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Name: __________________________________
Unit “5B”: Problem Solving –Calculations using Chemical Formulas. Introduction to the Mole.
I. A “mole” is a unit: Much like “dozen” represents a certain number; “mole” represents a certain number.
A. 1 mole of any substance is that amount which contains ~6.02 x 10 23 “representative particles”.
1. While it is not important to know the background to this number, many students ask me
about it, so if you are interested, here is the background
The unit we use for the “atomic mass” of an atom (the mass that is written on the periodic
table, the mass of 1 atom) is the “unified atomic mass unit” or the “dalton” (its symbol might be
either u, da, or amu).
2. The amu was designed to give a proton and a neutron a mass very close to 1. This then
gave our smallest atoms, hydrogen atoms, an average mass very close to 1, which people back in the
day thought made a lot of sense. On the periodic table on the back wall of our classroom, we
routinely round atomic mass to 2 decimal places, so the average mass of a H atom is 1.01 amu.
3. The mole is the number of particles that are calculated to be in a substance (element or
compound) when you measure out the atomic mass of that substance in grams. If we measured 1.01
grams of hydrogen atoms, we would have 1 mole (6.02 x 10 23) atoms. However, if we wanted 1
mole of helium atoms, since each atom of helium weights 4x more than an atom of hydrogen
(helium’s atomic mass is 4.00 amu), we would need to measure 4.00 grams of helium atoms, in order
to have 1 mole (6.02 x 10 23 ) atoms. Since we can’t see individual atoms, whenever we need a
specific amount (number) of atoms, we weigh them instead of counting them.
4. The number of atoms in a mole (6.022141… x 10 23) was first determined by Robert
Millikan, from his measurement of the charge of a single electron, compared to the measured charge
of a mole of electrons. More recently, scientists use X-rays and other devices to improve on the
estimation of a mole. In class, we will round to 2 decimal places (6.02 x 1023).
5. A representative particle is the smallest unit of a substance, that still has the properties of
that substance. The term “representative particle” is used so that the definition can apply to anything
you are interested in. If you want to know how many pretzels are in a barrel of pretzels, the
representative particle is a pretzel; but, in chemistry, a representative particle is likely to be an atom,
an ion, a molecule, an electron, a formula unit, etc.
II.
Chemical formulas are either “empirical” formulas or “molecular” formulas.
A. Previously, when you determined a correct formula for an ionic compound (metal cation, nonmetal
anion) you were told to drop/swap/simplify. An “empirical” formula is a formula that shows
only the lowest whole number ratio of atoms or ions. Formulas for ionic compounds are
always empirical formulas because it is the ratio (not the exact number) of ions that determines
the properties of the ionic compound.
1. NaCl is an empirical formula. There is 1 Na ion and 1 Cl ion; the ratio is 1:1
2. CaCl2 is an empirical formula. There is 1 Ca ion and 2 Cl ions; the ratio is 1:2
3. AlCl3 is an empirical formula. There is 1 Al ion and 3 Cl ions; the ratio is 1:3
4. A molecular formula is used to represent a molecular substance (nonmetal atoms joined by covalent
bonds consisting of shared pairs of electrons), and a molecular formula may be a whole number
multiple of a smaller empirical formula. And, several different compounds (with different
molecular formulas) may have the same empirical formula. For example:
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a. C6H12O6 is the molecular formula for “glucose” (blood sugar). CH2O is its empirical formula.
b. CH2O is the molecular formula for “formaldehyde” (embalming fluid). CH2O is its empirical formula.
Practice: Circle the empirical formulas in the list below:
Al2H6
III.
NH3
C 4 H8
CaSO4
P2O5
NaC2H3O2
N2O4
Molar Mass: You should (this year) round to 2 decimal places when calculating molar mass.
A. The molar mass is the mass in grams of 1 mole of a substance (element or compound). You will
find the molar mass for each element, listed in the element’s square on the periodic table. The
molar mass is identical to the element’s atomic mass (except that the unit is different).
1. The unit for “atomic mass” would be one of these: u, da, amu.
2. The unit for “molar mass” (the mass of 1 mole) is “grams” or “g”.
B. To calculate the molar mass when given a chemical formula, calculate the sum of the molar
masses of the atoms within the given formula. Maintain 2 decimal places.
Example: Calculate the molar mass of water
H2O = 2 H + 1 O
1.00 + 1.00 + 16.00 =
18.00 grams / mole.
C. It is important that you become used to writing the full unit for molar mass: “grams / mole”
(spoken: “grams per mole”). This will enable you to understand stoichiometry later this year.
D. Whenever you need to use molar mass as a factor: Every numerator and every denominator
should have a number, a unit, and a chemical formula. The number that should be placed in front
of “mole” is the whole number 1 (with infinite sig figs). The mass in grams is placed on the


1 mole H 2 O

opposite side of the division line: 
18.00
grams
H
O
2


 18.00 grams H 2 O 


1
mole
H
O
2


or
E. Calculating molar mass in a formula with parenthesis…. Multiply the number of each atom
within the parentheses by the subscript outside of the parentheses.
F. Example: calculate the molar mass of
Ca(NO3)2
There is 1 Ca ion and 2 (NO3) ions.
The “2” refers only to the (NO3) group; distribute the 2 over the (NO3) but not back to the Ca.
I’ve worked the calculation two different ways:
Ca(NO3)2
1 Ca + 2 N +
Ca(NO3)2
6O
14.01 x 2 = 28.02
= 40.00 + 28.02
+
OR: 1 Ca + 2 (NO3)
and 16.00 x 6 = 96.00
96.00
= 164.02 grams / mole
2 (14.01 + 48.00) = 124.02
40.00 + 124.02 = 164.02 g/mol
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IV.
Determining the “Percent Composition by Mass” of Each Element in a Compound.
A. When given the actual measured mass of the each element used to make a compound, you may do a
simple % calculation: (part / total) x 100 for each element.
(mass of first element / sum of the 2 masses) x 100 = % by mass for the first element
(mass of second element / sum of the 2 masses) x 100 = % by mass for the second element
Example: 23.0 grams of Na combines completely with 8.0 grams of O to form a compound. Determine
the percent composition of “O” in the compound
 part 

 x 100 
 total 


16.0 grams

 (23.0  8.0) grams 
 x 100 


 16.0 grams 

 31.0 grams 
  25.8 % O


Practice: 64.0 grams of Cu combines completely with 16.0 grams of S to form a compound. Determine the
percent composition of Cu in the compound.
B. When NOT given the actual measured mass of each element used to make a compound, you will
instead be given the compound’s formula. In this case, you will use the “molar masses” for each of the
elements (remember, the molar mass is the same as the average atomic mass, measured in grams, and
found on the periodic table). Again do a simple % calculation for each element; but, don’t forget to
double or triple (etc) the molar masses of the elements based on the subscripts that are given for each
element in the compound.
Ex: Determine the % composition by mass of Na2O
For each element: (part / total) x100
(mass from element / mass entire compound) x 100 = %
For sodium: 2 x Na = 22.99 x 2 = 45.98
For oxygen: 1 x O = 1 x 16.00 = 16.00
Entire compound: 45.98 + 16.00 = 61.98
 2 Na 

 x 100
 2 Na  1 O 
1O



 x 100
 2 Na  1 O 
 45.98 

 x 100  74.19 % Na
 61.98 
 16.00 

 x 100  25.81 % Oxy
 61.98 
Practice: What is the percent composition of oxygen in the compound, H2SO4?
*The only things needed in order to determine the % composition of a compound (if you’re not given the actual masses of
each element) are the formula of the compound and the molar mass (or the “atomic mass”) of its elements!
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V.
Determining a compound’s empirical formula when given its % composition (by mass).
Example: The % composition by mass of a compound is: 76.54% C, 12.13% H, 11.33% O.
Calculate its empirical formula.
Step 1: As long as the percentages add up to 100 (which they will), you may immediately change the unit
for each element from % to “grams”. Essentially, you assume that your sample contains 100 grams.
Then, for instance, 76.54% of a 100 gram sample = 76.54 grams. The math becomes very easy this way.
Step 1 is shown for you below, combined directly with step 2.
Step 2:
Multiply by a “dimensional analysis” factor in which you use the molar mass (atomic mass
from the periodic table) in grams in the denominator of the factor.
 1 mol 
  6.378 mol C
76.54 % C  76.54 g C x 
 12.00 g C 
 1 mol 
  12.0 mol H
12.13 % H  12.13 g H x 
 1.01 g H 
 1 mol 
  0.7081 mol O
11.33 % O  11.33 g O x 
 16.00 g O 
Step 3: Determine the “smallest whole number ratio of moles” for each element:
Just divide each of the previous answers in moles by the smallest of the answers in moles
6.378 mol C  0.7081  9.000 mol C 
9 mol C
12.0 mol H  0.7081  17.0 mol H 
17 mol H
0.7081 mol O  0.7081  1.00 0mol O 
1 mol O
As long as each answer is a whole number (or within + or – 0.1 mole of a whole number), you may use
those answers as the subscripts for an empirical formula.
Step 3: Write an empirical formula from the mole ratio:
C9H17O1
*Note: BUT, IF/WHEN you have one subscript that is a decimal or fraction, then proceed to step 4.
Step 4: Multiply the non-whole number subscript by the smallest integer (between 2 and 10) necessary to
generate a whole number. Once that integer is identified, then multiply the other subscripts by this same
integer.
Example: You are given a % composition for a compound, do everything correctly, and get this empirical
formula: C1.5 H3 O1 How should you adjust this to become the correct empirical formula?
Since 1.5 x 2 = 3 You now should multiply each of the remaining subscripts by 2 as well.
Correct formula: C3H6O2
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Practice question: The % composition by mass of a compound is: 82.4% N, 17.6% H. Calculate its
empirical formula.
Practice question: You use a given % composition for a compound and determine this to be its empirical
formula: KCrO3.5.
Determine the correct empirical formula.
Working With “Molecular Formulas”
VI.
A. Molecular formulas are whole number multiples of empirical formulas. A molecular formula will always
be either equal to or larger than its empirical formula; and, there may be several molecular formulas
that have the same empirical formula.
1. CH2O is an empirical formula. Some possible molecular formulas for it are:
C1H2O1
x1
CH2O
x2
C2H4O2
x3
C3H6O3
x4
C4H8O4
x5
x6 =
C5H10O5
C6H12O6
B. Calculating a compound’s molecular formula when given the molecule’s empirical formula and its
“molecular mass”. Molecular mass is a term that means: the mass of the entire molecule. Think of it as
the molar mass of a molecular substance. You are given a molecule’s simplified (empirical) formula,
and are being asked to calculate the whole number that the formula needs to be multiplied by in order to
get the larger formula. You are given the mass of the larger formula.
Example: Determine the molecular formula of a compound having P2O5 as an empirical formula. The
compound’s molecular mass is 283.89 g/mole.
Step 1: Calculate molar mass of the empirical formula:
(2 x 30.97)  (5 x 16.00)  141.94 g / mole
Step 2: Divide the (given) molar mass of the compound by the (calculated) molar mass of the empirical formula:
283.89 g/mole  141.94 g/mole  2
Step 3: Multiply each subscript of the given empirical formula by the whole number answer from step 2:
2
x (P2O5 ) = P4O10
Practice:
1. Calculate the molecular formula of a compound whose empirical formula is C3H7. The compound’s molar
mass is 86.0 g / mole.
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VII.
Determine the molecular formula of a substance when given the percent composition by mass
(instead of an empirical formula) and the molecular mass.
First determine an empirical formula for the molecule.
Second determine the molar mass of the empirical formula.
Third divide the given molecular mass by the calculated molar mass to arrive at a whole number.
Fourth multiply each subscript in the empirical formula by the whole number.
Example: The % composition by mass of a compound is: 76.54% C, 12.13% H, 11.33% O. The compound’s
molecular mass = 282.45 g/mole. Calculate the molecular formula.
Step 1: Change the unit from % to “grams”. Then multiply by a dimensional analysis factor with molar mass
(atomic mass from the periodic table) in grams in the denominator of the factor. Then divide each answer in
moles by the smallest answer in moles.
 1 mol 
  6.378 mol C
76.54 % C  76.54 g C x 
 12.00 g C 
 0.7081 = 9 mol C
 1 mol 
  12.0 mol H
12.13 % H  12.13 g H x 
 1.01 g H 
 0.7081 = 17 mol H
 1 mol 
  0.7081 mol O
11.33 % O  11.33 g O x 
 16.00 g O 
 0.7081 = 1 mol O
Write an empirical formula from the mole ratio: C9H17O1
If each subscript had not been a whole number, then you would have needed to multiply by consecutive integers
until you arrived at the smallest whole number formula.
Step 2: Calculate the molar mass of empirical formula:
(9 x 12.01) + (17 x 1.01) + (1 x 16.00)
= 141.26 g / mol
Step3: Molecular mass (given) ÷ molar mass of empirical formula (calculated):
282.45 ÷ 141.26 = 2
Step 4: Multiply each of the subscripts in your empirical formula by the whole number answer:
(C9H17O1 ) x 2
= C18H34O2 = molecular formula
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VIII. Nomenclature and Formulas of Hydrated Ionic compounds.
A. Some ionic compounds are known to absorb water from their environment very easily. When this occurs,
we make note of the water both in the name of the compound, and in the formula of the compound.
B. Naming a hydrated compound: We follow the usual name the ionic compound with a prefix prior to the
word, “hydrate”. The prefix represents both how many molecules of water bind to 1 formula unit of the
compound, and how many moles of water bind to 1 mole of the ionic compound. For example, in a
“pentahydrate”, there are 5 moles of water for each 1 mole of compound. We would indicate the amount of
water that is bound to this compound by attaching a dot, a coefficient and “H2O” to the compound’s formula,
in this manner: CuSO4 5 H2O for the compound, copper(II) sulfate pentahydrate. Careful, the dot does
NOT mean to multiple…It means, to add!
Prefix list:
1 = mono
monohydrate
7 = hepta
heptahydrate
2 = di
dehydrate
8 = octa
octahydrate
3 = tri
trihydrate
9 = nona
nonahydrate
4 = tetra
tetrahydrate
10 = deca
decahydrate
5 = penta
pentahydrate
11 = undeca undecahydrate
6 = hexa
hexahydrate
12 = dodeca dodecahydrate
Ex: CaCl2 H2O would be called calcium chloride dihydrate, because of the “2” after the large dot, connecting 2 moles of
water to the 1 mole of CaCl2.
C. Calculating the molar mass of a hydrated ionic compound – ADD the mass of the number of moles
of water as indicated by the coefficient in front of “H2O”.
Ex: CaCl2
2 H2O =
1 Ca + 2 Cl
+ 2 (H2O) =
40.08 + 2 (35.45) + 2 (1.01 + 1.01 + 16.00)
= 147.02 g / mol
XII. Determining the Percent by Mass of Ionic Compound and/or the Percent by Mass of Water in a given
hydrated ionic compound.
A. Calculate the molar mass of the ionic compound part.
B. Calculate the molar mass of the water part.
C. Calculate the molar mass of the entire hydrated ionic compound (compound + water).
D.
(mass ionic compound / mass entire hydrated compound) x 100 = % ionic compound by mass.
And/or
(mass of water / mass of entire hydrated compound) x 100 = % water by mass
Ex: determine the % by mass of NaI and the % by mass of water in NaI 2H2O
NaI
= 22.99 + 126.90
2(H2O) = 2 (18.02)
NaI + 2(H2O) = 149.89 + 2 (18.02)
= 149.89
= 36.04
= 185.93


Na  I

 x 100
(Na

I)

2
(H
O)
2


= (149.89 / 185.93 ) x 100 = 80.616 % NaI


2 (H 2 O)

 x 100
(Na

I)

2
(H
O)
2


= (36.04 / 185.93) x 100 = 19.39% H2O