Chapter 10
“Chemical Quantities”
Yes, you will need a
calculator for this chapter!
Section 10.1
Da Mole: A
Measurement of
Matter
How do we measure items?
 Measure mass in
 Measure volume in
 Measure amount in
.
.
.
What is the mole?
We’re not talking about this
kind of mole!
Moles (mol)
 It is
.
 Defined as the number of carbon atoms in
exactly 12 grams of carbon-12.

of representative particles.
 6.02 x 1023 is called:
number.
 1 mole = 6.02 x 1023 = Avogadro’s number
Similar Words for amount
mol muffins =
muffins
1 dozen muffins =
 2 mol puppies =
muffins
puppies
2 dozen puppies =
 2.4 x 1024 m&m stuffed
puppies
donuts =
mol
36 m&m stuffed donuts =
donuts
dozen donuts
Dozen:



1
What are Representative Particles?

The smallest pieces of a substance:

based on what we’re looking at
1) For a molecular compound: it is the
.
2) For an ionic compound: it is the formula unit (ex:
NaCl, MgS).
3) For an element: it is the
.
• 1 mol of CO2, 1 mol of NaCl, and 1 mol of H
equal 6.02 x 1023 of that thing.
Measuring Moles
 The
mass on the PT is
also the
mass (g/mol)
– mass (in grams) of 1 mole of that atom

=
Find the Molar Mass (g/mol)
of the following:
• Nitrogen
• Aluminum
• Zinc
What about compound mass?
In 1 mole of H2O, there are
moles of H atoms and
mole of O atoms
 To find the mass of a compound:
– Determine
of
each element present
–
the number times
their mass (from the periodic table)
–
them up for the
Calculating Compound Mass
Calculate the mass of magnesium
carbonate, MgCO3.
Practice Problem:
• What is the mass of one mole of CH4?
1 mole of C = 12.01 g/mol
4 mole of H x 1.01 g = 4.04g/mol
1 mole CH4 =
10.1 Review
1. 1 mol =
particles
2. 2.5 mol MgCl2 = ? particles MgCl2
=
3. 7.2 x 1024 Fe particles = ? mol Fe
=
4. Molar mass of BeF2 =
Be = 9.0 g/mol
F = 19.0 g/mol
5. Molar mass of C6H12O6 =
C = 12.0 g/mol
H = 1.0 g/mol
O= 16.0 g/mol
Section 10.2
Mole-Mass and
Mole-Volume
Relationships
Since Molar Mass is…
 Number of
in 1 mole.
 grams per mole (g/mol)
 Use to make
_
factors from these.
- Use molar mass to convert to
or
of a substance.
For example
• How many moles is 5.69 g of NaOH?
=
• How many grams in .53 mol of NaOH?
=
Practice Problems:
 How much would 2.34 moles of carbon weigh?
 How many moles of magnesium is 24.31 g of Mg?
The Mole-Volume Relationship
 Under different circumstances,
can
.
 Two things effect the volume of a gas:
a) Temperature and b)
**We need to compare all gases at the same
temperature and pressure.
Standard Temperature and Pressure
• abbreviated
• (273K) and
pressure
• At STP 1 mole of gas occupies
22.4 L= molar volume
•
of any gas at STP
Practice Problems:
 What is the volume of 4.59 mole of CO2 gas at STP?
 How many moles is 5.67 L of O2 at STP?
• What is the volume of 8.8 g of CH4 gas at STP?
=
Summary:
• These four items are all
:
a) 1 mole
b) molar mass (in grams/mol)
c)
particles
(atoms, molecules, or formula units)
d)
of a gas at
**Thus, we can make
factors from them.
Avog. # Practice problems:
 How many molecules of CO2 are in 4.6 moles of CO2?
 How many moles of water is in 5.87 x 1022 molecules?
 How many atoms of carbon are in 1.230 moles of Carbon?
 How many moles is 7.78 x 1024 formula units of MgCl2?
Mixed Practice Problems:
 How many atoms of lithium is 1.0 g of Li?
 How much would 3.45 x 1022 atoms of U weigh?
 What is the volume of 10.0 g of CH4 gas at STP?
Section 10.3
Percent
Composition and
Chemical Formulas
Percentage Composition
• Percentage by
of
element in a compound
mass of element
% composition 
 100
total mass compound
Calculating Percent Composition
of a Compound
 Like all percent problems:
part
x 100 % =
_
whole
1) Find mass of
(from the periodic table)
2) Next, divide by total mass of
; then
_
Percentage Composition
• Find the % composition of Cu2S.
%Cu =
%S =
Percentage Composition
• Find the mass percentage of water in
calcium chloride dihydrate,
CaCl2•2H2O?
%H2O =
Empirical Formula (EF)
•
whole number
of atoms in a compound
C2H6
simplify subscripts
Empirical Formula (EF)
 Just find
whole number ratio
C6H12O6
CH4N = already lowest ratio.
 Formula is not just ratio of atoms,
it is also ratio of
.
 In 1 mole of CO2 there is
mole of
carbon and
moles of oxygen.
 In one molecule of CO2 there is 1 atom of C
and 2 atoms of O.
Formulas (continued)
Formulas for
compounds are
ALWAYS empirical
(the lowest whole number ratio =
be reduced).
Examples:
NaCl
MgCl2
Al2(SO4)3
*Remember, we simplify beforehand
K2CO3
Formulas (continued)
Formulas for molecular compounds
be empirical
(lowest whole number ratio).
Molecular:
(Correct formula)
Empirical:
(Lowest whole
number ratio)
H2O
C6H12O6
C12H22O11
Formulas
• Empirical Formula (EF) =
lowest whole number ratio
of elements in a compound.
• Molecular Formula (MF) =
the actual ratio
of elements in a compound.
• The two can be the same.
• CH2 =
C2H4 =
• C3H6 =
H2O =
Calculating EF
• We can get a ratio from the percent composition.
1.) Assume you have 100 g.
-the percentage becomes grams (75.1% = 75.1 grams)
2.) Convert grams to moles.
3.) Find lowest whole number ratio by dividing by
the smallest # of moles.
**4.) If not a whole #, use a multiplier.
Example
• Calculate the empirical formula of a compound
composed of 38.67 % C, 16.22 % H, and 45.11
%N.
• Assume 100 g so:
• 38.67 g C x 1mol C = 3.220 mole C
12.01 gC
• 16.22 g H x 1mol H = 16.06 mole H
1.01 gH
• 45.11 g N x 1mol N = 3.220 mole N
14.01 gN
Now divide each value by the smallest value
Example
• 3.220 mol C = 1 mol C
• 16.06 mol H = 5 mol H
• 3.220 mol N = 1 mol N
3.220
EF = C1H5N1 = CH5N
3.220
3.220
Empirical Formula (EF)
• Find the empirical formula for a sample
of 25.9% N and 74.1% O.
25.9 g 1 mol
= 1.85 mol N
=1N
1.85 mol
14.01 g
74.1 g 1 mol
= 4.63 mol O
= 2.5 O
16.00 g
1.85 mol
Empirical Formula
N1O2.5
Need to make the subscripts
whole numbers  multiply by 2
N2O5
EF Practice
Find the empirical formula for a sample of
43.64% P and 56.36% O.
Molecular Formula (MF)
• “True Formula” - the actual number of
atoms in a compound
empirical
formula
CH3
?
molecular
formula
C2H6
Empirical to Molecular
• Since the empirical formula is the lowest
ratio, the actual molecule would weigh
more.
By a whole number multiple.
• **Divide the actual molar mass by the
empirical formula mass.
MF mass
n
EF mass
EF n  MF
Molecular Formula
• The empirical formula for ethylene is CH2. Find the
molecular formula if the molecular mass is
28.1 g/mol?
empirical mass = 14.03 g/mol
28.1 g/mol
14.03 g/mol
=2
(CH2)2  C2H4
Practice Problem:
• Caffeine (EF= C4H5N2O) has a molecular
mass of 194 g. What is its molecular
formula?