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Transcript
Chapter 10 – Chemical
Quantities
I
II
III
IV
What is the Mole?

A unit of measurement used in chemistry.

A counting number like – a dozen eggs, a
ream of paper, a bushel of apples, etc.

A mole describes an amount of matter.
Definition:
A
mole of any element is
defined as the number of atoms
of that element equal to the
number of atoms in exactly
12.0 grams of carbon-12.
Atomic Mass

Mass of a single atom  10-23 grams

Atomic Mass Unit (amu) is used to
describe the atomic mass of an atom.

A scale was devised to express the mass
of all the different atoms.

Based on the mass of an atom relative to
carbon-12.

Atomic Masses are listed in the
periodic table.
What it means:

1 atom of 12C has a mass of 12.01 amu

1 mole of 12C has a mass of 12.01 grams
What it means:


1 atom of 35Cl has a mass of 35.45 amu
1 mole of 35Cl has a mass of 35.45 grams
Formula Mass

Is calculated using the atomic mass of
each element in the compound, times
how many atoms of each element are in
the compound.
Example:

Find the formula mass of H2O
H = 2 x 1.0 = 2.0 grams
O = 1 x 16.0 = 16.0 grams
18.0 grams/mole
Example:

Find the formula mass of NaOH
Na = 1 x 23.0 = 23.0 grams
O = 1 x 16.0 = 16.0 grams
H = 1 x 1.0 = 1.0 grams
40.0 grams/mole
The Atomic Mass or Formula Mass

Tells the mass in grams of 1 mole
of a substance

There are 6.02 x 1023
objects in one mole.

602,000,000,000,000,000,000,000
A. What is the Mole?

1 mole of hockey pucks would
equal the mass of the moon!

1 mole of basketballs would fill a
bag the size of the earth!

1 mole of pennies would cover
the Earth 1/4 mile deep!
A mole of water
A mole of sugar
A mole of copper
Example:
16.0 g of oxygen =
1 mole of oxygen =
6.02 x 1023 atoms of oxygen
Example:
18.0 g of water
=
1 mole of water
=
6.02 x 1023 molecules of water
Example:
40.0 g of NaOH
dissolved in water
=
1 mole of Na+ ions and
=
1 mole of OH ions
6.02 x 1023 Na+ ions and
6.02 x 1023 OH- ions
Molar Mass
 Mass
in grams of 1 mole of a
substance.
Molar Mass Examples

water
 H2O
 2(1.01) + 16.00 = 18.02 g/mol

sodium chloride
 NaCl
 22.99 + 35.45 = 58.44 g/mol
Molar Mass Examples

sodium bicarbonate
 NaHCO3
 22.99 + 1.01 + 12.01 + 3(16.00)
= 84.01 g/mol

sucrose
 C12H22O11
 12(12.01) + 22(1.01) + 11(16.00)
= 342.34 g/mol
Molar Conversions
MASS
molar
mass
6.02  1023
÷
x
MOLES
IN
GRAMS
NUMBER
OF
x
÷
(g/mol)
(particles/mol)
PARTICLES
Molar Conversion Examples
 How
many moles of carbon are
in 26 g of carbon?
26 g C 1 mol C
12.01 g C
= 2.2 mol C
Molar Conversion Examples
 How
many molecules are in
2.50 moles of C12H22O11?
6.02  1023
2.50 mol molecules
1 mol
= 1.51  1024
molecules
C12H22O11
Molar Conversion Examples
the mass of 2.1  1024
molecules of NaHCO3.
 Find
2.1  1024
molecules
1 mol
84.01 g
6.02  1023 1 mol
molecules
= 293 g NaHCO3
Chapter 10 – % Composition,
Empirical and Molecular
Formulas
Suggested
Reading: Pages
305 - 313
I
II
III
IV
Percentage Composition
 the
percentage by mass of each
element in a compound
mass of element
% composition 
 100
total mass
Percentage Composition
 Find
%Cu =
%S =
the % composition of Cu2S.
128g Cu
160 g Cu2S
32 g S
160 g Cu2S
 100 =
80% Cu
 100 =
20% S
Percentage Composition
 Find
the percentage composition
of a sample that is 28 g Fe and
8.0 g O.
28 g
 100 = 78% Fe
%Fe =
36 g
%O =
8.0 g
36 g
 100 = 22% O
Percentage Composition
 How
many grams of copper are in
a 38.0-gram sample of Cu2S?
Cu2S is 80% Cu
(38.0 g Cu2S)(0.80) = 30.4 g Cu
Percentage Composition
 Find
the mass percentage of
water in calcium chloride
dihydrate, CaCl2•2H2O?
%H2O =
36 g
146 g
 100 = 25%
H2O
Empirical Formula
 Smallest
whole number ratio of
atoms in a compound
C 2H 6
reduce subscripts
CH3
Empirical Formula
1. Find mass (or %) of each element.
2. Find moles of each element.
3. Divide moles by the smallest # to
find subscripts.
4. When necessary, multiply
subscripts by 2, 3, or 4 to get
whole #’s.
Empirical Formula Song
(to help you remember steps)
1. Percent to Mass
2. Mass to Mole
3. Divide by Small
4. Multiply ‘til Whole
Empirical Formula
 Find
the empirical formula for a
sample of 25.9% N and 74.1% O.
25.9 g 1 mol
14 g
74.1 g 1 mol
16 g
= 1.85 mol N
=1N
1.85 mol
= 4.63 mol O
= 2.5 O
1.85 mol
Empirical Formula
N1O2.5
Need to make the subscripts whole
numbers  multiply by 2
N2O5
Molecular Formula
 “True Formula” - the actual number
of atoms in a compound
empirical
formula
CH3
?
molecular
formula
C2H6
Molecular Formula
1. Find the empirical formula.
2. Find the empirical formula mass.
3. Divide the molecular mass by the
empirical mass.
4. Multiply each subscript by the
answer from step 3.
MF mass
n
EF mass
EF n
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 g/mol
= 2.00
(CH2)2  C2H4