Chemical Composition
Chemical Quantities
Chapters 9&10
Pg. 221-293
1
Chemical Composition
• Atomic Masses-
– Because the mass of one atom is so small- on
the order of 10-23 gram- a special unit – the
atomic mass unit- is used to described the
mass of an atom.
– A system has been devised for expressing the
masses of atoms on a relative scale- that is, a
scale based on experimental comparison with a
standard. The scale for expressing atomic
masses is based on the mass of carbon- 12. 1
carbon 12 atom weighs 12 amu.
– Masses on the periodic table are weighted
averages of all the naturally occurring isotopes
for an element.
2
Atomic Mass = the mass of an atom expressed
relative to the mass of carbon –12
– Na =
– K=
– Cl =
• Notice that we will be rounding to the 1/10th place for all
calculations.
Formula Mass / Molecular Mass = the sum of
the atomic masses for all the atoms in a
compound.
H2O
H: 2 atoms X 1.0 amu = 2.0 amu
O: 1 atom X 16.0 amu = 16.0 amu
= 18.0 amu
HC2H3O2 = ?
3
The Mole- Three Faces
1. Avogadro’s Number
One mole refers to Avogadro’s
number ___________________
• 1 mole of silver =
• 1 mole of CO2 =
Just like a dozen represents 12 of
anything, and a gross represents 144
of anything.
4
5
2. Molar Mass
One mole of particles has a special
mass associated with it: the formula
mass or molecular mass of the
substance expressed in grams.
Substance
Formula /
Molar Mass
Molecular Mass
Calcium (Ca)
40.1 amu
1 mole = 40.1 grams
40.1 g / mol
Hydrogen
Perioxide
(H2O2)
34.0 amu
1 mole = 34.0 grams
34.0 g / mol
6
Question: Find the molar mass of the
following substances:
• Hydrogen gas
• Carbon Monoxide
• Glucose (C6H12O6)
7
3. Molar Volume of an Ideal Gas
Suppose the mole of particles happens to be
a gas. In this case, the term mole has a
special volume associated with it. If the
mole of gas particles behaves ideally, the
gas will occupy a volume of approximately
22.4 L at STP.
• 1 mole of He gas occupies 22.4 L
• 1 mole of hydrogen gas (H2) occupies 22.4 L
• 1 mole of carbon dioxide gas (CO2) occupies
22.4 L
8
Mole Conversions
• Because the mole measures mass, number
of particles, and volume of a gas, it is the
central unit in converting the amount of a
substance from one type of measurement
to another.
• These conversions will require use of the
Factor Label Method we learned at the
start of the year.
9
FLM
• How many apples are there in 3.75
dozen?
Unit you want
3.75 dozen X 12 apples = 45 apples
1 dozen
Unit you are trying to
get rid of
10
Mass and Moles
• If you know the mass of a given
amount of substance, you can
calculate the number of moles of the
substance.
• Your fact you will use in this
conversion is the molar mass of the
substance.
11
• Example: Suppose a laboratory
procedure calls for 0.65 moles of
NaCl to be added to a solution.
• Question: Suppose you produced 45.0
g of magnesium sulfate in the lab.
How many moles of magnesium sulfate
does this represent?
12
Particles and Moles
• This conversion requires only one fact
that does not change from atom to atom
or compound to compound:
Avagadro’s Number = 6.02 X 1023 “stuff” = 1 mole of “stuff”
Stuff could be atoms, molecules, particles,
electron, protons, ions, ….anything!
13
• Example: Determine the number of
molecules of water in 2.55 moles of
water.
• Question: How many moles of C6H12O6
contain 7.9 X 1024 molecules?
14
Multi- Step Conversions
• Using a mole map and the FLM, link your
conversion facts together and solve.
• Question: You need 250.g of table sugar
or sucrose (C12H22O11) to bake a cake. How
many sucrose molecules will be in the cake?
Note: There is no direct conversion
between grams and molecules. You must
first convert grams to moles, and then
moles to molecules.
15
• Question: A student fills a 1.0 L flask
with carbon dioxide at standard
temperature and pressure. How many
molecules of gas are in the flask?
16
Percent Composition by Mass
Percent = part per hundred
• Percentages are calculated by dividing
the quantity under consideration by
the total quantity involved and
multiplying the result by 100. For
example, if 7 persons in 20 wear
glasses, then the percentage of
persons wearing glasses is
17
• Similarly, we can calculate the
percentage composition of an element
in a compound by dividing the mass of
the element by the formula mass of
the substance and multiplying the
result by 100.
• Question: Calculate the percentage
of oxygen by mass in CuSO4.
18
Empirical Formula from Percent Composition
Empirical Formula =
• Example: The empirical formula for N2O4
is NO2, and the empirical formula for C6H6
is CH, but the empirical formula for H2O is
still H2O.
• Since we calculate percent composition
from chemical formulas, we are able to
reverse the process and determine
formulas from percent composition.
However, when we use this process, we are
able to determine only the empirical
formula.
19
• Question: Calculate the empirical
formula of a substance that contains
50.00 % sulfur and 50.00% oxygen by
mass.
• Solution: Assume we have 100 grams
of the substance. If so, then based
on the percentages, we would have
50.00g of S and 50.00g of O. If we
divide the given masses by the atomic
masses of the two elements, we
transform the mass into moles.
20
• As strange as the previous formula
looks, it is accurate because it shows
the relative number of moles of each
element. Our task is to make the
formula look like a formula, that is, to
convert it to one with whole numbers.
Usually this is accomplished by
dividing each element by the smallest
number of moles.
21
Determining Molecular Formula
• The empirical formula for a compound
indicates the simplest ratio of the
atoms in the compound. However, it
does not tell the actual numbers of
atoms in each molecule of the
compound.
• Molecular Formula =
22
• Question: The empirical formula for
glucose is CH2O. Its empirical
formula mass is 30.0 g / mol.
Experiments show that the molar
mass of glucose is 180 g / mol. What
is the molecular formula of glucose?
23
Mole Problems Involving
Chemical Equations
Consider the following equation:
N2 + 3H2  2NH3
How do we interpret the coefficients?
• Molecules OR
• Moles of molecules
1 mole of N2 reacts with 3 moles of H2
to form 2 moles of NH3
24
We can construct a mole diagram for solving
mole problems involving equations just as
we did for single substances. Note that
the diagram represents simply the joining
of two single mole maps. It does not
matter whether A and B are both
reactants, products, or one of each of a
given chemical reaction. The important
point is that (moles of ) A and (moles of ) B
are connected by the coefficients the two
substances have in the BALANCED
chemical reaction.
25
Mass
of A
Gas volume
of A at STP
Gas volume
of B at STP
Moles
of A
Moles
of B
Number of
Particles
of A
Mass
of B
Number of
Particles
of B
26
All equation problems are solved using the steps
listed below:
1. Examine the chemical equation and make sure
that it is balanced.
2. Refer to the preceding diagram and prepare a
solution map as follows:
Quantity of A  moles of A  moles of B  Quantity of B
given in problem from rxn
from rxn
calculate
3. Set up the problem, using FLM and the solution
map
4. Perform the calculations on the numbers and
units.
27
Mole Problems
• Question: In the equation
N2 + H2  NH3,
how many moles of N2 are needed to
produce 5.0 moles of NH3?
Tip- BALANCE EQUATION FIRST!
28
Mass – Mass Problems
• Question: In the equation
CH4 + O2  CO2 + H2O
how many grams of CO2 are formed
when 8.0 grams of CH4 reacts?
Tip- BALANCE EQUATION FIRST
29
Mass-Volume Problems
• Question: In the equation
CO (g) + O2 (g)  CO2 (g)
how many liters of CO2 (g) at STP is
produced by the reaction of 64.0
grams of O2 (g)?
Tip- BALANCE EQUATION FIRST
30
Volume – Volume Problems
• Question: In the equation
NH3 (g) + O2 (g)  NO (g) + H2O
(l)
how many liters of NH3 (g) at STP are
needed to react with 200. liters of
O2 (g) at STP?
Tip- BALANCE EQUATION FIRST
31
Problems Involving Numbers
of Particles
• Question: In the equation
C2H6 + O2  CO2 + H2O
How many molecules of C2H6 are
needed to produce 27 grams of H2O?
Tip- BALANCE EQUATION FIRST
32
Limiting Reactants and
Percent Yield
Consider the recipe for tollhouse
cookies. The number of batches of
cookies you can bake depends on the
amount of each of the starting
materials you have.
33
Original Nestlé Toll House Chocolate
Chip Cookies
Ingredients:
•2 1/4 cups all-purpose flour
•1 teaspoon baking soda
•1 teaspoon salt
•1 cup (2 sticks) butter or margarine, softened
•3/4 cup granulated sugar
•3/4 cup packed brown sugar
•1 teaspoon vanilla extract
•2 large eggs
•2 cups (12-ounce package) NESTLÉ TOLL HOUSE
Semi-Sweet Chocolate Morsels
•1 cup chopped nuts
34
If you are in short supply of any of the
ingredients, then the recipe must be
modified in order to make the
cookies.
For example, say you only had 1 1/4
cups of semi sweet chocolate chips.
– How would that affect the rest of your
recipe?
– How many cups of flour would you wind
up using?
35
When the quantities of reactants are
available in the exact ratio described by
the balanced equation, the chemists say
that the reactants are in stoichiometric
proportions. When this is the case, all the
reactants will take part in the reaction and
there will be no reactants left over one the
reaction is complete.
More often than not, one of the reactants is
available in limited quantities. The
reactants are then said to be in nonstoichiometric proportions.
36
Limiting Reactant
• The reactant that limit the amount of
product formed in a chemical
reaction. It is completely used up in
the chemical reaction.
37
To determine the limiting reactant in an
equation, you must complete two separate
mass- mass problems.
• First, write the balanced equation for the
reaction.
• Then, calculate the mass of one of the
reactants needed based on the other
reactant.
• Compare the calculated needed value to
what you actually are given.
• If the calculated needed value is more than
what is given, the substance is the limiting
reactant. If not, the other reactant will
by default be the limiting reactant.
38
• Question: Identify the limiting
reactant when 10.0 g of water reacts
with 4.5 g of sodium to produce
sodium hydroxide and hydrogen gas.
39
• Question: Identify the limiting
reactant when 12.5 L H2S at STP is
bubbled through a solution containing
24.0 g of KOH to from K2S and H2O.
40
Once the limiting reactant is identified,
it is easy to calculate the theoretical
mass or volume of the products.
Remember, the limiting reactant will
be used up completely in the chemical
reaction. The other reactant will be
in excess meaning there will be
reactant left over.
41
• Question: If 3.5 g of Zn and 3.5 g of
S are mixed together and heated,
what mass of ZnS will be produced?
42
• Question: What mass of barium
nitride (Ba3N2) is produced from the
reaction between 22.6 g barium and
4.2 g nitrogen gas?
43
Percent Yield
• Expected yield- the amount of a product
that should be produced based on
calculations.
• Actual yield- the amount of product that is
really obtained from a chemical reaction.
Why would the expected yield be different
than the actual yield?
– Loss of product in the chemical reaction
– Error in measuring the starting materials
– Side reactions take place between the
reactants leading to a different product than
was expected.
– Reaction did not occur because of temperature,
pressure, etc. was not ideal for the reaction. 44
Percent yield = percent of the
expected yield that was obtained.
Percent yield = Actual yield X 100
Expected yield
• Question: Determine the percent
yield for the reaction between 2.80 g
Al(NO3)3 and excess NaOH if 0.966 g
Al(OH)3 is recovered.
45
• Question: Determine the percent
yield for the reaction between 15.0 g
N2 and 15.0 g H2 if 10.5 g NH3 is
produced.
46