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TOPIC 1
STOICHIOMETRIC
RELATIONSHIPS
1.2
THE MOLE CONCEPT
ESSENTIAL IDEA
The mole makes it possible to
correlate the number of particles
with the mass that can be
measured.
NATURE OF SCIENCE (2.3)
Concepts – the concept of the mole
developed from the related concept of
“equivalent mass” in the early 19th century.
UNDERSTANDING/KEY IDEA
1.2.A
The mole is a fixed number of
particles and refers to the
amount, n, of substance.
A chemical species may be an atom, a
molecule or an ion.
This is the same as using the terminology
that “particles” refer to atoms, molecules and
formula units. (Note that formula units are
made up of ions.)
The exact definition of a mole is the amount
of a substance that contains the same
number of chemical species as there are
atoms in exactly 12g of the isotope
carbon-12.
Avogadro’s constant (L) has the value
6.02x1023 particles per mole.
The mole is the chemist’s counting unit.
It contains 6.02 x 1023 particles of a
substance.
Example:
1 dozen = 12 of something
1 mole = 6.02 x 1023 of something
What are species (particles)?



Atoms – smallest species of an element
Molecules – smallest species of a covalent cmpd
Ions – smallest species of an ionic cmpd
When we want to count atoms or molecules,
we use Avogadro’s # as a conversion factor.
There are 6.02 x 1023 number of atoms in
one mole of an element.
There are 6.02 x 1023 number of molecules
in a covalent compound.
UNDERSTANDING/KEY IDEA
1.2.B
Masses of atoms are
compared on a scale relative
to 12C and are expressed as
relative atomic mass (Ar) and
relative formula/molecular
mass (Mr).
1 The relative atomic mass (Ar) is the
average mass of an atom, taking into
account the relative abundances of all the
naturally occurring isotopes of the element.
2 The relative molecular mass (Mr) of a
covalent or molecular compound is the sum
of all the atomic masses of the elements in
the compound.
3 The relative formula mass (Mf) of an ionic
compound is the sum of all the atomic
masses of the ions in the formula.
 Notice that Ar, Mr and Mf have no units
because they are “relative” terms.
UNDERSTANDING/KEY IDEA
1.2.C
Molar mass (M) has the
units g/mol.
The mass of one mole of a species is
called the molar mass (M) and it has the
units of g/mol or g mol-1.
APPLICATION/SKILLS
Be able to calculate the molar
masses of atoms, ions, molecules
and formula units.
To calculate molar mass, add up the
atomic masses of the separate atoms
from the periodic table.
CO2 1-C = 1 x 12 = 12
2-O = 2 x 16 = 32
44 g/mol
CH4 1-C = 1 x 12 = 12
4-H = 4 x 1 = 4
16 g/mol
FIND THE FOLLOWING MOLAR MASSES
1. NaHCO3
2. Sr(CN)2
3. Al2(SO3)3
4. C12H22O11
5. (NH4)2CO3
APPLICATION/SKILLS
Be able to solve problems involving
the relationships between the
number of particles, the amount of
substance in moles and the mass
in grams.
Converting from moles to particles
or particles to moles
To convert from moles to particles or
particles to moles, use Avogadro’s number
as the conversion factor.
MASS
(g)
PARTI
CLES
MULTIPLY BY
6.02X1023 PART
MOL
ATOMS
MOLECULES
FORMULA UNITS
DIVIDE BY
6.02X1023 PART
MOL
MOLE
VOLUME
(L)
Mole/Particle Conversion Problems

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1. How many atoms are in 3.0 mol of Sn?
2. How many formula units are in 0.40 mol KCl?
3. How many molecules are in 7.50 mol SO2?
4. How many formula units are in 4.80x10-3 mole NaI?
5. How many moles are in 1.50x1023 molecules of NH3?
6. How many moles are in 1.0x109 molecules of O2?
7. How many moles are in 6.02x1023 molecules of Br2?
8. How many moles are in 4.81x1024 atoms of Li?
Converting from moles to mass or
mass to moles
To convert from moles to mass or mass to
moles, use the molar mass as the
conversion factor.
MASS
(g)
PARTI
CLES
MULTIPLY BY
MOLAR MASS
DIVIDE BY
MOLAR MASS
MOLE
VOLUME
(L)
ATOMS
MOLECULES
FORMULA UNITS
MASS/MOLE CONVERSIONS

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1. What is the mass of 2.40 mol N2?
2. How many moles is 5.00g H2?
3. What is the mass of 3.32 mol K?
4. How many moles is 11.0g CH4?
5. What is the mass of 0.160 mol H2O2?
6. How many moles is 847g (NH4)2CO3?
7. What is the mass of 5.08 mol Ca(NO3)2?
8. How many moles is 333g SnF2?
9. What is the mass of 10.0 mol Cr?
10. How many moles is 0.000264g Li2HPO4?
UNDERSTANDING/KEY IDEA
1.2.D
The empirical formula and
molecular formula of a
compound give the simplest
ratio and the actual number of
atoms present in a molecule
respectively.
Empirical formula – the simplest whole
number ratio of the atoms it contains.
Molecular formula – multiple of the empirical
formula showing the actual number of atoms
of each element present.
APPLICATION/SKILLS
Be able to determine the empirical
formula when given the percentage
composition by mass.
Rules for empirical formula
1. Find the % composition.
2. Assume 100 g and convert %’s to g’s.
3. Convert the grams to moles by dividing
by the molar mass.
4. Divide all mole quantities by the smallest
number of moles found in step 3.
5. If the numbers found in step 4 are not
whole numbers, multiply by a factor to make
whole numbers.
Sample Problem
A hydrocarbon contains 85.7% by mass of carbon. What is
the empirical formula?

Step 1: Find %’s
• 85.7 % C
• 100-85.7 = 14.3% H (Hydrocarbons contain H and C)

Step 2: Convert to grams
• 85.7 g C
• 14.3 g H

Step 3: Convert to moles
• 85.7 g C x mol/12.01 g C = 7.14 mol C
• 14.3 g H x mol/1.01 g H = 14.16 mol H

Step 4: Divide by the smallest number of
moles.
• 7.14 mol / 7.14 mol = 1
• 14.16 mol / 7.14 mol = 1.98 = 2

Step 5: The numbers in Step 4 are whole
numbers so this is the end of the problem.
• The empirical formula is CH2
APPLICATION/SKILLS
Be able to determine the molecular
formula of a compound from its
empirical formula and molar mass.
 Hint:
You have to be given the
experimental molar mass in order to solve
a molecular formula problem.
Rules for molecular formula
1. Solve for empirical formula. (It may be
given to you or you may have to solve for it.)
2. Find the empirical molar mass.
3. Divide the given molecular or molar mass
by the empirical molar mass – this answer
should be a whole number.
4. Multiply the empirical formula by the
number found in step 3.
Sample problem
The compound with the empirical formula CH2 is analyzed
by a mass spectrometer and found to have a relative
molecular mass of 42.09. Deduce its molecular formula.


Step 1: The empirical formula is CH2.
Step 2: Find the empirical molar mass.
• CH2 has a molar mass of 14.03 g/mol.

Step 3: Divide the given molar mass by the
empirical molar mass.
• 42.09 / 14.03 = 3

Step 4: Multiply the empirical formula by the
number found in Step 3.
• (CH2)3 = C3H6
APPLICATION/SKILLS
Be able to obtain and use
experimental data for deriving
empirical formulas from reactions
involving mass changes.
EMPIRICAL FORMULAS FROM
COMBUSTION
1. Sometimes you are not given the percentages,
but have to calculate them from a combustion
problem.
2. The problems usually give you an amount of
original substance and amounts of CO2 and H2O
that are formed from the combustion reactions.
3. Make these assumptions:
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All of the carbon in the original sample is converted
completely to CO2.
All of the hydrogen in the original sample is converted
completely to H2O.
EXAMPLE
You have a compound composed of carbon, hydrogen and
nitrogen. When .1156g of this compound is burned, .1638g
of CO2 and .1676g of H2O is produced. Determine the
empirical formula of the compound.
Citations
Brown, Catrin, and Mike Ford. Higher Level
Chemistry. 2nd ed. N.p.: Pearson Baccalaureate,
2014. Print.
Most of the information found in this power point
comes directly from this textbook.
The power point has been made to directly
complement the Higher Level Chemistry textbook by
Catrin and Brown and is used for direct instructional
purposes only.