Do Now:
If one stick of
Juicy Fruit gum
weighs 3.0g, what
percent of the
total mass of the
gum is sugar?
Hydrate Lab – for tomorrow
 Pre-lab
 Title
 Purpose
 Materials
 Procedure
 Data table
Hydrate
Anhydrous
Hydrates
• Water molecules are incorporated
into the crystalline structure.
Methane hydrate
 Hydrates, like zinc acetate
dihydrate, Zn(C2H3O2)2 * 2H2O
are commonly found in skin care
products such as moisturizer,
shampoo and lip balm.
What is the % H2O in nickel chloride
dihydrate, NiCl2 * 2H2O?
Element
#
g/mol
(molar mass)
TOTAL
Ni
1
58.69 g/mol
58.69g
Cl
2
35.45 g/mol
70.90g
H2O
2
18.02 g/mol
36.04g
MOLAR
MASS=
165.63g/mol
NiCl2 * 2H2O


36.04 g H 2O
%H2O = 
 = 21.76% H2O
165.63 g NiCl 2 * 2H2O 
Percentage Composition
(by mass...not atoms)
24.305
35.453
12
17
magnesium
chlorine
Mg
Cl
part
% = whole
24.31 g
% Mg =
x 100
95.21 g
25.52% Mg
Mg2+
Cl174.48% Cl
MgCl2
It is not 33% Mg and 66% Cl
1 Mg @ 24.31 g= 24.31 g
2 Cl @ 35.45 g= 70.90 g
95.21 g
x 100
percentage composition: the mass % of each
element in a compound
% of element =
g element
molar mass of compound
Find % composition.
PbO2
207.2 g Pb
32.0 g O
(NH4)3PO4
42.0 g N
12.0 g H
31.2 g P
64.0 g O
x 100
(see calcs above)
: 239.2 g = 86.6% Pb
: 239.2 g = 13.4% O
: 149.0 g
: 149.0 g
: 149.0 g
: 149.0 g
=
=
=
=
28.2% N
8.1% H
20.8% P
43.0% O
Empirical and 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
Different Types of Formulas
 Molecular Formula – shows the real # of atoms in one molecule
or formula unit
C6H6
 Empirical Formula – shows smallest whole
number mole ratio
CH
**Sometimes the empirical &molecular formula can be the same
 Structural Formula- molecular formula info PLUS bonding electron
and atomic arrangement
Calculating Empirical formula
Percent to mass (assume 100g)
Mass to mole (molar mass)
Divide by the smallest (ratio)
Multiply til’ whole*
Empirical Formula
Quantitative analysis shows that a compound contains 32.38% sodium,
22.65% sulfur, and 44.99% oxygen.
sodium sulfate
Find the empirical formula of this compound.
32.38% Na
32.38 g Na
 1mol Na 

 = 1.408 mol Na
 23 g Na 
/ 0.708 mol
= 2 Na
22.65% S
22.65 g S
 1mol S 


32
g
S


= 0.708 mol S
/ 0.708 mol
=1S
44.99% O
44.99 g O
 1 mol O 


16
g
O


= 2.812 mol O
/ 0.708 mol
=4O
Step 1) %  g
Step 2) g  mol
Step 3) mol
mol
Na2SO4
Empirical Formula
Quantitative analysis shows that a compound contains 66.75% copper, 10.84% phosphorus
and 22.41% oxygen.
Find the empirical formula of this compound.
copper (I) phosphate
66.75% Cu
66.75g Cu
 1 mol Cu 

=
63.55 g Cu 
10.84 % P
10.84gP
 1 mol P 


30.97 g P 
= 0.3500 mol P / 0.3500 mol = 1 P
22.41 % O
22.41gO
1 mol O 


 16 g O 
= 1.401 mol O
Step 1) %  g

1.050 mol Cu
Step 2) g  mol
/ 0.3500 mol =3 Cu
/ 0.3500 mol = 4 O
Step 3) mol
mol
Cu3PO4