Transcript PPT Empirical & Molecular Formulas

Percent Composition &
Empirical & Molecular Formulas
Molecular Formulas
The molecular formula
• Shows the actual number of each type of
atom in a compound
• Not always the simplest formula
Ex: C6H12O6 = glucose
Contains 6 atoms of carbon, 12 atoms of
hydrogen and 6 atoms of oxygen in 1
molecule
Empirical Formulas
The empirical formula
• Is the simplest whole number ratio of the
atoms.
• Is calculated by dividing the subscripts in the actual
(molecular) formula by a whole number to give the
lowest ratio.
Glucose
C6H12O6  6 = C1H2O1 = CH2O
molecular formula
empirical formula
Some Molecular and Empirical Formulas
• The molecular formula is the same or a multiple of the
empirical.
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You Try…
1. What is the empirical formula for C4H8?
2. What is the empirical formula for C8H14?
3. Which is a possible molecular formula for CH2O?
a) C4H4O4
b) C2H4O2
c) C3H6O3
4. A compound has an empirical formula SN. If there are 4
atoms of N in one molecule, what is the molecular
formula? Explain.
3 Steps for determining Chemical Formulas
1. Determine the percent composition of all elements.
2. Convert this information into an empirical formula
3. Find the true number of atoms/ elements in the
compound (Molecular Formula)
Percent Composition Review
Percent composition
• Is the percent by mass of each element in a formula.
Example: Calculate the percent composition of CO2.
CO2 = 1(12.01g/mol) + 2(16.00 g/mol) = 44.01 g/mol)
12.01 g C
x 100
44.01 g CO2
2(16.00) g O
44.01 g CO2
=
27.29 % C
x 100 =
72.71 % O
100.00 %
You Try…
5. What is the percent composition of
lactic acid, C3H6O3, a compound that
appears in the blood after vigorous
activity?
6. The chemical isoamyl acetate C7H14O2
gives the odor of pears. What is the
percent carbon in isoamyl acetate?
Solution 5
STEP 1: Calculate Molar Mass (Mr)
3C(12.01) + 6H(1.008) + 3O(16.00) = 90.08 g/mol
STEP 2: Calculate % Composition for each element
%C = 36.03 g C x 100
90.08 g
= 40.00% C
%H = 6.048 g H x 100 = 6.714% H
90.08 g
%O = 48.00 g O x 100 = 53.29% O
90.08 g
Solution 6
Molar mass C7H14O2 =
7C(12.01) + 14H(1.008) + 2O(16.00) = 130.18 g/mol
Total C = 7C(12.01) = 84.07g
%C
= total g C
total g
x 100
%C
= 84.07 g C x 100 = 64.58 % C
130.18 g
BRAIN BREAK: Hands
This Brain Break seems simple. However, you will find out soon that you will have a hard time
mastering it.
1. Stand Up.
2. Start by waving your right hand in front of you left to right. Your palm should be
facing away from you while keeping your hand with your fingers pointing up.
3. Now stop that hand and have your left hand in front of you waving it up and
down.
4. Now practice moving them at the SAME TIME. Do not move your hands going
diagonally.
5. Now switch to have your right hand up and down and your left hand left and
right. Do this faster and switch often to make it more difficult.
6. Lastly, to increase the difficulty, have your arms crossed while doing this.
Calculating Empirical Formula
1.
2.
3.
4.
If given % composition, assume 100 g of sample
Convert mass of each element to moles (gmol)
Divide each of these numbers by the smallest number
If necessary, multiply by the smallest number possible to
make each a whole number
5. These whole numbers are the subscripts in the empirical
formula called mole ratio
The percentage composition of a compound is found
to be 32.4% sodium, 22.5% sulfur, 45.1 % oxygen.
Determine the empirical formula.
Step 1: Assume 100 g of Sample
32.4% Na 
32.4 g Na
22.5% S

22.5 g S
45.1% O 
45.1 g O
Step 2: Convert grams to moles
The percentage composition of a compound is found
to be 32.4% sodium, 22.5% sulfur, 45.1 % oxygen.
Determine the empirical formula.
Step 3: Divide by smallest mol
1.41 mol Na
0.702 mol S
2.82 mol O
Step 4: If necessary, multiply by the smallest
number possible to make each a whole number
Step 5: Write Empirical Formula
Na2SO4
The percentage composition of acetic acid is found
to be 39.9% C, 6.7% H, and 53.4% O. Determine the
empirical formula of acetic acid.
Step 1. Assume 100 g of Sample:
39.9% C
6.7% H
53.4% O



39.9 g C
6.7 g H
53.4 g O
Step 2. Convert grams to moles
The percentage composition of acetic acid is found
to be 39.9% C, 6.7% H, and 53.4% O. Determine the
empirical formula of acetic acid.
Step 3: Divide by smallest mol
3.32 mol C
6.65 mol H
3.34 mol O
Step 4: If necessary, multiply by the smallest
number possible to make each a whole number
Step 5: Write Empirical Formula
CH2O
You Try
Phosphoric acid is found in some soft drinks. A sample
of phosphoric acid contains 0.086 g of hydrogen, 3.161
g of phosphorus, and 6.531 g of oxygen. What is the
empirical formula for phosphoric acid?
H3PO4
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BRAIN BREAK: The Crab
• Directions:
1. Stand Up
2. Put your arms out in front of you and touch your fingers and
thumbs together.
3. Now put lower your middle fingers so that the knuckles touch.
Keep them flat against each other.
4. Now un-touch and retouch your thumbs.
5. Now un-touch and retouch your index fingers.
6. Now un-touch and retouch your ring fingers.
7. Lastly, un-touch and retouch your pinkies.
Good luck. This one was difficult.
• https://www.youtube.com/watch?v=dp0sa7Z72R8
Converting Decimals to Whole Numbers
When the number of moles for an element is a
decimal, all the moles are multiplied by a small
integer to obtain whole number.
Aspirin is 60.0% C, 4.5 % H and 35.5 % O.
Calculate its empirical formula.
Step 1. Assume 100 g of Sample:
60.0% C
4.5% H
35.5% O



60.0 g C
4.5 g H
35.5 g O
Step 2. Convert grams to moles
Aspirin is 60.0% C, 4.5 % H and 35.5 % O.
Calculate its empirical formula.
Step 3: Divide by smallest mol
5.00 mol C
4.46 mol H
2.22 mol O
Step 4: If necessary, multiply by the smallest
number possible to make each a whole number
Step 5: Write Empirical Formula
C9H8O4
Your Turn
• A compound consists of 72.2% magnesium and
27.8% nitrogen by mass. What is the empirical
formula of the compound?
Mg3N2
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Calculating Molecular Formula From
Empirical Formula
•
Find the empirical formula mass by adding up the
weights of the atoms in the empirical formula.
•
Divide the molecular mass by the formula mass. This
number tells you how many times the empirical formula is
repeated to make the molecular formula.
Calculating Molecular Formula From
Empirical Formula
Your Turn
• Ibuprofen, a common headache remedy, has an
empirical formula of C7H9O and a molar mass of
approximately 215 g/mol. What is the molecular
formula for ibuprofen?
C14H18O2
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Calculating Molecular Formula & Empirical
Formula
A mixture of cyclopropane gas and oxygen is used in
an anaesthetic. Cyclopropane contains 85.7% C, and
14.3% H by mass. The molecular mass is 42.0 g/mol.
What is the empirical and molecular formula of the
cyclopropane?
=1C
=2H
Empirical
CH2
Calculating Molecular Formula & Empirical
Formula
A mixture of cyclopropane gas and oxygen is used in an anaesthetic.
Cyclopropane contains 85.7% C, and 14.3% H by mass. The molecular
mass is about 42.0 g/mol. What is the empirical and molecular
formula of the cyclopropane?
•
Find the empirical formula mass
• For CH2, the formula weight is 12.01 + 2 × 1.01 = 14.03 g/mol.
•
Divide the molecular weight by the formula weight.
In this case, the molecular formula weight divided by the
empirical formula weight is 42.0/14.03 ≈ 3.
•
3 x CH2  C3H6.
YOUR TURN
Epinephrine (adrenaline) is a hormone secreted into
the bloodstream in times of danger and stress. It is
59.0% carbon, 7.1% hydrogen, 6.2% oxygen, and 7.7%
nitrogen by mass. Its molar mass is about 180 g/mol.
Find the empirical and molecular formulas of
epinephrine.
Empirical
C9H13O3N
Molecular
C9H13O3N
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