Problem: The lead in my pencil is graphite-an almost pure
carbon compound. I wrote a really long English essay and used up
most of a pencil, how many carbon atoms did I use? Can I figure
this out.
Slide 4
What is the mass of 1 carbon atom? 1.99 * 10 -23 g Is this a
practical number to use for calculations? Scientists have defined
atomic mass units or amus to avoid this. 1 amu = 1.66 * 10 -24 g
The average atomic mass of 1 carbon atom is 12.01 amu. Any sample
of carbon from nature can be treated as though it were composed of
identical carbon atoms, each with a mass of 12.01 amu.
Slide 5
Conversion Relationship #1 1 amu = 1.66 * 10 -24 g This
equivalency allows conversion between these two measurements of
mass. Its just like using 2.2046 lbs = 1 Kg
Slide 6
Practice: g amu What is the mass of 5.00 g Carbon in atomic
mass units? What is the mass of 5 million amus of Silver in
grams?
Slide 7
Conversion Relationship #2 The mass of a single atom of an
element equals its mass on the Periodic Table in atomic mass units.
For Carbon:1 C atom = 12.0107 amu For Hydrogen:1 H atom = 1.00795
amu For Chlorine:1 Cl atom = 35.453 amu
Slide 8
Conversion Relationship #2 1 atom = mass on PT in amu The
periodic table contains equivalency statements that can be used to
convert between mass in amu and number of atoms. For example, What
is the mass of 12 Carbon atoms? 1 C atom = 12.01 amu 12 C atoms
(12.01 amu / 1 C atom) = 144.1 amu
Slide 9
Practice: # atoms amu Calculate the mass, in amu, of a sample
of aluminum that contains 75 aluminum atoms. 1 Al atom = 26.981538
amu
Slide 10
Practice: # atoms amu Calculate the mass of a sample that
contains 23 nitrogen atoms. 1 N atom = 14.0067 amu
Slide 11
Practice: # atoms amu Calculate the number of sodium atoms
present in a sample that has a mass of 1172.49 amu. 1 Na Atom =
22.98977 amu
Slide 12
Practice: # atoms amu Calculate the number of oxygen atoms in a
sample that has a mass of 288 amu. 1 O Atom = 15.9994 amu
Slide 13
Try these Calculate the mass of a sample that contains 2493
Lithium atoms. Which weighs more, 25 Platinum atoms, or 5 iron
atoms? How many atoms are in 10462.692 amu of oxygen?
Slide 14
THE MOLE
Slide 15
What is a Dozen? A Dozen is a word that represents a number. 1
Dozen = 12 1 dozen apples = ? = 12 apples 2 dozen cookies = ? = 24
cookies 3.2 dozen eggs = ? = 38.4 eggs (38 eggs)
Slide 16
What is a Mole? A Mole is a word that represents a number. 1
mole = 1 mol = 6.022 x 10 23 This number is called Avogadros
number. Avogadros Number = 1 mol = 6.022 x 10 23 A Mole is a lot
bigger than a Dozen! 1 mol apples = ? = 6.022 x 10 23 apples 2 mol
cookies = ? = 1.2044 x 10 24 cookies 3.2 mol eggs = ? = 1.92704 x
10 24 eggs = 1,927,040,000,000,000,000,000,000 eggs
Slide 17
Conversion Relationship #3 1 mol = 6.022 x 10 23 atoms How many
calcium atoms are present in a sample that contains 3.12 mol Ca?
How many moles are present in a sample of gold that contains 2.6 x
10 52 atoms?
Slide 18
Why is the mole necessary? Comparing equal masses of chemicals
does not compare equal numbers of particles. Consider the synthesis
of barium sulfide from its elements: Ba(s) + S(s) BaS If there are
100 amu of each element, how many molecules of BaS will form?
Calculate the number of atoms in each 100 amu sample. Are they the
same? It is much more efficient to use moles to compare quantities
of chemicals. If there are 100 moles of Ba and 100 moles of S, how
many moles of BaS will form? 100 moles is the same number of atoms
no matter what kind of atom it is. It is a basis for an equal
comparison. That is a much easier way of thinking of things.
Slide 19
The Mole is the Most Important Quantity in Chemistry!!!!
Comparing quantities in chemistry using moles allows an equal basis
of comparison. Comparing quantities in chemistry using grams does
not.
Slide 20
Conversion Relationship #4 The Mass of a Mole The Periodic
Table of Elements tells us the mass of 1 mole of each of the atoms.
1 mole of an element = mass on Periodic Table in grams. To Clarify
between the mass of 1 atom and the mass of one mole of atoms: 1
atom = atomic mass in amu The mass of 1 carbon atom is 12.0107 amu
1 mole of atoms = atomic mass in grams The mass of 1 mole of carbon
atoms is 12.0107 g
Slide 21
Practice: g-mol & mol-#atoms How many moles Al is contained
in 10.0 g Al? How many atoms does this sample contain?
Slide 22
Practice: grams-moles-#atoms A silicon chip used in an
integrated circuit of a microcomputer has a mass of 5.68 mg. How
many Si atoms are present in this chip
Slide 23
Practice: #atoms-moles-grams Calculate both the number of moles
in, and the mass of, a sample of chromium containing 5.00 x 10 20
atoms.
Slide 24
8.4 MOLAR MASS Goals: 1. To define Molar Mass. 2. To determine
the molar mass of a polyatomic substance. 3. To convert between
moles and mass of a substance.
Slide 25
The Molar Mass of Atoms The molar mass of an atom is its atomic
mass from the Periodic Table in grams. What is the molar mass of an
atom of Oxygen? 15.9994 g/mol What is the molar mass of elemental
oxygen? O 2 (g) 2 x 15.9994 g/mol = 31.9988 g/mol
Slide 26
The Molar Mass of Polyatomic Substances A polyatomic substance
is a collection of atoms. The molar mass of a polyatomic substance
is simply the sum of the molar masses of all the atoms of which the
substance is made. Example: Find the molar mass of methane (CH 4 )
1.Determine the chemical composition: CH 4 = 1 C + 4 H 2.Then add
up the masses of all the atoms. CH 4 = 1(12.0107) + 4(1.00794) =
16.04246 The molar mass is the mass of 6.022 x 10 23 molecules of
CH 4, in grams. The molar mass of methane = 16.0425 g/mol CH 4
Slide 27
Summary of Molar Mass Molar Mass The Molar Mass of a substance
is the mass (in grams) of 1 mole of the substance. 1.For Elements,
the molar mass is equal to the atomic mass on the periodic table
(in grams). 2.For Diatomic & Polyatomic Substances, the molar
mass is equal to the sum of the atomic weights of all the atoms in
the substance (in grams).
Slide 28
Practice: Calculating Molar Mass Calculate the molar mass of
sulfur dioxide. 1.Chemical Composition: SO 2 = S + 2O 2.Add the
molar mass of all the atoms: SO 2 = 1(32.065) + 2(15.9994) Answer:
64.0638 = 64.064 g/mol SO 2
Slide 29
Practice: Calculating Molar Mass Polyvinyl chloride (called
PVC) which is widely used for floor coverings (vinyl) and for
plastic pipes in plumbing systems, is made from a molecule with the
formula C 2 H 3 Cl. Calculate the molar mass of this substance.
1.Chemical Composition: C 2 H 3 Cl = 2C + 3H + 1Cl 2.Sum of molar
masses: C 2 H 3 Cl = 2(12.0107) + 3(1.00794) + 1(35.453) Answer:
62.49822 = 62.498 g/mol C 2 H 3 Cl
Slide 30
Conversion Relationship #4 The Mass of a Mole (Revisited) Once
the molar mass of a substance is known, it can be used to convert
between moles and masses of the substance. The molar mass of
methane (CH 4 ) is 16.04246 g/mol. This gives an equivalency
statement that can be used as a conversion factor: 1 mol CH 4 =
16.04246 g CH 4 1.What is the mass of 3.2 moles of methane? 2.How
many moles are in 5.93 g methane?
Slide 31
Practice: grams-moles (with molecules this time) Calcium
Carbonate, CaCO 3 (also called calcite), is the principal mineral
found in limestone, marble, chalk, pearls, and the shells of marine
animals such as clams. a.Calculate the molar mass of calcium
carbonate. b.What is the mass in grams of 4.86 mol CaCO 3 ?
Slide 32
Calculate the molar mass for sodium sulfate, Na 2 SO 4. A
sample of sodium sulfate with a mass of 300.0 g represents what
number of moles of sodium sulfate? a.Calculate the molar mass of
Sodium Sulfate. b.Convert mass Na 2 SO 4 moles Na 2 SO 4 Practice:
grams-moles
Slide 33
Practice: moles-grams Juglone, a dye known for centuries, is
produced from the husks of black walnuts. It is also a natural
herbicide (weed killer) that kills off competitive plants around
the black walnut tree but does not affect grass and other
noncompetitive plants. The formula for juglone is C 10 H 6 O 3.
a.Calculate the molar mass of Juglone: b.How many moles are
contained in a 1.56 g sample of juglone?
Slide 34
Conversion Relationship #3 (Revisited) Remember Avogadros
Number? 6.022 x 10 23 Avogadros number represents 1 mole of
anything. 1 mole of any substance contains 6.022 x 10 23 units of
that substance. This relationship is an equivalency factor that can
be used to make conversions between moles and # of molecules. 1 mol
= 6.022 x 10 23 molecules
Slide 35
Practice: mass-moles-#molecules Isopentyl Acetate, C 7 H 14 O
2, the compound responsible for the scent of bananas, can be
produced commercially. Interestingly, bees release about 1 g (1x10
-6 g) of this compound when they sting. This attracts other bees,
which then join the attack. How many moles and how many molecules
of isopentyl acetate are released in a typical bee sting?
a.Determine the molar mass of C 7 H 14 O 2 : b.Convert mass C 7 H
14 O 2 moles C 7 H 14 O 2 # of molecules C 7 H 14 O 2 :
Slide 36
Practice: mass-moles-#molecules The substance Teflon, the
slippery coating on many frying pans, is made from the C 2 F 4
molecule. Calculate the number of C 2 F 4 units present in 135 g of
Teflon. a.Calculate the molar mass of C 2 F 4 : b.Convert mass
moles # molecules
Slide 37
MOLAR RELATIONSHIPS
Slide 38
RELATIONSHIP #1 PARTICLES IN A MOLE
Slide 39
Particles in a Mole A chemical mole is defined as the number of
atoms present in a sample of Carbon- 12 that weighs exactly
12.0107g. A Mole is a word that represents a Number, and that
number is also known as Avogadros Number. 1 mole = 6.022 x 10 23
particles Particles are either atoms or molecules.
Slide 40
Practice: Moles - # Particles How many carbon atoms are in a
sample that contains 5.16 moles C? How many moles are present in a
sample of nitric acid that contains 5 billion molecules?
Slide 41
RELATIONSHIP #2 ATOMS IN A MOLECULE
Slide 42
Atoms in Molecules Consider a single water molecule: What is
the composition by type of atom? Using moles, 1 mole H 2 O contains
2 moles H and 1 mole O. The following equivalencies can be made: 1
mol H 2 O = 2 mol H 1 mol H 2 O = 1 mol O How many moles of
hydrogen are in 2.53 mole H 2 O?
Slide 43
Practice: Atoms in Molecules How many moles of carbon are in a
6.57 mol sample of Aluminum Acetate?
Slide 44
PERCENT COMPOSITION BY MASS 8.5 Pages 226 - 228
Slide 45
Mass Percent
Slide 46
Calculating Mass Percent Consider the compound Ethanol (C 2 H 5
OH) Follow these steps to determine the mass percent of Carbon in
Ethanol: 1.Determine the number of moles of C in one mole of
ethanol: There are 2 moles of Carbon, so n = 2. 2.Determine the
mass of C in ethanol: 2 mol C x (12.0107 g C / 1 mol C) = 24.0214 g
C 3.Determine the molar mass of ethanol: 2(12.0107) + 6(1.00794) +
1(15.9994) = 46.0684 g C 2 H 5 OH 4.Divide molar mass C in ethanol
by molar mass ethanol and multiply by 100: (24.0214 g)/(46.0684 g)
x 100 = 52.14 % C
Slide 47
Practice: Calculating Mass Percent Carvone is a substance that
occurs in two forms, both of which have the same molecular formula
(C 10 H 14 O) and molar mass. One type of carvone gives caraway
seeds their characteristic smell; the other is responsible for the
smell of spearmint oil. Compute the mass percent of each element in
Carvone.
Slide 48
Practice: Calculating Mass Percent Penicillin F has the formula
C 14 H 20 N 2 SO 4. Calculate the mass percent of each element in
Penicillin F.
Slide 49
Using Mass Percent A sample of Calcium Carbonate has a mass of
5.69 g. What is the mass of Oxygen present in the sample?
Slide 50
Using Mass Percent What mass of carbon is present in 15.39 mol
C 6 H 12 O 6 ?
Slide 51
THE EMPIRICAL FORMULA 8.6 Pages 228 - 230
Slide 52
Combustion Analysis When a compound containing only C, H, &
O is burned in the presence of excess oxygen, the composition by
mass of C, H, & O can be determined by measuring the mass of CO
2, and H 2 O produced. 0.2015 g of a substance containing C, H,
& O is found to contain 0.0806 g C, 0.01353 g H, and 0.1074 g
O. What is the ratio of C:H:O atoms in the compound?
Slide 53
Step 1: Convert mass to moles 0.0806 g C (1 mol C / 12.0107 g
C) = 0.006710683 mol C =.00671 mol C 0.01353 g H (1 mol H / 1.00794
g H) = 0.0134234181 mol H = 0.01342 mol H 0.1074 g O (1 mol O /
15.9994 g O) = 0.0067127517 mol O = 0.006713 mol O
Slide 54
Step 2: Convert Moles to Atoms 0.00671 mol C (6.022 x 10 23
atoms C / 1 mol C) = 4.04 x 10 21 atoms C 0.01342 mol H (6.022 x 10
23 atoms H / 1 mol H) = 8.084 x 10 21 atoms H 0.006713 mol O
((6.022 x 10 23 atoms H / 1 mol H) = 4.042 x 10 21 atoms O
Slide 55
Step 3: Determine the Ratio of C, H, & O C:H:O (4.04 x 10
21 atoms C) : (8.084 x 10 21 atoms H) : (4.042 x 10 21 atoms O)
1:2:1 Notice that the mole ratio is the same (0.00671 mol C) :
(0.01342 mol H) : (0.006713 mol O)
Slide 56
Is this ratio the molecular formula? The ratio of C:H:O is
1:2:1 Does this mean the molecular formula is CH 2 O? What about
the molecular formulas of the following compounds? C 2 H 4 O 2, C 3
H 6 O 3, C 5 H 10 O 5 All of the above have a ratio of C:H:O of
1:2:1.
Slide 57
Empirical Formulas The Empirical Formula of a compound is the
smallest whole-number (integer) ratio of atoms. All four of the
compounds CH 2 O, C 2 H 4 O 2, C 3 H 6 O 3, & C 5 H 10 O 5 have
the same empirical formula; CH 2 O. The Empirical Formula of a
compound is the most simplified formula.
Slide 58
Determining Empirical Formulas Determine the empirical formula
for each compound: 1.C 6 H 6 2.C 12 H 4 Cl 4 O 2 3.C 6 H 16 N 2
Answers: 1.CH 2.C 6 H 2 Cl 2 O 3.C 3 H 8 N
Calculating Empirical Formulas 1. Determine the number of moles
of each element present in the compound. (This can be done 1 of 2
ways) a)If the amounts are given in grams, simply convert them to
moles using the molar masses of the elements from the periodic
table of elements. b)If the amounts are give in percent composition
by mass, simply base your calculations on a sample of mass = 100. g
of the compound. The percent mass of each element will represent
the mass in grams of that element. 2. Divide each value of the
number of moles by the smallest of the values. If each resulting
number is a whole number (after appropriate rounding) these numbers
represent the subscripts of the elements in the empirical formula.
3. If the numbers obtained in the previous step are not whole
numbers, multiply each number by an integer so that the results are
all whole numbers, which are the subscripts of the elements in the
empirical formula.
Slide 61
Practice: Calculating Empirical Formulas An oxide of aluminum
is formed by the reaction of 4.151 g Al with 3.692 g O. Calculate
the empirical formula of the compound. 1. Determine the mass of
each element: Given 2. Determine the moles of each: 4.151 g Al (1
mol Al / 26.981538 g Al) = 0.1538 mol Al 3.692 g O (1mol O /
15.9994 g O) = 0.2308 mol O 3. Divide each by the least number of
moles: Al: 0.1538 / 0.1538 = 1 O: 0.2308 / 0.1538 = 1.5 4. Multiply
to get whole numbers to determine the formula: Al:O = (1:1.5) x 2 =
2:3 Answer: Al 2 O 3 Aluminum Oxide
Slide 62
Practice: Calculating Empirical Formulas When a 0.3546 g sample
of vanadium metal is heated in air, it reacts with oxygen to
achieve a final mass of 0.6330 g. Calculate the empirical formula
of this vanadium oxide. 1. Determine the mass of each element
present: mV = 0.3546 g mO = 0.6330 0.3546 = 0.2784 g 2. Determine
the moles of each element present: molV = 0.3546 g (1 mol / 50.9415
g) = 0.0069609258 mol V mol O = 0.2784 g (1 mol / 15.9994 g) =
0.0174006525 mol O 3. Determine the mole:mole ratio of elements: V:
0.0069609258 / 0.0069609258 = 1 O: 0.0174006525 / 0.0069609258 =
2.499761242 2.5 4. Determine the lowest whole number ration of
elements: V:O = (1:2.5) x 2 = 2:5 Answer: V 2 O 5 Vanadium (V)
Oxide
Slide 63
Practice: Calculating Empirical Formulas A sample of lead
arsenate, an insecticide used against the potato beetle, contains
1.3813 g of lead, 0.00672 g of hydrogen, 0.4995 g arsenic, and
0.4267 g oxygen. Calculate the empirical formula for lead arsenate.
1. Determine the mass of each element: given 2. Determine moles of
each element: 1.3813 g Pb (1 mol / 207.2 g) = 0.0066665058 mol Pb
0.00672 g H (1 mol / 1.00794 g) = 0.0066670635 mol H 0.4995 g As (1
mol / 74.9216 g) = 0.0066669692 mol As 0.4267 g O (1 mol / 15.9994
g) = 0.0266697501 mol O 3. Determine the mole ratios: Pb:
0.0066665058 / 0.0066665058 = 1 H: 0.0066670635 / 0.0066665058 =
1.000083657 1 As: 0.0066669692 / 0.0066665058 = 1.000069512 1 O:
0.0266697501 / 0.0066665058 = 4.000559048 4 4. Multiply the mole
ratios to get the lowest whole # integers: Pb:H:As:O = 1:1:1:4
already done Answer: PbHAsO 4
Slide 64
Practice: Calculating Empirical Formulas Analysis of the
chemical carbamic acid finds that it is composed of 0.8007 g C,
0.9333 g N, 0.2016 g H, and 2.133 g O. Determine the empirical
formula of carbamic acid. 1. Determine the mass of each element:
given 2. Determine moles of each element: 0.8007 g C (1 mol /
12.0107 g) = 0.0666655565 mol C 0.9333 g N (1 mol / 14.0067 g) =
0.0666323974 mol N 0.2016 g H (1 mol / 1.00794 g) = 0.2000119055
mol H 2.133 g O (1 mol / 15.9994 g) = 0.1333174994 mol O 3.
Determine the mole ratios: C: 0.0666655565 / 0.0666323974 =
1.00497642 1 N: 0.0666323974 / 0.0666323974 = 1 H: 0.2000119055 /
0.0666323974 = 3.001721584 3 O: 0.1333174994 / 0.0666323974 =
2.000790976 2 4. Multiply the mole ratios to get the lowest whole #
integers: C:N:H:O = 1:1:3:2 already done Answer: CNH 3 O 2
Slide 65
Practice: Calculating Empirical Formulas Cisplatin, the common
name for a platinum compound that is used to treat cancerous
tumors, has the composition (mass percent) 65.02 % Pt, 9.34 % N,
2.02 % H, and 23.63 %Cl. Calculate the empirical formula for
cisplatin. 1. Determine the mass of each element: given in % so use
a 100.g sample: 65.02 g Pt + 9.34 g N + 2.02 g H + 23.63 g Cl =
100.00 g sample 2. Determine moles of each element: 65.02 g Pt (1
mol / 195.08 g) = 0.3332991593 mol Pt 9.34 g N (1 mol / 14.0067 g)
= 0.6668237344 mol N 2.02 g H (1 mol / 1.00794 g) = 2.004087545 mol
H 23.63 g Cl (1 mol / 35.453 g) = 0.6665162328 mol Cl 3. Determine
the mole ratios: Pt: 0.3332991593 / 0.3332991593 = 1 N:
0.6668237344 / 0.3332991593 = 2.000676317 2 H: 2.004087545 /
0.3332991593 = 6.012879088 6 Cl: 0.6665162328 / 0.3332991593 =
1.999753717 2 4. Multiply the mole ratios to get the lowest whole #
integers: Pt:N:H:Cl = 1:2:6:2 already done Answer: PtN 2 H 6 Cl
2
Slide 66
Practice: Calculating Empirical Formulas The most common form
of nylon (nylon 6) is 63.68% C, 12.38% N, 9.80% H, and 14.14% O.
Calculate the empirical formula of nylon 6. 1. Determine the mass
of each element: given in % so use a 100.g sample: 63.68 g C +
12.38 g N + 9.80 g H + 14.14 g O = 100.00 g sample 2. Determine
moles of each element: 63.68 g C (1 mol / 12.0107 g) = 5.301939104
mol C 12.38 g N (1 mol / 14.0067 g) = 0.8838627228 mol N 9.80 g H
(1 mol / 1.00794 g) = 9.7228096 mol H 14.14 g O (1 mol / 15.9994 g)
= 0.883731419 mol O 3. Determine the mole ratios: C: 5.301939104 /
0.883731419 = 5.999491463 6 N: 0.8838627228 / 0.883731419 =
1.000148579 1 H: 9.7228096 / 0.883731419 = 11.00199607 11 O:
0.883731419 / 0.883731419 = 1 4. Multiply the mole ratios to get
the lowest whole # integers: C:N:H:O = 6:1:11:1 already done
Answer: C 6 NH 11 O
Slide 67
8.8 CALCULATION OF MOLECULAR FORMULAS Pages 235 237
Slide 68
Calculating Molecular Formulas Knowing the percent composition
or knowing the mass ratios of a compound allows us to calculate the
empirical formula of the compound. But this does not determine the
identity of the compound In order to calculate the molecular
formula of a compound, the molar mass must also be known. n = molar
mass of compound molar mass of empirical formula If n can be found,
then the identity of the compound can be determine as [n x
(empirical formula)].
Slide 69
Practice: Calculating Molecular Formulas A white powder is
analyzed and found to have an empirical formula of P 2 O 5. The
compound has a molar mass of 283.88 g. What is the compounds
molecular formula? 1. Determine molar mass of the empirical
formula: P 2 O 5 = 141.94452 g/mol 2. Determine the value of n: n =
283.88 / 141.94452 = 1.999936313 2 3. Determine the molecular
formula: n(P 2 O 5 ) = 2(P 2 O 5 ) Answer: P 4 O 10
Slide 70
Practice: Calculating Molecular Formulas A compound used as an
additive for gasoline to help prevent engine knock shows the
following percentage composition: 71.65% Cl, 24.27% C, 4.07% H. The
molar mass is known to be 98.96 g. Determine the empirical and the
molecular formula for this compound. Determine the empirical
formula: Mass of each, moles of each, mole ratio, lowest whole
number mole ratio = empirical formula The empirical formula is ClCH
2 Determine the molar mass of the empirical formula: ClCH 2 =
49.47958 g/mol Determine the value of n: 98.96 g / 49.47958 g =
2.000016977 2 Determine the molecular formula: n(ClCH 2 ) = 2(ClCH
2 ) Answer: Cl 2 C 2 H 4