1
2
The MoleThe Mole
4
The mass of a single atom is too small to measure on a balance.
mass of hydrogen atom = 1.673 x 10-24 g
5
This is an
infinitesimal
mass
1.673 x 10-24 g
6
• Chemists have chosen a unit for counting atoms.
• That unit is the
• Chemists require a unit for counting which can express large numbers of atoms using simple numbers.
MOLE
7
1 mole = 6.02 x 1023 objects
8
LARGE6.02 x 1023
is a very
number
9
6.02 x 1023
is
number
Avogadro’s Number
10
Amadeo (Amedeo) Avogadro
11
If 10,000 people started to count Avogadro’s number and counted at the rate of 100 numbers per minute each minute of the day, it would take over 1 trillion years to count the total number.
12
1 mole of any element contains
6.02 x 1023
particles of that substance.
13
The atomic weight in grams
of any element23
contains 1 mole of atoms.
14
This is the same number of particles6.02 x 1023
as there are in exactly 12 grams of
C126
15
ExamplesExamplesExamplesExamples
16
Species
Quantity
Number of H atoms
H
1 mole
6.02 x 1023
17
Species
Quantity
Number of H2 molecules
H2
1 mole
6.02 x 1023
18
Species
Quantity
Number of Na atoms
Na
1 mole
6.02 x 1023
19
Species
Quantity
Number of Fe atoms
Fe
1 mole
6.02 x 1023
20
Species
Quantity
Number of C6H6 molecules
C6H6
1 mole
6.02 x 1023
21
1 mol of atoms = 6.02 x 1023 atoms
6.02 x 1023 molecules
6.02 x 1023 ions
1 mol of molecules =
1 mol of ions =
22
• The mole weight of an element is its atomic weight in grams.
• It contains 6.02 x 1023 atoms (Avogadro’s number) of the element.
23
ElementAtomic mass
Mole weightNumber of
atoms
H 1.008 amu 1.008 g 6.02 x 1023
Mg 24.31 amu 24.31 g 6.02 x 1023
Na 22.99 amu 22.99 g 6.02 x 1023
24
ProblemsProblemsProblemsProblems
25
Convert ToMoles!
26
Atomic weight iron = 55.85
How many moles of iron does 25.0 g of iron represent?
Conversion sequence: grams Fe → moles Fe
1 mol Fe(grams Fe)
55.85 g Fe
1 mol Fe(25.0 g Fe)
55.85 g Fe
0.448 mol Fe
Set up the calculation using a conversion factor between moles and grams.
27
Atomic weight iron = 55.85
Conversion sequence: grams Fe → atoms Fe
236.02 x 10 atoms Fe(grams Fe)
55.85 g Fe
How many iron atoms are contained in 40.0 grams of iron?
236.02 x 10 atoms Fe(40.0 g Fe)
55.85 g Fe
234.31 x 10 atoms Fe
Set up the calculation using a conversion factor between atoms and grams.
28
Mole weight Na = 22.99 g
Conversion sequence: atoms Na → grams Na
23
22.99 g Na(atoms Na)
6.02 x 10 atoms Na
What is the mass of 3.01 x 1023 atoms of sodium (Na)?
2323
22.99 g Na(3.01 x 10 atoms Na)
6.02 x 10 atoms Na
11.5 g Na
Set up the calculation using a conversion factor between grams and atoms.
29
Atomic weight tin = 118.7
What is the mass of 0.365 moles of tin?
Conversion sequence: moles Sn → grams Sn
mole weight Sn(moles Sn)
1 mole Sn
118.7 g Sn(0.365 moles Sn)
1 mole Sn
43.3 g Sn
Set up the calculation using a conversion factor between grams and atoms.
30
2(2.00 mol O )23
2
2
6.02 x 10 molecules O1 mol O
2
2 atoms O1 molecule O
Conversion sequence: moles O2 → molecules O2 → atoms O
232
2
6.02 x 10 molecules O1 mol O
How many oxygen atoms are present in 2.00 mol of oxygen molecules?
Two conversion factors are needed:
2
2 atoms O1 molecule O
24= 2.41 x10 atoms O
31
Mole Weight of Mole Weight of CompoundsCompounds
32
The mole weight of a compound can be determined by adding the mole weights of all of the atoms in its formula.
33
2 C = 2(12.01 g) = 24.02 g
6 H = 6(1.01 g) = 6.06 g
1 O = 1(16.00 g) = 16.00 g
46.08 g
Calculate the mole weight of C2H6O.
34
1 Li = 1(6.94 g) = 6.94 g1 Cl = 1(35.45 g) = 35.45 g4 O = 4(16.00 g) = 64.00 g
106.39 g
Calculate the mole weight of LiClO4.
35
Calculate the mole weight of (NH4)3PO4 .
3 N = 3(14.01 g) = 42.03 g12 H = 12(1.01 g) = 12.12 g
1 P = 1(30.97 g) = 30.97 g4 O = 4(16.00 g) = 64.00 g
149.12 g
36
Calculate the mole weight of NaC2H3O2 · 3 H2O
1 Na = 1(22.99 g) = 22.99 g 2 C = 2(12.01 g) = 24.02 g
9 H = 9(1.01 g) = 9.09 g 5 O = 5(16.00 g) = 80.00 g
136.10 g
Solids that contain water molecules as part of their structure are called Hydrates.
37
Avogadro’s Number of Particles
6.02 x 1023 Particles
Mole Weight
1 MOLE
38
1 MOLE Ca
Avogadro’s Number ofCa atoms
6.02 x 1023 Ca atoms
40.078 g Ca
39
1 MOLE H2O
Avogadro’s Number of
H2O molecules
6.02 x 1023 H2O
molecules
18.02 g H2O
40
H Cl HCl
6.02 x 1023 H atoms
6.02 x 1023 Cl atoms
6.02 x 1023 HCl molecules
1 mol H atoms 1 mol Cl atoms1 mol HCl molecules
1.008 g H 35.45 g Cl 36.46 g HCl
1 mole weight H atoms
1 mole weight Cl atoms
1 mole weight HCl molecules
These relationships are present when hydrogen combines with chlorine.
41
In dealing with diatomic elements (H2, O2, N2, F2, Cl2, Br2, and I2), distinguish between one mole of atoms and one mole of molecules.
42
Calculate the mole weight of 1 mole of H atoms.
1 H = 1(1.01 g) = 1.01 g
Calculate the mole weight of 1 mole of H2 molecules.
2 H = 2(1.01 g) = 2.02 g
43
ProblemsProblemsProblemsProblems
44
How many moles of benzene, C6H6, are present in 390.0 grams of benzene?
Conversion sequence: grams C6H6 → moles C6H6
6 6
6 6
78.12 grams C HUse the conversion factor:
1 mole C H
6 6
6 6
1 mole C H 78.12 g C H
6 6(390.0 g C H ) 6 6= 5.000 moles C H
The mole weight of C6H6 is 78.12 g.
45
How many grams of (NH4)3PO4 are contained in 2.52 moles of (NH4)3PO4?
Conversion sequence: moles (NH4)3PO4
→ grams (NH4)3PO4
4 3 4
4 3 4
149.12 grams (NH ) POUse the conversion factor:
1 mole (NH ) PO
4 3 4(2.52 mol (NH ) PO )) 4 3 4
4 3 4
149.12 g (NH ) PO1 mol (NH ) PO
4 3 4= 376g (NH ) PO
The mole weight of (NH4)3PO4 is 149.12 g.
46
2(56.04 g N ) 2
2
1 mol N28.02 g N
232
2
6.02 x 10 molecules N
1 mol N
56.04 g of N2 contains how many N2 molecules?
The mole weight of N2 is 28.02 g.
Conversion sequence: g N2 → moles N2 → molecules N2
Use the conversion factors
2
2
1 mol N 28.02 g N
232
2
6.02 x 10 molecules N
1 mol N
242= 1.20 x 10 molecules N
47
2
2
1 mol N28.02 g N
2(56.04 g N )23
2
2
6.02 x 10 molecules N
1 mol N
56.04 g of N2 contains how many N atoms?
The mole weight of N2 is 28.02 g.
Conversion sequence: g N2 → moles N2 → molecules N2
→ atoms NUse the conversion factors
2
2
1 mol N 28.02 g N
232
2
6.02 x 10 molecules N
1 mol N 2
2 atoms N 1 molecule N
24= 2.41 x 10 atoms N2
2 atoms N1 molecule N
48
Percent CompositionPercent Compositionof Compoundsof Compounds
49
Percent composition of a compound is the mass percent of each element in the compound.
H2O11.19% H by mass 88.79% O by mass
50
Percent Composition Percent Composition From FormulaFrom Formula
51
If the formula of a compound is known, a two-step process is needed to calculate the percent composition.
Step 1 Calculate the mole weight of the formula.
Step 2 Divide the total mass of each element in the formula by the mole weight and multiply by 100.
52
total mass of the element x 100 = percent of the element
mole weight
53
Step 1 Calculate the mole weight of H2S.2 H = 2 (1.01 g) = 2.02 g
1 S = 1 (32.07 g) = 32.07 g34.09 g
Calculate the percent composition of hydrosulfuric acid H2S(aq) .
54
Calculate the percent composition of hydrosulfuric acid H2S.
Step 2 Divide the mass of each element by the mole weight and multiply by 100.
H
5.93%
S
94.07%
32.07g SS: (100) 94.07%
34.09g
2.02 g HH: (100) = 5.93%
34.09 g
55
Percent Composition Percent Composition From Experimental DataFrom Experimental Data
56
Percent composition can be calculated from experimental data without knowing the composition of the compound.
Step 1 Calculate the mass of the compound formed.
Step 2 Divide the mass of each element by the total mass of the compound and multiply by 100.
57
Step 1 Calculate the total mass of the compound1.52 g N
3.47 g O
4.99 g
A compound containing nitrogen and oxygen is found to contain 1.52 g of nitrogen and 3.47 g of oxygen. Determine its percent composition.
= total mass of product
58
Calculate the percent composition of hydrosulfuric acid H2S.
Step 2 Divide the mass of each element by the total mass of the compound formed.
3.47g O(100) = 69.5%
4.99g
1.52 g N(100) = 30.5%
4.99 g
N
30.5%
O
69.5%
59
Empirical Formula versus Empirical Formula versus Molecular FormulaMolecular Formula
60
• The empirical formula or simplest formula gives the smallest whole-number ratio of the atoms present in a compound.
• The empirical formula gives the relative number of atoms of each element present in the compound.
61
• The molecular formula is the true formula of a compound.
• The molecular formula represents the total number of atoms of each element present in one molecule of a compound.
62
ExamplesExamplesExamplesExamples
63
C2H4Molecular Formula
CH2Empirical Formula
C:H 1:2Smallest Whole Number Ratio
64
C6H6Molecular Formula
CHEmpirical Formula
C:H 1:1Smallest Whole Number Ratio
65
H2O2Molecular Formula
HOEmpirical Formula
H:O 1:1Smallest Whole Number Ratio
66
67
Two compounds can have identical empirical formulas and different molecular formulas.
68
69
CalculatingCalculatingEmpirical FormulasEmpirical Formulas
70
Step 1 Assume a definite starting quantity (usually 100.0 g) of the compound, if not given, and express the mass of each element in grams.
Step 2 Convert the grams of each element into moles of each element using each element’s mole weight.
71
Step 3 Divide the moles of atoms of each element by the moles of atoms of the element that had the smallest value
– If the numbers obtained are whole numbers, use them as subscripts and write the empirical formula.
– If the numbers obtained are not whole numbers, go on to step 4.
72
Step 4 Multiply the values obtained in step 3 by the smallest numbers that will convert them to whole numbers
Use these whole numbers as the subscripts in the empirical formula.
FeO1.5
Fe1 x 2O1.5 x 2 Fe2O3
73
• The results of calculations may differ from a whole number.
– If they differ ±0.1 round off to the next nearest whole number.
2.93– Deviations greater than 0.1 unit from a
whole number usually mean that the calculated ratios have to be multiplied by a whole number.
75
ProblemsProblemsProblemsProblems
76
The analysis of a salt shows that it contains 56.58% potassium (K); 8.68% carbon (C); and 34.73% oxygen (O). Calculate the empirical formula for this substance.Step 1 Express each element in grams. Assume 100
grams of compound.
K = 56.58 g
C = 8.68 g
O = 34.73 g
77
The analysis of a salt shows that it contains 56.58% potassium (K); 8.68% carbon (C); and 34.73% oxygen (O). Calculate the empirical formula for this substance.Step 2 Convert the grams of each element to moles.
K: 56.58 g K1 mol K atoms
39.10 g K
1.447 mol K atoms
C: 8.68 g C1 mol C atoms
12.01 g C
0.723 mol C atoms
O: 34.73 g O1 mol O atoms
16.00 g O
2.171 mol O atoms
C has the smallest number of moles
0.723 mol C atoms
78
The analysis of a salt shows that it contains 56.58% potassium (K); 8.68% carbon (C); and 34.73% oxygen (O). Calculate the empirical formula for this substance.Step 3 Divide each number of moles by the smallest
value.1.447 mol
K = = 2.000.723 mol
0.723 molC: = 1.00
0.723 mol
2.171 molO = = 3.00
0.723 mol
The simplest ratio of K:C:O is 2:1:3
Empirical formula K2CO3
C has the smallest number of moles
0.723 mol C atoms
79
The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance.
The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance.
Step 1 Express each element in grams. Assume 100 grams of compound.
N = 25.94 g
O = 74.06 g
80
Step 2 Convert the grams of each element to moles.
N: 25.94 g N1 mol N atoms
14.01 g N
1.852 mol N atoms
O: 74.06 g O1 mol O atoms
16.00 g O
4.629 mol C atoms
The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance.
81
Step 3 Divide each number of moles by the smallest value.
1.852 molN = = 1.000
1.852 mol4.629 mol
O: = 2.5001.852 mol
The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance.
This is not a ratio of whole numbers.
82
Step 4 Multiply each of the values by 2.
The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance.
Empirical formula N2O5
N: (1.000)2 = 2.000 O: (2.500)2 = 5.000
83
Calculating the Molecular Formula Calculating the Molecular Formula from the Empirical Formulafrom the Empirical Formula
84
• The molecular formula can be calculated from the empirical formula if the mole weight is known.
• The molecular formula will be equal to the empirical formula or some multiple n of it.
• To determine the molecular formula evaluate n.
• n is the number of units of the empirical formula contained in the molecular formula.
mole weightn = =
empirical weight formula units number of empirical
85
What is the molecular formula of a compound which has an empirical formula of CH2 and a mole weight of 126.2 g?
The molecular formula is (CH2)9 = C9H18
Let n = the number of formula units of CH2.
Calculate the mass of each CH2 unit
1 C = 1(12.01 g) = 12.01g
2 H = 2(1.01 g) = 2.02g
14.03g126.2 g
n 9 (empirical formula units)14.03 g
86
A compound made of 30.4% N and 69.6% O has a mole weight of 138 grams/mole. Find the molecular formula.
Old Way:
1mole30.4g 2.17mole of N
14.0g
1mole69.6g 4.35mole of O
16.0g
2.17 24.352.17 2.17
N O NO 46g/ mole
2 3 6
1383units 3(NO ) N O
46
87
A compound made of 30.4% N and 69.6% O has a mole weight of 138 grams/mole. Find the molecular formula.
New Way:
138gcomp'd 30.4gN 1moleN3moleN
1molecomp'd 100gcomp'd 14.0g
138gcomp'd 69.6gO 1moleO6moleO
1molecomp'd 100gcomp'd 16.0g
N3O6
88
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