THE MOLE
The Chemist’s Package
HOW DO WE GROUP THINGS?
Pairs,
 Dozen,
 ???


Chemists group chemicals into moles.

A mole of any chemical is the same amount.

Just like a dozen is always 12…
FROM ONE TO ANOTHER…
MOLES TO MOLES

We can change from one substance to another
using a ratio.


Usually a part to a whole
This ratio is a mole ratio.

It converts from moles of one thing to moles of
another.

____moles X = ____moles Y

Ex. H2O : Part to Whole


2 moles H = 1 mole H2O
1 mole O = 1 mole H2O
HOW MANY ARE THERE?
MOLES TO PARTICLES

Moles can also be converted to tell us how many
particles there are.


Particles can be atoms (atom), ions (ion),
molecules(molec), and formula units (fmu).
There are always 6.022 x 1023 particles in a mole.

This number is known as Avagadro’s Number.
Ex. H2O: 1 mole H2O = 6.022 x 1023 molec H2O
 Ex. Na:
1 mole Na = 6.022 x 1023 atoms Na

HOW MUCH IS IT?
MOLES TO GRAMS
The mass of one mole is a chemical substance’s
molar mass.
 Each substance has its own molar mass that can
be determined using the periodic table.
 The molar mass of an element is equal to its
atomic mass.



Ex. Na: 1 mol Na = 22.99 g Na
The molar mass of a compound is equal to the
sum of the atomic masses of the elements that
are in the compound.

Ex. H2O: 1 mole H2O = 18.02 g H2O
(2H + 1O)
HOW BIG IS IT?
MOLES TO VOLUME

The volume of a mole of gas under standard
conditions is always the same.


Standard Conditions of Temperature and Pressure
(STP) are:


This is not true under nonstandard conditions or for
liquids and solids.
273 K and 101.3 kPa
At STP, 1 mole of gas has a volume of 22.4 L.

This is called molar volume.
Ex. O2 at STP: 1 mole O2 = 22.4 L O2
 Ex. CO2 at STP: 1 mole CO2 = 22.4 L CO2

CONVERSION FACTORS

These are all conversion factors.
Mole to Mole – Mole Ratio
 Mole to Particle – Avagadro’s Number
 Mole to Mass – Molar Mass
 Mole to Volume – Molar Volume

VOLUME
(L)
Molar
Volume
Avagadro’s
Number
MOLE
(mol)
PARTICLES
(atoms, ions, molec, fmu)
MOLES OF
ANOTHER
Mole Ratio
Molar
Mass
MASS
(g)
PRACTICE PROBLEMS
MOLE TO MOLE

How many moles of carbon are in 2.0 moles of
sucrose, C12H22O11?
Known: 2.0 moles sucrose Unknown: ? moles carbon
 Conversion Factor: Mole Ratio 1 mol C12H22O11 = 12 mol C

2.0 mol C12H22O11
12 mol C
=
24 mol C
1 mol C12H22O11

How many moles of sucrose will contain 12 moles
of hydrogen?
K: 12 mol H
UK: ? Mol C12H22O11
 CF: Mole Ratio - 22 mol H = 1 mol C12H22O11
12 mol H
1 mol C12H22O11 =
0.55 mol C12H22O11
22 mol H

PRACTICE PROBLEMS
MOLE TO PARTICLE

How many moles are equal to 2.41 x 1024 formula units of
sodium chloride?
K: 2.41 x 1024 fmu NaCl
UK: ? mol NaCl
 CF: Avagadro’s Number - 1 mol NaCl = 6.022 x 1023 fmu NaCl


How many moles are equal to 9.03 x 1024 atoms of mercury?



How many atoms are equal to 4.50 moles of copper?



K: 9.03 x 1024 atoms Hg
UK: ? mol Hg
CF: 1 mol Hg = 6.022 x 1023 atoms Hg
K: 4.50 mol Cu
UK: ? atoms Cu
CF: 1 mol Cu = 6.022 x 1023 atoms Cu
How many molecules are equal to 100.0 moles of carbon
dioxide?


K: 100.0 moles CO2
UK: ? molec CO2
CF: 1 mol CO2 = 6.022 x 1023 molec CO2
PRACTICE PROBLEMS
MOLES TO PARTICLES (2 STEPS)

How many atoms of oxygen are in 3.65 moles of
sucrose?
K: 3.65 mol C12H22O11
 CF: Mole Ratio
 CF: Avagadro’s Number

3.65 mol C12H22O11
UK: ? atoms O
1 mol C12H22O11 = 11 mol O
1 mol O= 6.022 x 1023 atoms O
11 mol O
1 mol C12H22O11
6.022 x 1023 atoms O
=
1 mol O
= 2.42 x 1025 atoms O

How many atoms are in 1.00 mole of sucrose?
PRACTICE PROBLEMS
MOLAR MASS

Determine the molar mass of the following
compounds.
 Carbon dioxide
CO2 : 1 Carbon + 2 Oxygen = 12.01g + 2(16.00 g) = 44.01 g CO2

Sulfur trioxide
SO3 : 1 Sulfur + 3 Oxygen =

Bromine
Br2 : 2 Bromine =

Sodium hydroxide
NaOH : 1 Sodium + 1 Oxygen + 1 Hydrogen =

Barium nitrate
Ba(NO3)2 : 1 Barium + 2 Nitrogen + 6 Oxygen =
PRACTICE PROBLEMS
MOLES TO MASS
Calculate the mass in grams of 0.250 moles of the following
compounds.
K: 0.250 moles
UK: ? grams
 Sucrose


Sodium chloride

Potassium permanganate

Calcium sulfide

Lithium chlorate
PRACTICE PROBLEMS
MOLES TO MASS
Calculate the number of moles in 100.0 grams of the following
compounds.
K: 100.0 grams
UK: ? moles
 Sucrose
CF:


Sodium chloride
CF:

Potassium permanganate

Calcium sulfide
CF:

Lithium chlorate
CF:
CF:
PRACTICE PROBLEMS
MOLES TO VOLUME

What is the volume in liters of 8.35 moles of sulfur trioxide gas at
STP?
K: 8.35 mol SO3
CF: 1 mol SO3 = 22.4 L SO3

What is the volume in liters of 0.750 moles of carbon dioxide gas
at STP?
K: 0.750 mol CO2
CF: 1 mol CO2 = 22.4 L CO2

UK: ? L CO2
How many moles are there in 52.5 liters of oxygen gas at STP?
K: 52.5 L O2
CF: 1 mol O2 = 22.4 L O2

UK: ? L SO3
UK: ? mol O2
How many moles are there in 15.0 liters of nitrogen gas at STP?
K: 15.0 L N2
CF: 1 mol N2 = 22.4 L N2
UK: ? mol N2
MULTI-STEP PROBLEMS


Same conversions…
just more than one at a time…
Don’t solve them any differently…
Identify your known and unknown.
 Identify the conversion factors you need.



Let the Mole Map do the work for you…
Solve.
MULTI-STEP PRACTICE

How many grams of carbon dioxide are there in 4.20
L of carbon dioxide at STP?

What is the volume of 245 grams of water vapor at
STP?

How many formula units are there in 63.4 grams of
barium hydroxide?
MULTI-STEP PRACTICE

What is the volume of 6.21 x 1031 molecules of sulfur
trioxide gas at STP?

What is the mass of 5.23 x 1018 atoms of gold?

How many molecules of silicon dioxide gas does it
take to fill a 6.0 L container at STP?
MULTI-STEP PRACTICE

How many atoms of carbon are there in 4.24 grams of
carbon tetrabromide?

How many liters of carbon dioxide gas are there if you have
4.7 x 1020 atoms of oxygen?

If you have 3.62 x 1024 atoms of nitrogen, then what is the
maximum number of grams of dinitrogen pentoxide that
can be formed?
EMPIRICAL FORMULAS

Empirical Formulas are the LOWEST whole
number ratio of elements in a compound.


Sometimes this is the same as the chemical formula
Examples:
C2H6 → CH3
 C6H12O6 → CH2O


Empirical Formulas can be determined
experimentally from the percent composition of a
compound.
EMPIRICAL FORMULA DETERMINATION
STEPS

Assume you have 100 g of the substance (makes
the math easier because everything is a straight
percent).



Consider the amounts you are given as being in units
of grams.
Convert the grams to moles for each element.
Find the smallest whole number ratio of moles
for each element by dividing all values by the
smallest value.
EXAMPLE: FIND THE EMPIRICAL FORMULA FOR A
COMPOUND CONSISTING OF
63% MN AND 37% O
Solution for Finding the Empirical Formula
 Assuming 100 g of the compound, there would be 63 g Mn
and 37 g O
 Convert grams to moles. (Use molar mass from the
periodic table.)
 63 g Mn × (1 mol Mn)/(54.94 g Mn) = 1.1 mol Mn
 37 g O × (1 mol O)/(16.00 g O) = 2.3 mol O
 Find the smallest whole number ratio by dividing the
number of moles of each element by the number of moles
for the element present in the smallest molar amount. (In
this case there is less Mn than O, so divide by the number
of moles of Mn)
 1.1 mol Mn/1.1 = 1 mol Mn
 2.3 mol O/1.1 = 2.1 mol O
 The best ratio is Mn:O of 1:2 and the formula is MnO2
The empirical formula is MnO2
EMPIRICAL FORMULA CALCULATION
WORKSHEET
1. 88.8% Cu and 11.2% O
 Assume you have 100 grams.



88.8 g Cu and 11.2 g O
Convert grams to moles

88.8g Cu

11.2g O
1 mol Cu
63.55g Cu
1 mol O
16.00g O
= 1.40 mol Cu
= 0.700 mol O
Find the smallest whole number ratio.
1.40 mol Cu / 0.700 = 2
 0.70 mol O /0.700 = 1


So the empirical formula is Cu2O.
MOLECULAR FORMULA
FROM EMPIRICAL FORMULA


The empirical formula is the lowest ratio of elements
and not always the actual ratio of elements in a
compound.
The actual ratio (molecular formula) can be
determined using the molar mass of the actual
compound and the molar mass of the empirical
formula.
The molar mass of the molecular formula is experimentally
determined (Given).
 The molar mass of the empirical formula can be
determined from the periodic table.



The molecular mass is divided by the empirical mass
and equals a whole number.
The whole number is multiplied by the subscripts in
the molecular formula to give the empirical formula.
EXAMPLE: FIND THE MOLECULAR FORMULA FOR A
COMPOUND THAT HAS AN EMPIRICAL FORMULA OF
CH2 AND A MOLECULAR MASS OF 41.5G.

The molecular mass is 41.5 grams.


Given
The empirical mass is 14.03 grams.

C + 2H = 12.01 + 2(1.008) = (14.026)14.03

41.5 / 14.03 = (2.96) 3

3 (CH2) = C3H6

The molecular formula is C3H6 .
MOLECULAR FORMULA PRACTICE
Use Empirical Formulas from Worksheet with the
following Molecular Masses.
1. 290 g/mol
2. 178 g/mol
3. 26 g/mol
4. 160 g/mol
5. 34 g/mol
6. 160 g/mol
7. 138 g/mol
8. 240 g/mol
9. 245 g/mol
10. 230 g/mol