Transcript The Mole - Phillips Scientific Methods

THE MOLE
Magnet Chemistry
Phillips 2016
Measurements





Dozen = 12
Can you
Baker’s dozen = 13
imagine a
mole of
Gross= 144
donuts?
23
Mole (mol) = 6.02x10
A mole of water molecules has a volume of only 18
mL!


Molecules are unbelievably small, so a lot of them doesn’t
take up that much space.
The mole is useful in chemistry because it links the
microscopic world of atoms, molecules, and ions, to the
macroscopic world
Avogadro’s Number


6.02x1023 is so important in chemistry it’s given its
own name…Avogadro’s Number
In numerical form it looks like this:
 602,000,000,000,000,000,000,000
A
mole of paper would reach past Pluto
 A mole of basketballs is the size of Earth
 A mole of rice would cover the land masses of Earth to
a depth of 75 meters!
Scientific Notation Practice


Using the mole requires that we are comfortable
using scientific notation
Write the following in scientific notation
6000
0.0067
78,000,000
698,700
0.000009
0.0090076
6.00x103
6.70x10-3
7.80x107
6.99x105
9.00x10-6
9.01x10-3

Write the following in standard notation
6.02x104
9.03x109
7.77x10-2
9.21x10-7
60,200
9,030,000,000
0.0777
0.000000921
The mole

The mole establishes a relationship between the
atomic mass and the gram. Individual atoms and
molecules are too small to be directly measured, so we refer to
them using the unit amu. Using moles allows direct measurement.

The mass in grams of 1 mole of a substance is equal
to its atomic mass.
 6.02x1023
atoms Cu = 63.5 g
 6.02x1023 atoms H
= 1.001 g
 6.02x1023 atoms Fe = 55.8 g
 Mole was based on Carbon. How many g of C are in 1
mol?
What’s in a Mole?
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A mole of particles in an element is usually talking about
atoms.
The number of molecules in a mole of any molecular compound
is 6.02x1023
How many atoms are in a molecule of ammonia (NH3)?

4 atoms

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1 N atom and 3 H atoms
Each molecule has 4 atoms
So 1 mole of ammonia gas contains 1 mole of NH3 molecules,
but four times as many atoms, or 4 moles of atoms (1 mol N
atoms and 3 mol H atoms)
Mole Conversions
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Chemists measure amounts of substances by volume
or mass.
Because the mole measures both a mass and a
number of particles (and volume, but that will be
covered later), it is the central unit in converting the
amount of substance from one type of measurement
to another.
Moles to Molecules
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The number of particles in 1 mole of any substance
is always the same- Avogadro’s number.
6.02x1023 particles
or
1 mol
6.02x1023 particles

Molecules OR atoms
1 mol
How many atoms are in 3 mol of elemental Ne?
3 mol Ne
1
6.02x1023 atoms
1 mol
1.81x1024 atoms Ne
Unit Cancellation/ Dimensional Analysis
Method (sometimes called ‘unit analysis’)
1.
2.
3.
Write down units asked for in answer to the right
Write down the given value over 1 on the left
Apply one or more unit factors to cancel units
It’s as easy
as 1-2-3!
Moles to Molecules (cont.)
PARTICLES
Atoms
6.02x1023 particles
1 mol
Molecules?
Use the subscript
1 mol
6.02x1023 particles
MOLES
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How many atoms of oxygen are in 6 mol of O2 molecules?
6 mol O2
1

6.02x1023 molecules O2
1 mol O2
2 atoms O
1 molecule O2
7.22x1024 atoms O
How many moles of I2are in 8.02x1020 molecules of I2?
8.02x1020 molec. I2
1
1 mol I2
6.02x1023 molecules I2
1.33x10-3 mol I2
Molar Mass
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The atomic mass of any substance expressed in grams
is the molar mass (MM) of that substance.
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The atomic mass of iron is 55.85 amu.
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Therefore, the molar mass of iron is 55.85 g/mol.
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Since oxygen occurs naturally as a diatomic, O2, the
molar mass of oxygen gas is 2 times 16.00 g or 32.00
g/mol.
Calculating Molar Mass
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The molar mass of a substance is the sum of the molar
masses of each element and expressed as g/mol.

What is the molar mass of magnesium nitrate,
Mg(NO3)2?
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The sum of the atomic masses is:
24.31 + 2(14.01 + 16.00 + 16.00 + 16.00) =
24.31 + 2(62.01) = 148.33 amu

The molar mass for Mg(NO3)2 is 148.33 g/mol.
Molar Mass Practice
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Ne = 20.18 g/mol
O2 = 16.00*2 = 32.00 g/mol
U = 238.03 g/mol
NaOH = 22.99 + 16.00 + 1.008 = 40.00 g/mol
CO2 = 12.01 + 16.00*2 = 44.01 g/mol
Al2(CO3)3 = 26.98*2 + (12.01*3) + (16.00*9) =
233.99 g/mol
Moles to Grams Conversions
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If you know the mass of a substance, you can
calculate the number of moles.
You have 11.2 g of NaCl. How many moles is that?
 1.
determine the molar mass of NaCl (using the
periodic table)
 Na=
23.0 g/mol
Cl= 35.5 g/mol
 23.0 g/mol + 35.5 g/mol = 58.5 g/mol (get these #’s
from the periodic table)
Moles to Grams Conversions
1.
2.
Molar mass of NaCl = 58.5 g/mol
Set up a conversion factor
(a fraction whose value is equal to 1)
1.
The units we need should be in the numerator and the
units you already know in the denominator.
Moles to Grams Conversions
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You can also determine the mass of a sample if you
know the number of moles
2.50 mol of NaCl is how many grams?
Set up a conversion factor 
Moles to Grams Conversions
MASS
moles x
g = g
mol
g x mol = moles
g
MOLES
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How many moles are in 14 g LiOH?
14 g LiOH
1
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1 mol LiOH
23.95 g LiOH
0.58 mol LiOH
How many moles are in 15 g N2?
15 g N2
1
1 mol N2
28.02 g N2
0.54 mol N2
Moles to Grams Conversions

How many grams are in 4 mol H2O2?
4 mol H2O2
1

34.016 g H2O2
1 mol H2O2
136.06 g H2O2
How many grams are in 56 mol CaCO3?
56 mol CaCO3
1
100.09 g CaCO3
1 mol CaCO3
5605.04 g CaCO3
Mole Calculations II
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What is the mass of 2.55 × 1023 atoms of lead?
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We want grams, we have atoms of lead.

Use Avogadro’s number and the molar mass of Pb
2.55 ×
1023
1 mol Pb
207.2 g Pb
atoms Pb ×
×
23
1 mole Pb
6.02×10 atoms Pb
= 87.8 g Pb
Mole Calculations II
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How many O2 molecules are present in 0.470 g of
oxygen gas?
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We want molecules O2, we have grams O2.
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Use Avogadro’s number and the molar mass of O2
1 mol O2
6.02×1023 molecules O2
0.470 g O2 ×
×
1 mole O2
32.00 g O2
8.84 × 1021 molecules O2
Multistep Conversions
MASS
PARTICLES
Avogadro’s
Number
Use molar
mass
MOLES
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You want to impress your date by boasting you know how many molecules
of table sugar are in the cake you just made. You need 250 g of sugar
(C12H22O11). How many sucrose molecules will be in the cake?
Plan: Convert the mass to moles using the molar mass and then convert to
moles using Avogadro’s number.
250 g C12H22O11
1
1 mol C12H22O11
342.3 g C12H22O11
6.02x1023 molecules
1 mol C12H22O11
4.4x1023 molec. C12H22O11
Multistep Conversions
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If you burned 4.0x1024 molecules of methane (CH4) during a
laboratory experiment, what mass of methane did you use?
Plan: Convert your given # of molecules to moles using
Avogadro’s number, then convert the moles to grams using the
molar mass of methane. (12.00 + 1.004*4)
4.0 x 1024 molec. CH4
1
1 mol CH4
6.02x1023 molec. CH4
16.016 g CH4
1 mol CH4
106.42 g CH4
Moles and Gases
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At the same temperature and pressure, equal
volumes of gases contain the same number of gas
particles.
1 mole of any gas at 0oC and 1 atm (Standard
Temperature and Pressure; STP) has a volume of
22.4 L.
This volume, 22.4 L/mol, is called molar volume.
Gas Density- not on this test
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The density of gases is much less than that of liquids.
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We can calculate the density of any gas at STP easily.
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The formula for gas density at STP is:
molar mass in grams
molar volume in liters
= density, g/L
Calculating Gas Density not on this test
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What is the density of ammonia gas, NH3, at STP?

First we need the molar mass for ammonia;
 14.01
+ 3(1.01) = 17.04 g/mol
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The molar volume NH3 at STP is 22.4 L/mol.
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Density is mass/volume:
17.04 g/mol
22.4 L/mol
= 0.761 g/L
Molar Mass of a Gas
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We can also use molar volume to calculate the molar
mass of an unknown gas.
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1.96 g of an unknown gas occupies 1.00L at STP. What
is the molar mass?

We want g/mol, we have g/L.
1.96 g 22.4 L
×
= 43.9 g/mol
1.00 L 1 mole
Mole Unit Factors
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
We now have three interpretations for the mole:
1
mol = 6.02 × 1023 particles
1
mol = molar mass
1
mol = 22.4 L at STP for a gas
This gives us 3 unit factors to use to convert between
moles, particles, mass, and volume.
Mole-Volume Calculation
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A sample of methane, CH4, occupies 4.50 L at STP.
How many moles of methane are present?
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We want moles, we have volume.
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Use molar volume of a gas: 1 mol = 22.4 L
1 mol CH4
4.50 L CH4 ×
= 0.201 mol CH4
22.4 L CH4
Mass-Volume Calculation
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What is the mass of 3.36 L of ozone gas, O3, at STP?
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We want mass O3, we have 3.36 L O3.
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Convert volume to moles then moles to mass:
1 mol O3
48.00 g O3
3.36 L O3 ×
×
22.4 L O3
1 mol O3
= 7.20 g O3
Molecule-Volume Calculation
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How many molecules of hydrogen gas, H2, occupy
0.500 L at STP?
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We want molecules H2, we have 0.500 L H2.

Convert volume to moles and then moles to molecules:
1 mol H2
6.02×1023 molecules H2
×
0.500 L H2 ×
22.4 L H2
1 mole H2
= 1.34 × 1022 molecules H2
Moles and Gases: A Summary
MASS
PARTICLES
Avogadro’s
Number
Molar
Mass
MOLES
Molar Volume
(22.4 L/mol)
VOLUME
of gases at STP
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You try it: A student fills a 1.0-L flask with CO2 at STP. How
many molecules of gas are in the flask?
Plan:
1) Convert from volume to moles using molar volume.
2) Then convert from moles to molecules using Avogadro’s
number.
Moles and Gases
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Let’s see how you did…
1.0-L CO2
1
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1 mol CO2
22.4 L CO2
6.02 x 1023 molec CO2
1 mol CO2
2.7 x 1022 molec. CO2
A chemical reaction produces 0.82 mole of oxygen gas. What
volume will that gas occupy at STP?
0.82 mol O2
1
22.4 L O2
1 mol O2
18 L O2
Law of Definite Composition
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The law of definite composition states that
“Compounds always contain the same elements in a
constant proportion by mass”.

Sodium chloride is always 39.3% sodium and 60.7%
chlorine by mass, no matter what its source.

Water is always 11.2% hydrogen and 88.8% oxygen by
mass.
Law of Definite Composition
A drop of water, a glass of water, and a lake of water all contain
hydrogen and oxygen in the same percent by mass.
Percent Composition
• The percent composition of a compound lists the
mass percent of each element.
• For example, the percent composition of water,
H2O is:
– 11% hydrogen and 89% oxygen
• All water contains 11% hydrogen and 89%
oxygen by mass.
Calculating Percent Composition
• There are a few steps to calculating the percent
composition of a compound. Lets practice using
H2O.
– Assume you have 1 mole of the compound.
– One mole of H2O contains 2 mol of hydrogen
and 1 mol of oxygen.
– 2(1.01 g H) + 1(16.00 g O) = molar mass H2O
– 2.02 g H + 16.00 g O = 18.02 g H2O
Calculating Percent Composition
• Next, find the percent composition of water by
comparing the masses of hydrogen and oxygen in
water to the molar mass of water:
2.02 g H
× 100% = 11.2% H
18.02 g H2O
16.00 g O
× 100% = 88.79% O
18.02 g H2O
Percent Composition Problem
• TNT (trinitrotoluene) is a white crystalline
substance that explodes at 240°C. Calculate the
percent composition of TNT, C7H5(NO2)3.
Solution Percent Composition Problem
• TNT (trinitrotoluene) is a white crystalline substance that explodes
at 240°C. Calculate the percent composition of TNT, C7H5(NO2)3.
• First, find molar mass of TNT:
• 7(12.01 g C) + 5(1.01 g H) + 3 (14.01 g N + 32.00 g O)
= g C7H5(NO2)3
• 84.07 g C + 5.05 g H + 42.03 g N + 96.00 g O
= 227.15 g C7H5(NO2)3.
Percent Composition of TNT
84.07 g C
× 100% = 37.01% C
227.15 g TNT
1.01 g H
× 100% = 2.22% H
227.15 g TNT
42.03 g N
× 100% = 18.50% N
227.15 g TNT
96.00 g O
× 100% = 42.26% O
227.15 g TNT
Try one more: Percent
Composition
What is the percent carbon in C5H8NO4 (the
glutamic acid used to make MSG
monosodium glutamate), a compound used
to flavor foods and tenderize meats?
a) 8.22 %C
b) 24.3 %C
c) 41.1 %C
Chemical Formulas of Compounds
• Formulas give the relative numbers of atoms or
moles of each element in a formula unit - always a
whole number ratio (the law of definite
proportions).
NO2
2 atoms of O for every 1 atom of N
1 mole of NO2 : 2 moles of O atoms to every 1
mole of N atoms
• If we know or can determine the relative number
of moles of each element in a compound, we can
determine a formula for the compound.
Types of Formulas
• Empirical Formula
The formula of a compound that
expresses the smallest whole number
ratio of the atoms present.
*Ionic formula is always empirical formula
• Molecular Formula
The formula that states the actual
number of each kind of atom found in one
molecule of the compound.
To obtain an Empirical Formula
1. Determine the mass in grams of each
element present, if necessary.
2. Calculate the number of moles of each
element.
3. Divide each by the smallest number of
moles to obtain the simplest whole
number ratio.
4. If whole numbers are not obtained* in
step 3), multiply through by the smallest
number that will give all whole numbers
* Be
careful! Do not round off numbers prematurely
A sample of a brown gas, a major air pollutant, is found
to contain 2.34 g N and 5.34g O. Determine a formula
for this substance.
Sol’n: require mole ratios, so convert grams to moles
moles of N = 2.34g of N = 0.167 moles of N
14.01 g/mole
moles of O = 5.34 g = 0.334 moles of O
16.00 g/mole
Formula:
N 0.167 O 0.334  NO 2
N0.167 O0.334
0.167
0.167
(HONORS only)
Calculation of the Molecular Formula
A compound has an empirical formula
of NO2. The colorless liquid, used in
rocket engines has a molar mass of
92.0 g/mole. What is the molecular
formula of this substance?
Empirical Formula from % Composition
A substance has the following composition by
mass: 60.80 % Na ; 28.60 % B ; 10.60 % H
What is the empirical formula of the substance?
Consider a sample size of 100 grams
This will contain 28.60 grams of B and
10.60 grams H
Determine the number of moles of each
Determine the simplest whole number ratio
More Empirical & Molecular Formulas
• The empirical formula of a compound is the
simplest whole number ratio of ions in a formula
unit or atoms of each element in a molecule.
• The molecular formula of benzene is C6H6
– The empirical formula of benzene is CH.
• The molecular formula of octane is C8H18
– The empirical formula of octane is C4H9.
Another Exp: Empirical Formulas
• We can calculate the empirical formula of a
compound from its composition data.
• We can determine the mole ratio of each element
from the mass to determine the formula.
Ex: radium oxide, Ra?O?.
• A 1.640 g sample of radium metal was heated to
produce 1.755 g of radium oxide. What is the
empirical formula?
• We have 1.640 g Ra and 1.755-1.640 = 0.115 g O.
Calculating Empirical Formulas
• The molar mass of radium is 226.03 g/mol and the
molar mass of oxygen is 16.00 g/mol.
1 mol Ra
= 0.00726 mol Ra
1.640 g Ra ×
226.03 g Ra
1 mol O
= 0.00719 mol O
0.115 g O ×
16.00 g O
• We get Ra0.00726O0.00719. Simplify the mole ratio
by dividing by the smallest number.
• We get Ra1.01O1.00 = RaO is the empirical formula.
Another Exp: Empirical Formulas from
Percent Composition
• We can also use percent composition data to
calculate empirical formulas.
• Assume that you have 100 grams of sample.
• Benzene is 92.2% carbon and 7.83% hydrogen,
what is the empirical formula.
• If we assume 100 grams of sample, we have
92.2 g carbon and 7.83 g hydrogen.
Empirical Formulas from
Percent Composition
• Calculate the moles of each element:
1 mol C
= 7.68 mol C
92.2 g C ×
12.01 g C
7.83 g H ×
1 mol H
= 7.75 mol H
1.01 g H
• The ratio of elements in benzene is C7.68H7.75.
Divide by the smallest number to get the formula.
C
7.68
7.68
H
7.75
7.68
= C1.00H1.01 = CH
Another Exp: Molecular Formulas
• The empirical formula for benzene is CH. This
represents the ratio of C to H atoms of benzene.
• The actual molecular formula is some multiple of
the empirical formula, (CH)n.
• Benzene has a molar mass of 78 g/mol. Find n to
find the molecular formula.
(CH)n
78 g/mol
=
CH
13 g/mol
n = 6 and the molecular
formula is C6H6.
Conclusions
• Avogadro’s number is 6.02 × 1023 and is one mole
of any substance.
• The molar mass of a substance is the sum of the
atomic masses of each element in the formula.
• At STP, 1 mole of any gas occupies 22.4 L.
Conclusions Continued
• We can use the following flow chart for mole
calculations:
Conclusions Continued
• The percent composition of a substance is the
mass percent of each element in that substance.
• The empirical formula of a substance is the
simplest whole number ratio of the elements in the
formula.
• The molecular formula is a multiple of the
empirical formula.