Chapter 12: Chemical Quantities
SECTION 1: COUNTING PARTICLES OF
MATTER
The Mole
 Mr. Wiedeman, What is a mole?
 A number: 6.02X1023
 Just another number right? How BIG is it though?
 To answer this, lets first look at money

Here is a hundred bucks:
The Mole
 Heres 10,000 $
The Mole
 Heres 1 million dollars:
 (100 packets of 10,000$)
The Mole
 Heres 100 million dollars:
The Mole
 Heres 1 billion dollars:
The Mole
 Heres 1 trillion dollars:
 If you had a 1 mole of money…
 You could make a stack from the earth to the moon
The Mole
 Avogadro’s number (the mole)
 A quantity for counting particles/atoms/molecules
 6.02X1023 particles/atoms/molecules
 How long, if placed length wise, would a mole of
paper clips reach into space?

2 million light years
Stoichiometry
 The study of quantitative relationships between
reactants and products in a chemical reaction

For example, a sample’s mass or volume can be converted to a
count of the number of its particles

Such as atoms, ions, or molecules
 When counting particles, it is easier to group them
instead of dealing with large numbers
Using the Mole
 Example: counting pennies in a
large barrel

Hard way:


Count one by one
Easy way:
Weigh one penny,
 Weigh the barrel,
 divide by barrel’s weight by one penny


Or:

Place the pennies into groups and
weigh the groups
 Ex: 1000 pennies = 2890.7g
Avogadro’s number
 For counting atoms, you need groupings much larger
than 1000

Atoms are so small that even counting by the thousands would
be unmanageable, even grouping by the millions
 The unit of measurement used to count numbers of
atoms, molecules, ions, or formula units of
substances is the mole (mol)

6.02 X 1023
Avogadro’s number
 All kinds of subatomic particles can be conveniently
counted using this number



1 mole of water molecules = 6.02 x 1023 water molecules
1 mole of copper atoms = 6.02 x 1023 copper atoms
1 mole of sodium ions = 6.02 x 1023 sodium ions
Molar mass
 Like most materials, things come in packages or
groups

A ream of paper, a dozen eggs, etc.
 Chemicals in the storeroom do not come in
convenient bundles of moles all set for counting

But, you can use a balance to measure the mass of a sample of
matter
Molar mass
 A dozen limes does not weight the same as a dozen eggs
 Similarly, different elements have different masses
 Recall from last semester, the average atomic masses of the
elements are given on the periodic table
Molar mass
 How does the mass of one atom relate to the mass of
one mole of that atom?


The atomic mass unit (u) is defined so that the atomic mass of
an atom of the most common carbon isotope is exactly 12u
Conveniently enough, the mass of one mole of the most
common isotope of carbon atoms is exactly 12g
 Simple definition of molar mass:
 The number underneath an element on the periodic table
stated in grams (g) rather than atomic mass units (u)
Rounded to the hundredths place
 The unit for just the mass is: gram (g)
 The unit for molar mass is: grams per mole (g/mol)

Molar mass
 The molar mass of Oxygen = __________
 The molar mass of Iron = __________
Molar mass for methanol (covalent) and calcium
chloride (ionic)
Practice Molar Mass on Homework
Conversions
 Conversions (using dimensional analysis) allow us to:
 Move between moles of a substance and grams of a substance
 Moles  Grams and Grams  Moles
 Background info:
 If you are going to do any converting at all you need…

The molar mass of whatever compound you are working with
Conversions
 How many moles are in 16.8 grams of iron(III) oxide
(Fe2O3)?


Find molar mass first
Use molar mass just like the pressure conversions
Practice Mol to Gram Conversions on homework
Conversions
 How many grams are in 1.55 mol of HCl?
Practice Gram to Mol Conversions on homework
How many atoms are in a sample of an element?
Molar mass and stoichiometry
 Molar mass makes it easy to determine the number
of particles in a sample of a substance by measuring
the mass of the sample
Next section: convert between moles of reactants and moles of
products
Chapter 12: Chemical Quantities
SECTION 2: USING MOLES
Real-World Chem. Application
 Pancakes!



Situation: You just mixed enough pancake mix and ingredients for
yourself
Problem: brother/sister just offered you 5$ to make enough for them
too
Solution to make pancakes for two people:
Double the mix?
 Add water?
 2 cups (pancakes) + 1 cup (water) = 3-6 pancakes (enough for you)
 3 cups (pancakes mix) + 2 cups (water) = 6-9 pancakes (Success!)


Chemical reactions are similar, today, relating reactants and products
to moles
Using Molar masses in stoichiometeric problems
 Important step in this section
 You cannot jump from one mass of a substance to another
mass of a substance
 You first need to convert the given mass to moles
 Recall that coefficients are used to make equations
balanced
 They also tell you the number of moles of each
chemical in the reaction
Using Molar masses in stoichiometeric problems
 Once you have the number of moles for any reactant
or product…

Use the coefficients in the equation to convert to moles of the
other reactants and products
(Find the molar mass)
( Use the coefficients
from the formula)
(molar mass of H2)
Ideal Gas Law
 PV=nRT
Objectives
 Review ‘gram’ to ‘gram’ stoichiometry problems
 Attach Theoretical Yield and Actual Yield
 Practice calculating Mass Percentages of compounds
 Apply stoichiometry skills to ‘Left-over aluminum
lab’
Theoretical Yield and Actual Yield
 The amount of a chemical reaction predicted by
stoichiometry is called theoretical yield

This just a different way of looking at a ‘gram’ to ‘gram’
problem
2Al + 3CuCl2  2AlCl3 + 3Cu
 Problem #1
 A student reacts 21g of Al with CuCl2 and obtains 65.3g of Cu.
What is the theoretical yield of Cu?
 What is the actual yield of Cu?

 Part 1: Solve theoretical yield using ‘gram’ to ‘gram’
stoichiometry

Answer = 74.20
 Part 2: Solve the actual yield
 Actual is given, the student got 65.3g of Cu
 Part 3: Find the percent yield
 Percent Yield = (Actual yield/Theoretical Yield) X 100
 Answer = 88% Yield
CaO + H2O  Ca(OH)2
 Problem #2
 A student reacts 7.5g of CaO with water and obtains 9.5g of
Ca(OH)2
 Part 1: Solve theoretical yield using ‘gram’ to ‘gram’
stoichiometry

Answer = 9.91g of Ca(OH)2
 Part 2: Solve the actual yield
 Actual is given, the student got 9.5g of Ca(OH)2
 Part 3: Find the percent yield
 Percent Yield = (Actual yield/Theoretical Yield) X 100
 Answer = 95.9% Yield
Objectives
 1. Review percent yield
 2. Determining mass percents
 3. Practice determining chemical formulas
Obj 1. Review
 CaO + H2O  Ca(OH)2
 - Using the above equation, solve the following problem
 A student reacts 8.5g of CaO and obtains 10.8g of Ca(OH)2. What is
the percent yield of this reaction?
 Theoretical Yield
= 11.23g
 Actual Yield = 10.8g
 Percent Yield = Actual/Theoretical yield X 100
 Answer = 96.17%
Obj 2: Determining Mass Percents
Geraniol: C10H18O
(on page 421)
 As you know…
 10 carbon atoms
 18 hydrogen atoms
 1 oxygen atom
 But, which is the major element by mass?
 By looking at the masses of the elements and the compound they
are in you can find the percent composition of a compound
Percent Composition:
C10H18O
 Molar mass of this geraniol = 154g/mol
 One mole of geraniol has
10 mol of carbon
 18 mol of hydrogen
 1 mol of oxygen

 Step 1: convert mol to g (use molar mass)
 Answer for C = 120.g C
 Answer for H = 18.0g H
 Answer for O = 16.0g O
C10H18O
 From step 1:
 The following are the masses for
the elements using the numbers
from the compound
C = 120g C
 H = 18.0g H
 O = 16.0g O

 Step 2: Use the molar mass of
geraniol (154g/mol) to solve
for the percent composition



Mass percent for C = 77.9%
Mass percent for H = 11.7%
Mass percent for O = 10.4%
Obj 3: Determining Chemical Formulas, p426
% to Empirical Formula
 Empirical Formula:
 The formula of a compound having the smallest whole-number
ratio of atoms in the compound
Obj 3: Determining Chemical Formulas, p426
% to Empirical Formula
 Unknown compound has…



18.8% sodium
29.0% chlorine
52.2% oxygen

These numbers equal 100%, so if we has 100g of this
substance:

18.8g sodium
29.0g chlorine
52.2g oxygen


% to Empirical Formula
 Step 1: use the molar mass of each element to find
the number of moles



18.8g sodium
29.0g chlorine
52.2g oxygen
 Solutions:
 Answers:
0.817 mol Na
 0.817 mol of Cl
 3.26 mol of O

% to Empirical Formula
 From part 1:
0.817 mol Na
 0.817 mol of Cl
 3.26 mol of O

 Part 2: select the smallest number and divided
everything by it

Smallest number is 0.817 mol
 Solutions
 0.817 Na/0.817 = 1.00 mol Na
 0.817 mol Cl/0.817 = 1.00 mol Cl
 3.26 mol O/0.817 = 3.99 mol O
Use these numbers to
find the empirical
formula:
NaClO4
On the Quiz
 You may use your:
 Gram to gram worksheet
 Ideal gas law worksheet
 Topics on the quiz
 Gram to Gram


Ideal gas law


Gram of one reactant to gram of product
PV=nRT
Questions about the lab
Percent yield question
 Theoretical and actual, use to find percent yield
