Empirical Formulas &
Percent Composition
CO2 Carbon dioxide
BCl3 boron trichloride
Molar Mass
How many moles of alcohol are
there in a “standard” can of beer
if there are 21.3 g of C2H6O?
(a)Molar mass of C2H6O = 46.08 g/mol
(b)Calc. moles of alcohol
1 mol
21.3 g •
= 0.462 mol
46.08 g
How many molecules of alcohol are
there in a “standard” can of beer if
there are 21.3 g of C2H6O?
We know there are 0.462 mol of C2H6O.
6.022 x 1023 molecules
0.462 mol •
1 mol
= 2.78 x 1023 molecules
How many atoms of C are there in
a “standard” can of beer if there are
21.3 g of C2H6O?
There are 2.78 x 1023 molecules.
Each molecule contains 2 C atoms.
Therefore, the number of C atoms is
23
2.78 x 10
2 C atoms
molecules •
1 molecule
= 5.57 x 1023 C atoms
Empirical & Molecular
Formulas
A pure compound always consists of the
same elements combined in the same
proportions by weight.
Therefore, we can express molecular
composition as PERCENT BY
WEIGHT
Ethanol, C2H6O
52.13% C
13.15% H
34.72% O
Percent Composition
Consider some of the family of nitrogenoxygen compounds:
NO2, nitrogen dioxide
Structure of NO2
What is the weight percent of N and of O?
Percent Composition
Consider NO2, Molar mass = 46 g/ mol
Percent of N and of O?
Wt. % N =
14.0 g N
• 100% = 30.4 %
46.0 g NO2
Wt. % O  2 (16 .0 g O per mole ) x 100%  69 .6%
46 .0 g
What are percentages of N and O in NO?
Determining
Formulas
In chemical analysis we determine
the % by weight of each element in a given
amount of pure compound and derive the
EMPIRICAL or SIMPLEST formula.
PROBLEM: What is its empirical
formula of a compound of B and H
if 81.10% is boron?
Determination of empirical formulas:
mass of each
%B + %H = 100%
100% - 81.10% B = 18.90% H.
If there were 100.0 g of the compound:
there are 81.10 g of B and 18.90 g of H.
Calculate the number of MOLES of each.
Determination of empirical formulas:
moles of each
Calculate the number of moles of each
element assuming 100.0 g total.
1 mol
81.10 g B •
= 7.502 mol B
10.81 g
1 mol
18.90 g H •
= 18.75 mol H
1.008 g
Determination of empirical formulas:
moles ratio
Take the ratio of moles of B and H.
atoms combine in small whole number ratios
Always
divide
18.75 mol H
by the
and
7.502 mol
2.5 mol H
smaller
number.
7.502 mol B
7.502 mol
to
1.0 mol B
EMPIRICAL FORMULA =
B2H5
But we need a whole number ratio.
2.5 mol H
1.0 mol B
or
5 mol H
2 mol B
If the empirical formula is B2H5,
what is its molecular formula?
We need to do an
EXPERIMENT to find
the MOLAR MASS.
MS gives 53.3 g/mol
Compare empirical
mass
= 26.66
g/unit
Find the ratio of these
masses.
53.3
2
=
26.66
Molecular
formula =
B4H10
Determine the formula of a
compound of Sn and I using
the following data.
• Reaction of Sn and I2 is done using
excess Sn.
• Mass of Sn in the beginning = 1.056 g
• Mass of Sn remaining
= 0.601 g
• Mass of iodine (I2) used
= 1.947 g
Tin and Iodine Compound
Find the mass of Sn that combined
with 1.947 g I2.
Mass of Sn initially =
1.056 g
Mass of Sn recovered = 0.601 g
Mass of Sn used =
0.455 g
Find moles of Sn used:
1 mol
-3
0.455 g Sn •
= 3.83 x 10 mol Sn
118.7 g
Tin and Iodine Compound
Now find the number of moles of I2 that
combined with 3.83 x 10-3 mol Sn. Mass
of I2 used was 1.947 g.
1 mol
1.947 g I2 •
= 7.671 x 10-3 mol I2
253.81 g
How many mol of iodine atoms?
7.671 x 10
-3
2 mol I atoms 
mol I2 

 1 mol I2

= 1.534 x 10-2 mol I atoms
Tin and Iodine Compound
Now find the ratio of number of moles of
moles of I and Sn that combined.
-2
1.534 x 10 mol I
4.01 mol I
=
-3
1.00 mol Sn
3.83 x 10 mol Sn
Empirical formula is
SnI4
Practice Formula
Determination
Homework:
Worksheets pgs 17-20
Pg 13-14 good review for
mole calculations
Prelab questions