Ch. 10 – The Mole
II. Formula
Calculations
I
II
III
Hydrates

Some compounds contain H2O in their
structure. These compounds are called
hydrates.

This is different from (aq) because the
H2O is part of the molecule (not just
surrounding it).

The H2O can usually be removed if
heated.
Hydrates

A dot separates water in the formula
 CuSO4•5H2O

When naming a Greek prefix indicates
the # of H2O groups.
 copper(II) sulfate pentahydrate
Naming Hydrates




Na2SO4•10H2O
NiSO4•6H2O
CoCl2•6H2O
MgSO4•7H2O




Sodium sulfate
decahydrate
Nickel (II) sulfate
hexahydrate
Cobalt (II) chloride
hexahydrate
Magnesium sulfate
heptahydrate
Writing Formulas of Hydrates




sodium carbonate
monohydrate
barium chloride
dihydrate
Tin (IV) chloride
pentahydrate
Barium hydroxide
octahydrate

Na2CO3•H2O

BaCl2•2H2O

SnCl4•5H2O

Ba(OH)2•8H2O
What is Percent Composition?
A. Percent Composition

the percentage by mass of each
element in a compound
total mass of element
% mass of element 
 100
total mass of compound
mass of element in 1 mol compound
% mass of element 
 100
molar mass of compound
A. Percent Composition
Find the percent composition of a sample
that is 28 g Fe and 8.0 g O.
Known:
Unknown:
 Mass of Fe = 28 g
 % Fe = ?
 Mass of O = 8.0 g
 %O= ?
 Total Mass = 28 + 8.0 g = 36 g

%Fe =
%O =
28 g
36 g
8.0 g
36 g
 100 = 78% Fe
 100 = 22% O
Check:
78% + 22%
= 100%

A. Percent Composition


Find the % composition of Cu2S.
Use this formula when finding % composition from a
chemical formula: % mass of element  mass of element in 1 mol compound  100
molar mass of compound
Known:
Unknown:
 Mass of Cu in 1 mol Cu2S =
 % Cu = ?
2(63.55g) = 127.10 g Cu
 %S= ?
 Mass of S in 1 mol Cu2S = 32.07 g S
 Molar Mass = 127.10 g + 32.07 g = 159.17 g/mol
%Cu =
%S =
127.10 g Cu
159.17 g Cu2S
32.07 g S
159.17 g Cu2S
 100 = 79.85% Cu
 100 = 20.15% S
A. Percent Composition
 How
many grams of copper are in
a 38.0-gram sample of Cu2S?
 Use answer from last question as
a conversion factor.
79.85 g Cu
Cu2S is 79.85% Cu =
100 g Cu2S
38.0 g Cu2S 79.85 g Cu
= 30.3 g Cu
100 g Cu2S
A. Percent Composition
 Find
the percent composition of
Cu2SO4.
Known:
 Mass of Cu in 1 mol Cu2SO4 = 2(63.55 g) = 127.10 g Cu
 Mass of S in 1 mol Cu2SO4 = 32.07 g S
 Mass of 0 in 1 mol Cu2SO4 = 4(16.00 g) = 64.00 g O
 Molar Mass = 127.10 g + 32.07 g + 64.00 g = 223.17 g/mol
Unknown:
 % Cu = ?
 %S= ?
 %O= ?
A. Percent Composition
 Find
the percent composition of
Cu2SO4.
%Cu =
%S =
%O =
127.10 g
223.17 g
32.07 g
223.17 g
64.00 g
223.17 g
 100 = 56.95% Cu

 100 = 14.37% S
 100 = 28.68% O
Check:
56.95% +
14.37% +
28.68% =
100%
A. Percent Composition

Find the mass percentage of water in
calcium chloride dihydrate, CaCl2•2H2O.
Unknown:
Known:
 % H2 O = ?
 Mass of H2O in 1 mol compound =
2(2(1.01g) + 16.00 g) = 36.04 g H2O
 Molar Mass = 40.08 g + 2(35.45 g) +
36.04 g = 147.02 g
36.04
g/mol
 100 = 24.51%
%H2O =
H2O
147.02 g/mol
B. Empirical Formula
 Smallest
whole number ratio of
atoms in a compound
C 2H 6
reduce subscripts
CH3
B. Empirical Formula
1. Find mass (or %) of each element.
2. Find moles of each element.
3. Divide moles by the smallest # to
find subscripts.
4. When necessary, multiply
subscripts by 2, 3, or 4 to get
whole #’s.
B. Empirical Formula
 Find
the empirical formula for a
sample of 25.9% N and 74.1% O.
25.9 g N 1 mol N = 1.85 mol N
=
1
N
14.01 g N 1.85 mol
N1.85O4.63
74.1 g O 1 mol O = 4.63 mol O
=
2.5
O
16.00 g O 1.85 mol
B. Empirical Formula
N1O2.5
Need to make the subscripts whole
numbers  multiply by 2
N2O5
B. Empirical Formula
 Find
the empirical formula for a
sample of 94.1% O and 5.9% H.
94.1 g O 1 mol O = 5.88 mol O
=
1
O
16.00 g O 5.84 mol
5.9 g H 1 mol H = 5.84 mol H
=
1
H
1.01 g H
5.84 mol
EF = OH
C. Molecular Formula

“True Formula” - the actual number of atoms
in a compound, either the same as or a
whole-number multiple of the empirical
formula
empirical
formula
CH3
?
molecular
formula
C2H 6
C. Molecular Formula
1. Find the empirical formula.
2. Find the empirical formula mass.
3. Divide the molar mass by the
empirical formula mass.
4. Multiply each subscript in your EF
by the answer from step 3.
molar mass
n
EF mass
EF n
C. Molecular Formula
 The
empirical formula for ethylene
is CH2. Find the molecular formula
if the molar mass is 28.1 g/mol?
empirical formula mass = 14.03 g/mol
n=
28.1 g/mol
14.03 g/mol
= 2.00
(EF)n = (CH2)2  C2H4
D. Put it all together!!

1,6-diaminohexane is 62.1% C, 13.8% H,
and 24.1% N. What is the empirical
formula? If the molar mass is 116.21 g/mol,
what is the molecular formula?
62.1 g C
1 mol C
12.01 g C
13.8 g H 1 mol H
1.01 g H
24.1 g N
1 mol N
14.01 g N
= 5.17 mol C
1.72 mol
=3C
= 13.7 mol H
=8H
1.72 mol
= 1.72 mol N
1.72 mol
=1N
EF = C3H8N
D. Put it all together!!
C3H8N
empirical formula mass = 58.12 g/mol
n=
116.21 g/mol
58.12 g/mol
(C3H8N)2 
= 2.00
C6H16N2