Chapter 7
The Mole: A Measurement of
Matter
At the end of this chapter, you should
be able to:
Describe how Avogadro’s number
is related to a mole of any substance
Calculate the mass of a mole of
any substance
The Mole (aka Avagadro’s Number):
6.02 x
23
10
The Mole and Avogadro’s Number
SI unit that measures the amount of
substance
1 mole = 6.02 x 1023 representative
particles
Representative particles are usually
atoms, molecules, or formula units
(ions)
But Why the Mole?
Just as 12 = 1 dozen, or 63,360 inches = 1 mile,
the mole allows us to count microscopic items
(atoms, ion, molecules) on a macroscopic scale.
So, 1 mole of any substance is a set number of
Items, namely: 6.02 x 1023.
Chemistry = awesome
Examples:
Substance
Representative
Particle
Chemical
Formula
Representative
Particles in 1.00
mol
Atomic
nitrogen
Atom
N
6.02 x 1023
Water
Molecule
H2O
6.02 x 1023
Calcium
ion
Ion
Ca2+
6.02 x 1023
Solve
Substance
Nitrogen
gas
Calcium
Fluoride
Sucrose
Carbon
Representative
Particle
Chemical
Formula
Representative
Particles in
1.00 mol
Answers
Nitrogen gas-molecule-N2
Calcium fluoride-formula unit-CaF2
Sucrose-molecule-C12H22O11
Carbon-atom-C
All have 6.02 x 1023 representative
particles in 1.00 mol
How many atoms are in a mole?
Determined from the chemical formula
List the elements and count the atoms
Solve for CO2
C - 1 carbon atom
O - 2 oxygen atoms
Add: 1 + 2 = 3
Answer: 3 times Avogadro’s number of
atoms
Solve: How many atoms are in a
mole of
1.
2.
3.
4.
Carbon monoxide – CO
Glucose – C6H12O6
Propane – C3H8
Water – H2O
How many moles of magnesium is
1.25 x 1023 atoms of magnesium?
Refer to page 174 in text
Divide the number of atoms or
molecules given in the example by
6.02 x 1023
Divide (1.25 x 1023) by (6.02 x 1023)
Express in scientific notation
Answer = 2.08 x 10-1 mol Mg
Objectives
Use the molar mass to convert
between mass and moles of a
substance
Use the mole to convert among
measurements of mass, volume, and
number of particles
Molar mass
Mass (in grams) of one mole of a
substance
Broad term (can be substituted) for
gram atomic mass, gram formula
mass, and gram molecular mass
Can be unclear: What is the molar
mass of oxygen?
O or O2 ? - element O or molecular
compound O2 ?
Molar Mass
Gram atomic mass (gam) – atomic
mass of an element taken from the
periodic table
Gram molecular mass (gmm) – mass
of one mole of a molecular compound
Gram formula mass (gfm) – mass of
one mole of an ionic compound
Can use molar mass instead of gam,
gmm, or gfm
Calculating the Molar Mass of
Compounds (Molecular and Ionic)
1. List the elements
2. Count the atoms
3. Multiply the number of atoms of
the element by the atomic mass of the
element (atomic mass is on the
periodic table)
4. Add the masses of each element
5. Express to tenths place
What is the molar mass (gfm) of
ammonium carbonate (NH4)2CO3?
N 2 x 14.0 g = 28.0 g
H 8 x 1.0 g = 8.0 g
C 1 x 12.0 g = 12.0 g
O 3 x 16.0 g = 48.0 g
Add
________
Answer
96.0 g
Practice Problems
1. How many grams are in 9.45 mol
of dinitrogen trioxide (N2O3) ?
a. Calculate the grams in one mole
b. Multiply the grams by the number
of moles
2. Find the number of moles in 92.2 g
of iron(III) oxide (Fe2O3).
a. Calculate the grams in one mole
b. Divide the given grams by the
grams in one mole
Answers
1. 718 g N2O3 (one mole is 76.0g)
2. 0.578 mol Fe2O3 (one mole is
159.6 g)
Volume of a Mole of Gas
Varies with a change in temperature
or a change in pressure
At STP, 1 mole of any gas occupies a
volume of 22.4 L
Standard temperature is 0°C
Standard pressure is 101.3 kPa
(kilopascals), or 1 atmosphere (atm)
22.4 L is known as the molar volume
22.4 L of any gas at STP contains
6.02 x 1023 representative particles of
that gas
One mole of a gaseous element and
one mole of a gaseous compound
both occupy a volume of 22.4 L at
STP (Masses may differ)
Study Figure 7.13 on page 186
Molar mass (g/mol) = Density (g/L) x
Molar Volume (L/mol)
Objectives
Define the terms
Calculate the percent composition of
a substance from its chemical formula
or experimental data
Derive the empirical formula and the
molecular formula of a compound
from experimental data
Terms to Know
Percent composition – relative
amounts of each element in a
compound
Empirical formula – lowest wholenumber ratio of the atoms of an
element in a compound
An 8.20 g piece of magnesium
combines completely with 5.40 g of
oxygen to form a compound. What
is the percent composition of this
compound?
1. Calculate the total mass
2. Divide each given by the total mass
and then multiply by 100%
3. Check your answer: The
percentages should total 100%
Answer
The total mass is 8.20 g + 5.40 g =
13.60 g
Divide 8.2 g by 13.6 g and then
multiply by 100% = 60.29412 = 60.3%
Divide 5.4 g by 13.6 g and then
multiply by 100% = 39.70588 = 39.7%
Check your answer: 60.3% + 39.7% =
100%
Calculate the percent composition
of propane (C3H8)
1. List the elements
2. Count the atoms
3. Multiply the number of atoms of
the element by the atomic mass of the
element (atomic mass is on the
periodic table)
4. Express each element as a
percentage of the total molar mass
5. Check your answer
Answer
Total molar mass = 44.0 g/mol
36.0 g C = 81.8%
8.0 g H = 18.2%
Calculate the mass of carbon in
82.0 g of propane (C3H8)
1. Calculate the percent composition
using the formula (See previous
problem)
2. Determine 81.8% of 82.0 g
Move decimal two places to the
left (.818 x 82 g)
3. Answer = 67.1 g
Calculating Empirical Formulas
Microscopic – atoms
Macroscopic – moles of atoms
Lowest whole-number ratio may not
be the same as the compound
formula
Example: The empirical formula of
hydrogen peroxide (H2O2) is HO
Empirical Formulas
The first step is to find the mole-to-mole
ratio of the elements in the compound
If the numbers are both whole numbers,
these will be the subscripts of the elements
in the formula
If the whole numbers are identical,
substitute the number 1
Example: C2H2 and C8H8 have an empirical
formula of CH
If either or both numbers are not whole
numbers, numbers in the ratio must be
multiplied by the same number to yield
whole number subscripts
What is the empirical formula of a
compound that is 25.9% nitrogen
and 74.1% oxygen?
1. Assume 100 g of the compound, so that
there are 25.9 g N and 74.1 g O
2. Convert to mole-to-mole ratio:
Divide each by mass of one mole
25.9 g divided by 14.0 g = 1.85 mol N
74.1 g divided by 16.0 g = 4.63 mol O
3. Divide both molar quantities by the
smaller number of moles
4. 1.85/1.85 = 1 mol N
4.63/1.85 = 2.5 mol O
5. Multiply by a number that converts
each to a whole number (In this case,
the number is 2 because 2 x 2.5 = 5,
which is the smallest whole number )
2 x 1 mol N = 2
2 x 2.5 mol O = 5
Answer: The empirical formula is
N 2O 5
Determine the Empirical Formulas
1. H2O2
2. CO2
3. N2H4
4. C6H12O6
5. What is the empirical formula of a
compound that is 3.7% H, 44.4% C,
and 51.9% N?
Answers
Compound
1. H2O2
2. CO2
3. N2H4
4. C6H12O6
5. HCN
Empirical Formula
HO
CO2
NH2
CH2O
Calculating Molecular Formulas
The molar mass of a compound is a
simple whole-number multiple of the
molar mass of the empirical formula
The molecular formula may or may
not be the same as the empirical
formula
Calculate the molecular formula of
the compound whose molar mass
is 60.0 g and empirical formula is
CH4N.
1. Using the empirical formula, calculate
the empirical formula mass (efm)
(Use the same procedure used to calculate
molar mass.)
2. Divide the known molar mass by the
efm
3. Multiply the formula subscripts by this
value to get the molecular formula
Answer
Molar mass (efm) is 30.0 g
60.0 g divided by 30.0 g = 2
Answer: C2H8N2
The G.U.S. Method
Allows students to organize data in an easy
way. More importantly, makes my grading job
easier. Works like so:
G: the given. Write down all data given in the
problem WITH proper units.
U: the unknown. Write down what you are
looking for AND the unit.
S: solve. Show ALL work WITH units.