The Mole
Molar Mass


Molar mass relates moles to grams
The molar mass is the mass in grams of 1 mole of
an element or compound

Use the mass numbers from the periodic table, instead of
amu’s, the units are grams
Round all masses from the periodic table to the nearest tenth place

The chemical formula gives the ratio of atoms in a
compound

The chemical formula can also give the ratio of ions
to one another
Molar masses are computed in
the typical way
Volume-Mass Relationships of
Gases

Gay-Lussac’s law of combining volumes of
gases states that at constant temperature
and pressure, the volumes of gaseous
reactants and products can be expressed
as ratios of small whole numbers
 Example:
2 L of H2 reacts with 1 L of O2 to
produce 2L of H2O

Avogadro’s law states that equal
volumes of gases at the same
temperature and pressure contain
equal numbers of molecules
 Also
indicates that volume is directly
proportional to the amount of gas, at
a given temperature and pressure

V=kn


According to Avogadro’s
law, one mole of any
gas will occupy the
same volume as any
other gas at the same
conditions
The standard molar
volume of gas had been
determined to be
22.41410 L (22.4L)
The Ideal Gas Law

The ideal gas law is the mathematical
relationship among pressure, volume,
temperature, and the number of moles of each
gas
1
V
P

V T
The ideal gas law is
derived by combining
Boyle’s law and
Charles’s law and the
logic of Avogadro’s
law
V n
1
V  T  n
P
By adding a constant R we get:
1
V  R  T  n
P
Or PV= nRT

The constant R is known as the
ideal gas constant
 Its
value depend on the units
chosen
mRT
PV 
M

The ideal gas law can
be manipulated to
solve for molar mass
and density
 Given
that m = mass
and M = molar mass
 n=m/M and D=m/V
mRT
M
PV
MP
D
RT
Empirical formula
1.
2.
3.
4.
5.
Choose an arbitrary sample size (100g).
Convert masses to amounts in moles.
Write a formula.
Convert formula to small whole numbers.
Multiply all subscripts by a small whole
number to make the subscripts integral.
Example 3-5
Determining the Empirical and Molecular Formulas of
a Compound from Its Mass Percent Composition.
Dibutyl succinate is an insect repellent used against
household ants and roaches. Its composition is 62.58%
C, 9.63% H and 27.79% O. Its experimentally
determined molecular mass is 230 u. What are the
empirical and molecular formulas of dibutyl succinate?
Step 1: Determine the mass of each element in a 100g
sample.
C 62.58 g
H 9.63 g
O 27.79 g
Example
3-5
Step 2: Convert masses to amounts in moles.
1 mol C
 5.210 mol C
12.011 g C
1 mol H
nH  9.63 g H 
 9.55 mol H
1.008 g H
1 mol O
nO  27.79 g O 
 1.737 mol O
15.999 g O
nC  62.58 g C 
Step 3: Write a tentative formula.
C5.21H9.55O1.74
Step 4: Convert to small whole numbers. C2.99H5.49O
Slide 17 of 37
Prentice-Hall © 2002
General Chemistry:
Chapter 3
Example
3-5
Step 5: Convert to a small whole number ratio.
Multiply x2 to get C5.98H10.98O2
The empirical formula is C6H11O2
Step 6: Determine the molecular formula.
Empirical formula mass is 115 u.
Molecular formula mass is 230 u.
The molecular formula is C12H22O4
Slide 18 of 37
Prentice-Hall © 2002
General Chemistry:
Chapter 3
Concentration & Molarity

The ratio of the solute to the solvent is the solution
concentration



Molarity is the moles of solute per liter of solution
It is abbreviated with an M
5.0 M HCl is read as a 5 molar solution of HCl
molesofsolute
molarity 
litersofsolution
Molality is the concentration of a solution
expressed in moles of solute per kilogram
of solvent
 Symbol is m
