Download The Mole - Bakersfield College

Chapter 4
Chemical Composition
• Mole Quantities
• Moles, Masses, and Particles
• Determining Empirical and Molecular
Formulas
• Chemical Composition of Solutions
4-1
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
The Mole
• The unit that acts as a bridge between the
microscopic world and the macroscopic
world
• Contains 6.022 x 1023 particles (molecules,
atoms, ions, formula units, etc.)
– This number is called Avogadro’s number.
• The amount of substance that contains as
many basic particles (atoms, molecules,
or formula units) as there are atoms in
exactly 12 g of carbon-12
4-2
Moles of Various Elements and
Compounds
Figure 4.8
4-3
1
Molar Mass
• Describes the mass of 1 mole of a substance
• We obtain the Molar Mass (MM) from the
periodic table by assigning different units to the
atomic mass.
– Instead of assigning the atomic mass units of amu,
we assign the atomic mass units of grams per 1
mole.
• Molar mass is the conversion factor between
mass and moles.
4-4
Practice - Molar Mass
• Complete the table.
Element or
Atomic Mass
Compound
Molar Mass
H
O
Na
Cl
H2O
NaCl
4-5
Practice Solutions – Molar Mass
Element or
Compound
Atomic Mass
Molar Mass
H
1.008 amu
1.008 g/mol
O
16.00 amu
16.00 g/mol
Na
22.99 amu
22.99 g/mol
Cl
35.45 amu
35.45 g/mol
H2O
18.02 amu
18.02 g/mol
NaCl
58.44 amu
58.44 g/mol
4-6
2
Percent Composition by Mass
• An expression of the
portion of the total
mass contributed by
each element
• To find the percent
composition of E (E
is any element):
%E =
mass of E
x 100%
mass of sample
4-7
Practice - Percent Composition
1. What are the percent iron and the
percent sulfur in an 8.33-g sample of
chalcopyrite that contains 2.54 g Fe and
2.91 g S?
2. A 4.55-g sample of limestone (CaCO3)
contains 1.82 g of calcium. What is the
percent Ca in limestone?
4-8
Practice Solutions - Percent
Composition
1. What are the percent iron and the percent
sulfur in an 8.33-g sample of chalcopyrite
that contains 2.54 g Fe and 2.91 g S?
% Fe =
%S=
2.54 g Fe
8.33 g sample
2.91 g S
8.33 g sample
x 100% = 30.5% Fe in sample
x 100% = 34.9% S in sample
4-9
3
Practice Solutions - Percent
Composition
2. A 4.55-g sample of limestone (CaCO3)
contains 1.82 g of calcium. What is the
percent Ca in limestone?
% Ca =
1.82 g Ca
x 100% = 40.0% Ca in limestone
4.55 g limestone
4 - 10
Practice – Conversions with
Molar Mass
1. How many moles of sulfur are in the
1.28 g of sulfur (S) found in a
sample of chalcopyrite?
4 - 11
Practice Solutions – Conversions with
Molar Mass
1. How many moles of sulfur are in the 1.28 g of
sulfur (S) found in a sample of chalcopyrite?
The conversion factor needed is the MM of sulfur:
1 mole S = 32.07 g S
We can write this equality as a ratio as well:
1 mole S
or
32.07 g S
32.07 g S
1 mole S
To solve, we need to cancel out the grams of S. Therefore:
1.28 g S x
1 mol S
= 0.0399 mol S
32.07 g S
4 - 12
4
Extra Practice – Conversions with
Molar Mass
1. How many moles of aspartame (C14H18N2O5)
are found in 40. mg of aspartame? How many
molecules of aspartame are found in this
mass?
2. If one aspirin tablet contains 0.324 g of
acetylsalicylic acid (C9H8O4), then how many
molecules of acetylsalicylic acid are in 2
aspirin tablets?
4 - 13
Extra Practice Solutions – Conversions with
Molar Mass
1. How many moles of aspartame (C14H18N2O5) are found
in 40. mg of aspartame? How many molecules of
aspartame are found in this mass?
1. We first need to convert from mg to g:
1 g C14H18N2O5
40. mg C14H18N2O5 x
= 0.040 g C14H18N2O5
1000 mg C14H18N2O5
2. Next, we need to find the MM of C14H18N2O5:
(14 moles C x 12.01 g C) + (18 moles H x 1.01 g H) +
1 mol C
1 mol H
(2 moles N x 14.01 g N) + (5 moles O x 16.00 g O)
1 mol N
1 mol O
= 294.34 g/mol C14H18N2O5
4 - 14
Extra Practice Solutions Continued
– Conversions with Molar Mass
1. How many moles of aspartame (C14H18N2O5) are
found in 40. mg of aspartame? How many
molecules of aspartame are found in this
mass?
3. Next, convert from grams to moles:
0.040 g C14H18N2O5 x
•
1 mol C14H18N2O5
= 1.4 x 10-4 mol C14H18N2O5
294.34 g C14H18N2O5
Finally, we convert from moles to molecules:
1.4 x 10 -4 mol C14H18N2O5 x
6.022 x 10 23 molecules C14H18N2O5
= 8.2 x 1019 molecules C14H18N2O5
1 mol C14H18N2O5
4 - 15
5
Extra Practice Solutions –
Conversions with Molar Mass
2.
If one aspirin tablet contains 0.324 g of acetylsalicylic acid
(C9H8O4), then how many molecules of acetylsalicylic acid
are in 2 aspirin tablets?
0.324 g C9H8O4
1 mol C9H8O4
6.022 x 10 23 molecules C9H8O4
x
x
1 aspirin tablet
180.17 g C9H8O4
1 mol C9H8O4
= 1.08 x 1021 molecules C9H8O4
1 aspirin tablet
2 aspirin tablets x
1.08 x 1021 molecules C9H8O4
1 aspirin tablet
= 2.16 x 1021 molecules C9H8O4
4 - 16
Summary – Conversions with Molar
Mass and Avogadro’s Number
• To convert from moles to grams or from grams to
moles, use a molar mass (MM) as your conversion
factor.
• To convert from moles to particles (molecules,
atoms, ions, or formula units) or from particles to
moles, use Avogadro’s number as your conversion
factor.
4 - 17
Determining Empirical and
Molecular Formulas
• Empirical formula
– Expresses the simplest ratios of atoms in a
compound
– Written with the smallest whole-number
subscripts
• Molecular formula
– Expresses the actual number of atoms in a
compound
– Can have the same subscripts as the empirical
formula or some multiple of them
4 - 18
6
Determining Empirical and
Molecular Formulas
4 - 19
Practice – Empirical and
Molecular Formulas
• For which of these substances is the empirical
formula the same as the molecular formula?
4 - 20
Empirical or Molecular Formulas
Table 4.1 Some Empirical and Molecular Formulas
Substance
Molecular Formula
Empirical Formula
cyclopentane
C5H10
CH2
cyclohexane
C6H12
CH2
ethylene
C2H4
CH2
hydrogen
sulfide
H2S
H2S
calcium
chloride
This compound does
not have a molecular
formula
CaCl2
4 - 21
7
Empirical Formulas from Percent
Composition
4 - 22
Finding Empirical Formulas
•
To find the empirical formula:
1. If starting with a percent composition, find
the mass of the element by assigning the
percent composition (which has no units
but a % instead) the units of grams.
– If starting with another set of units, then
convert the units to masses if
necessary.
2. Convert from mass to moles using the MM
of the element.
4 - 23
Finding Empirical Formulas
•
To find the empirical formula cont’d:
3. Repeat for all elements in the compound.
4. Find whole number subscripts by:
1. Dividing the moles of the each element by
the smallest number of moles. The
quotients will give whole numbers which
are now the subscripts for the empirical
formula.
2. If #1 does not give whole numbers, then
multiply all numbers by a multiplier that
will resolve the quotients into whole
numbers.
4 - 24
8
Practice – Finding Empirical
Formulas
1. Determine the empirical formula for the
mineral covellite, which has the percent
composition 66.5% Cu and 33.5% S.
2. Shattuckite is a fairly rare copper mineral.
It has the composition 48.43% copper,
17.12% silicon, 34.14% oxygen, and 0.31%
hydrogen. Calculate the empirical
formula of shattuckite.
4 - 25
Practice Solutions – Finding Empirical
Formulas
1. Determine the empirical formula for the mineral
covellite, which has the percent composition 66.5% Cu
and 33.5% S.
First, reassign the percentages units of grams: 66.5 g Cu
and 33.5 g S.
Then, convert to moles and divide both numbers by the
lowest number.
66.5 g Cu x 1 mol Cu = 1.05 mol Cu
1.05 mol Cu = 1
63.55 g Cu
1.05 mol
33.5 g S x 1 mol S = 1.05 mol S
1.05 mol S = 1
32.07 g S
1.05 mol
The whole numbers in purple then become our subscripts.
The empirical formula is therefore: CuS
4 - 26
Practice Solutions Continued – Finding
Empirical Formulas
2.
Shattuckite is a fairly rare copper mineral. It has the
composition 48.43% copper, 17.12% silicon, 34.14% oxygen,
and 0.31% hydrogen. Calculate the empirical formula of
shattuckite.
48.43 g Cu x 1 mol Cu = 0.7621 mol Cu = 2.5 mol Cu x 2 = 5 mol Cu
63.55 g Cu
0.3069
17.12 g Si x 1 mol Si = 0.6095 mol Si = 2 mol Si x 2 = 4 mol Si
28.09 g Si
0.3069
34.14 g O x 1 mol O = 2.134 mol O = 7 mol O x 2 = 14 mol O
16.00 g O
0.3069
0.31 g H x 1 mol H = 0.3069 mol H = 1 mol H x 2 = 2 mol H
1.01 g H
0.3069
Therefore, the empirical formula is: Cu5Si4O14H2
4 - 27
9
Molecular Formulas
•
•
To determine a molecular
formula, the problem must
give a piece of
experimental data, such as
a molar mass, MM.
To find the molecular
formula:
1. Find the empirical
formula first.
2. Divide the empirical
formula’s molar mass by
the experimental molar
mass (which is given).
4 - 28
Practice - Molecular Formulas
1. Potassium persulfate is a strong
bleaching agent. It has a percent
composition of 28.93% potassium,
23.72% sulfur, and 47.35% oxygen.
The experimental molar mass of
270.0 g/mol. What are the empirical
and molecular formulas of
potassium persulfate?
4 - 29
Practice Solutions - Molecular Formulas
1. Potassium persulfate is a strong bleaching agent. It
has a percent composition of 28.93% potassium,
23.72% sulfur, and 47.35% oxygen. The experimental
molar mass of 270.0 g/mol. What are the empirical and
molecular formulas of potassium persulfate?
First, find the empirical formula:
28.93 g K x 1 mol K = 0.7399 mol K = 1 mol K
39.10 g K
0.7396
23.72 g S x 1 mol S = 0.7396 mol S = 1 mol S
32.07 g S
0.7396
47.35 g O x 1 mol O = 2.959 mol O = 4 mol O
16.00 g O
0.7396
Thus, the empirical formula is KSO4.
4 - 30
10
Practice Solutions Continued –
Molecular Formulas
1. Potassium persulfate is a strong bleaching agent. It
has a percent composition of 28.93% potassium,
23.72% sulfur, and 47.35% oxygen. The experimental
molar mass of 270.0 g/mol. What are the empirical and
molecular formulas of potassium persulfate?
To find the molecular formula:
Calculate the MM of KSO4 = (1 mol K x 39.10 g/mol K) + (1
mol S x 32.07 g/mol S) + (4 mol O x 16.00 g/mol O) =
135.17 g/mol KSO4
270.0 g/mol (Experimental MM)
= 2
135.17 g/mol (Empirical Formula’s MM)
Thus, the molecular formula is K2S2O8.
4 - 31
Determining Percent Composition
Using Molar Mass
• To determine the percent composition of an
element (E) in a compound using molar mass
(MM):
%E=
[ MM (E) x # of moles E in compound]
Total MM of compound
4 - 32
Practice – Percent Composition
• Calculate the percent composition of
each element in the following
compounds:
[HINT: you must determine the compound formula 1st]
1. iron(II) chloride
2. dinitrogen tetroxide
3. sodium phosphate
4 - 33
11
Practice Solutions –
Percent Composition
Calculate the percent composition of each element in the
following compounds:
1. iron(II) chloride – FeCl2
Find the MM of the compound 1st:
(1 mol Fe x 55.85 g/mol Fe) + (2 mol Cl x 35.45 g/mol Cl)
= 126.75 g/mol FeCl2
•
% Fe = (1 mol Fe x 55.85 g/mol Fe) = 44.06% Fe in FeCl2
126.75 g/mol FeCl2
% Cl = (2 mol Cl x 35.45 g/mol Cl) = 55.94% Cl in FeCl2
126.75 g/mol FeCl2
NOTE: % Cl = 100.00% FeCl2 - 44.06% Fe = 55.94% Cl
4 - 34
Practice Solutions Continued
Percent Composition
Calculate the percent composition of each element in the
following compounds:
2. Dinitrogen tetroxide – N2O4
Find the MM of the compound 1st:
(2 mol N x 14.01 g/mol N) + (4 mol O x 16.00 g/mol O)
= 92.02 g/mol
•
% N = (2 mol N x 14.01 g/mol N) = 30.45% N in N2O4
92.02 g/mol N2O4
% O = (4 mol O x 16.00 g/mol O) = 69.55% O in N2O4
92.02 g/mol N2O4
NOTE: % O = 100.00% N2O4 – 30.45% N = 69.55% O
4 - 35
Practice Solutions Continued
Percent Composition
Calculate the percent composition of each element in the
following compounds:
3. Sodium phosphate – Na2PO4
Find the MM of the compound 1st:
(2 mol Na x 22.99 g/mol Na) + (1 mol P x 30.97 g/mol P) + (4 mol O x
16.00 g/mol O) = 140.95 g/mol Na2PO4
•
% Na = (2 mol Na x 22.99 g/mol Na) = 32.62% Na in Na2PO4
140.95 g/mol Na2PO4
% P = (1 mol P x 30.97 g/mol P) = 21.97% P in Na2PO4
140.95 g/mol Na2PO4
% O = (4 mol O x 16.00 g/mol O) = 45.41% O in Na2PO4
140.95 g/mol Na2PO4
4 - 36
12
Chemical Composition of Solutions
• Solutions
– are any homogeneous mixture at the
molecular or ionic scale
– are composed of solutes and solvents
• Solutes
– Are present in a lesser amount
– The substances that are dissolved (can be
either wet or dry)
• Solvents
– Are present in the larger amount
– The substances that dissolve
4 - 37
Making a Solution
4 - 38
Solution Concentration
• Concentration
– Is the relative amounts of solute and
solvent in a solution
– When compared with one another,
solutions are classified as dilute or
concentrated.
• Dilute solution
– A solution that contains a relatively small
amount of solute
• Concentrated solution
– A solution that contains a relatively large
amount of solute
4 - 39
13
Concentration of Solutions
4 - 40
Determining Concentration
Insert diagram at bottom of pg. 138
(or pg. 140)
• Percent by Mass
– Expresses concentration via percentage
% mass = mass solute x 100%
mass solution
• Molarity (M)
– The moles of solute dissolved in 1 L of solution
– The most common units of concentration
M = moles solute
L solution
4 - 41
Practice – Solution Concentration
1. A solution is prepared from 22.5 g of H2S
dissolved in sufficient water to give 250.0 mL
of solution. What is the molarity of the
solution?
2. Bluestone is copper(II) sulfate pentahydrate,
CuSO4•5H2O, with a molar mass of 249.7 g/mol.
A sample of pond water was found to have a
concentration of 6.2 x 105 M copper(II) sulfate.
If the pond has a volume of 1.8 x 107 L, then
what mass of bluestone did the farmer add to
the pond?
4 - 42
14
Practice Solutions – Solution Concentration
1. A solution is prepared from 22.5 g of H2S dissolved in
sufficient water to give 250.0 mL of solution. What is
the molarity of the solution?
M = moles of solute
L solution
H2S is the solute, so
Moles of H2S = 22.5 g H2S x 1 mole H2S = 0.660 moles H2S
34.09 g H2S
To find L of solution:
= 0.2500 L
250.0 mL x 1 L
1000 mL
M = 0.660 moles H2S = 2.64 M H2S solution
0.2500 L solution
4 - 43
Practice Solutions Continued –
Solution Concentration
2. Bluestone is copper(II) sulfate pentahydrate, CuSO4•5H2O,
with a molar mass of 249.7 g/mol. A sample of pond
water was found to have a concentration of 6.2 x 105 M
copper(II) sulfate. If the pond has a volume of 1.8 x 107 L,
then what mass of bluestone did the farmer add to the
pond?
Start with the volume as it only has one set of units:
1.8 x 107 L x 6.2 x 105 mol x 249.7 g = 2.8 x 1015 g CuSO4•5H2O
1L
1 mol
4 - 44
Dilution and the Dilution
Equation
• Dilution
– The process of adding more solvent to solution
• The dilution equation:
Molescon = MconVcon
MconVcon = MdilVdil
Moles1 = M1V1
M1V1 = M2V2
4 - 45
15
Diluting a More Concentrated
Solution
4 - 46
Practice – Dilution
1. If 42.8 mL of 3.02 M H2SO4 solution
is diluted to a final volume 500.0 mL,
what is the molarity of the diluted
solution of H2SO4?
2. What is the concentration of a
solution prepared by diluting 35.0
mL of 0.150 M KBr to 250.0 mL?
4 - 47
Practice Solutions – Dilution
1. If 42.8 mL of 3.02 M H2SO4 solution is
diluted to a final volume 500.0 mL, what
is the molarity of the diluted solution of
H2SO4?
MconVcon = MdilVdil
(3.02 M)(42.8 mL) = Mdil(500.0 mL)
Mdil = (3.02 M)(42.8 mL) = 0.259 M
500.0 mL
4 - 48
16
Practice Solutions – Dilution
2. What is the concentration of a
solution prepared by diluting 35.0
mL of 0.150 M KBr to 250.0 mL?
MconVcon = MdilVdil
(0.150 M)(35.0 mL) = Mdil(250.0 mL)
Mdil = (0.150 M)(35.0 mL) = 0.0210 M
250.0 mL
4 - 49
17