Transcript Section 7.3 Day 1

Monday, April 21st: “A” Day
Tuesday, April 22nd: “B” Day
Agenda
Go over Sec. 7.2 quiz
Begin 7.3: “Formulas & Percentage Composition”
In-Class Assignment:
Practice pg. 243: #1-4
Homework:
Pg. 56 Worksheet: 1 a-d
Concept Review
Must SHOW WORK!
Sec. 7.2 Quiz:
“Relative Atomic Mass and
Chemical Formulas”
This quiz gave some of you trouble, so I
wanted to go over it before we continue with
section 7.3…
7.3: “Formulas and Percentage
Composition”
The percentage composition is the percentage by
mass of each element in a compound.
Percentage composition helps verify
a substance’s identity.
Percentage composition can also be used to
compare the ratio of masses contributed by the
elements in two different substances.
Percent Composition of Iron Oxides
Empirical Formula
An actual formula shows the actual ratio of
elements or ions in a single unit of a compound.
Empirical formula: a chemical formula that shows
the simplest ratio for the relative numbers and
kinds of atoms in a compound.
For example, consider the empirical formula and
actual formulas for hydrogen peroxide:
HO
Empirical Formula
H2O2
Actual formula
Rules for Determining Empirical
Formulas
 You can use the percentage composition of a
compound to determine its empirical formula.
1. Change the percentage of each element in the
compound to grams.
%
grams
2. Use the molar mass to change grams
moles
3. Compare the amounts in moles to find the
simplest whole-number ratio.
Rules for Determining Empirical
Formulas
To find the simplest whole-number ratio,
divide each amount of moles by the smallest
number of moles you found.
This will give a subscript of 1 for the atoms
present in the smallest amount.
Finally, you may need to multiply all of the
amounts of moles by a number to convert all
subscripts to small, whole numbers.
The final numbers of moles you get are the
subscripts in the empirical formula.
Determining an Empirical Formula form
Percentage Composition
(Sample Problem G, pg. 242)
Chemical analysis of a liquid shows that it is
60.0% C, 13.4% H, and 26.6% O by mass. Calculate
the empirical formula of this substance.
1. Change %
grams:
Assume that you have a 100 g sample so that each
percentage is the same as the amount in grams:
C: 60.0% = 60.0 g C
H: 13.4% = 13.4 g H
O: 26.6% = 26.6 g O
Sample Problem G, continued…
2. Use the molar mass to change grams
(remember sig figs!)
moles:
Sample Problem G, continued…
3. Divide each number of moles by the smallest
number of moles found. (1.66 mol O)
Carbon: 5.00 mol = 3.01 mol C
1.66 mol
Hydrogen: 13.3 mol= 8.01 mol H
1.66 mol
Oxygen: 1.66 mol = 1.00 mol O
1.66 mol
 These numbers are within experimental error to
be considered whole numbers and become the
subscripts, so the empirical formula is: C3H8O
Example #1
Find the empirical formula given the following
percentage composition: 32.37% Na, 22.58% S,
45.05% O.
1. Assume 100 g sample and change %
32.37 g Na
22.58 g S
45.05 g O
grams:
Example #1 cont…
2. Use the molar mass to change grams
moles:
Na: 32.37 g Na X 1 mol Na
22.99 g Na
= 1.408 mol Na
S:
= .7041 mol S
O:
22.58 g S X 1 mol S
32.07 g S
45.05 g O X 1 mol O
16.00 g O
= 2.816 mol O
Example #1 cont…
3. Divide each number of moles by the smallest
number of moles found (.7041 mol)
Na: 1.408 mol Na
=
2.000 mol Na
.7041 mol
S: .7041 mol S
=
1.000 mol S
.7041 mol
O: 2.816 mol O
=
3.999 mol O
.7041 mol
These ARE whole numbers, so the empirical
formula is:
Na2SO4
Additional Practice
Find the empirical formula given the following
percentage composition:
26.58% K, 35.35% Cr, and 38.07% O
1.Assume 100 g sample and change %
26.58 g K
35.35 g Cr
38.07 g O
grams:
Additional Practice, cont…
2. Use the molar mass to change grams
K:
moles:
26.58 g K X 1 mole K = .6798 mol K
39.10 g K
Cr: 35.35 g Cr X 1 mole Cr = .6798 mol Cr
52.00 g Cr
O:
38.07 g O X 1 mole O = 2.379 mol O
16.00 g O
(remember sig figs!)
Additional Practice, cont…
3. Divide each number of moles by the smallest
number of moles found (.6798 mol)
K:
.6798 mol K = 1 mol K
.6798 mol
Cr:
.6798 mol Cr = 1 mol Cr
.6798 mol
O:
2.379 mol O = 3.5 mol O
.6798 mol
Additional Practice, cont…
4. Since 3.5 mol of oxygen is not a whole number,
multiply each number of moles by 2 to get
whole numbers:
K: (2) 1 mol K =
2 mol K
Cr: (2) 1 mol Cr = 2 mol Cr
O: (2) 3.5 mol O = 7 mol O
 These ARE whole numbers, so the empirical
formula is:
K2Cr2O7
In-Class Assignment/Homework
In-Class Assignment:
Practice pg. 243: 1-4
Homework:
Worksheet pg. 56: 1a-d (side 56 only)
Concept Review: “Formulas Percentage
Composition”: #1-5
Must Show Work!
Next time: You will have a sub
Good luck to all juniors taking the ACT!