Stoichiometry
Chapters 2, 3, 4
Significant Figures - A.K.A. Sig. Fig.’s
•
•
•
•
•
In almost every experiment, scientists measure data.
These measured numbers help determine many things in
life - as such, we need a way to recognize the error in
each measurement.
We do this by using significant figures.
Numerically 3.0, 3.00, 3.000 are of the same value, but
3.000 shows that it was measured with the more precise
instrument.
The zeroes in all three numbers are considered
"significant figures".
They are shown to indicate the precision of the
measurements.
If we take away the zeroes, the value does not change.
Types of Numbers
Exact Numbers
Approximate Numbers
• numbers obtained
through counting.
• numbers obtained
through measurement.
• These stand as is.
• With these numbers we
must determine how
precise they are when
we use them in
calculations.
Determining Sig. Figs
•
All non-zero integers are significant.
Ex. 256.5
4 sig. figs
• Captive zeros are always significant.
Ex. 206
3 sig. figs
• Leading zeros are NEVER significant.
Ex. 0.0056
2 sig. figs
• Trailing zeros in a decimal are significant.
Ex. 25.00
4 sig. figs
• Assumed decimals are not significant.
Ex. 500
1 sig. figs
Rules For Calculations:
Addition and Subtraction
Multiplication and Division
• Do the math.
• Do the math.
• The answer should contain
the same number of Sig.
Figs as the number with
the least number of Sig.
Figs
• The answer should contain
the same number of
decimal places as the
number with the least
decimal places.
Ex. 25.46 + 12.5 + 82.91 =
Ex. 0.14 x 98.54 x 12.5 =
120.87
172.445
only one dec. place
 120.9
only 2 sig. fig
 170
Chapter 2.1
Isotopes and
Average Atomic Mass
Isotopes
Recall that the mass number tells the total number of
protons and neutrons.
• All neutral atoms have the same number of protons and
electrons, however, the number of neutrons may vary.
• Isotopes - atoms of an element that have the same
number of protons but different number of neutrons.
atoms of the same element do not necessarily have
the same mass number.
Ex.
Oxygen naturally occurs in 3 different forms:
Carbon – exists as 3 isotopes...
carbon-12 carbon-13 carbon-14
6 p+, 6n
6p+, 7n
6p+, 8n
The Relative Mass of An Atom
• The mass of an atom is expressed in atomic
mass units, u.
• Atomic masses are relative - all are based on
the mass of carbon-12, 12 u.
 the mass of all other elements are defined
in comparison to carbon-12.
Isotopic Abundance
• An element may have several naturally occurring
isotopes, but not all of its isotopes will exist in the
same amount.
Ex. Magnesium exists in 3 isotopes –
Mg-24, Mg-25, and Mg-26
• All sources of magnesium occur in the following
amounts:
Mg-24 79%, Mg-25 10%, Mg-26 11%
• Isotopic Abundance - the relative amount in which
each isotope is present in an element.
• Determined with a sophisticated machine called a
mass spectrometer. It uses a magnetic field to
separate the different masses in a sample.
Complete #1-4 Page 45
Average Atomic Mass
And the Periodic Table
• Average atomic mass the average of all the
masses of all the
element’s isotopes –
accounting for the
abundance of all the
isotopes.
• the average atomic mass
is the mass given in the
Periodic Table.
To Calculate the Average
Atomic Mass
• 1. Multiply the mass of
each isotope by the
relative abundance (you
must change the
abundance to a decimal).
• 2. Add all these totals
together.
Average Atomic Mass = 106.9 u (.518) + 108.9 u(.482)
= 107.9 u
Note: This is the mass number given on the Periodic
Table. No atom of silver actually has a mass of
107.9 u - it is just an average.
The Mole
• chemists need a unit of measure that works
for many things – they use the mole.
Mole
• A quantity, an amount.
• symbol – mol
▫
▫
▫
▫
1 mol element= 6.02 x 1023 atoms
1 mol compound = 6.02 x 1023 molecules
1 mol gas = 22.4 L @STP
1 mol = mass of substance
2.3 Molar Mass
• the mass of one mole of a substance.
• expressed in g/mol.
Molar Mass of an Element
• a mass expressed in grams that is
numerically equivalent to the element’s
average atomic mass.
• 1 mol Na = 22.99 g
• 1 mol Ar = 39.95 g
Molar Mass Of Compounds
• the mass of one mole of a compound is equal to
the molar mass of all the elements it contains.
• 1 mole MgO = mass Mg + mass of O
= 24.31 g + 16.00 g
= 40.31 g
• 1 mole Ca3(PO4)2
= 3(Ca) + 2(P) + 8(O)
= 3(40.08) + 2(30.97) + 8(16.00)
= 310.18 g
• 1 mole Ca3(PO4)2 = 310.18 g
or molar mass of Ca3(PO4)2 = 310.18 g/mol
Conversion Flashback
• When 2 things are equal – we can set up
conversion factors...
1 hour = 60 minutes
1hour
60 min
or
60 min
1hour
How many minutes are in 8.49 hours?
8.49 hours x 1hour 60 min = 509 min
60 min
1hour
Converting Mass to Moles
• When given the mass of a compound, we can convert
to moles using the molar mass.
Steps:
1. Determine molar mass.
2. Start with what is given.
3. Set up factors so they cancel.
How many moles are in 43.38 grams of NaOH?
Molar Mass NaOH = 40.00 g/mol
43.38 g NaOH x
1molNaOH
40.00 g
= 1.085 mol
Converting Moles to Mass
• When given the amount of moles of a compound, we
can convert to mass using the molar mass.
Steps:
1. Determine molar mass.
2. Start with what is given.
3. Set up factors so they cancel.
How many grams are in 5.48 moles of NaOH?
Molar Mass NaOH = 40.00 g/mol
5.48 mol NaOH x
40.00 g
1molNaOH
= 219 g
Converting Moles To Particles
• 1 mole contains 6.02 x 1023 particles
When talking about particles we must look at the substance to
see what the particles are.
An Element
– smallest particle is the atom.
1 mole of an element = 6.02 x 1023 atoms
A Compound
• smallest particle for a covalent compound is a molecule.
H2O - one molecule of H2O
- also 2 atoms of H and 1 atom of O
- or 3 atoms
• smallest particle for an ionic compound is the formula unit.
NH4Cl – one form. unit of NH4Cl
- also one ion of NH4+ and one ion of Cl- also 1 N atom, 4 H atoms, and 1 Cl atom
- or 6 atoms
Conversion Factors for Elements
•
if 1 mole = 6.02 x 1023 atoms then
1mole
6.02 x1023 atoms
OR
6.02 x1023 atoms
1mole
How many moles are present in a sample of carbon that is
made up of 5.82 x 1024 atoms of C.
5.82 x 1024 atoms x
1moleC
6.02 x1023 atoms
= 9.67 mol of C
Conversions for Compounds
Covalent Compounds
• 1 mole = 6.02x1023 molecules
1mole
6.02 x1023 molecules
OR
6.02 x1023 molecules
1mole
Then look at one molecule and
Example: C3H8
count:
1 mol C3H8 = 6.02x1023 molecules
• how many atoms are in one
molecule
1 molecule C3H8 = 11 atoms
• how many atoms of a certain 1 molecule C H = 3 carbon atoms
3 8
type.
1 molecule C3H8 = 8 hydrogen atoms
Conversions for Compounds
Ionic Compounds
• 1 mole = 6.02x1023 formula units
1mole
6.02 x10 23 F .U .
OR
6.02 x1023 F .U .
1mole
Then look at one formula unit and count:
•
•
•
•
how many ions are in one molecule
how many ions of a certain type
how many atoms are in one molecule
how many atoms of a certain type.
Example: Na2SO4
1 mol Na2SO4 = 6.02x1023 F.U.
1 F.U. Na2SO4 = 3 ions
1 F.U. Na2SO4 = 2 Na+ ions
1 F.U. Na2SO4 = 1 SO42- ions
1 F.U. Na2SO4 = 2 Na atoms
1 F.U. Na2SO4 = 1 S atom
1 F.U. Na2SO4 = 4 O atoms
The Mole Road Map
1 mole =
6.02 x 1023
molecules
For Compounds
grams
moles
1 mole =
what you
count
Molecules or
Formula Units
For Elements
1 mol=6.02x1023 atoms
STP 1 mol=22.4L
SATP 1 mol=24.4L
Volume
Atoms or
Ions
3.1 Percent Composition
• The Law of Definite Proportions:
Elements in a chemical compound are always present in
the same proportions by mass.
• the mass of an element in a compound, expressed as a
percent of the total mass of a compound, is the element’s
mass percent.
Whether water, H2O, comes from a lake, a bottle, or a
fountain, it will contain 11.2 % hydrogen and 88.8% oxygen.
Mass Percent =
Mass % Oxygen =
massofelement x 100%
massofcompound
16.00 g x 100%  88.8%
18.02 g
Percent Composition
• a statement of all the mass percents of
every element in a compound.
• Vanillin is C8H8O3
▫ 63.1% C, 5.3% H, 31.6% O
Page 81
Practice Problems #1-4 Page 82
Percent Composition From a Formula
• When you are not given masses in a problem, you can still
determine the percent composition.
• Assume you have one mole of the sample
• Use molar masses and masses from PT to get percents.
Page 84
Practice Problems #5-8 Page 85
Empirical Formula of a Compound
• The empirical formula (also simplest formula) of a
compound shows the lowest whole number ratio of the
elements in a compound.
• The molecular formula (also the actual formula)
shows the number of atoms of each element in a
molecule.
• Many compounds can have molecular formulas that are the
same as their empirical formulas.
Ex.
NH3, H2O, CO2 etc.
• Note: Ionic compounds do not have molecular formulas
because they do not exist as molecules. Because of the
ions involved in the compound, there is only one way that
they can combine - therefore the chemical formula is the
empirical formula.
Write the Empirical Formula for Each of the Following:
a. P4O6
b. C6H9
c. CH2OHCH2OH
d. BrCl2
e. C6H8O6
f. C10H22
g. Cu2C2O4
h. Hg2F2
Determining the Empirical Formula
“Percent to mass,
mass to mole,
divide by small,
multiply until whole”
Page 88
Page 90
Practice Problems #9-12 Page 89
Practice Problems #13-16 Page 91
1. A compound composed of: 72% iron (Fe) and 27.6% oxygen (O) by
mass.
Fe3O4
2. A compound composed of: 9.93% carbon (C), 58.6% chlorine (Cl),
and 31.4% fluorine (F).
CCl2F2
3. A compound composed of: .556g carbon (C) and .0933g hydrogen
(H).
CH2
3.3 Molecular Formula
• the molecular formula is a whole number ratio of the
empirical formula
Name
Formaldehyde
Molecular
Formula
Multiple
Information
CH2O
1
disinfectant
Acetic Acid
C2H4O2
2
Vinegar
Lactic Acid
C3H6O3
3
in muscles during exercise
Erythrose
C4H8O4
4
formed from metabolizing sugar
Ribose
C5H10O5
5
in nucleic acid and Vit.B
Glucose
C6H12O6
6
sugar
Determining Molecular Formula
1. Determine the empirical formula.
2. molar mass of molecular formula = ratio
molar mass of empirical formula
3. Multiply the ratio by the empirical formula.
Page 97
1. NutraSweet is 57.14% C, 6.16% H, 9.52% N, and 27.18%
O. Calculate the empirical formula of NutraSweet and
find the molecular formula. (The molar mass of
NutraSweet is 294.30 g/mol)
C14H18N2O5
2. An oxide of nitrogen, NxOy, contains 30.43% N. Its
molecular formula is determined to be 138 g/mol. What
is the molecular formula?
N3O6
2. A butane is a common fuel for lighters. When analyzed
the compound if found to contain 82.63% carbon and
17.37% hydrogen. The molar mass of butane is 58.14
g/mol. What is the molecular formula?
C4H10
Practice Problems # 17-20 Page 97
Hydrated Ionic Compounds
Many ionic compounds crystallize from a water solution
with water molecules incorporated into their crystal
structure, forming a hydrate.
• Hydrate - a compound with a specific number of water
molecules chemically bonded to each formula unit.
Ex. MgSO4 7H2O
• every formula unit of MgSO4 has 7 water molecules
chemically bonded to it.
• a raised dot in a chemical formula denotes a hydrated
compound.
▫ It represents a weak bond between the ionic compound
and the water molecules.
When a hydrate is heated – we destroy the weak
bonds between the ionic compound and evaporate the
water molecules. This will form an anhydrous compound.
Anhydrous Compound or Anhydrate
• a compound that has no water molecules incorporated
into their chemical formula.
CaSO4 - anhydrous calcium sulfate
CaSO4 2H2O - calcium sulfate dihydrate
Calculations Involving Hydrates
“Mass to Moles,
Divide by small.”
• Sample Problem - page 102
•
A hydrate of barium hydroxide, Ba(OH)2
∙xH2O, is used to make barium salts and to
prepare certain organic compounds. Since it
reacts with CO2 from the air to yield BaCO3,
it must be tightly stored.
•
A) A 50.0 g sample of the hydrate
contains 27.2 g of Ba(OH)2. Calculate the
percent, by mass, of water.
•
B) Find the formula of the hydrate.
• Percent By Mass
•
Ba(OH)2 = 27.2 g
•
water = 50.0g - 27.2 g = 22.8 g
• % water = 22.8 g water
x 100%
•
50.0 g compound
•
= 45.6 %
• Formula
•
27.2 g Ba(OH)2 x 1 mole Ba(OH)2 = 0.159 mol Ba(OH)2 =
1
•
171.3 g Ba(OH)2
0.159
•
22.8 g H2O x 1 mole H2O
= 1.27 mol H2O = 8
•
18.02 g H2O
0.159
• Therefore: Ba(OH)2 ∙8H2O
Stoichiometry Continued...
• Balanced chemical equations are EXTREMELY important in
chemistry.
• The coefficients in front of the formulas and symbols for
the compounds and elements tell you the ratio of particles
involved in a chemical reaction.
• The coefficients also tell us the ratio of the number of
molecules in the equation.
• The coefficients tell us the number of moles involved in
the chemical equation.
the mole ratio or the mole-mole relationship.
Page 118
Mole-Mole Conversions
•
when starting with one compound and making calculations
about another compound you need to use the mole-mole
relationship in conversion factors...
Common Everyday Example:
2 eggs + 75 mL oil + 125 mL water → 24 muffins
If you had 14 eggs, how many muffins could you make?
14 eggs x 24 muffins
2 eggs
= 168 muffins
Consider the following chemical rxn:
How many moles of ammonia, NH3, is formed from 13.3
moles of nitrogen gas reacts with excess hydrogen?
13.3 mol N2 x 2 mol NH3
1 mol N2
= 26.6 mol NH3
How many moles of hydrogen are needed to react
completely with 0.589 moles of nitrogen gas?
0.589 mol N2 x 3 mol H2 = 1.77 mol H2
1 mol N2
Practice Problems #4-6 Page 115
Page 116
Practice Problems # 8-10 Page 117
Mass To Mass Conversions
• We can also convert from the mass of one compound to
the mass of another compound through mole-mole
relationships...
Consider the following reaction:
1 Co(NO3)2 + 2 NaOH → 2 NaNO3 + 1 Co(OH)2
How many grams of Co(OH)2 will be produced when 25.47 g
of NaOH reacts with excess Co(OH)2?
25.47 g NaOH x 1 mol NaOH x 1 mol Co(OH)2 x 92.95 g Co(OH)2
40.00 g NaOH
2 mol NaOH
1 mol Co(OH)2
= 29.59 g Co(OH)2
Mixed Stoichiometry
The Limiting Reagent
•
Theoretical Yield & Percent Yield
• Stoichiometry can be used to predict the amount of
product that could be expected if an experiment went
perfectly to completion.
theoretical yield
competing reactions
Reasons why we do not
ever achieve the
theoretical yield
faulty or poor equipment
impure reactants
poor technique
•
When a chemists actually does an experiment, they can
measure the amount of product that they actually
produced.
This is the Actual Yield.
• The Actual Yield is usually less than the theoretical yield.
% Yield =
Actual Yield
Theoretical Yield
x 100%
Ex. When 2.3 g of NaOH reacts with excess Co(NO3)2 a
precipitate is formed. A chemists completes this reaction and
produces 1.74 g of Co(OH)2. Calculate the percent yield.
2.3 g NaOH x 1 mole NaOH x 1 mol Co(OH)2 x 92.95 g Co(OH)2 = 2.67 g
40.0 g NaOH
2 mol NaOH
1 mol Co(OH)2
% Yield =
actual yield
= 1.74 g x 100% = 65.2 %
theoretical yield
2.67 g