal element, so the particles of
aluminum are atoms. There are Avogadro's number of
aluminum atoms in 27.0 grams of it. But 1.008 grams of hydrogen
is NOT a mol of hydrogen! Why not? Remember that
hydrogen is one of the diatomic gases. There is really no such
thing as loose hydrogen atoms. The total mass of a single hydrogen
diatomic molecule (H2) is 2. 016 AMU. A mol
of hydrogen gas has a mass of 2.016 grams. In that 2.016 gram
mass is Avogadro's number of H2 molecules
because that is the way hydrogen comes. A mol of water is 18.016 grams because
each water molecule has two hydrogen atoms and one oxygen atom. A
mol of water has in it Avogadro's number of water
molecules. Another way to view the same thing is that a formula weight
is the total mass of a formula in AMU expressed with units of grams per mol.
So Avogadro's number is just a number, like dozen or score or
gross or million or billion, but it is a very large number. You
could consider a mol of sand grains or a mol of stars. We are more
likely to speak of a mol of some chemical, for which we can
find the mass of a mol of the material by adding the atomic
weights of all the atoms in a formula of the chemical. The unit
of atomic weight or formula weight is grams/mol.
The chemical formula of a material should tell you; (a) which
elements are in the material, (b) how many atoms of each
element are in the formula, (c) the total formula weight, and (d)
how the elements are attached to each other. The symbols
of the elements tell you which elements are in the material. The
numbers to the right of each symbol tells how many atoms
of that element are in the formula. The type of atoms and their
arrangement in the formula will tell how the elements are
attached to each other. A metal and a nonmetal or negative
polyatomic ion shows an ionic compound. A pair of non-metals
are bonded by covalent bonds. Some crystals have water of
hydration loosely attached in the crystal. This is indicated by
the dot such as in blue vitriol, Cu(SO4)
· 5H2O, showing five molecules
of water of hydration to one formula of cupric sulfate.
The unit of the formula weight or molecular weight or
atomic weight is "grams per mol," so it provides a relationship
between mass in grams and mols of material.
nFw = m
'n' is the number of mols, 'Fw is the formula weight, and 'm' is the mass.
Back to the beginning of Mols, Percents, and Stoichiometry.
PERCENTS BY WEIGHT
All men weigh 200 pounds. All women weigh 125 pounds. What is
the percent by weight of woman in married couples? A married couple is one man and one
woman. (No political implications intended.) The total weight is 325
pounds. The formula for percent is:
In this case the woman is the target, so the weight of the woman
goes on top, and the total weight goes on the bottom of the fraction.
325 # | x||100%||=||38.461%||=
Notice that the units of pound cancel to make the percent a
pure number of comparison.
The weights of atoms are the atomic weights. What is the
percentage of chloride in potassium chloride? The atomic weight
of potassium is 39.10 g/mol. The atomic weight of chlorine is
35.45 g/mol. So the formula weight of potassium chloride is
74.55 g/mol. The chloride is the target and the potassium
chloride is the total. 35.45 g/mol / x 100% = 47.55198 % or
47.6 % to three significant figures.
74.55 g/mol |x||100%||=||47.5520%||=||47.6%|
You can do that with any part of a compound. What is the
percentage of sulfate in beryllium sulfate tetrahydrate?
Notice that the examples here are done to two decimal points
of the atomic weights. The problems in the practice bunch at
the end of this chapter are done to one decimal point of the
Back to the beginning of Mols, Percents, and
Pronounce stoichiometry as "stoy-kee-ah-met-tree," if you want to sound like you know what you are talking about, or "stoyk," if you want to sound like a real geek.
Stoichiometry is just a five dollar idea dressed up in a fifty
dollar name. You can compare the amounts of any materials in the
same chemical equation using the formula weights and the
coefficients of the materials in the equation. Let's
consider the equation for the Haber reaction, the combination of
nitrogen gas and hydrogen gas to make ammonia. N2 + 3 H2 —> 2 NH3
for nitrogen is N2 and the formula for
hydrogen is H2.
They are both diatomic gases. The formula for ammonia is NH3. The balanced equation requires one nitrogen molecule and
three hydrogen molecules to make two ammonia molecules, meaning that one
nitrogen molecule reacts with three hydrogen molecules to make two ammonia
molecules or one MOL of nitrogen and three MOLS of hydrogen make two MOLS
of ammonia. Now we are getting somewhere. The real way we measure amounts
is by weight (actually, mass), so 28 grams (14 g/mol times two atoms of
nitrogen per molecule) of nitrogen and 6 grams of hydrogen (1 g/mol times
two atoms of hydrogen per molecule times three mols) make 34 grams of ammonia.
Notice that no mass is lost or gained, since the formula weight for ammonia is
17 (one nitrogen at 14 and three hydrogens at one g/mol) and there are two mols
of ammonia made. Once you have the mass proportions, any
mass-mass stoichiometry can be done by good old proportionation.
What is the likelihood you will get just a simple mass-mass
stoich problem on your test? You should live so long. Well, you
should get ONE.
Rather than thinking in terms of proportions, think in mols
and mol ratios, a much more general and therefore more useful
type of thinking. A mol ratio is just the ratio of one material
in a chemical equation to another material in the same equation.
The mol ratio uses the coefficients of the materials as they
appear in the balanced chemical equation. What is the mol ratio
of hydrogen to ammonia in the Haber equation? 3 mols of hydrogen
to 2 mols of ammonia. Easy. In the standard stoichiometry
calculations you should know, ALL ROADS LEAD TO MOLS. You can
change any amount of any measurement of any material in the same
equation with any other material in any measurement in the same
equation. That is powerful. The setup is similar to Dimensional
Analysis, and the calculations can include portions of DA.
1. Start with what you know (GIVEN), expressing it as a fraction.
2. Use definitions or other information to change what you know to mols of that material.
3. Use the mol ratio to exchange mols of the material given to the mols of material you want to find.
4. Change the mols of material you are finding to whatever other measurement you need.
How many grams of ammonia can you make with 25 grams of
hydrogen? (Practice your mol math rather than doing this by
proportion. Check it by proportion in problems that permit it.)
You are given the mass of 25 grams of hydrogen. Start there.
25 g H2/1 Change to mols of hydrogen by
the formula weight of hydrogen 1 mol of H2 =
2.0 g. (The 2.0 g goes in the denominator to cancel with the gram units in the material given.)
Change mols of hydrogen to mols of ammonia by the mol
ratio. 3 mols of hydrogen = 2 mols of ammonia. (The mols of
hydrogen go in the denominator to cancel with the mols of
hydrogen. You are now in units of mols of ammonia.) Convert the
mols of ammonia to grams of ammonia by the formula weight of
ammonia, 1 mol of ammonia = 17 g. (Now the mols go in the
denominator to cancel with the mols of ammonia.) Cancel the
units as you go.
The math on the calculator should be the last thing you do.
2 5 ÷ 2 . 0
x 2 ÷ 3 x 1
7 = and the number you get (141.66667)
will be a number of grams of ammonia as the units in your
calculations show. Round it to the number of significant digits
your instructor requires (often three sig. figs.) and put into
scientific notation if required. Most professors suggest that
scientific notation be used if the answer is over one thousand or
less than a thousandth. The answer is 142 grams of
The calculator technique in the preceding paragraph
illustrates a straightforward way to do the math. If you include
all the numbers in order as they appear, you will have
less chance of making an error. Many times students have been observed
gathering all the numbers in the numerator, gathering all the
numbers in the denominator, presenting a new fraction of the