1. science
2. chemistry

Topic 3
Mass Relations and
Stoichiometry
Mass and Moles of a Substance
Chemistry requires a method for determining
the numbers of molecules in a given mass of a
substance which has lead to the development
of the mole (quantity of substance to be
discussed later).
– This allows the chemist to carry out “recipes” for
compounds based on the relative numbers of
atoms involved.
– stoichiometry - the calculation involving the
quantities of reactants and products in a chemical
equation.
2
Molecular Weight and Formula
Weight
The molecular weight (for molecular
substances) of a substance is the sum of the
atomic weights of all the atoms in a molecule of
the substance.
– For, example, a molecule of H2O contains 2
hydrogen atoms (at 1.01 amu each) and 1 oxygen
atom (16.00 amu), giving a molecular weight of
18.02 amu.
H: 2 x 1.01 amu = 2.02 amu
O: 1 x 16.00 amu = 16.00 amu
18.02 amu (mass of 1 molecule of H2O)
3
Molecular Weight and Formula Weight
The formula weight of a substance is the sum of the
atomic weights of all the atoms in one formula unit of the
compound, whether molecular or not.
one formula unit (FU) of NaCl:
Na: 1 x 22.99 amu = 22.99 amu
Cl: 1 x 35.45 amu = 35.45 amu
58.44 amu
(mass of 1 FU NaCl)
iron (III) sulfate,
Fe2(SO4)3
Fe: 2 x 55.85 amu = 111.70 amu
S: 3 x 32.07 amu = 96.21 amu
O: 12 x 16.00 amu = 192.00 amu
399.91 amu
(mass of 1 FU Fe2(SO4)3)
Glucose, C6 H12 O6
Note: molecular wt and
formula wt are calculated the
same way.
C: 6 x 12.01 amu = 72.06 amu
H: 12 x 1.01 amu = 12.12 amu
O: 6 x 16.00 amu = 96.00 amu
180.18 amu
4
(mass of 1 molecule)
Mass and Moles of a Substance
The Mole Concept
– A mole is defined as the quantity of a given
substance that contains as many molecules or
formula units as the number of atoms in exactly 12
grams of carbon–12.
– The number of atoms in a 12-gram sample of
carbon–12 is called Avogadro’s number (Na). The
value of Avogadro’s number is 6.022 x 1023.
1 mole = 6.022 x 1023 X X = ions, particles, atoms,
molecules, items, etc.
1 mole Na2CO3  6.022 x 1023 FU Na2CO3
1 mole CO2  6.022 x 1023 molecules CO2
5
Mass and Moles of a Substance
The molar mass (Mm) of a substance is the mass of one mole
(6.022 x 1023 FU or molecules) of a substance. This is the
term we will use the most in the course and is done the same
way as formula wt. except using gram instead of amu.
– For all substances, molar mass, in grams per mole, is numerically
equal to the formula weight in atomic mass units.
– That is, one mole of any element weighs its atomic mass in grams.
1 molecule of H2O - MW = 18.02 amu  1 mole of H2O - Mm = 18.02 g H2O
6.022 x 1023 molecules
Note: we can use Mm as a
conversion factor between
grams and mols.
Mm of 1 mole of
MgSO4 . 7H2O
Mg: 1 x 24.31
S: 1 x 32.07
O: 4 x 16.00
H:14 x 1.01
O: 7 x 16.00
g=
g=
g=
g=
g=
24.31 g
32.07 g Note: must account for
64.00 g mass of water in hydrates
14.14 g
112.00 g
6
.
246.52 g/mol MgSO4 7H2O
How is it possible that Mm and FW/MW are the same value but different
units?Avogadro’s number and amu are both based on carbon-12 and inverse values of each other.
Let’s convert 1 Na atom  g/mol
A.)What is the mass in grams of one Cl atom?
B.)in one HCl molecule?
Notes:
Converting mass to mols or vice versa will require using molar mass.
Converting mass to atoms, molecules, ions, etc. or vice versa will require going
through mols by using Avogadro’s number.
7
Mass and Moles of a Substance
Mole calculations
– Converting the number of moles of a given substance
into its mass, and vice versa, is fundamental to
understanding the quantitative nature of chemical
equations.
mass of " A"
moles of " A" 
molar mass of " A"
Converting mass to mols or vice
versa will require using molar mass.
Mm NaCl = 58.44 g/mol
58.44 g NaCl
1 mol NaCl
1 mol NaCl
58.44 g NaCl
Note: we can use Mm as a
conversion factor between
grams and mols with g in
numerator or denominator.
8
Mass and Moles of a Substance
Mole calculations
Suppose we have 5.75 moles of magnesium. What is
its mass?
Note: we use Mm as a conversion factor between
grams and mols with mol in this case being placed in
the denominator to cancel with the 5.75 mols we are
converting to g.
9
Mass and Moles of a Substance
Mole calculations
Suppose we have 100.0 grams of H2O. How many
moles does this represent?
10
Mass and Moles and Number of
Molecules or Atoms
The number of molecules or atoms in a sample is related
to the moles of the substance by Avogadro’s #:
1 mole HCl  6.02  10 HCl molecules
23
1 mole Fe  6.02  10 23 Fe atoms
How many molecules are there in 56 mg HCN?
HW 20-22
code: moles
11
Determining Chemical Formulas
The percent composition of a compound is the mass
percentage of each element in the compound.
– We define the mass percentage of “A” as the parts of “A” per
hundred parts of the total, by mass. That is,
mass of " A" in whole
mass % " A" 
 100%
mass of the whole
20 g A
20 % A 
100 g sample
If given the %A by mass, it is useful to put the percentage in
ratio form for dimensional analysis calculations.
12
Mass Percentages from Formulas
Let’s calculate the percent composition (%C,
%H) of one molecule of butane, C4H10.
First, we need the molecular wt of C4H10.
4 carbons @ 12.01 amu/atom  48.04 amu
10 hydrogens @ 1.01 amu/atom  10.10 amu
1 molecule of C 4 H 10  58.14 amu
Now, we can calculate the percents (basically, part/whole).
%C
%H
48.04 amu C
58.14 amu total
10.10 amu H
58.14 amu total
 100 %  82 .63 % C
 100 %  17 .37 % H
What if we wanted the %C & %H for 1 mol of butane?
Same calculation and results except we would use molar
mass, g/mol, instead of amu/atom.
Note: % is a unit like g,
etc. Don’t use the %
button on calculator.
13
How many grams of carbon are there in 83.5 g of
formaldehyde, CH2O, (40.0% C, 6.73% H, 53.3% O)?
rewrite % into ratio
= 33.4 g C
14
An unknown acid contains only C, H, O. A 4.24 mg sample of acid is
completely burned. It gives 6.21 mg of CO2 and 2.54 mg H2O. What
is mass% of each element in the unknown acid?
goal of problem
O2
%C, H, O in sample
unknown acid  CO2 + H2O
C:
has C, H, O
4.24 mg
6.21 mg
2.54 mg
mol to mol ratio needed
to convert from one
species to another
Note: We did the same calculation for H as we did for C except we did it in terms of mg and mmol
to demonstrate how taking advantage of the prefix can simplify your work by adding m to the mols
and g of the molar mass conversion factor (mmol/mg) and mol to mol ratio (mmol/mmol). This
eliminates the useless conversion from mg to g and back again.
H:
O:
Total mass = 4.24 mg = mass C + mass H + mass O = 1.69 mg + 0.285 mg + mass O
mass O = 4.24 mg – 1.69 mg – 0.285 mg = 2.26 mg O
%C:
%H:
%O:
100% - 39.9% - 6.72% = 53.4% O
Alternatively, we could have just subtracted the 15
percent C and H from 100% to determine the %O.
Determining Chemical Formulas
We can determine the formula of a compound
from the percent composition.
– The percent composition of a compound leads
directly to its empirical formula.
– An empirical formula (or simplest formula) for a
compound is the formula of the substance written
with the smallest integer (whole number)
subscripts.
– From the empirical formula, we can find the
molecular formula of a substance which will be a
multiple of the empirical formula based on its
16
molar mass.
Determining Chemical Formulas
Benzoic acid is a white, crystalline powder used as a
food preservative. The compound contains 68.8% C,
5.0% H, and 26.2% O by mass. What is its empirical
formula?
In other words, give the smallest whole-number ratio of
the subscripts in the formula based on mols (C:H:O)
Cx HyOz
mols
17
Determining Chemical Formulas
For the purposes of calculations of this type, we will
assume we have 100.0 grams of sample; therefore,
benzoic acid at 68.8% C means
which means the mass of each element equals the
numerical value of the percentage.
Since x, y, and z in our formula represent mole-mole
ratios, we must first convert these masses to moles.
Cx HyOz
18
Determining Chemical Formulas
Our assumed 100.0 grams of benzoic acid (68.8%C,
5.0%H, 26.2%O) would contain:
1 mol C
68.8 g C 
 5.72(8) mol C
12.01 g
1 mol H
5 .0 g H 
 4.9(5) mol H
1.01 g
1 mol O
26.2 g O 
 1.63(7 ) mol O
16.00 g
This isn’t quite a
whole number ratio,
but if we divide each
number by the
smallest of the three,
a better ratio will
emerge. We are
dividing by the
smallest number to
get the species all on
the same scale –
“normalizing” the
values.
Note: 68.8g + 5.0 g + 26.2 g = assumed 100.0 g sample
19
Determining Chemical Formulas
Next step is to get all species on the same scale,
basically normalizing to moles of smallest species (in this
case, 1.63 mols of O) for the purpose of getting values
that can easily be converted to whole number ratios.
1 .63( 7 ) m ol O  1.63(7)  1.00
5 .728 m ol C  1.63(7)  3.50
4 .95 m ol H  1.63(7)  3.0
2 x C3.5 H3 O1
 C7H6O2
now it’s not too difficult
to see that the smallest
whole number ratio is
7:6:2 (multiple
everything by 2 to get
whole number.
The empirical formula
is C7H6O2 .
This is the empirical as well as molecular formula of benzoic acid because the
molar mass of the empirical formula is the same as the molecular formula of 20
benzoic acid.
Determining Chemical Formulas
An empirical formula gives only the smallest whole-number ratio of
atoms in a formula.
The molecular formula should be a multiple of the empirical formula
(since both have the same percent composition).
C 2 H 3 O2
empirical formula (lowest whole # ratio)
C 4 H 6 O4
molecular formula (multiple of 2 x emp)
C8H12O8
molecular formula (multiple of 4 x emp)
Which is not an empirical formula meaning not lowest whole #?
CH4
CH4O
C2H4O2
C2H6O
C2H4O2 is divisible by 2; CH2O would be its empirical formula
To determine the molecular formula, we must know the molecular
weight (molar mass) of the compound.
21
Determining Chemical Formulas
Determining the molecular formula from the
empirical formula.
molar mass = n x empirical formula mass
where n is the multiple factor
n = molar mass
empirical mass
22
Acetic Acid contains 39.9% C, 6.7% H, 53.4% O.
Determine empirical formula. The molecular mass of
acetic acid is 60.0 g/mol. What is the molecular formula?
We assume 100.0 g sample meaning we have 39.9 g C, 6.7 g H,
53.4 g O and calculate the mols of each. Next, we get them on same scale by dividing
each by the smallest mols, 3.33 mols
empirical formula = CH2O Mm= 30.03 g/mol
2x
HW 23
code: formula
CH2O  C2H4O2
23
Stoichiometry: Quantitative Relations
in Chemical Reactions
Stoichiometry is the calculation of the quantities
of reactants and products involved in a chemical
reaction.
– It is based on the balanced chemical equation and
on the relationship between mass and moles (molar
mass is an important concept here; g  mol,
mol  g) and mol to mol ratios.
– Such calculations are fundamental to most
quantitative work in chemistry.
What we are discussing deals with mol to
mol ratios (very important concept) and is
the basis of many calculations.
24
Molar Interpretation of a Chemical Equation
The balanced chemical equation can be interpreted in
numbers of molecules, but generally chemists interpret
equations as “mole-to-mole” relationships. Lets look at the
following reaction:

N 2 (g)
3H 2 (g)

2 NH 3 ( g )
1 molecule N2
+
3 molecules H2
1 mol N 2

3 mol H 2
28.02 g N2 +
3(2.02 g) H2
 2 (17.04 g) NH3
28.02 g
6.06 g

34.08 g
=
34.08 g
+
34.08 g
2 molecules NH3

2 mol NH 3
If the number of each particular atom is the same on both sides,
the Law of Conservation of Mass will be preserved and the mass
of reactants will be equal to the mass of the products.
25
Molar Interpretation of a Chemical Equation
Suppose we wished to determine the number of moles of
NH3 we could obtain from 4.8 mol H2. We will assume N2 is
in excess which means H2 is limited and will dictate the
amount of product that will be formed; once the H2 is gone,
the reaction will stop.
N 2 (g)  3H 2 (g)  2 NH 3 ( g )
We should realize that we have conversion factors that exist from the mol to mol
ratios available from the balanced equation to go from one species to another.
26
Mass Relationships in Chemical
Equations
How many grams of HCl are required to react with
5.00 grams MnO2 according to this equation?
4 HCl(aq)  MnO 2 (s )  2 H 2O(l)  MnCl 2 (aq)  Cl 2 ( g )
?g
5.00 g
Note: To get from one substance to another requires
the use of mol to mol ratios.
27
Mass Relationships in Chemical
Equations
How many grams of CO2 gas can be produced
from 1.00 kg Fe2O3?
Fe2O3 (s) + 3 CO (g)  2 Fe (s) + 3 CO2 (g)
1.00 kg
assume excess
?g
HW 24
code: stoich
28
Limiting Reagent
• The limiting reactant (or limiting reagent) is the reactant
that is entirely consumed when the reaction goes to
completion.
• The limiting reagent ultimately determines how much product
stops when one component ,
can be obtained.
R + R  P Reaction
limiting reagent, is exhausted
– For example, bicycles require one frame and two wheels. If you have
20 wheels but only 5 frames, how many bicycles can be made?
oo oo oo oo oo oo oo oo oo oo
20 wheels
10 wheels in excess; therefore, frames are limiting factor
1 frame + 2 wheels  1 bike
Smaller value is correct answer or you can
determine which is the limiting reagent and
only calculate that value.
Basically, we must find out which species will
run out first because it will dictate how much
29
product will form.
Limiting Reagent
If 0.30 mol Zn is added to hydrochloric acid containing
0.52 mol HCl, how many grams of H2 are produced?
Zn(s)  2 HCl(aq)  ZnCl 2 (aq)  H 2 ( g )
0.30 mol
0.52 mol
?g
First, we calculate how many mols of H2
is possible for each reactant.
limiting reagent HCl
We use the mols of H2 possible from the limiting
reagent, HCl, to determine the mass of H2.
The maximum product
possible is based on the
limiting reagent; smaller
mols of product is
maximum possible
based on the limiting
reagent which is HCl in
this case.
Note: The limiting reagent is not necessarily the species with the smaller mass or 30
#mols; you must account for the mol-to-mol ratio as demonstrated in this problem.
If 7.36 g Zn was heated with 6.45 g sulfur, what amount
of ZnS was produced?
8 Zn(s)  S 8  8 ZnS
7.36 g
mols produced by
each reactant:
6.45 g
?g
smaller mols; limiting reagent
This is known as
the theoretical
yield.
31
Theoretical and Percent Yield
The theoretical yield of product is the
maximum amount of product that can be
obtained from given amounts of reactants.
The percentage yield is the actual yield
(experimentally determined) expressed as a
percentage of the theoretical yield (calculated like in
previous example).
actual yield (exp)
%Yield 
100%
theoretical yield (calc) Once again,
% is a unit.
32
Theoretical and Percent Yield
To illustrate the calculation of percentage yield, recall that
the theoretical yield of ZnS in the previous example was
11.0 g ZnS.
If the actual yield experimentally for the reaction is 9.32 g
ZnS, what is the %yield?
experimental value
9.32 g ZnS
% Yield 
 100%  84.7%
11.0 g ZnS
% is unit, not button
theoretical value;
calculated value
on calculator
33
If 11.0 g CH3OH are mixed with 10.0 g CO, what is the theoretical
yield of HC2H3O2 in the following reaction? If the actual yield was
19.1 g, what is the %yield of HC2H3O2?
CH3OH (l) + CO (g)  HC2H3O2 (l)
11.0 g
10.0 g
mols produced by
each reactant:
?g
smaller mols; limiting reagent
theoretical yield
experimental value
theoretical value;
calculated value
HW 25
code: limiting
34