Math with
Chemical Formulas
Honors Chemistry Unit 5
Unit Objectives
 Be
able to perform math
functions with and without your
calculator using correct scientific
notation.
 Be able to find
molar/molecular/formula mass
using the periodic table.
Unit Objectives, cont.
 Be
able to calculate Molarity.
 Be able to calculate percent
composition.
 Be able to determine empirical
and molecular formulas using lab
data.
Unit Objectives, cont.
 Understand
the mole and
Avogadro’s number.
 Be able to convert to/from
atoms, ions, molecules, moles
and grams
What is a “Mole” (mol)?
A
mole is a counting unit
 just
like a dozen
 other examples…..
What is a Mole, cont.?
“
Official Definition”
 the
amount of a substance that
contains as many particles as
there are atoms in exactly 12g
of carbon-12
Avogadro’s Number
 Constant
 the
number of particles in
exactly one mole of a pure
substance
6.02 X 1023
Memorize this Number
Conversion Factors – 1 mol
1
mol = 6.02 X1023 of anything
Conversion Factors – 1 mol
Examples…
1 mol
6.02 X 1023
atoms
1 mol
6.02 X 1023
ions
1 mol
6.02 X 1023
molecules
Molar Mass
Mass
in g of 1 mole of
anything
For elements,
the molar mass is equal to
the atomic mass
Molar Mass Examples…..
1 mol
1 mol C
12.01 g C
atomic wt. (g)
1 mol Li
6.94 g Li
You try……
1 mol Ca
? g Ca
1 mol Fe
? g Fe
Now you have Two conversion
Factors for a mole…..
1 mol
6.02 X 1023
atoms, ions or molecules
and
1 mol
______(g)
Example 1
 How
many g in 2.0 mol of He?
2.0 mol He
4.0g He = 8.0g He
1.0 mol He
Example 2
many moles in 3.01 X 1023
atoms Ag?
 How
3.01 X1023 atoms Ag
1mol Ag
6.02 X 1023 atoms Ag
0.500 mol Ag
Oct. 16-22, 2005
=
Example 3

What is the mass of 1.20 X 108 atoms of Cu?
1.20 X108 atoms Cu
1mol Cu
6.02 X 1023 atoms Cu
63.6g Cu
1 mol Cu
1.27 X 10-14 g Cu
=
You try:
 Convert
11.5 g B to moles B
1.06 mol B
19
 Convert 8.0 X 10 atoms of
Ag to g
0.014 g Ag
Formula Mass/Molecular
Mass/Molar Mass
 Sum
of masses in a compound
 Molar Mass of sodium chloride
NaCl
Na
Cl
Total:
1 mol X 22.99g/mol = 22.99g
1 mol X 35.45g/mol = 35.45g
58.44g
Example 2
 Molar
Mass of magnesium chloride
MgCl2
Mg
Cl
Total:
1 mol X 24.31g/mol = 24.31g
2 mol X 35.45g/mol = 70.90g
95.21g
Example 3 - Calcium Nitrate
Molar Mass of Ca(NO3)2
Ca
N
O
Total:
1 mol X 40.08 g/mol = 40.08g
2 mol X 14.01 g/mol = 28.02g
6 mol X 16.00 g/mol = 96.00g
164.10g
You try:
Calculate the molar mass of
sodium phosphate
Na3PO4
Na
3 mol X 22.99 g/mol =
P
1 mol X 30.97 g/mol =
O
4 mol X 16.00 g/mol =
Total:

68.97g
30.97g
64.00g
163.94g
Conversion Factors using Formula
Mass, Molecular Mass/Molar Mass
1 mol NaCl
58.44g NaCl
1 mol MgCl2
95.21g MgCl2
1 mol Ca(NO3)2
164.1g Ca(NO3)2
can be used as a conversion factors
You try….
 How
many mol in 127g barium
chloride?
(set up on board)
 Answer:
0.610 mol BaCl2
Percent Composition
 Percent
composition is the
percent by mass of each element
in a compound.
 Percent
composition is the same,
regardless of the size of the
sample.
% Composition Calculations
% comp = mass of element X 100%
molar mass of cpd
= % element in the compound
Examples
 Find
the % composition of Cu2S
 First, find the molar mass:
2 mol Cu = 2 X 63.55g/mol = 127.1g Cu
1 mol S = 1 X 32.06g/mol = 32.06g S
Total:
159.15g/mol
Example, cont.
Next, find the % of each element
For Cu:
% Cu = 127.1g X 100% = 79.86% Cu
159.15g
For S:
% S = 32.06g X 100% = 20.14% S
159.15g
Next,
check your work – do the %s add up to 100?

You try:
 barium
choride
 sodium
phosphate
Answers:

barium chloride
barium
chloride

65.90%
34.10%
sodium phosphate
sodium
42.07%
phosphorus 18.89%
oxygen
39.04%
Back to
Objectives
Determining Formulas

Empirical Formula =
Simplest Formula
To find the empirical formula
from data:
1.
2.
3.
4.
Assume 100% sample; change % to grams
for each element
Find moles from the grams of each element
Find the smallest whole # ratio by dividing
by the smallest number of moles
If necessary, multiply to get rid of fractions.
Example
A compound is 78% B and 22% H. What is
the empirical formula?
First, change % to grams and find moles:
78g B
1mol B
= 7.22 mol
10.81g B

22g H
1mol H
1.01g H
=
21.78 mol
Example, cont.
Next, divide all mole numbers by the smallest
number of moles:
B: 7.22 mol =
1
7.22 mol
H:
21.78 mol =
7.22 mol
3.02 = 3
Example, cont.

Finally, use these whole numbers as the
number of each individual element. They are
the subscripts.
Empirical Formula = BH3
Example 2


Analysis shows a compound to contain
26.56% K, 35.41% Cr, and 38.03% O. Find
the empirical formula of this compound:
First (always!) assume 100g sample, convert
% to g and then find moles of each element
Example 2 cont.

Next, Conversion to moles:
26.56g K 1mol K
=
39.10g K
35.41g Cr 1mol Cr =
52.00g Cr
38.03g O 1mol O
=
16.00g O
0.6793 mol K
0.6810 mol Cr
2.377 mol O

Next, divide all numbers by the smallest whole
number to find the smallest whole number ratios:
0.6793 mol K = 1.00 mol K
0.6793
0.6810 mol Cr = 1.003 mol Cr ~ 1.00 mol Cr
0.6793
2.377 mol O
= 3.499 mol O – can’t be rounded
0.6793
 So,
if you have:
.25 or .75
.33 or .66
.50
multiply all by:
4
3
2
For our example:
2 X 1.00 mol K =
2 X 1.00 mol Cr =
2 X 3.499 mol O=

2 mol K
2 mol Cr
7 mol O
Empirical Formula = K2Cr2O7
You try:

What is the empirical formula if we have
a sample containing 66.0% Ca and
34.0% P?

Answer:
Ca3P2
You try:
 Find
the empirical formula of a
compound with 32.38% Na; 22.65%
S; and 44.99% O.
 Answer:
Na2SO4
Molecular Formula
Molecular Formula = Actual Formula
Example:
C2H6
CH3
molecular
empirical

MF = (EF)x
where X = Molecular mass
Empirical mass
Example

The empirical formula of a compound was
found to be P2O5. Experimentation shows
that the molar mass of this compound is
283.89 g/mol. What is the compound’s
molecular formula?
Back to
Objectives
Moles in Solution




Molarity is the term used for moles dissolved
in solution
Symbol for Molarity = M
Definition – moles of solute per liter of
solution
Formula
M = moles solute (mol)
liter solution (L)
Example

What is the molarity of a 0.5L solution
containing 2 moles of NaCl?

M=
2moles NaCl = 4 M NaCl
0.5 L
Example 2

What is the molarity of a 250 mL solution
containing 12.7 g of lithium bromide?
M = moles = 12.7g LiBr 1mol LiBr
1000mL
L
86.8g LiBr 250mL
1L
= 0.59 M LiBr
Example 3

How would you make 500mL of a 0.32M
solution of LiBr and water?
Given: M=0.32M, L=0.500
M= #moles/liters
.32M= #moles
.500L
#moles= 0.16 molesLiBr

0.16 molesLiBr






X 86.841g
1 mole
= 13.894gLiBr in .500L of Water
You try:

Calculate the M of a 700. mL solution of
23.2g calcium chloride

How would you make a 0.2 L solution of
0.50 M CaCl2 solution?
Back to
Objectives