
is one where the substance retains its identity.


Examples of physical reactions:
◦
◦
◦
◦
◦
melting / freezing
boiling / condensing
subliming
subdivision
dissolving (solvation)

Stoichiometry


the atoms of one or more substances are rearranged
to produce new substances with new properties.
the atoms remain the same, but their arrangement
changes:


H2O  H2 + O2



water is broken down into two substances, oxygen
gas and hydrogen gas.
The two new substances bear no resemblance to the
water of which they are made.
Stoichiometry
a)
b)
c)
d)
a colour change
gas production
an energy change
precipitate formation
Stoichiometry


refers to balancing chemical equations and the
associated mathematics.
is possible because of the law of conservation
of mass.
Stoichiometry
Concise representations of chemical reactions
Stoichiometry
CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (g)
Stoichiometry
CH4 (g) + 2 O2 (g)
Reactants appear on the
left side of the equation.
CO2 (g) + 2 H2O (g)
Stoichiometry
CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (g)
Products appear on the right side of the equation.
Stoichiometry
CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (g)
The states of the reactants and products are written
in parentheses to the right of each compound. Stoichiometry
CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (g)
Coefficients are inserted to balance the equation.
Stoichiometry

Subscripts tell the number of atoms of each element
in a molecule
Stoichiometry

Coefficients tell the number of molecules
Stoichiometry

Matter cannot be lost in any chemical reaction.
◦ all the atoms must balance in the chemical
equation.
 same number and type on each side of the
equation
◦ can only change the stoichiometric coefficients
in front of chemical formulas.
◦ Subscripts in a formula are never changed
when balancing an equation.
Stoichiometry

Balance the following equation:
Na(s) + H2O(l)  H2 (g) + NaOH(aq)

count the atoms of each kind on both sides of the
arrow:
◦ The Na and O atoms are balanced (one Na and one O on each
side)
◦ there are two H atoms on the left and three H atoms on the
right.
Stoichiometry

Balance the following equation:
Na(s) + 2 H2O(l)  H2 (g) + NaOH(aq)



the 2 increases the number of H atoms among
the reactants.
O is now unbalanced.
have 2 H2O on the left; can balance H by putting
a coefficient 2 in front of NaOH on the right:
Stoichiometry

Balance the following equation:
Na(s) + 2 H2O(l)  H2 (g) + 2 NaOH(aq)



H is balanced
O is balanced
Na is now unbalanced, with one on the left but
two on the right.
◦ put a coefficient 2 in front of the reactant:
Stoichiometry

Balance the following equation:
2 Na(s) + 2 H2O(l)  H2 (g) + 2 NaOH(aq)

Check our work:
◦ 2 Na on each side
◦ 4 H on each side
◦ 2 O on each side

Balanced !!
Stoichiometry

CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g)
Stoichiometry

Pb(NO3)2 (aq) + 2 NaI (aq)
 PbI2 (aq) + 2 NaNO3 (aq)
Stoichiometry

Remember:
◦ when changing the coefficient of a compound, the
amounts of all the atoms are changed.
◦ do a recount of all the atoms every time you change a
coefficient.
◦ do a final check at the end.
◦ fractions are not allowed. Multiply to reach whole
numbers.
Stoichiometry

(s) (l) (g) (aq) -

(ppt) -



solid
liquid
gas
aqueous (meaning that the compound is
dissolved in water.)
precipitate (meaning that the reaction
produces a solid which falls out of solution.)


Reaction conditions occasionally appear above or
below the reaction arrow
◦ e.g., "Δ" is often used to indicate the addition of heat).
Stoichiometry

Complete 3.11 to 3.14
Stoichiometry


Two or more
substances
react to form
one product
Examples:
N2 (g) + 3 H2 (g)  2 NH3 (g)
C3H6 (g) + Br2 (l)  C3H6Br2 (l)
2 Mg (s) + O2 (g)  2 MgO (s) Stoichiometry
Stoichiometry


One substance breaks
down into two or more
substances
Examples:
CaCO3 (s)  CaO (s) + CO2 (g)
2 KClO3 (s)  2 KCl (s) + O2 (g)
2 NaN3 (s)  2 Na (s) + 3 N2 (g)
Stoichiometry



Rapid reactions that
produce a flame
Most often involve
hydrocarbons
reacting with
oxygen in the air
Examples:
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g)
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g)
Stoichiometry

Complete Problems 3.15 to 3.20
Stoichiometry


Sum of the atomic weights for the atoms in a
chemical formula
So, the formula weight of calcium chloride, CaCl2,
would be
Ca: 1(40.08 amu)
+ Cl: 2(35.45 amu)
110.98 amu

These are generally reported for ionic
compounds
Stoichiometry


Sum of the atomic weights of the atoms in a
molecule
For the molecule ethane, C2H6, the molecular
weight would be
C: 2(12.01 g/mol)
+ H: 6(1.01 g/mol)
30.08 g/mol
Stoichiometry




formula and molecular weights are often called
molar mass.
The formula weight (in amu) is numerically
equal to the molar mass (in g/mol).
Always report molar mass in g/mol.
Always round the atomic weights to 2 decimal
places.
Stoichiometry

Complete questions 3.21 and 3.22
Stoichiometry
Percentage composition is obtained by dividing
the mass contributed by each element (number
of atoms times AW) by the formula weight of
the compound and multiplying by 100.
(number of atoms)(atomic weight)
% element =
(molar mass of the compound)
x 100
Stoichiometry
So the percentage of carbon in ethane (C2H6)
is…
%C =
(2)(12.01 g/mol)
(2(12.01) + 6(1.01))g/mol
= 24.02 g/mol x 100
30.08 g/mol
= 79.85%
Stoichiometry
• the formula weight is 98.09 amu (do calculation)
• % H = 2(1.01 amu) x 100 = 2.06 % H
98.09 amu
• % S = 32.07 amu x 100 = 32.69 % S
98.09 amu
• % O = 4(16.00 amu) x 100 = 65.25 % O
98.09 amu
the sum of the percent compositions will always
equal 100 %
Stoichiometry


Complete questions 3.24 and 3.26
Calculate the % composition of each element in
each compound.
Stoichiometry


6.022 x 1023
1 mole of 12C has a
mass of 12 g
Stoichiometry

By definition, these are the mass of 1 mol of a
substance (i.e., g/mol)
◦ is the atomic weight on the periodic table
◦ The formula weight (in amu’s) will be the same number
as the molar mass (in g/mol)
Stoichiometry
Moles provide a bridge from the molecular
scale to the real-world scale
Stoichiometry

Using the mass and molar mass to calculate the
number of MOLES of a compound or element.
◦ m
◦ n
◦ M
Mass
amount of matter
Molar mass
number of
moles
n
=
=
grams
moles
grams per mole
g
mol
g/mol
mass
molar mass
m
M
Stoichiometry

If you have 20.2 grams of Potash, how many
moles do you have?
Molar Mass KCl (calculated):
74.55 g/mol
Mass KCl:
20.2 g

20.2 g
= 0.271 moles KCl
74.55 g/mol
if the units work out, you are correct.


Stoichiometry




How many moles are present in 0.750 g of
ammonium phosphate?
Ammonium phosphate:
(NH4)3PO4
Molar mass:
(3(14.01) + 12(1.01) + 30.98 + 4(16.00)) g/mol
= 149.13 g/mol
Number of moles = 0.750 g
= 5.03 x 10-3 mol
149.13 g/mol
(NH4)3PO4
Stoichiometry

What is the mass of 12.31 mol of NaOH ?
◦ Step 1. Calculate molar mass
 1 x 22.00 g/mol
 1 x 16.00 g/mol
 1 x 1.01 g/mol

40.00 g/mol
◦ Step 2. Calculate mass
 (12.31 mol)(40.00 g/mol) = 492.4 g NaOH
Stoichiometry




What is the mass of 2.1 x 10-4 mol of diarsenic
trioxide?
Diarsenic trioxide: As2O3
Molar mass: (2(74.92) + 3(16.00)) g/mol =
197.84 g/mol
Mass = (2.1 x 10-4 mol)(197.84 g/mol)
= 0.042 g As2O3
Stoichiometry
How many moles of caffeine (C8H10N4O2 ) is
required to reach the 12.0 g lethal dose in a 140
pound female?
Step 1. Molar Mass





Step


8
10
4
2
x 12.01 g/mol
x 1.01 g/mol
x 14.01 g/mol
x 16.00 g/mol
194.22 g/mol
2. Calculate Moles
12.00 g
194.22 g/mol
= 0.06179 mol caffeine
Stoichiometry




We can also find the number of atoms of a
substance using Avogadro’s Number:
6.022 x 1023 atoms/mol
Example:
Calculate the number of atoms in 0.500 moles of
silver?
0.500mol Ag x 6.022x1023 atoms = 3.01 x 1023 atoms
mol of Ag
Stoichiometry



For molecules!
Avogadro’s number represents the number of
particles, whether atoms or molecules.
Eg. Calculate the number of molecules in 4.99mol
of methane.
4.99 mol x 6.022 x 1023 molecules = 3.00 x 1024 molecules
mol of CH4
Stoichiometry


How many molecules are present in 0.045 mol of
H2O?
Number of atoms =
(0.045 mol)(6.022 x 1023 particles/mol)
= 2.7 x 1022 molecules H2O
Stoichiometry


How many moles are present in 8.75 x 1026
molecules of Vitamin C?
Moles = 8.75 x 1026 molecules = 1450 mol
6.022 x 1023 particles/mol
Vitamin C
Stoichiometry

What is the mass of 2.1 x 1022 molecules of N2O?

Step 1. Convert molecules to moles:
◦ Moles =

2.1 x 1022 molecules
6.022 x 1023 particles/mol
= 0.0349 mol N2O
Step 2. Convert moles to mass:
◦ Mass
= (0.0349 mol)((2(14.01) + 16.00) g/mol)
= 1.54 g N2O
Stoichiometry

How many molecules are present in 4.65 t of CO2?

Step 1: Convert mass to moles
◦ Moles =

(4.65 t)(1000 kg/t)(1000 g/kg) = 1.06 x 105
(12.01 + 2(16.00)) g/mol
mol CO2
Step 2: Convert moles to molecules
◦ Molecules = (1.06 x 105 mol)(6.022 x 1023 particles/mol)
= 6.36 x 1028 molecules CO2
Stoichiometry

Complete questions 3.32 to 3.42, even
(leave 3.32 for last)
Stoichiometry
One can calculate the empirical formula from
the percent composition
Stoichiometry
1. Assume we start with 100 g of sample.
2. The mass percent then translates as the number of
grams of each element in 100 g of sample.
3. From these masses, calculate the number of moles
(use atomic weight from Periodic Table)
4. The lowest number of moles becomes the divisor for
the others. (gives a mole ratio greater than 1)
5. Adjust mole ratios so all numbers are whole (1, 2, etc)
6. The lowest whole-number ratio of moles is the
empirical formula.
Stoichiometry
The compound para-aminobenzoic acid (you may have
seen it listed as PABA on your bottle of sunscreen) is
composed of carbon (61.31%), hydrogen (5.14%),
nitrogen (10.21%), and oxygen (23.33%). Find the
empirical formula of PABA.
Assumptions:
1. These are % by mass; in 100g of the compound,
the percentages represent masses in grams.
2. Mass divided by molar mass gives moles of each
atom.
Stoichiometry
Assuming 100.00 g of para-aminobenzoic acid, each
percentage converts into a mass:
C:
61.31 g x
H:
5.14 g x
N:
10.21 g x
O:
23.33 g x
1 mol
12.01 g
1 mol
1.01 g
1 mol
14.01 g
1 mol
16.00 g
= 5.105 mol C
= 5.09 mol H
= 0.7288 mol N
= 1.456 mol O
Stoichiometry
Calculate the mole ratio by dividing by the smallest number
of moles:
C:
5.105 mol
0.7288 mol
= 7.005  7
H:
5.09 mol
0.7288 mol
= 6.984  7
N:
0.7288 mol
0.7288 mol
= 1.000
O:
1.458 mol
0.7288 mol
= 2.001  2
Stoichiometry
These are the subscripts for the empirical formula:
C7H7NO2
Stoichiometry

Complete questions 3.43 to 3.46
Stoichiometry


Once we know the composition of a compound
the next step is to determine the molar mass.
From these two pieces of information the actual
molecular formula can be determined:
◦ take the ratio of the actual molar mass divided by the
molar mass determined by the empirical formula
◦ multiply the atoms in the empirical formula by the results
of the ratio.
Stoichiometry

Complete questions 3.47 to 3.50
Stoichiometry

Compounds containing C, H and O are routinely
analyzed through combustion in a chamber like this
◦ C is determined from the mass of CO2 produced
◦ H is determined from the mass of H2O produced
◦ O is determined by difference after the C and H have been
determined
Stoichiometry
Compounds
containing other
elements are
analyzed using
methods analogous
to those used for C,
H and O
Stoichiometry

Complete questions 3.51 & 3.52
Stoichiometry
The coefficients in the balanced equation give
the ratio of moles of reactants and products
Stoichiometry
From the mass of
Substance A you can
use the ratio of the
coefficients of A and
B to calculate the
mass of Substance B
formed (if it’s a
product) or used (if
it’s a reactant)
Stoichiometry
C6H12O6
+
6 O2

6 CO2
+
6 H2O
Stoichiometry


Work on questions 3.57 to 3.66
Copy examples done in class.
Stoichiometry


You can make cookies
until you run out of
one of the ingredients
Once this family runs
out of sugar, they will
stop making cookies
(at least any cookies
you would want to eat)
Stoichiometry

In this example the
sugar would be the
limiting reactant,
because it will limit the
amount of cookies you
can make
Stoichiometry
The limiting reactant
is the reactant
present in the
smallest
stoichiometric
amount
Stoichiometry

The limiting reactant is the reactant you’ll run out
of first (in this case, the H2)
Stoichiometry
In the example below, the O2 would be the
excess reagent
Stoichiometry

Complete questions 3.69 to 3.76
Stoichiometry

The theoretical yield is the amount of product that
can be made
◦ In other words it’s the amount of product possible as
calculated through the stoichiometry problem

This is different from the actual yield, the amount
one actually produces and measures
Stoichiometry
A comparison of the amount actually obtained
to the amount it was possible to make
Actual Yield
Percent Yield =
Theoretical Yield
x 100
Stoichiometry

Complete questions 3.77 to 3.80
Stoichiometry