Download The Mole - Solon City Schools

It’s time to learn about . . .
Happy Friday!

Please have out your work for the molar
conversions on Page 4.

We will have a brief review before the
quiz.
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
% Composition – Page 5

Is the percent by mass that each element
contributes to the molar mass.
1)
Find the molar mass of the compound.
Element Mass/Molar Mass * 100
2)
EXAMPLE: What is the percent composition of
hydrogen and oxygen in water?
1) What is the molar mass of water?
2) Element Mass/Molar Mass * 100
Try: CuBr2
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
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 – Page 5

Empirical Formula
The formula of a compound that expresses the
smallest whole number ratio of the atoms present.
It is the simplest ratio possible.
Ex –
P4O6
C6H9
CH2OHCH2OH
BrCl2
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
Page 6, #9 & #11
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
2.64 mol Na
2.65 mol B
10.50 mol H
Determine the simplest whole number ratio
NaBH4
Molecular Formula – Page 7

1)
2)
Molecular Formula
The formula that states the actual
number of each kind of atom found in one
molecule of the compound.
Molar Mass/Empirical Mass
Empirical Formula * Answer
Molecular Formula, Page 7 #1
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?
Empirical Mass = 14.01 + 32.00 = 46.01g/mol
92.0 g/mol
46.01 g/mol = ~2
N2O4
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 compositions of a compound
to determine the empirical and/or molecular
formula
Empirical Formula from % Composition

A substance has the following composition by
mass: 12.8% C, 2.1% H, and 85.1% Br (by mass).
Calculate the empirical formula and the molecular
formula of this compound given that the molar
mass is 188 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
12.8 g/12.01 g/mol =
1.07 mol C
2.1 g/1.01 g/mol =
2.08 mol H
85.1 g/79.90 g/mol =
1.07 mol Br
CH2Br
Molecular Formula from Empirical
Formula
CH2Br
Empirical Molar Mass =
93.93 g/mol
Molecular Molar Mass = 188 g/mol
93.93 g/mol
~2
C2H4Br2