The Mole
Chapters 10 & 11
What is a mole?
http://www.topnews.in/health/scientists-identify-protein-marks-difference-between-mole-and-melanoma-2990
What is a mole?
http://scienceblogs.com/neurophilosophy/2009/08/the_star_nosed_moles_amazing_appendages.php
http://animals.nationalgeographic.com/animals/mammals/naked-mole-rat/
What is a mole in CHEMISTRY?
• A unit of measurement for molecules
• # of particles of any chemical substance
• 1 mole = 6.02x1023 particles
This is known as Avagodro’s #
•
1mole of Carbon = 6.02x1023 atoms of Carbon
•
1mole of water = 6.02x1023 molecules of H2O
•
1mole of glucose = 6.02x1023 molecules of C6H12O6
•
1mol of popcorn kernels = 6.02x1023 kernels
(enough to bury the entire US under 9 feet of popcorn kernels!)
http://www.holidayforeveryday.com/?p=451
Practical Application of the Mole
= mass of 1 mole of a substance in grams
= average atomic mass given in periodic table
Substance
Molar Mass
Carbon
12.0 g
Magnesium
24.3 g
Copper
63.5 g
Lead
207.2 g
Practical Application of the Mole
= sum of the atomic masses of each element
represented in formula
Compound
Formula
Formula Mass
Water
H2O
2(1) + 1(16) = 18g
Calcium Hydroxide
Ca(OH)2
1(40) + 2(16) + 2(1) = 74g
Iron (III) Chloride
FeCl3
1(55.8) + 3(35.4) = 162.0g
Potassium Nitrate
KNO3
1(39) + 1(14) + 3(16) = 101g
Learning Check
• Prozac, C17H18F3NO, is a widely used
antidepressant that inhibits the uptake
of serotonin by the brain. Find its
molar mass.
17 x 12 =
204g
18 x 1 =
18g
3 x 19 =
57g
1 x 14 =
14g
1 x 16 =
16g
309g
Calculations with Molar Mass
molar mass
Grams
Moles
Calculations with Molar Mass
Practice Example #1
Aluminum is often used for the structure of lightweight bicycle frames. How many grams of Al are in
3.00 moles of Al?
Given
Solving For
3.00 moles of Al
_____ g Al
3.00 moles of Al
27 g of Al
1 mole of Al
= 81g of Al
Learning Check
The artificial sweetener aspartame (NutraSweet) formula C14H18N2O5 is used to
sweeten diet foods, coffee and soft drinks.
How many moles of aspartame are present
in 225 g of aspartame?
225g aspartame
1 mole aspartame
294g aspartame
= 0.76 mol aspartame
Calculations with Molar Mass
molar mass
Avogadro’s #
Grams
Moles
Particles
Multistep Practice Problem #1
How many atoms of Cu are present in 35.4 g of Cu?
35.4 g Cu 1 mol Cu
63.5 g Cu
6.02 X 1023 atoms Cu
1 mol Cu
= 3.4 X 1023 atoms Cu
Multistep Practice Problem #2
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
39.0 g K
1 mol K
= 1.21 x 1024 atoms K
Multistep Practice Problem #3
What is the mass (in grams) of 1.20 X 1024 molecules of
glucose (C6H12O6)?
1.20x1024 molecules of C6H12O6
1 mol glucose
180 g glucose
6.02 X 1023 molecules
1 mole glucose
= 358 g glucose
Multistep Practice Problem #4
How many atoms of O are present in 78.1 g of oxygen
gas?
78.1 g O2
1 mol O2
32.0 g O2
6.02 X 1023 molecules O2
1 mol O2
2 atoms O
1 molecule O2
= 2.94 x 1024 atoms of O
Calculations with Molar Volume
Gases at STP (Standard
Temperature & Pressure)
with the same number of
particles will occupy the
same volume
One mole of any gas at STP = 22.4L
Calculations with Molar Volume
Molar Volume Practice Problem
A student fills a 1.0L flask with carbon dioxide (CO2) at
standard temperature and pressure. How many
molecules of gas are in the flask?
1.0 L CO2
1 mol CO2
6.02 x 1023 molecules of CO2
22.4 L CO2
1 mol CO2
= 2.69 x 1022 molecules CO2
Empirical & Molecular Formulas
• The mass of each element in a compound
compared to the total mass
Percent Hydrogen of
1 mole of H2O
%H=
2.0 g H
18 g H2O
= 11%
% Composition Using Experimental Data
Knowns
Mass Mg = 8.2 g
Unknowns
% Mg =
8.2 g
= 60.3% Mg
13.6 g
Mass O = 5.4 g
Total Mass =
13.6 g
5.4 g
%O=
13.6 g = 39.7% O
Calculating Empirical Formulas
What is the empirical formula of a compound
containing 25.9% N and 74.1% O?
Step 1: Assume 100 g of the
compound (only necessary if given
percents to start)
25.9% N = 25.9 g N
74.1% O = 74.1 g O
Step 2: Convert grams to moles
25.9 g N
1 mol N
14.0 g N
74.1 g O
1 mol O
16.0 g O
Step 3: Divide both mole amounts
by the smaller of the 2 numbers
1.85 mol N
1.85
4.63 mol O
= 1.85 mol N
= 4.63 mol O
= 1 mol N
= 2.50 mol O
1.85
Step 4: Multiply both to obtain whole
numbers
1 mol N x 2 = 2 mol N
2.5 mol O x 2=5 mol O
N2O5
Calculating Molecular Formulas
Step 1: Calculate the empirical
formula
Step 2: Based on the empirical
formula, calculate the
formula mass
Step 3: Divide the given molar
mass by the calculated
formula mass
Step 4: Multiply
the
subscripts in the
empirical formula by the
answer found in #3
N2O5
(answer from previous problem)
2x14 = 28
5x16 = 80
108g
If the molar mass of the compound is 324g,
what is the molecular formula?
324g/108g = 3
N6O15
Stoichiometry: Mole-Mole
A balanced formula indicates the number of moles of each compound.
Ex: 2HCl + Zn  ZnCl2 + H2
2 moles HCl+1 mole Zinc yield 1 mole Zinc Chloride +1 mole hydrogen gas
Based on this formula, how many moles of HCl are needed to react with 2.3
moles of Zn?
2.3 mol Zn
2 mol HCl
1 mol Zn
= 4.6 mol HCl
Stoichiometry: Mass-Mass
In a thermite reaction, powdered aluminum reacts with
iron (III) oxide to produce aluminum oxide and molten
iron. What mass of aluminum oxide is produced when
2.3g of aluminum reacts with iron (III) oxide?
Balanced Equation
2 Al +
Fe2O3  Al2O3 + 2 Fe
Stoichiometry
2.3g Al
1 mol Al
1 mol Al2O3
102g Al2O3
27g Al
2 mol Al
1 mol Al2O3
= 4.3g Al2O3
Stoichiometry: Mass-Volume
Sodium Bicarbonate (NaHCO3) can be used to
extinguish a fire. When heated, it decomposes into
sodium carbonate, carbon dioxide and water. If a
sample contains 4.0g NaHCO3, what volume of CO2 gas
is produced?
Balanced Equation
2 NaHCO3 
Na2CO3 + H2O + CO2
Stoichiometry
4.0g NaHCO3
1 mol NaHCO3 1 mol CO2
84g NaHCO3
22.4L CO2
2 mol NaHCO3 1 mol CO2
= .53L CO2
Limiting Reactants
The limiting reactant in any equation
• would produce a smaller amount of product
• is completely used up first
• determines the amount of product made
Which reactant in an equation is the limiting reactant?
Identify the limiting reactant when 1.7 g of sodium reacts with 2.6L of
chlorine gas at STP to produce sodium chloride.
Step 1
Write the balanced equation
Step 2
Determine the mass of product that would be produced
from each given reactant quantity.
2 Na + Cl2  2 NaCl
1.7g Na
2.6L Cl2
1mol Na
2 mol NaCl
58g NaCl
23g Na
2 mol Na
1 mol NaCl
1 mol Cl2
2 mol NaCl
58g NaCl
22.4L Cl2
1 mol Cl2
1 mol NaCl
= 4.3g NaCl
= 13g NaCl
Percent Yield
Percent Yield =
Actual yield
Expected yield
X 100
Expected yield is determined from a stoichiometry problem.
Actual yield is the amount of product measured or recovered.
A piece of copper with a mass of 5.00g is placed in a solution of silver (I) nitrate
containing excess AgNO3. The silver metal produced has a measured mass of
15.2g. What is the percent yield for this reaction?
Step 1
Write balanced equation
Step 2
Find expected mass of product
Cu + 2 AgNO3  2 Ag + Cu(NO3)2
5.00g Cu
1mol Cu
2 mol Ag
107.9g Ag
63.5g Cu
1 mol Cu
1 mol Ag
Percent Yield= Actual yield =
Expected yield
15.2g Ag
17.0g Ag
X 100
= 17.0g Ag
= 89.4%