The Mole
Chapter 10
1
Objectives
Use the mole and molar mass to make conversions among moles, mass, and number of particles
Determine the percent composition of the components of a compound
Calculate empirical and molecular formulas for compounds
Determine the formulas for hydrates Examine relationship between volume
and amount of gas (Avogadro’s Law –
Ch. 13.2) 2
Dimensional Analysis
Conversion Factors:
– What is 8a*5b/2a=
- Real Example: 1 mile = 5280 ft – Convert 5.5 miles to feet. – 5.5 miles x 5280 ft =
1 mi 20 b
29040 ft
3
The Mole
Mole (mol)
measures the number of particles of a substance
(atom, molecule, formula unit)
Using “mole” is just a shorthand way for saying 6.02 x 1023 particles
6.02 x 1023 particles = Avogadro’s number
4
How much mass is in one
atom of
carbon-12
?
Exactly 12 amu (by definition)
Molar Mass of Atoms
Definition:
12.0000g of Carbon-12 is exactly 1 mol (6.02x1023) of Carbon-12
atoms. Therefore:
Finding Molar Mass of Other
Elements
All other elements are
determined with respect to
Carbon-12 by mass
spectroscopy.
For example: the He-4 has 4/12
or 1/3 the mass of Carbon-12
7
Finding Molar Mass of Other
Elements
Another element: Fe-56 has 55.935 amu
– So it is 55.935/12 or 4.67 x the mass of C-12
– If we put all the iron isotopes together, the ‘average’ Fe atom is 55.85 amu – Each amu has a mass of 1.66 x 10-24g, so:
8
Finding the Molar Mass of an
Element
9
.
x
.=
9.27 x 10-23g/atom.
x
.=
55.85 g/mol FeMass of Atoms
If you have two 1 g samples of different substances, what do they have in common?
•••••••••••••••••••••••• •••••••••••••••••••••••• •••••••••••••••••••••••• •••••••••••••••••••••••• •••••••••••••••••••••••• ••••••••••••••••••••••••
10
Mass of Atoms
•• •• •• •• •• •• •• •• •• •• •• •• •• •• •• •• •• •• •• •• •• •• •• •• •••••• •••••• •••••• •••••• •••••• •••••• •••••• ••••••
The atomic mass of ONE iron
atom is 55.85 AMU (atomic
The number of grams in a mole
is called the
molar mass
.
13
Why does a mole of iron weigh
more than a mole of carbon?
For the same reason that a dozen bowling balls weighs more than a
dozen golf balls!
Bowling balls are heavier than golf
balls.
14
Mass of Iron on Balance
55.85 g of iron (iron’s molar mass) on scale is:
1 mole of iron (1 mol Fe = 55.85 g Fe)
6.02 x 1023atoms of iron
15
A conversion you already know.
How many eggs are in 14 dozen eggs?
14 dozen eggs x 12 eggs = 168 eggs 1 dozen eggs
How many dozens of eggs would you need
to buy if you were going to feed 504 people 1 egg each?
504 eggs x 1 dozen eggs = 42 dozen eggs 12 eggs
16
Moles and Number of Particles
Examples
If have 2 mole of iron how many atoms of
iron do you have?
2 moles Fe x 6.02x1023atoms = 12.04x1023atoms
1 mole
If have 12.04 x 1023atoms of iron how
many moles of iron have you?
12.04x1023atoms x 1 mole____ = 2.0 moles 6.02x1023atoms
Mass and Moles Examples
What is the mass in grams of 2.00 moles of Cu?
2.00 moles Cu x 63.5 grams = 127 grams Cu 1 mole Cu
How moles is 190. g of Cu? 190. g Cu x 1 mole Cu = 3.00 mol Cu
Mass
↔
Particles Examples
In order to go from mass to number of particles, you have to go through moles.
How many atoms are in 46.0g of Na?
46.0 g x
.
2.00
2.00 mol Na x .
12.04 10
19
Mass
↔
Particles Examples
What is the mass in grams of 15.05 x 1023 atoms of Al?
15.05 x 1023atoms x
.
2.50
2.50 mol Al x
.
67.5
20
Flowchart
Atoms or
Molecules
Moles
Mass (grams)
Divide by 6.02 X 1023
Multiply by 6.02 X 1023
Multiply by molar mass from periodic table
Divide by molar mass from periodic table
21
Mass, Moles, and Number of
Particles Practice
How many moles are in 25.5 g of Ag? How many grams are 42.60 mol of silicon
(Si)?
How many atoms are in 15.0 mol of Xe? How many moles are in 7.23 x 1024atoms
of Xe?
How many atoms are in 6.50 g of B? How many grams are in 5.53 x 1022atoms
of Mg?
22
Practice
How many moles are in 25.5 g of Ag?
Practice
How many grams are 42.60 mol of silicon (Si)?
Practice
How many atoms are in 15.0 mol of Xe?
25
Practice
How many moles are in 7.23 x 1024 atoms of Xe?
26
Practice
How many atoms are in 6.50 g of B?
27
Practice
How many grams are in 5.53 x 1022 atoms of Mg?
28
One Mole of
Four
Elements
One mole each of helium, sulfur, copper, and mercury. How many atoms of helium are present? Of sulfur? Of copper? Of mercury?
One-Mole Quantities of Some
Elements & Compounds
We can also calculate the molar mass
of compounds like
carbon dioxide.
The formula for carbon dioxide is
CO
2.
That means there is one carbon
atom and two oxygen atoms in
every molecule of CO
2.
31
We can also calculate the molar mass
of compounds like carbon dioxide.
Carbon weighs 12.01 grams/mole
Oxygen weighs 16.00 grams/mole.
Therefore CO
2has a molar mass of:
C + O + O
= CO
212.01 + 16.00 + 16.00 = 44.01 g 32
Finding Molar Mass of CaCl
2(assume 1 mole compound)
Atom # mol Atoms in 1 mol of compound
Molar Mass (g/mol)
Total (g) Ca
Cl CaCl2
33
Let’s find out the molar mass of
glucose (C
6H
12O
6).
Use the molar masses from the
periodic table:
Carbon – 12.0 grams/mole
Hydrogen – 1.0 gram/mole
Oxygen – 16.0 grams/mole
34
How many grams are in a mole of
glucose (C
6H
12O
6)?
Carbon: 6 x 12.0 g/mol = 72.0 g Hydrogen: 12 x 1.0 g/mol =12.0 g Oxygen: 6 x 16.0 g/mol = 96.0 g
180.0 g/mol (For consistency, round all molar masses
(elements & compounds) to 0.1 g.)
Moles to Mass of Compound
How many grams of KCl are in 2.30 mol of KCl?
First, find molar mass.
K: 1 K x 39.1 g/mol = 39.1 g Cl: 1 Cl x 35.5 g/mol = 35.5 g
Moles to Mass of Compound
How many grams of KCl are in 2.30 mol of KCl?
2.30 mol KCl x 74.6 g KCl = mol KCl
=171.6 g KCl = 172 g KCl (proper Sig Figs)
# of moles Molar mass
37
Mass to Moles in a Compound
How many moles of KCl are present in 253.6 grams KCl
?
First, determine the molar mass of KCl.
(Same as before, which is 74.6 g/mol) Then, divide mass by molar mass to get moles.
253.6 g KCl x 1 mol =3.40 mol KCl
74.6 g KCl
38
Mass to Moles in a Compound
Can determine the number of moles of each of the atoms/ions that make up the compound. Multiply by
ion/compound conversion factor. Conversion factor: mole of specific atom
1 mol of compound Example: 2 mol Cl
-1 mol CaCl2
39
Mass to Moles in a Compound
How many moles of Cl-ions are there 5.5 mol of CaCl2?
5.5 mol CaCl2x 2 mol Cl- = 11.0 mol Cl
-1 mol CaCl2
40
Mass, Moles & Particles
Use the flowchart from the elemental calculations and use them for
compounds.
Moles are still centralGet to mass (g) by multiplying by molar mass Get to # molecules (covalent
compounds) or formula units (ionic compounds) by multiplying by Avogadro’s number.
Mass to Moles in a Compound
How many moles of Na+are there in 561.8 g of Na2CO3?
First, find Molar Mass of Na2CO3:
Na: 2 x 23.0 g/mol = 46.0 g C: 1 x 12.0 g/mol = 12.0 g O: 3 x 16.0 g/mol = 48.0 g
Mass to Moles in a Compound
How many moles of Na+are there in 561.8 g of Na2CO3?
Next, find the number of moles of Na2CO3.
561.8 g x 1 mol = 5.300 mol Na2CO3
106.0 g
Use conversion of moles of ion/mole of
compound.
5.300 mol Na2CO3x 2 mol Na+ = 10.6 mol Na+
1 mol Na2CO3
43
Mass to Number of Particles
How many Na+ ions are in 561.8 g of Na2CO3?
Note: Looking for individual # of ions. Will be a very large number. First, find # of moles of the ion or
element you are interested in. In this case it was 10.6 mol Na+.
44
Mass to Number of Particles
Finally, multiply # of mol by Avogadro’s Constant.
10.6 mol Na+x 6.02 x 1023 ions 1 mol
= 6.38 x 1024 Na+ions
45
Mass to Number of Particles
How many formula units of Na2CO3are there in the 5.30 mol of Na2CO3?
5.30 mol Na2CO3 x 6.02 x 1023 Frm Unts 1 mol of Na2CO3 = 31.9 x 1023 formula Units
46
Practice Problems
Determine the # of formula units, the number of moles of each ion, and the number of each ion in:
a) 2.50 mol ZnCl2 b) 623.7 g of Fe2S3
Determine the mass in grams of 2.11x1024 formula units of Na
2S.
Practice Problems
Determine the # of formula units in 2.50 mole of ZnCl2.
Practice Problems
Determine the # of moles Zn2+ ions in 2.50 mol of ZnCl2.
49
Practice Problems
Determine the # of moles Cl- ions in 2.50 mol of ZnCl2.
50
Practice Problems
Determine the # of Zn2+ions in 2.50 mol of ZnCl2.
51
Practice Problems
Determine the # of Cl-ions in 2.50 mol of ZnCl2.
52
Practice Problems
Determine the # of formula units in in 623.7g of Fe2S3.
Practice Problems
Determine the # of formula units in in 623.7g of Fe2S3.
Practice Problems
Determine the # of moles of Fe3+ions in 623.7g of Fe2S3.
55
Practice Problems
Determine the # of moles of S2-ions in 623.7g of Fe2S3.
56
Practice Problems
Determine the # Fe3+ions ions in 623.7g of Fe2S3.
57
Practice Problems
Determine the # S2-ions in 623.7g of Fe2S3.
58
Practice Problems
Determine the # Fe3+ions and S2-ions in 623.7g of Fe2S3.
Practice Problems
Determine the mass in grams of 2.11x1024 formula units of Na
2S.
Practice Problems
Determine the mass in grams of 2.11x1024 formula units of Na
2S.
61
Ch. 10.4 - Percent Composition
Percent composition is the
percent, by mass, of each element in a compound.
In general, it’s the mass of the element/mass of the formula:
Mass of element x 100 = %mass Mass of compound of element
62
Percent Composition
Example: If a 50.0 g sample of H2O contains 5.6 g of H and 44.4 g of O the percent composition is:
5.6 g H x 100% = 11.1% H 50.0 g H2O
44.5 g O x 100% = 88.9% O 50 g H2O
63
Percent Composition
You can calculate the percent composition by finding the mass of each element in 1 mole of a
compound.
Example: Water’s formula is H2O which means there are 2 mol of hydrogen and 1 mol oxygen in one mol of water.
64
Percent Composition
So, if you have 1 mol of water, you have 18.0 g of water (molar mass) The 18.0 g of water is made up of 2
mol Hydrogen (2 g) and 1 mol oxygen (16 g).
H: 2 g x 100% = 11.1% 18.0 g
O: 16 g x 100% = 88.9% 18 g
Percent Composition Example
Calculate the percent composition of each element in Ca(OH)2.
1. Determine the mass of each element present in 1 mol of cmpd.
Ca: 1 mol x 40.1 g/mol = 40.1 g O: 2 mol x 16.0 g/mol = 32.0 g H: 2 mol x 1.0 g/mol = 2.0 g 2. Determine mass in g of one mole
Percent Composition Example
(Continued)
3. Calculate percentage of each element
Ca: 40.1 g/74.1 g * 100 = 54.1% H: 2.0 g/74.1 g * 100 = 2.7% O: 32.0 g/74.1 g * 100 = 43.2%
67
Percent Composition Practice
Determine the %
Comp of each element in: 1. KBr 2. Fe2O3
3. Barium nitrate
68
Empirical Formulas
An empirical formula for a compound is the formula of a substance written with the smallest integer subscripts. In other words, the simplest mole ratios. You can use the percent composition of a
compound to determine is empirical formula.
69
Empirical Formula
Steps to determine formula:
1. Consider 100g of compound
2. Convert percentages of elements to grams
3. Divide each element’s respective mass by its molar mass to obtain moles
4. Divide each mole value by the smallest mole value. This give the mole ratio. 5. Multiply by appropriate number to get
whole number subscripts.
70
Empirical Formula Example
Determining the Empirical Formula from the Masses of Elements.
We have determined the mass percentage composition of calcium chloride: 36.0% Ca and 64.0% Cl. What is the empirical formula of calcium chloride?
Empirical formula Calcium Chloride
(36% Calcium; 64% Chlorine)
Atom Mass % In grams Molar Mass
Moles Ratio Ca
Cl
Empirical Formula Practice
Determine the empirical formula of a compound that is 36.8% nitrogen and 63.2% oxygen.
73
More Empirical Formula Practice
Determining The Empirical Formula from Percentage Composition. (General)
Benzene is a widely used industrial solvent. This compound has been analyzed and found to contain 92.26% carbon and 7.74% hydrogen by mass. What is its empirical formula? Hint: Consider a 100 g sample.
The empirical formula is:
74
Compounds with different molecular formulas can have the same empirical formula, and such substances will have the same percentage composition.
Remember that the molecular formula has the actual number of atoms of each element that make one molecule of that compound.
Eg. Compound A = C2H2 Compound B = C6H6 both have the empirical formula =
Molecular Formulas
75
Molecular Formula from Empirical
Formula
The molecular formula of a compound is a multiple of its empirical formula.
Molecular mass = n x empirical formula mass where n = number of empirical formula units in the molecule.
76
Determining the Molecular Formula from the Percent Composition and Molar Mass. We have already determined the mass
composition and empirical formula of benzene (CH). In a separate experiment, the molar mass of benzene was determined to be 78.1. What is the molecular formula of benzene
Mass of empirical formula =
Molecular Formula Example
Molar mass benzene _ = Mass of empirical formula
Molecular Formula Example
What is the molecular formula of a compound that has an empirical formula of CH3and a molar mass of 30.0 g/mol.
What is the mass of each empirical unit? C: 1 x 12.0 = 12.0
H: 3 x 1.0 = 3.0 15.0
How many times units? 30/15 = 2 So molecular formula is…C H
Molecular Formula Practice
1. Empirical Formula isNO2; molar mass is
92.0 g/mol
2. A compound contains 26.76% C, 2.21%H, 71.17% O and has a molar mass of 90.04 g/mol. Determine its molecular formula.
79
Ch. 10.5 - Salt Hydrates
Water molecules bound to a compound
For example, CaCl2•2H2O
– Each molecule of calcium chloride has two water molecules bound to it
See “conceptual” picture next slide
80
CaCl
2•2H
2O
Ca+2 Cl
-Cl- Cl
-Water
Water
-
-81
Example of Percent
Composition of a Hydrate
Determine the percent salt and percent water (by mass) in 1 mol of CaCl2•2H2O.
1. Find the mass of 1 mole of the compound.
Ca: 1 x 40.1 = 40.1
Cl: 2 x 35.5 = 71.0 = 147.1 g/mol H2O: 2 x 18.0 = 36.0
82
Example of Percent
Composition of a Hydrate
2. Find the percent of salt.
111.1 g CaCl2 = 75.5% CaCl2
147.1 g CaCl2•2H2O 3. Find the percent of water.
36.0 g H2O = 24.5% H2O
147.1 g CaCl2•2H2O
Or you can say 100% – 75.5% CaCl
Determination of Formula
Example
A nickel(II) cyanide hydrate,
Ni(CN)2•XH2O, contains 39.4% water by mass. What is the formula of the hydrated compound?
1. Assume 100 g of the compound. If it is 39.4% water, it is 100 - 39.4 or 60.6% Ni(CN)2. So there are
Determination of Formula
Example
2. Determine the number of moles of the salt and the water.
60.6 g Ni(CN)2 = 0.547mol Ni(CN)2 110.7 g/mol
39.4 g H2O = 2.19 mol H2O 18.0 g/mol
Ni: 1 x 58.7 = 58.7 C: 2 x 12.0 = 24.0 N: 2 x 14.0 = 28.0 110.7
85
Determination of Formula
Example
3. Determine the mol ratio of the water to the salt.
2.19 mol H2O 0.547 mol Ni(CN)2
= 4 mol H2O/mol Ni(CN)2
So the formula is Ni(CN)2•4H2O 86
Moles and Gases
(Ch. 13.1-13.2)
Remember Boyle’s Law, Charles’s Law and Combined Gas Law? Boyle: P1V1=P2V2
Charles: V1/T1=V2/T2 Combined: P1V1 = P2V2
T1 T2
There is another law that relates volume to moles.
87
The Gas Laws – Avogadro’s Law
Amedeo
Avogadro
(1776 – 1856)
studied the
relationship
between
volume and
amount of gas.
88
The Gas Laws – Avogadro’s Law
Avogadro’s Principle: Equal volumes of
gas at the same P & T contain equal numbers of particles (molecules or atoms).
Avogadro’s Law – The volume of a gas
at constant P & T is directly proportional to the number of moles of gas.
Volume αn; hold P & T constant
V = k x n
Avogadro’s Principle
As an extension of Avogadro’s Principle, 1 mol of a gas at 0˚C (273 K) and 1 atm of pressure occupies a volume of 22.4 Liters (Molar
Volume)
0˚C (273 K) and 1 atm of pressure are standard temperature and
Avogadro’s Principle
Since volume is related to moles, if you know the volume a gas occupies at STP, you know the # of moles.
22.4 L 1 mol
Example: If you have 11.2 L of a gas at STP, how many moles are there? 11.2 L x 1 mol = 0.500 mole
22.4 L
91
Avogadro’s Principle Practice
Given the conditions
at STP find:
Volume of 0.881 mol
of gas
# of moles of N2(g) in
a 2.0 L flask
# of atoms of Kr in
28.5 L
92
Ch. 13.2 - Ideal Gas Law
The gas laws that we covered all have one thing in common: they relate the volume of the gas to one of the other variables.
Boyle: V 1/P Charles: V T Avogadro: V n
We can put them all together to get…
93
Ideal Gas Law
V
nT
P
To change from a
proportionality,
, to an
equation, we introduce, R,
the proportionality constant
or
Universal Gas Constant.
94Ideal Gas Law
This makes the previous proportionality relationship into
V = R*nT/P Or more commonly
PV = nRT
the IDEAL GAS
EQUATION
Ideal Gas Law
The most common value of R is 0.0821 L-atm
Mol-K
There are other values of R with different units that are listed in your book, but we’ll use the one above. Note: they all are equivalent, it just
Relationship to Other Gas Laws
P
1
V
1
= P
2
V
2
n
1
T
1
n
2
T
2
Both equations are equal to a constant, which is…R (the gas
constant)
97
Ideal Gas Equation Example
A deodorant can has a volume of 0.175 L and a pressure of 3.80 atm at 22ºC. How many moles of gas are contained in the can?
Given: P=3.80 atm; V = 0.175 L; T=22+273=295 K; R = 0.0821 L-atm/mol-K
PV=nRTLooking for n (moles) So… n = PV = (3.80 atm)(0.175 L)
RT (0.0821L*atm/mol*K)*295K
=0.0275 mol 98
Ideal Gas Equation Practice
Calculate the volume that a 0.323 mol sample of gas will occupy at 265 K and 0.900 atm.
A 47.3 L container containing 1.62 mol of He is heated until the pressure reaches 1.85 atm. What is the
temperature in ⁰C?
Kr gas in a 18.5 L cylinder exerts a pressure of 8.61 atm at 24.8ºC What is the mass of Kr?
99
Gas Density and Molar Mass
Let Mstand for molar mass (g/mol) Where n = m/M ;m = mass in grams, n
is moles
And we know PV=nRT So then PV= (m/M)RT Rearranging gives: M = mRT
V P Finally we get: M = dRT/P
100 Mass/volume =
Density
Gas Density & Molar Mass
Example 1
What is molar mass of a gas that has a density of 1.40 g/L at STP?
Given: T=273; P=1.00 atm; d=1.40 g/L; R=0.0821 L-atm/mol-K
Know M = DRT/P so…
=(1.40 g/L)(0.0821 L-atm/mol-K)(273K) 1.0 atm
= 31.4 g/mol
Dalton’s Law of Partial
Pressures
Dalton’s Law of Partial Pressure
(end of Ch. 12.1) states that the total pressure of a mixture of gases is equal to the sum of the pressures of all the gases in the mixture. In other words…
Dalton’s Law of Partial
Pressures
John Dalton 1766-1844
103
Dalton’s Law of Partial
Pressures
partial pressure depends only on:
– number of moles of gas – container volume – temperature of the gas.
It does not depend on the identity of the gas.
Because gas molecules are so far apart, they don’t interact with each other.
104
Dalton’s Law Example 1
We have a mixture of O2, CO2, and N2in a vessel at STP. The partial
pressure of CO2 is 0.70 atm and the partial pressure of N2is 0.12 atm. What is the partial pressure of O2?
Answer: What is total pressure? 1 atm, so…
1.00 = 0.70 + 0.12 + P(O2)
P(O2) = 1.00 – 0.70 -0.12 = 0.18 atm
105
Partial Pressure & Mole
Fraction
The partial pressure of a gas
component in a mixture is dependent on how much (moles) are there: Pa = nART/V
The total pressure of a gas mixture is Ptotal = ntotalRT/V
106
Partial Pressure & Mole
Fraction
Ratio of pressure of component to total gas pressure:
=
/
/
RT & V are constants so…
=
Partial Pressure & Mole
Fraction
=
This means that the partial
pressure of a gas is equal to its mole
fraction multiplied by the total
pressure:
Pa= na * Ptotal
Mole Fraction Example 1
A mixture of O2, N2, and He gases are in an enclosed tank. The total pressure in the tank is 12.3 atm. If there are 10.6 mol of N2, 3.3 mol O2 and 1.2 mol He in the tank, what is the partial pressure of each gas?
109
Mole Fraction Example 1
Find total Number of moles:
10.6 + 3.3 + 1.2 = 15.1 mol gas total. For N2: PN2= .
.
∗ 12.3
= 8.6 atmFor O2: PO2= .
.
∗ 12.3
= 2.7 atmFor He: PHe= .
.
∗ 12.3
= 1.0 atmThe sum of the pressures is 12.3 atm.110
Mole Fraction Example 2
A gas mixture containing 0.538 mol He, 0.315 mol Ne and 0.103 mol Ar is confined to a 7.00 L vessel at 25°C.a) Calculate the partial pressure of each gas in the mixture
b) Calculate the total pressure of the mixture. P(He) = 0.538 mol(0.0821 L-atm/mol-K)(298K)/7.0 L
= 1.88 atm
P(Ne) = 0.315 mol(0.0821 L-atm/mol-K)(298K)/7.0 L = 1.10 atm
P(Ar) = 0.103 mol(0.0821 L-atm/mol-K)(298K)/7.0 L = 0.36 atm
111
Mole Fraction Example 2
Can do previous problem because the gases are at same T & V. Therefore, their mole ratios will be
proportional.
Greater moles means greater partial pressure!
112