Do Now
Crime scene investigators recovered a sample of
poison from a corpse.
They suspected it was
strychnine (C21H22N2O2) and got the following
elemental analysis results. Is the poison strychnine?
symbol
unknown
C
72.30%
H
6.00%
N
11.30%
O
10.40%
Put homework
into red basket
slide 1
slide 2
Do Now
• What is the percent composition of C21H22N2O2?
symbol number
atomic
mass
C
21
12.011
H
22
1.008
N
2
14.007
O
2
15.999
mass
percent
comp
slide 3
Do Now
• What is the percent composition of C21H22N2O2?
symbol number
atomic
mass
mass
C
21
12.011
252.231
H
22
1.008
22.176
N
2
14.007
28.014
O
2
15.999
31.998
percent
comp
334.419
slide 4
Do Now
• What is the percent composition of C21H22N2O2?
symbol number
atomic
mass
mass
percent
comp
C
21
12.011
252.231
75.424%
H
22
1.008
22.176
6.631%
N
2
14.007
28.014
8.377%
O
2
15.999
31.998
9.568%
334.419
100%
slide 5
Do Now
Do they match?
symbol strychnine
unknown
C
75.424%
72.30%
H
6.631%
6.00%
N
8.377%
11.30%
O
9.568%
10.40%
No!
The poison found in the corpse is not strychnine
slide 6
Last Class
slide 7
Identifying Unknowns 2
SWBAT use percent composition to find an
empirical formula, a molecular formula and a
hydrate formula.
Write this in your notes
slide 8
Percent Composition to Empirical Formula
• Elemental analysis can provide the percent
composition (percent by mass) of a compound.
• We are going to learn a technique to calculate the
empirical formula using the percent composition
slide 9
Percent Composition to Empirical Formula
Elemental analysis of a sample of
potassium carbonate shows it to have
the following percent composition:
K 56.6%
C 8.7%
O 34.7%
What is the empirical formula for
potassium carbonate?
We are going to learn a 4-step process
to solve this type of problem
slide 10
Write this in your notes
Percent Composition to Empirical Formula
1) Assume a 100 g sample and assign masses to
each element based upon percent composition
K 56.6 g
C 8.7 g
O 34.7 g
slide 11
Write this in your notes
Percent Composition to Empirical Formula
1) Assume a 100 g sample and assign masses to
each element based upon percent composition
K 56.6 g
C
8.7 g
O 34.7 g
slide 12
Write this in your notes
Percent Composition to Empirical Formula
2) Divide each mass by the atomic mass to
determine the moles in the sample
K 56.6 g ÷ 39.1 = 1.45
C
8.7 g ÷ 12.0 = 0.73
O 34.7 g ÷ 16.0 = 2.17
slide 13
Write this in your notes
Percent Composition to Empirical Formula
3) Divide the moles of each element by the
smallest value (the moles of C in this case)
K 56.6 g ÷ 39.1 = 1.45 ÷ 0.73 = 1.98
C
8.7 g ÷ 12.0 = 0.73 ÷ 0.73 = 1.00
O 34.7 g ÷ 16.0 = 2.17 ÷ 0.73 = 2.97
slide 14
Write this in your notes
Percent Composition to Empirical Formula
4) Convert the ratio to whole numbers by
multiplication and/or rounding
K 56.6 g ÷ 39.1 = 1.45 ÷ 0.73 = 1.98 ≈ 2
C
8.7 g ÷ 12.0 = 0.73 ÷ 0.73 = 1.00 = 1
O 34.7 g ÷ 16.0 = 2.17 ÷ 0.73 = 2.97 ≈ 3
The empirical formula is K2CO3
slide 15
Check for Understanding
Rubbing alcohol was found to contain 60.0%
carbon, 13.4% H, and the remaining mass was
due to oxygen. What is the empirical formula?
slide 16
Check for Understanding
Rubbing alcohol was found to contain 60.0%
carbon, 13.4% H, and the remaining mass was
due to oxygen. What is the empirical formula?
C 60.0 g ÷ 12.0 = 5.0
H 13.4 g ÷ 1.0 = 13.4
O 26.6 g ÷ 16.0 = 1.7
÷ 1.7
÷ 1.7
÷ 1.7
= 2.9
= 7.9
= 1.0
≈ 3
≈ 8
= 1
The empirical formula is C3H8O
slide 17
Getting the Molecular Formula
• Other techniques
provide the molar
mass of an unknown
sample
• On the left is a mass
spectrum that
provides such
information
• With the molar mass
and the empirical
formula, the
molecular formula
can be determined
slide 18
Write this in your notes
The empirical formula may or may not be
the same as the molecular formula.
Empirical
formula
CH2O
CH2O
Molecular
formula
CH2O
Formaldehyde
C6H12O6 Glucose
slide 19
Write this in your notes
The molecular mass will always
be equal to, or a multiple of,
the empirical formula mass.
Empirical
formula
CH2O
30.03 g/mole
Molecular
formula
x6
x6
C6H12O6
Glucose
180.16 g/mole
slide 20
Write this in your notes
A molecule has an empirical formula of CH,
and a molar mass of 78.0 g/mole. What is its
molecular formula?
Empirical
formula
Molecular
formula
CH
n
x g/mole
n
calculate this
CnHn
n(x) g/mole
78.0 g/mole
slide 21
Write this in your notes
C 1 x 12.0 = 12.0
H 1 x 1.0 = 1.0
13.0
Empirical
formula
Molecular
formula
CH
n
x g/mole
n
calculate this
CnHn
n(x) g/mole
78.0 g/mole
slide 22
Write this in your notes
C 1 x 12.0 = 12.0
H 1 x 1.0 = 1.0
13.0
Empirical
formula
Molecular
formula
CH
n
x g/mole
n
13.0 g/mole
CnHn
n(x) g/mole
78.0 g/mole
slide 23
Write this in your notes
𝒘𝒉𝒂𝒕 𝒊𝒔 𝒕𝒉𝒆 𝒗𝒂𝒍𝒖𝒆 𝒐𝒇 𝒏?
Empirical
formula
Molecular
formula
CH
n
x g/mole
n
13.0 g/mole
CnHn
n(x) g/mole
78.0 g/mole
slide 24
Write this in your notes
𝟕𝟖. 𝟎
𝐧=
=𝟔
𝟏𝟑. 𝟎
Empirical
formula
Molecular
formula
CH
n
x g/mole
n
13.0 g/mole
CnHn
n(x) g/mole
78.0 g/mole
slide 25
Write this in your notes
replace every n with 6
Empirical
formula
Molecular
formula
CH
n
x g/mole
n
13.0 g/mole
CnHn
n(x) g/mole
78.0 g/mole
slide 26
Write this in your notes
replace every n with 6
Empirical
formula
Molecular
formula
CH
6
13.0 g/mole
6
13.0 g/mole
C6H6
6(13.0) g/mole
78.0 g/mole
slide 27
Write this in your notes
The molecular formula is
C6H6
Empirical
formula
Molecular
formula
CH
6
13.0 g/mole
6
13.0 g/mole
C6H6
6(13.0) g/mole
78.0 g/mole
slide 28
SECTION
10.4
Check for Understanding
Find the molecular formula for the following:
empirical molar
formula mass
1)
C2OH4
88.0 g/mole
2)
CFBrO
254.4 g/mole
3)
C2H8N
46.0 g/mole
molecular
formula
slide 29
SECTION
10.4
Check for Understanding
Find the molecular formula for the following:
empirical molar
formula mass
molecular
formula
1)
C2OH4
88.0 g/mole
C4O2H8
2)
CFBrO
254.4 g/mole
3)
C2H8N
46.0 g/mole
slide 30
SECTION
10.4
Check for Understanding
Find the molecular formula for the following:
empirical molar
formula mass
molecular
formula
1)
C2OH4
88.0 g/mole
C4O2H8
2)
CFBrO
254.4 g/mole
C2F2Br2O2
3)
C2H8N
46.0 g/mole
slide 31
SECTION
10.4
Check for Understanding
Find the molecular formula for the following:
empirical molar
formula mass
molecular
formula
1)
C2OH4
88.0 g/mole
C4O2H8
2)
CFBrO
254.4 g/mole
C2F2Br2O2
3)
C2H8N
46.0 g/mole
C2H8N
slide 32
Write this in your notes
Hydrates
• Some molecules like to absorb water
• A hydrate is a compound that has a specific
number of water molecules bound to its
atoms.
• The number of water molecules associated
with each formula unit of the compound is
written following a dot.
• Sodium carbonate decahydrate =
Na2CO3 • 10H2O
slide 33
Examples of Hydrates
slide 34
Analyzing a Hydrate
• When heated, water molecules are released
from a hydrate leaving an anhydrous
compound.
• To determine the formula of a hydrate, find
the number of moles of water associated with
1 mole of hydrate.
slide 35
Write this in your notes
Determining Hydrate
• Sodium carbonate forms hydrates. A 100.0 g
sample of an unknown sodium carbonate
hydrate is dried to a constant mass of 37.1 g.
What is the formula of the hydrate?
100.0 g Na2CO3 ● nH2O
37.1 g Na2CO3
62.9 g nH2O
slide 36
Write this in your notes
slide 37
Write this in your notes
62.9 g
slide 38
Write this in your notes
3.49
9.97
slide 39
Check for Understanding
NiCl2 forms hydrates. A 5.00 g sample of an
unknown NiCl2 hydrate is dried to a constant
mass of 2.73 g. What is the formula of the
hydrate?
NiCl2 ● 6H2O
slide 40
Use of Hydrates
• Anhydrous forms of hydrates are often used
to absorb water, particularly during shipment
of electronic and optical equipment.
• In chemistry labs, anhydrous forms of
hydrates are used to remove moisture from
the air and keep other substances dry.
slide 41
Identifying Unknowns 2 - Homework
• pg 346, # 58, 59 (empirical formula from % composition)
• pg 350, # 64, 65 (molecular formula from % composition)
• pg 353, # 74, 75 (calculate hydrate formula)
Determine formula only for 74 & 75
slide 42
Backup Slides
slide 43
Not Covered
slide 44
The periodic table gives us atomic masses
and molar masses for each element
slide 45