1. Dimensional Analysis (a.k.a. Factor Labeling)
Many problems in chemistry involve using relationships to convert one unit of measurement to another.
Conversion factors are relationships between two units.
For our purposes conversion factors are exact.
Hence conversion factors have unlimited significant figures.
Conversion factors generated from equivalence statements.
e.g., 1 inch = 2.54 cm can give or

2. Dimensional Analysis (a.k.a. Factor Labeling)
Arrange conversion factors so the starting unit cancels.
Arrange conversion factor so the starting unit is on the bottom of the conversion factor.
May string conversion factors.
So we do not need to know every relationship, as long as we can find something else the starting and desired units are related to :

3. ( ) ______ How many cm are in 1.32 meters? equality: 1 m = 100 cm
(or 0.01 m = 1 cm)
applicable conversion factors:
______
1 m
100 cm
______
1 m
100 cm or ( )
______
1 m
100 cm
132 cm
X cm = 1.32 m =
We use the idea of unit cancellation
to decide upon which one of the two
conversion factors we choose.

4. Again, the units must cancel.
How many meters is 8.72 cm?
equality:
1 m = 100 cm
applicable conversion factors:
______
1 m
100 cm
______
1 m
100 cm or ( )
______
1 m
100 cm m
X m = 8.72 cm =
Again, the units must cancel.

5. Again, the units must cancel.
How many feet is inches?
equality:
1 ft = 12 in
applicable conversion factors:
______
1 ft
12 in
______
1 ft
12 in or ( )
____
1 ft
12 in
3.28 ft
X ft = in =
Again, the units must cancel.

6. ( ) ( ) ____ ______ How many kilometers is 15,000 decimeters? 10 dm
1 km
1.5 km
X km = 15,000 dm =

7. How many seconds is 4.38 days?
____ ( ) ( )
_____ ( )
____
24 h
1 d
1 h
60 min
1 min
60 s
X s = 4.38 d
378,432 s =
If we are accounting for significant figures, we would change this to…
3.78 x 105 s

8. Counting Atoms
Chemistry is a quantitative science - we need a "counting unit."
The MOLE
1 mole is a sample of a substance that contains as many particles as there are in 12.0 g of C-12. A mole equates a gram to an amu.
How many particles are in a mole?

9. Amedeo Avogadro (1776 – 1856) Particles in a Mole
A more precise value of Avagadro’s number = E23.
How could you determine the number of sugar crystals in a bowl of sugar?
Is it possible to get a reasonably close estimate?
quadrillions
thousands
trillions
billions
millions
1 mole =
or x 1023
There is Avogadro's number of particles in a mole of any substance.

10. How Big is a Mole? One mole of marbles would cover the entire Earth
(oceans included) for a depth of two miles.
One mole of $1 bills stacked
one on top of another would
reach from the Sun to Pluto
and back 7.5 million times.
It would take light 9500 years to travel from the bottom
to the top of a stack of 1 mole of $1 bills.

11. Visualizing a Chemical Reaction2
Na Cl NaCl 2
___ mole Na 10 10
___ mole Cl2 5 5
___ mole NaCl 10 10 ?

12. Molecular Weight and Molar Mass
Molecular weight (or Formula mass) is the sum of atomic weights
of all atoms in the molecule or formula unit.
example: NaCl has a molecular weight of 58.5 amu.
this is composed of a single formula unit of NaCl
Molar mass = molecular weight in grams. Units = g/mol
example: NaCl has a molar mass of 58.5 g/mol
this is composed of a 6.02 x1023 formula units of NaCl; it is also composed of 6.02 x1023 sodium atoms and 6.02 x1023 chlorine atoms

13. Mole Diagram 1 mole = 6.022 x 1023 particles (atoms or molecules)
1 mole = molar mass (g)
Mass
Mole
Particles

14. Percentage Composition
(by mass...not atoms) Mg
magnesium
24.305 12 Cl
chlorine
35.453 17
% Mg = x 100
24 g
95 g
% = x 100
part
whole
25.52% Mg
Mg2+ Cl1-
74.48% Cl
MgCl2
It is not 33% Mg and 66% Cl
amu = amu
2 Cl @ amu = amu
amu

15. 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. The
simplest formula is called the Empirical Formula
Ethanol, C2H6O
52.13% C
13.15% H
34.72% O

16. Empirical Formula
Quantitative analysis shows that a compound contains 32.38% sodium,
22.65% sulfur, and 44.99% oxygen.
Find the empirical formula of this compound.
sodium sulfate
32.38% Na
22.65% S
44.99% O
32.38 g Na
22.65 g S
44.99 g O
= mol Na
/ mol
= 2 Na
Na2SO4
Na2SO4
= mol S
= 1 S
= mol O
= 4 O
Step 1) %  g
Step 2) g  mol
Step 3) mol
mol

17. Empirical Formula
A sample weighing g is analyzed and found to contain the following:
27.38% sodium
1.19% hydrogen
14.29% carbon
57.14% oxygen
27.38 g Na
1.19 g H
14.29 g C
57.14 g O
Assume sample is 100 g.
Determine the empirical formula of this compound.
Step 1) convert %  gram
Step 2) gram  moles
Step 3) mol / mol
NaHCO3
/ 1.19 mol = 1 Na
/ 1.19 mol = 1 H
/ 1.19 mol = 1 C
/ 1.19 mol = 3 O

18. Empirical & Molecular Formula
(contains only hydrogen + carbon)
(~17% hydrogen)
A 175 g hydrocarbon sample is analyzed and found to contain ~83% carbon.
The molar mass of the sample is determined to be 58 g/mol.
Determine the empirical and molecular formula for this sample.
Determine the empirical formula of this compound.
Step 1) convert %  gram
Step 2) gram  moles
Step 3) mol / mol
2 12 g = 24 g
5 1 g = 5 g
29 g
Assume sample is 100 g.
Then, 83 g carbon and 17 g hydrogen.
MMempirical = 29 g/mol
/ mol = 1 C
/ mol = 2.5 H
( H)
CH2.5
C2H5
MMmolecular = 58 g/mol
58/29 = 2
Therefore 2(C2H5) = C4H10
butane

19. Find the molar mass and percentage composition of zinc acetate
Zn2+
CH3COO1-
acetate = CH3COO1-
Zn(CH3COO)2
g/mol = g
/ g x 100% = % Zn
/ g x 100% = % C
/ g x 100% = 3.3 % H
/ g x 100% = % O
4 12 g/mol = 48 g
6 1 g/mol = 6 g
4 16 g/mol = 64 g
Zn(CH3COO)2
183.4 g

20. A compound is found to be 45.5% Y and 54.5% Cl.
Its molar mass (molecular mass) is 590 g.
Assume a 100 g sample size
a) Find its empirical formula
45.5 g Y
1 mol Y
= mol Y
/ mol
= 1 Y
88.9 g Y
YCl3
54.5 g Cl
1 mol Cl
= mol Cl
= 3 Cl
35.5 g Cl
g/mol = g
b) Find its molecular formula
g/mol = g
590 / 195.4
= 3
YCl3
195.4 g
3 (YCl3)
Y3Cl9

21. Molar Mass vs. Atomic Mass
6.02x1023
Molar Mass vs Atomic Mass
H2 = _____
2 g
H2 = _______
2 amu
H2O = _____
18 g
H2O = ________
18 amu
MgSO4 = _____
120 g
MgSO4 = ________
120 amu
(NH4)3PO4 = _____
149 g
(NH4)3PO4 = ________
149 amu
Percentage Composition
(by mass)
Empirical Formula
%  g
g  mol
mol
% = x 100 %
part
whole
Empirical vs. Molecular Formula
(lowest ratio)