Transcript Empirical formula - Morrison Community Unit District 6

Friday, April 15th: “A” Day
Monday, April 18th: “B” Day
Agenda
Homework questions/problems?
Quiz over section 7.2
Begin 7.3: “Formulas & Percentage Composition”
In-Class Assignments:
– Practice pg. 243: #1-4
– Practice pg. 245: #1-3
Quiz 7.2: “Relative Atomic Mass and
Chemical Formulas”
You can use your book and your guided notes
for this walk-talk quiz…
Remember: answer only the question that
corresponds to the month you were born in.
If you were born in November, answer
question #2. If you were born in December,
answer question # 7.
Once everyone has answered their question,
get up, walk around, and talk/compare
answers with others.
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, the empirical formula for
hydrogen peroxide is HO, while the actual
formula is H2O2
Determining Empirical Formulas
 You can use the percentage composition for a
compound to determine its empirical formula.
1. Convert the percentage of each element in
the compound to grams.
2. Convert from grams to moles using the molar
mass of each element as a conversion factor.
3. Compare these amounts in moles to find the
simplest whole-number ratio among the
elements.
Determining Empirical Formulas
To find the simplest whole-number ratio,
divide each amount in moles by the smallest
of all the amounts of moles.
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 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. Assume that you have a 100 g sample so that
each percentage is the same as the amount in
grams:
 for C: 60.0% = 60.0 g C
 for H: 13.4% = 13.4 g H
 for O: 26.6% = 26.6 g O
Sample Problem G, continued…
2. Use the molar mass to convert each
amount in grams to amount in moles:
Sample Problem G, continued…
3. Divide each number of moles found by the
smallest number of moles found. (1.66 moles 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 mol O
1.66 mol
 These numbers are within experimental error to
be considered whole numbers so the empirical
formula is: C3H8O
Additional Practice
Find the empirical formula given the following
composition:
26.58% K, 35.35% Cr, and 38.07% O
1.Assume 100 g sample:
26.58 g K
35.35 g Cr
38.07 g O
Additional Practice
2. Use molar mass to convert from grams to
moles.
 26.58 g K X 1 mole K = .6798 mol K
39.10 g K
 35.35 g Cr X 1 mole Cr = .6798 mol Cr
52.00 g Cr
 38.07 g O X 1 mole O = 2.379 mol O
16.00 g O
Additional Practice
3. Divide each number of moles found by the
smallest number of moles found (.6798 mol).
 .6798 mol K = 1 mol K
.6798 mol
 .6798 mol Cr = 1 mol Cr
.6798 mol
 2.379 mol O = 3.5 mol O
.6798 mol
Additional Practice
4. Since these are not whole numbers, multiply
each one by 2 to get whole numbers.
 1 mol K (2) =
2 mol K
 1 mol Cr (2) = 2 mol Cr
 3.5 mol O (2) = 7 mol O
 These ARE whole numbers, so the empirical
formula is:
K2Cr2O7
Molecular Formulas are Multiples of
Empirical Formulas
The formula for an ionic compound shows the
simplest whole-number ratio of the large numbers
of ions in a crystal of the compound.
A molecular formula is a whole-number multiple
of the empirical formula.
The molar mass of any compound is equal to the
molar mass of the empirical formula times a
whole number, n.
n (empirical formula) = molecular formula
Comparing Empirical and Molecular
Formulas
Compound
Formaldehyde
Acetic Acid
Empirical
Formula
CH2O
CH2O
Glucose
CH2O
Molecular Formula
CH2O
C2H4O2
2X the empirical
formula
n = 2
C6H12O6
6X the empirical
formula
n = 6
Determining a Molecular Formula from an
Empirical Formula
(Sample Problem H, pg. 245)
The empirical formula for a compound is P2O5.
Its experimental molar mass is 284 g/mol.
Determine the molecular formula of the
compound.
1. Use the periodic table to find the molar
mass of the empirical formula:
For P: 2(30.97) = 61.94 g/mol
For O: 5(16.00) = 80.00 g/mol
Molar mass of P2O5 = 141.94 g/mol
Sample Problem H, continued…
2. Find the multiplier, n:
n = experimental molar mass of compound
molar mass of empirical formula
 n = 284 g/mol = 2
141.94 g/mol
Hint: the bigger #
always goes on top!
3. To find the molecular formula, simply multiply
the empirical formula by 2:
2 (P2O5) =
P4O10
Additional Practice
Determine the molecular formula for the
following:
Molar mass: 232.41 g/mol
Empirical formula: OCNCl
1. Find molar mass of empirical formula:
 O = 16.00 g/mol
 C = 12.01 g/mol
 N = 14.01 g/mol
 Cl = 35.5 g/mol
=
77.52 g/mol
Additional Practice
2. Find the multiplier, n:
n = experimental molar mass of compound
molar mass of empirical formula
 n = 232.41 g/mol = 3
77.52 g/mol
3. To find the molecular formula, simply
multiply the empirical formula by 3:
3 (OCNCl) =
O3C3N3Cl3
In-Class Assignments
You Must SHOW WORK!
Practice pg. 243: #1-4
Practice pg. 245: #1-3
We will finish this section next time…