It’s time to learn about . . .
Interesting angle to a picture!!!
Stoichiometry: Percent Composition
At the conclusion of our time together,
you should be able to:
1. Determine the percent composition of
each of the elements in a compound
2. Use percent composition to determine the
mass of each element in a compound
Chemical Formulas of Compounds
Formulas give the relative numbers of atoms or
moles of each element in a formula unit - always a
whole number ratio (the law of definite
proportions).
NO2 molecule: 2 atoms of O for every 1 atom
of N
1 mole of NO2 : 2 moles of O atoms to every 1
mole of N atoms
 If we know or can determine the relative number of
moles of each element in a compound, we can
determine a formula for the compound.

Types of Formulas

Empirical Formula
The formula of a compound that expresses the
smallest whole number ratio of the atoms present.
Ionic formulas are always empirical formulas

Molecular Formula
The formula that states the actual number of
each kind of atom found in one molecule of the
compound.
What is the percent composition of hydrogen in
water?
Determine the grams in 2 moles of hydrogen
and divide that by the grams in 1 mole of H2O,
then multiply by 100.
2.02 g H2
18.02 g H2O
x 100
= 11.21 % H2
What Lies At The Bottom Of The
Ocean And Twitches?
A Nervous Wreck.
Stoichiometry: Percent Composition
Let’s see if you can:
1. Determine the percent composition of
each of the elements in a compound
2. Use percent composition to determine the
mass of each element in a compound
#1: What is the percent composition of Cu in
copper bromide?
Determine the grams in 1 mole of copper and
divide that by the grams in 1 mole of copper
bromide, then multiply by 100.
63.55 g Cu
223.35 g CuBr2
x 100
= 28.45 % Cu
#7: In 75.5 g of water, how many grams of
hydrogen are there?
Start with the grams of water and multiply by
the hydrogen to water ratio.
75.5 g H2O
x 2.02 g H2
18.02 g H2O
= 8.46 g H2
Lessons from your elders…
An old prospector named Ralph, shuffled into
town leading an old tired mule.
Old Ralph headed straight for the only saloon
to clear his parched throat.
He walked up and tied his old mule to the hitch
rail. As he stood there, brushing some of the
dust from his face and clothes, a young
gunslinger stepped out of the saloon with a gun
in one hand and a bottle of whiskey in the
other.
The young gunslinger looked at the old man and
laughed, saying, "Hey old man, have you ever
danced?"
Lessons from your elders…
Ralph looked up at the gunslinger and said, "No,
I never did dance.. never really wanted to.”
A crowd had gathered as the gunslinger grinned
and said, "Well, you old fool, you're gonna
dance now," and started shooting at the old
man's feet.
The old prospector - not wanting to get a toe
blown off - started hopping around like a flea
on a hot skillet. Everybody was laughing, fit to
be tied.
When his last bullet had been fired, the young
gunslinger, still laughing, holstered his gun and
turned around to go back into the saloon.
Lessons from your elders…
Old Ralph turned to his pack mule, pulled out a
double-barreled shotgun, and cocked both
hammers.
The loud clicks carried clearly through the desert
air.
The crowd stopped laughing immediately.
The young gunslinger heard the sounds too, and
he turned around very slowly. The silence was
almost deafening.
Lessons from your elders…
The crowd watched as the young gunman stared
at the old timer and the large gaping holes of
those twin barrels.
The barrels of the shotgun never wavered in Old
Ralph's hands, as he quietly said, "Son, have
you ever kissed a mule's butt?“
The gunslinger swallowed hard and said, "No
sir..... but... I've always wanted to!!!"
Lessons from your elders…
Here are a few of the lessons:
Never be a cocky punk.
Don't waste ammunition.
Alcohol makes you think you're smarter
than you are.
Don't mess with old men; they didn't get old
by being stupid.
Stoichiometry: Percent Composition
At the conclusion of our time together,
you should be able to:
1. Determine the percent composition of
each of the elements in a compound
2. Use percent composition to determine the
mass of each element in a compound
3. Use percent compositions of a compound
to determine the empirical and/or molecular
formula
Why Do Gorillas Have Big Nostrils?

Because They Have Big Fingers.
To Obtain an Empirical Formula
1. Determine the mass in grams of each element
present, if necessary.
2. Calculate the number of moles of each
element.
3. Divide each by the smallest number of moles to
obtain the simplest whole number ratio.
4. If whole numbers are not obtained in step 3,
multiply through by the smallest number that
will give all whole numbers
A sample of a brown gas, a major air pollutant, is found
to contain 2.34 g N and 5.34g O. Determine the
formula for this substance.
This requires mole ratios, so convert grams to moles
moles of N = 2.34 g of N
14.01 g/mole
= 0.167 moles of N
moles of O = 5.34 g
16.00 g/mole
= 0.334 moles of O
Formula:
N0.167 O0.334
N 0.167 O 0.334  NO 2
0.167
0.167
Empirical Formula from % Composition
A substance has the following composition by mass:
60.80 % Na ; 28.60 % B ; 10.60 % H
What is the empirical formula of the substance?
Consider a sample size of 100 grams
This would contain 60.80 g of Na,
28.60 grams of B and 10.60 grams H
Determine the number of moles of each
Determine the simplest whole number ratio
Empirical Formula from % Composition
Determine the number of moles of each
60.80 g Na x 1 mol/22.99 g =
2.64 mol Na
28.60 g B x 1 mol/10.81 g =
2.65 mol B
10.60 g H x 1 mol/1.01 g =
10.50 mol H
Determine the simplest whole number ratio
NaBH4
Calculation of the Molecular Formula
A compound has an empirical formula of NO2. The
colorless liquid, used in rocket engines has a molar
mass of 92.0 g/mole. What is the molecular formula
of this substance?
Formula mass = 14.01 + 32.00 =
46.01 g/mol
92.0 g/mol
46.01 g/mol = ~2
N2O4
Bill Gates' Rules
Here is a list of 11 things that many high school and
college graduates did not learn in school. In his book,
Bill Gates talks about how feel-good, politicallycorrect teachings created a full generation of kids
with no concept of reality and how this concept has
set them up for failure in the real world.
RULE 7
Before you were born, your parents weren't as boring as they are
now. They got that way from paying your bills, cleaning your
clothes and listening to you talk about how cool you are. So
before you save the rain forest from the parasites of your parents'
generation, try "delousing" the closet in your own room.
Stoichiometry: Percent Composition
Let’s see if you can:
1. Determine the percent composition of
each of the elements in a compound
2. Use percent composition to determine the
mass of each element in a compound
3. Use percent compositions of a compound
to determine the empirical and/or molecular
formula
Check out my latest invention…
Percent Composition
What is the percent carbon in C5H8NO4 (the
glutamic acid used to make MSG monosodium
glutamate), a compound used to flavor foods
and tenderize meats?
a)
b)
c)
d)
8.22 % C
24.3 % C
41.1 % C
not listed
Empirical Formulas #10:
0.556 g C and 0.0933 g H.
Determine the formula for this substance.
moles of C =
0.556 g C
12.01 g/mole
= 0.0463 moles of C
moles of H = 0.0933 g H
1.01 g/mole
= 0.0924 moles of H
Formula:
C0.0463 H0.0924
C 0.0463 H 0.0924  CH 2
0.0463
0.0463
Stoichiometry: Percent Composition
Let’s see if you can:
1. Determine the percent composition of
each of the elements in a compound
2. Use percent composition to determine the
mass of each element in a compound
3. Use percent compositions of a compound
to determine the empirical and/or molecular
formula
Law of Mechanical Repair After your hands become coated with grease,
your nose will begin to itch and you'll have to
pee.
Empirical Formulas #9:
72% iron and 28% oxygen
Determine the formula for this substance.
moles of Fe
72 g Fe
55.85 g/mole
= 1.29 moles Fe
moles of O =
28 g H
16.00 g/mole
= 1.75 moles O
Formula:
Fe1.29 O1.75
Fe1.29 O1.79  FeO1.36
1.29
1.29
Empirical Formulas #9:
72% iron and 28% oxygen
Determine the formula for this substance.
Fe1.29 O1.75
Fe 1 O1.36  Fe3/3O 4/3
Multiply subscripts by 3 to get
Fe3 O4
Example #4: Empirical Formula from
% Composition

A substance has the following composition by
mass: 40.0% C, 6.7% H, and 53.3% O(by mass).
Calculate the empirical formula and the molecular
formula of this compound given that the molar
mass is 60.00 g/mol.
 Consider a sample size of 100 grams
Determine the number of moles of each
Determine the simplest whole number ratio
Empirical Formula from % Composition
Determine the number of moles of each
40.0 g x 1 mol/12.01 g =
3.33 mol C
6.7 g x 1 mol/1.01 g =
6.63 mol H
53.3 g x 1 mol/16.00 g =
3.33 mol O
CH2O
Molecular Formula from Empirical
Formula
CH2O
Empirical Molar Mass =
30.03 g/mol
Molecular Molar Mass = 60.00 g/mol
30.03 g/mol
~2
C2H4O2
15 Helpful Hints On The Lab Report from
Mr. T’s Vast Lab Experience!!!
Hint #10. If an experiment works, you must
be using the wrong equipment.