UNIT 2 –
QUANTITATIVE
CHEMISTRY
(I.E. IN COMES THE MATHS!! BRING A CALCULATOR!!)
Scientific Notation… A (very) brief
introduction
Stoichiometry

The study of the relationships between the
quantities of reactants and products involved in
chemical reactions.

Stoikheion = element

metron = measure
Proportions in compounds

The composition of a compound depends on two
things

1. The elements they are composed of


2. The quantities of each element.
Law of definite proportions

A specific compound always contains the same elements in
definite proportions by mass.

This is always constant even if produced other ways.
Relative atomic mass and isotopic
abundance (p80)

Relative atomic mass = the mass of an element that would
react with a fixed mass of a standard element, currently
carbon-12.

One atomic mass unit (u) is described as 1/12 the mass of a
carbon-12 atom.
Atomic mass unit

Now we know that it weighs
1

u = 1.660 540 2 x 10-27 kg.
This is the standard used world wide
Isotopic abundance

Some elements have isotopes. In calculating the relative
atomic mass of an element with isotopes, the relative mass
and proportion of each is taken into account. For example,
naturally occurring carbon consists of atoms of relative
isotopic masses C-12 (98.89%) and-C 13 (1.11%). Its relative
atomic mass is 12.01 u.
Isotopic abundance

mc = (98.89/100 x 12) + (1.11/100 x 13) =

mc = 11.8668 + 0 .1443 = 12.01 u

No atom will contain 12.01 u. it is an average.
Example

Cl-35 = 75.53%

Cl-37 = 24.47%

(35 x .7553) + (37 x .2447) = 35.49 u

Therefore the average atomic mass is 35.49u

Page 81 # 1
Atomic Mass and Molecular Mass

Atomic Mass
 The
mass of one atom of an element
expressed in atomic units, u.

Molecular mass
 The
mass of one molecule, expressed in
atomic mass units, u.
Example

MNH3 = 1(mN) + 3(mH)

= 1(14.01 u) + 3(1.01 u)

= 17.04 u .

Therefore, one molecule of ammonia has a mass of 17.04
units.
Formula unit of mass

The mass of one formula unit of an ionic compound,
expressed in atomic mass units, u
Example

NaCl = 1(mNa) + 1(mCl)

= 1 (22.99 u) + 1(35.45 u)

= 58.44 u

Therefore, one formula unit of sodium chloride has a mass of
58.44 units.

Page 81 # 1-3
The Periodic Table and the Mass of
Elements

The atomic number of a substance is written near the top of
the symbol. This number tells the number of protons in the
element.

The Atomic mass is usually written near the bottom. This
number is given as a numerical number of the elements mass.
The mass of one mole of the element measured in grams.
The Mole
It's not a spy, a machine for digging tunnels, a burrowing
animal, or a spot of skin pigmentation - it's
- a unit of measurement containing about
6.02 x 1023
particles.

Chemists created the mole as their own counting unit.
1 mole of atoms is
602, 214,199, 000, 000, 000, 000, 000
atoms!
(This is the number of Carbon atoms there are in 12.01 g or
Carbon)
Avogadro’s Number
6.02 x 1023 particles / mole
When you say you have a mole, it’s kind of like
saying you have ‘a dozen’ of something…
A dozen donuts = 12 donuts
A dozen apples = 12 apples
A mole of CO2 molecules = 6.02 x 1023 molecules
A mole of Na atoms = 6.02 x 1023 atoms
Just How Big is a Mole?

Enough soft drink cans to cover the surface
of the earth to a depth of over 200 miles.


If you had Avogadro's number of unpopped
popcorn kernels, and spread them across
the United States of America, the country
would be covered in popcorn to a depth of
over 9 miles.

If we were able to count atoms at the rate
of 10 million per second, it would take
about 2 billion years to count the atoms in
one mole.
Molar mass (g/mol) = M

the mass, in grams per mole, of one mole of a substance. The
SI unit for a mol is g/mol

Molar mass is conventionally abbreviated with a capital M. For
elements it is numerically equal to the atomic mass and is
often called atomic mass units.
Steps to calculating molar mass
1.
Write the chemical formula of the substance
2.
Determine the number of atoms or ions of each
element in one formula unit of the substance.
3.
Us the atomic molar masses from the periodic table
and the amounts in moles to determine the molar
mass of the chemical
4.
Write the molar mass in units of grams per mol
(g/mol).
Example

Molar mass of copper(Cu) = mass of 1 mol of Cu atoms

M = 63.546 g/mol Cu

M = mass of 6.022 x 10 23 atoms
Example 2







Molar mass of glucose (C6H12O6) = mass of
exactly 1 mol of glucose molecules
M = (6 x C) + (12 x H) + ( 6 x O)
M = ((6 x 12.01 (C)) + (12 x 1.01(H)) + (6x 16.00
(O)) ) / mol
M = ( (72.06 g) + (12.12 g) + (96.00g) ) / mol
M = ( 180.18 g) / mol
M = 180.18 g/mol
Therefore, 1 mol glucose has a mass of 180.18 g
and contains 6.022 x 1023 glucose
molecules(C6H12O6)
Important things to note

Specify what is being measured

Ex. molar mass of oxygen atom = 16.00 g/mol
 Molar
mass of oxygen molecules = 32.00 g/mol

When talking about oxygen we are normally referring to
the diatomic molecule

Since electrons have negligible mass compared to
protons and neutrons, ions will essentially have the same
mass as the atoms.
Moles of Elements

Why that number?
1
mole element = (atomic mass) in grams of that element
1 mole C = 12.011 g C
1 mole B = 10.81 g B
1 mole Cu = 63.55 g Cu

i.e. atoms are teeny, teeny, tiny, teeny, itty, bitty things and it is not reasonable to
talk about them in their atomic weights, but if scale them up to moles, then
working with balanced chemical equations etc. becomes easier.
Determine the mass one mole of the
following in grams:
a.
CO2
b.
NaCl
c.
K2CrO4
d.
Fe(NO3)3

PAGE 85 # 4
The Mole and Molar Mass

The Mole = n

Avogadro’s constant (NA) = the number of entities in one
mole
= 6.02 x1023

Eg. 1 dozen oxygen atoms = 12 oxygen atoms
1
mole oxygen atoms = 6.022 x 1023 oxygen atoms
Calculations involving Molar mass

Amount in moles (n) =
mass(g)
Molar mass(g/mol)
The Magic Triangle
n=m/M
m=nxM
M=m/n
GRASP method

1 Given = What do you know?

2 Required = What do you want to know?

3 Analysis = What formula can you use?

4 Solution = Show your work

5 Paraphrase = State the answer in a sentence.
Example

Lets say that we are given 22 grams of graphite.
How many moles do we have?

Lets say we want to know how much CaCl2 we need for
a reaction that asks for 13.52 moles

You are given an unknown sample that has a mass of
32.04 grams, 2.000 moles. Find M.
Practice:
1.
Determine the number of mol in 13.7 g of CO
2.
What is the mass of 2.75 mol of C3H8
Homework

Page 86 # 5-7

Isotopic Abundance and Mole calculations Worksheet
Calculating Number
of Entities
Calculating Number of Entities (N)
from Mass

We are not able to count the number of atoms in reactions; we
must count them indirectly by measuring out a certain mass of
a substance and calculating the number of moles.

In order to estimate the number of entities in a large sample
we need to take a small sample and find out the mass of a
small number of entities.
Calculating Number of Entities
and Avogadro’s number

There are times when need to know exactly how many atoms
are in something. This is especially true when dealing with
gases as they change in pressure with temperature.

To find the number of entities we use the formula

N = nNA

Number of entities = (# of mols) x(Avogadro's #)
Example

Find the number of atoms in 0.004 moles of water.

Find the number of molecules in 30.0 g of Cl2 (g)
Determining the number of particles in
a mole
Avogadro’s Number as
Conversion Factor
6.02 x 1023 particles
1 mole
or
1 mole
6.02 x 1023 particles
Note that a particle could be an atom OR a molecule!
Particles
Molecules
Compound
Formula
unit
Atom
Element
Equation to find number of particles!
number of moles x
6.02 x 1023 particles
= number of particles
1 mole
Example: How many particles of sucrose are in 3.50 moles?
Equation to find number of moles!
number of particles x
1 mole
= number of moles
6.02 x 1023 particles
Example: How many moles are contained in 4.50 x 1024 atoms of zinc?
Learning Check
1. Number of atoms in 0.500 mole of Al
2.Number of moles of S in 1.8 x 1024 S atoms
Molar Mass
Molar Mass

The Mass of 1 mole (in grams)

Molar mass is equal to the atomic mass (look on
the periodic table)
1 mole of C atoms = 12.01 g
1 mole of Au atoms = 196.97g
1 mole of Cu atoms = 63.55 g

Always round to the hundredths place.
Mole to Mass Conversion
Where will you
find the number
of grams per
mole?
Number of moles x number of grams = mass
1 mole
The
Periodic
Table
Example: Calculate the mass in grams of 0.0450
Look at your
moles of chromium.
Periodic Table
0.0450 moles Cr x 52.0 g Cr = 2.34 g Cr
1 mol Cr
Mass to Mole
Conversion
Mass x
1 mole
number of grams
Where will you
find the number
of grams per
mole?
= number of moles
The
Periodic
Table
Example: How many moles of calcium are there in
525 grams of calcium?
525 g Ca x
1 mol Ca = 13.1 mol Ca
40.1 g Ca
Look at your
Periodic Table
Mass to Atoms Conversion
Two steps:
1.
Use the mass to moles conversion
2.
Now calculate the number of atoms using the moles to
particles equation.
Example:
How many atoms of gold are in a pure gold nugget
having a mass of 25.0 g?
Atoms/Molecules and Grams
How many atoms of Cu are present in 35.4 g of
Cu?
35.4 g Cu
1 mol Cu
6.02 X 1023 atoms Cu
63.5 g Cu
1 mol Cu
= 3.4 X 1023 atoms Cu
Learning Check!
How many atoms of K are present in 78.4 g of
K?
78.4 g K
1 mol K
6.02 X 1023 atoms K
39.1 g K
1 mol K
= 1.20 X 1024 atoms K
Empirical and Molecular Formulas
11.4
Percent Composition
Mass of element
x 100 = percent by mass
Mass of compound
Example: Determine the percent composition of NaHCO3.
Determine the mass of each element first!
%mass element = mass of element in 1 mol compound x 100
molar mass of compound
Percent Na =
22.99 g Na
x 100 = 27.37% Na
84.01 g NaHCO3
You have to complete this step for each element in the compound!!!!
Determine the Percent Composition
from the Formula C2H5OH
l
Divide the mass of each element by the
molar mass of the compound and multiply
by 100% 24.02g
 100%  52.14%C
46.07g
6.048g
 100%  13.13%H
46.07g
16.00g
 100%  34.73%O
46.07g
Types of Formulas

Empirical Formula
The formula of a compound that expresses the
smallest whole number ratio of the atoms
present.

Molecular Formula
The formula that states the actual number of
each kind of atom found in one molecule of the
compound.
Determine the Empirical Formula of
Acetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
¬ Convert the percentages to grams by
assuming you have 100 g of the
compound
– Step can be skipped if given masses
47gC
100g 
 47gC
100g
47gO
100g 
 47gO
100g
6.0gH
100g 
 6.0gH
100g
Determine the Empirical Formula of
Acetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
-
Convert the grams to moles
1 mol C
47g C 
 3.9 mol C
12.01g
1 mol H
6.0 g H 
 6.0 mol H
1.008g
1 mol O
47 g O 
 2.9 mol O
16.00g
Determine the Empirical Formula of
Acetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
® Divide each by the smallest number of
moles
3.9 mol C  2.9  1.3
6.0 mol H  2.9  2
2.9 mol O  2.9  1
Determine the Empirical Formula of
Acetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
¯
If any of the ratios is not a whole number,
multiply all the ratios by a factor to make it a
whole number
– If ratio is ?.5 then multiply by 2; if ?.33 or ?.67 then
multiply by 3; if ?.25 or ?.75 then multiply by 4
Multiply all the
Ratios by 3
Because C is 1.3
3.9 mol C  2.9  1.3 x 3  4
6.0 mol H  2.9  2 x 3  6
2.9 mol O  2.9  1 x 3  3
Determine the Empirical Formula of
Acetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
°
Use the ratios as the subscripts in the
empirical formula
3.9 mol C  2.9  1.3 x 3  4
6.0 mol H  2.9  2 x 3  6
2.9 mol O  2.9  1 x 3  3
C4H6O3
Molecular Formulas
• The molecular formula is a multiple of
the empirical formula
• To determine the molecular formula you
need to know the empirical formula and
the molar mass of the compound
Determine the Molecular Formula of Benzopyrene
if it has a molar mass of 252 g and an
empirical formula of C5H3
u Divide the given molar mass of the
compound by the molar mass of the
empirical formula
–Round to the nearest whole number
252 g
4
63.07 g
Determine the Molecular Formula of Benzopyrene
if it has a molar mass of 252 g and an
empirical formula of C5H3
u Multiply the empirical formula by the
calculated factor to give the molecular
formula
(C5H3)4 = C20H12