The
Mole
Chemistry 6.0
Getting to know the terms…
MICROSCOPIC
Mass
MACROSCOPIC
Molar
Mass
Atom
Atomic mass
amu
Element
g/mol
Molecule
Molecular mass
amu
Molecular
Compound
g/mol
Formula Unit
Formula mass
amu
Ionic Compound
g/mol
The following elements exist in
nature as molecules:
H2 O2 F2 Br2 I2 N2 Cl2 S8 P4
MOLE RELATIONSHIPS
1 Mole = 6.02x1023 particles of substance
(atoms, formula units, molecules)
1 Mole = mass (g) of substance from PT
Also remember your formula information:
1 molecule = _________ atoms
1 formula unit = _________ ions or
_________ atoms
Mole Conversions
MUST use factor label!
A. Moles & Mass
1. How many grams in 3.0 moles of water?
know: 1 mole H2O = 18.0 g H2O
54 g H2O
2. How many moles in 60.0 g of copper?
know: 1 mole Cu = 63.5 g Cu
0.945 g Cu
B. Moles & Particles
1. How many atoms in 3.0 moles of copper?
know: 1 mole Cu = 6.02 x 1023 atoms of copper
1.8 x 1024 atoms Cu
2. How many atoms in 3.00 moles of water?
know: 1 mole H2O = 6.02 x 1023 molecules of H2O
know: 1 molecule H2O = 3 atoms
5.42 x 1024 atoms
Mole Conversions
MUST use factor label!
C. Mass & Particles
1. How many atoms in 100.0 g of copper?
63.5 g copper
know: 1 mole = _________
atoms of copper
1 mole = 6.02 x 1023 __________
9.480 x 1023 atoms Cu
2. How many oxygen atoms are in 75.0 g of sucrose,
C12H22O11?
342.0
know: 1 mole = __________
g of C12H22O11
molecules of C12H22O11
1 mole = 6.02 x 1023 _____________
atoms of oxygen
1 molecule of C12H22O11 = 11 ________
1.45 x 1024 atoms
Molar Volume of Gases at STP
Avogadro’s Law
Amount - Volume Relationship.
Equal volumes of gases at the same temperature and pressure
contain an equal number of particles.
volume
constant
4 He
222 Rn
molar mass
1 mole gas = 22.4 L = 6.02 x 1023 particles
at STP (273 K & 1 atm)
He
O2
Rn
Therefore because of Avogadro’s Law if these
three gases have the same number of particles
and are at the same temperature and pressure,
they must take up the same volume.
Molar Mass does not affect
volume of a gas
Avogadro’s Law
• At STP, the amount of gas is directly
proportional to the volume.
Problem #1: Which of the following samples
of gases occupies the largest volume,
assuming that each sample is the same
temp and pressure?
50.0 g Ne
50.0 g Ar
50.0 g Xe
Avogadro’s Law
V1 = V2
n1
n2
n ____ P ____ (more gas, ____________)
more collisions
Ideal Gas Law
Although no “ideal gas” exists, this law can be used to
explain the behavior of real gases under ordinary
conditions.
P = pressure (atm)
V = volume (L or dm3)
PV = nRT
n = number of moles
R = 0.08206 L•atm/mol•K
universal gas constant
T = Kelvin temperature
• Individual gas laws describe the relationships between
these variables.
• Ideal gas law relates all 4 variables that describe a gas
at one set of conditions.
Ideal Gas Law Problems
1. Calculate the volume of a gas balloon filled
with 1.00 mole of helium when the pressure
is 760. torr and the temperature is 0.oC.
22.4 L
2. Calculate the pressure, in atm, exerted by
54.0 g of xenon in a 1.00-L flask at 20.oC.
9.9 atm
3. Calculate the density of nitrogen dioxide, in
g/L, at 1.24 atm and 50.oC.
2.16 g/L
Empirical Formulas
1. Definition: always the smallest whole-number
ratio of the atoms, or ions, in a formula
2. Use experimental data to find the empirical
formula
3. Examples
a. Determine the empirical formula of a compound if a
2.500-g sample contains 0.900 g of calcium and 1.600
g of chlorine.
CaCl2
b. Determine the empirical formula for an iron oxide that
is 78% iron. Name the compound.
FeO
iron(II) oxide
C. Molecular Formula
1. Definition: the formula of a molecular
compound. The molecular formula shows the
actual number of atoms of each element
present in 1 molecule of a compound.
Molecular formula for benzene: C6H6
Empirical formula for benzene: CH
D. Molecular formula is always a wholenumber multiple of the empirical formula.
molecular formula = (empirical formula)n
n = molar mass molecular formula
molar mass empirical formula
Example
Find the molecular formula of a compound that
contains 42.5 g of palladium and 0.80 g of hydrogen.
The molar mass of the compound is 216.8 g/mol.
Empirical formula - PdH2
Molecular formula – Pd2H4
Concentration
• Percent concentration by mass
– (solute/solution) x 100% = % Concentration
• Molarity (M)
– Moles of solute/Liters of solution =
mol/L
• Molality (m)
– Moles of solute/mass of solvent =
mol/kg
Molarity or Concentration
a. Definition: number of moles of
solute per liter of solution
1 L = 1 dm3 = 103mL = 103cm3 = 103cc
b. Abbreviation: M
Units: mol/L
c. Preparation of solutions
Need to know the desired volume & calculate the
mass of needed solute.
Prepare 500. mL of 1.0 M NaCl
29 grams of NaCl to a 500Transfer ________
mL volumetric flask, and add water to the
line.
*Note: Always add acid to water.
Problems
1.
Calculate the molarity if 37 g of NaCl are dissolved in 150 mL
of solution.
4.2 M NaCl
2.
How many moles of HCl are present in 145 mL of a 2.25 M
HCl solution?
0.326 mol HCl
3.
How many grams of NaCl are contained in 2.5 L of a 1.5 M
solution?
220 g NaCl
Example Problems
1. Calculate the molarity of a solution
that contains 8.50 g of calcium nitrate
in 2.0 L.
2. Calculate the molality of a solution
that contains 8.50 g of calcium nitrate
in 125 g H2O.