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Transcript of Introductory Chemistry , 2 nd Edition Nivaldo Tro
Roy KennedyMassachusetts Bay Community College
Wellesley Hills, MA
Introductory Chemistry, 2nd EditionNivaldo Tro
Chapter 6Chemical
Composition
2006, Prentice Hall
Tro's Introductory Chemistry, Chapter 6
Why is Knowledge of Composition Important?
• everything in nature is either chemically or physically combined with other substances
• to know the amount of a material in a sample, you need to know what fraction of the sample it is
• Some Applications: the amount of sodium in sodium chloride for diet the amount of iron in iron ore for steel production the amount of hydrogen in water for hydrogen
fuel the amount of chlorine in freon to estimate ozone
depletion
Tro's Introductory Chemistry, Chapter 6
3
Counting Nails by the Pound
• I want to buy a certain number of nails for a project, but the hardware store sells nails by the pound!
• How do I know how many nails I am buying when I buy a pound of nails?
• AnalogyHow many atoms in a given
mass of an element?
Tro's Introductory Chemistry, Chapter 6
4
Counting Nails by the Pound
A hardware store customer buys 2.60 pounds of nails. A dozen of the nails has a mass of 0.150 pounds. How many nails did the customer buy?
Solution map:
Tro's Introductory Chemistry, Chapter 6
5
Counting Nails by the Pound
• The customer bought 2.60 lbs of nails and received 208 nails. He counted the nails by weighing them!
nails 208 doz. 1
nails 12
lbs. 0.150
nails doz. 1lbs. 2.60
Tro's Introductory Chemistry, Chapter 6
6
Counting Nails by the Pound
• What if he bought a different size nail? Would the mass of a dozen be 0.150 lbs?Would there be 208 nails in 2.60 lbs? How would this effect the conversion factors?
Tro's Introductory Chemistry, Chapter 6
7
Counting Atoms by Moles
• If we can find the mass of a particular number of atoms, we can use this information to convert the mass of a element sample to the number of atoms in the sample.
• The number of atoms we will use is 6.022 x 1023 and we call this a mole1 mole = 6.022 x 1023 things
Like 1 dozen = 12 things
Tro's Introductory Chemistry, Chapter 6
8
Chemical Packages - Moles• mole = number of particles equal to the
number of atoms in 12 g of C-12 1 atom of C-12 weighs exactly 12 amu1 mole of C-12 weighs exactly 12 g
• The number of particles in 1 mole is called Avogadro’s Number = 6.0221421 x 1023 1 mole of C atoms weighs 12.01 g and has
6.022 x 1023 atomsthe average mass of a C atom is 12.01 amu
Example 6.1Converting Between Moles and
Number of Atoms
Tro's Introductory Chemistry, Chapter 6
10
Example:• A silver ring contains 1.1 x 1022 silver atoms. How many
moles of silver are in the ring?
Tro's Introductory Chemistry, Chapter 6
11
Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring?
• Write down the given quantity and its units.
Given: 1.1 x 1022 Ag atoms
Tro's Introductory Chemistry, Chapter 6
12
• Write down the quantity to find and/or its units.
Find: ? moles
Information
Given: 1.1 x 1022 Ag atomsExample:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring?
Tro's Introductory Chemistry, Chapter 6
13
• Collect Needed Conversion Factors:
1 mole Ag atoms = 6.022 x 1023 Ag atoms
Information
Given: 1.1 x 1022 Ag atoms
Find: ? moles
Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring?
Tro's Introductory Chemistry, Chapter 6
14
• Write a Solution Map for converting the units :
Information
Given: 1.1 x 1022 Ag atoms
Find: ? moles
Conv. Fact.: 1 mole = 6.022 x 1023
Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring?
atoms Agatoms Ag moles Agmoles Ag
atoms Ag10022.6
Ag mole 123
Tro's Introductory Chemistry, Chapter 6
15
• Apply the Solution Map:
moles atoms Ag 106.022
Ag mole 1atoms Ag 101.1
2322
= 1.8266 x 1022 moles Ag
= 1.8 x 1022 moles Ag
• Sig. Figs. & Round:
InformationGiven: 1.1 x 1022 Ag atomsFind: ? molesConv. Fact.: 1 mole = 6.022 x 1023
Sol’n Map: atoms mole
Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring?
Tro's Introductory Chemistry, Chapter 6
16
• Check the Solution:
1.1 x 1022 Ag atoms = 1.8 x 10-2 moles Ag
The units of the answer, moles, are correct.The magnitude of the answer makes sense
since 1.1 x 1022 is less than 1 mole.
InformationGiven: 1.1 x 1022 Ag atomsFind: ? molesConv. Fact.: 1 mole = 6.022 x 1023
Sol’n Map: atoms mole
Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring?
Tro's Introductory Chemistry, Chapter 6
17
Relationship Between Moles and Mass
• The mass of one mole of atoms is called the molar mass
• The molar mass of an element, in grams, is numerically equal to the element’s atomic mass, in amu
Tro's Introductory Chemistry, Chapter 6
18
Mole and Mass RelationshipsSubstance Pieces in 1 mole Weight of 1 mole hydrogen 6.022 x 1023 atoms 1.008 g
carbon 6.022 x 1023 atoms 12.01 g
oxygen 6.022 x 1023 atoms 16.00 g
sulfur 6.022 x 1023 atoms 32.06 g
calcium 6.022 x 1023 atoms 40.08 g
chlorine 6.022 x 1023 atoms 35.45 g
copper 6.022 x 1023 atoms 63.55 g
1 moleSulfur32.06 g
1 moleCarbon12.01 g
Example 6.2Converting Between
Grams and Moles of Atoms
Tro's Introductory Chemistry, Chapter 6
20
Example:• Calculate the number of moles of sulfur in 57.8 g of
sulfur
Tro's Introductory Chemistry, Chapter 6
21
Example:Calculate the number of moles of sulfur in 57.8 g of sulfur
• Write down the given quantity and its units.
Given: 57.8 g S
Tro's Introductory Chemistry, Chapter 6
22
• Write down the quantity to find and/or its units.
Find: ? moles S
Information
Given: 57.8 g SExample:Calculate the number of moles of sulfur in 57.8 g of sulfur
Tro's Introductory Chemistry, Chapter 6
23
• Collect Needed Conversion Factors:
1 mole S atoms = 32.06 g
Information
Given: 57.8 g S
Find: ? moles S
Example:Calculate the number of moles of sulfur in 57.8 g of sulfur
Tro's Introductory Chemistry, Chapter 6
24
• Write a Solution Map for converting the units :
Information
Given: 57.8 g S
Find: ? moles S
Conv. Fact.: 1 mole S = 32.06 g
g Sg S moles Smoles S
S g 2.063
S mole 1
Example:Calculate the number of moles of sulfur in 57.8 g of sulfur
Tro's Introductory Chemistry, Chapter 6
25
• Apply the Solution Map:
S moles S g 2.063
S mole 1S g 57.8
= 1.80287 moles S
= 1.80 moles S
• Sig. Figs. & Round:
Information
Given: 57.8 g S
Find: ? moles S
Conv. Fact.: 1 mole S = 32.06 g
Sol’n Map: g moles
Example:Calculate the number of moles of sulfur in 57.8 g of sulfur
Tro's Introductory Chemistry, Chapter 6
26
• Check the Solution:
57.8 g sulfur = 1.80 moles sulfur
The units of the answer, moles, are correct.The magnitude of the answer makes sense
since 57.8 g is more than 1 mole.
Information
Given: 57.8 g S
Find: ? moles S
Conv. Fact.: 1 mole S = 32.06 g
Sol’n Map: g moles
Example:Calculate the number of moles of sulfur in 57.8 g of sulfur
Example 6.3Converting Between
Grams and Number of Atoms
Tro's Introductory Chemistry, Chapter 6
28
Example:• How many aluminum atoms are in an aluminum can with
a mass of 16.2 g?
Tro's Introductory Chemistry, Chapter 6
29
Example:How many aluminum atoms are in an aluminum can with a mass of 16.2 g?
• Write down the given quantity and its units.
Given: 16.2 g Al
Tro's Introductory Chemistry, Chapter 6
30
• Write down the quantity to find and/or its units.
Find: ? atoms Al
Information
Given: 16.2 g AlExample:How many aluminum atoms are in an aluminum can with a mass of 16.2 g?
Tro's Introductory Chemistry, Chapter 6
31
• Collect Needed Conversion Factors:
1 mole Al atoms = 26.98 g Al
1 mole = 6.022 x 1023
Information
Given: 16.2 g Al
Find: ? atoms Al
Example:How many aluminum atoms are in an aluminum can with a mass of 16.2 g?
Tro's Introductory Chemistry, Chapter 6
32
• Write a Solution Map for converting the units :
g Al mol Al atoms Al
Al g 26.98
Al mol 1
Al mol 1
Al atoms 10.0226 23
Information
Given: 16.2 g Al
Find: ? atoms Al
C F: 1 mol Al = 26.98 g
1 mol = 6.022 x 1023
Example:How many aluminum atoms are in an aluminum can with a mass of 16.2 g?
Tro's Introductory Chemistry, Chapter 6
33
Al mole 1
Al atoms 10022.6
Al g 26.98
Al mole 1Al g 6.21
23
• Apply the Solution Map:
= 3.6159 x 1023 atoms Al
= 3.62 x 1023 atoms Al
• Sig. Figs. & Round:
Information
Given: 16.2 g Al
Find: ? atoms Al
CF: 1 mol Al = 26.98 g
1 mol = 6.022 x 1023
SM: g mol atoms
Example:How many aluminum atoms are in an aluminum can with a mass of 16.2 g?
Tro's Introductory Chemistry, Chapter 6
34
• Check the Solution:
16.2 g Al = 3.62 x 1023 atoms Al
The units of the answer, atoms, are correct.The magnitude of the answer makes sense since 16.2 g is less than the mass of 1 mole, 26.98 g.
Information
Given: 16.2 g Al
Find: ? atoms Al
CF: 1 mol Al = 26.98 g
1 mol = 6.022 x 1023
SM: g mol atoms
Example:How many aluminum atoms are in an aluminum can with a mass of 16.2 g?
Tro's Introductory Chemistry, Chapter 6
35
Molar Mass of Compounds
• the relative weights of molecules can be calculated from atomic weights
Formula Mass = 1 molecule of H2O
= 2(1.01 amu H) + 16.00 amu O = 18.02 amu
• since 1 mole of H2O contains 2 moles of H and 1 mole of O
Molar Mass = 1 mole H2O
= 2(1.01 g H) + 16.00 g O = 18.02 g
Example 6.4Converting Between
Grams and Moles of Compound
Tro's Introductory Chemistry, Chapter 6
37
Example:• Calculate the mass (in grams) of 1.75 mol of water
Tro's Introductory Chemistry, Chapter 6
38
Example:Calculate the mass (in grams) of 1.75 mol of water
• Write down the given quantity and its units.
Given: 1.75 mol H2O
Tro's Introductory Chemistry, Chapter 6
39
• Write down the quantity to find and/or its units.
Find: ? g H2O
Information
Given: 1.75 mol H2OExample:Calculate the mass (in grams) of 1.75 mol of water
Tro's Introductory Chemistry, Chapter 6
40
• Collect Needed Conversion Factors:
Molar Mass H2O = 2(atomic mass H) + 1(atomic mass O)
= 2(1.01) + 1(16.00)
= 18.02 g/mol
1 mol H2O = 18.02 g H2O
Information
Given: 1.75 mol H2O
Find: ? g H2O
Example:Calculate the mass (in grams) of 1.75 mol of water
Tro's Introductory Chemistry, Chapter 6
41
• Write a Solution Map for converting the units :
Information
Given: 1.75 mol H2O
Find: ? g H2O
C F: 1 mole H2O = 18.02 g H2O
mol H2Omol H2O g H2Og H2O
OH mol 1
OH g 18.02
2
2
Example:Calculate the mass (in grams) of 1.75 mol of water
Tro's Introductory Chemistry, Chapter 6
42
• Apply the Solution Map:
OH g OH mol 1
OH g 18.02OH mol 1.75 2
2
22
= 31.535 g H2O
= 31.5 g H2O
• Sig. Figs. & Round:
Information
Given: 1.75 mol H2O
Find: ? g H2O
C F: 1 mole H2O = 18.02 g H2OSol’n Map: mol g
Example:Calculate the mass (in grams) of 1.75 mol of water
Tro's Introductory Chemistry, Chapter 6
43
• Check the Solution:
1.75 mol H2O = 31.5 g H2O
The units of the answer, g, are correct.The magnitude of the answer makes sense
since 31.5 g is more than 1 mole.
Information
Given: 1.75 mol H2O
Find: ? g H2O
C F: 1 mole H2O = 18.02 g H2OSol’n Map: mol g
Example:Calculate the mass (in grams) of 1.75 mol of water
Example 6.5Converting Between
Grams and Number of Molecules
Tro's Introductory Chemistry, Chapter 6
45
Example:
• Find the mass of 4.78 x 1024 NO2 molecules?
Tro's Introductory Chemistry, Chapter 6
46
Example:Find the mass of 4.78 x 1024 NO2 molecules
• Write down the given quantity and its units.
Given: 4.78 x 1024 NO2 molecules
Tro's Introductory Chemistry, Chapter 6
47
• Write down the quantity to find and/or its units.
Find: ? g NO2
Information
Given: 4.78 x 1024 molec NO2
Example:Find the mass of 4.78 x 1024 NO2 molecules
Tro's Introductory Chemistry, Chapter 6
48
• Collect Needed Conversion Factors:
Molar Mass NO2 = 1(atomic mass N) + 2(atomic mass O)
= 1(14.01) + 2(16.00)
= 46.01 g/mol
1 mole NO2 molec = 46.01 g NO2
1 mole = 6.022 x 1023
Information
Given: 4.78 x 1024 molec NO2
Find: ? g NO2
Example:Find the mass of 4.78 x 1024 NO2 molecules
Tro's Introductory Chemistry, Chapter 6
49
• Write a Solution Map for converting the units :
molec NO2 mol NO2 g NO2
2
2
NO mol 1
NO g 01.46
223
2
NO molec 10.0226
NO mol 1
Information
Given: 4.78 x1024 NO2 molec
Find: ? g NO2
C F: 1 mol NO2 = 46.01 g NO2
1 mol = 6.022 x 1023
Example:Find the mass of 4.78 x 1024 NO2 molecules
Tro's Introductory Chemistry, Chapter 6
50
2
2
223
22
23
NO mole 1
NO g 6.014
NO molec 10022.6
NO mole 1NO molec 10022.6
• Apply the Solution Map:
= 365.21 g NO2
= 365 g NO2 • Sig. Figs. & Round:
Information
Given: 4.78 x 1024 molec NO2
Find: ? g NO2
CF: 1 mol NO2 = 46.01 g
1 mol = 6.022 x 1023
SM: molec mol g
Example:Find the mass of 4.78 x 1024 NO2 molecules
Tro's Introductory Chemistry, Chapter 6
51
• Check the Solution:
4.78 x 1024 molecules NO2 = 365 g NO2
The units of the answer, g, are correct.The magnitude of the answer makes sense since
4.78 x 1024 molecules is more than 1 mole.
Information
Given: 4.78 x 1024 molec NO2
Find: ? g NO2
CF: 1 mol NO2 = 46.01 g
1 mol = 6.022 x 1023
SM: molec mol g
Example:Find the mass of 4.78 x 1024 NO2 molecules
Tro's Introductory Chemistry, Chapter 6
52
Chemical Formulas as Conversion Factors
• 1 spider 8 legs• 1 chair 4 legs• 1 H2O molecule 2 H atoms 1 O atom
Tro's Introductory Chemistry, Chapter 6
53
Mole Relationships inChemical Formulas
• since we count atoms and molecules in mole units, we can find the number of moles of a constituent element if we know the number of moles of the compound
Moles of Compound Moles of Constituents
1 mol NaCl 1 mole Na, 1 mole Cl
1 mol H2O 2 mol H, 1 mole O
1 mol CaCO3 1 mol Ca, 1 mol C, 3 mol O
1 mol C6H12O6 6 mol C, 12 mol H, 6 mol O
Example 6.7Converting Between
Grams of a Compound and Grams of a Constituent Element
Tro's Introductory Chemistry, Chapter 6
55
Example:
• Carvone, (C10H14O), is the main component in spearmint oil. It has a pleasant odor and mint flavor. It is often added to chewing gum, liquers, soaps and perfumes. Find the mass of carbon in 55.4 g of carvone.
Tro's Introductory Chemistry, Chapter 6
56
Example:Find the mass of carbon in 55.4 g of carvone, (C10H14O).
• Write down the given quantity and its units.
Given: 55.4 g C10H14O
Tro's Introductory Chemistry, Chapter 6
57
• Write down the quantity to find and/or its units.
Find: ? g C
Information
Given: 55.4 g C10H14OExample:Find the mass of carbon in 55.4 g of carvone, (C10H14O).
Tro's Introductory Chemistry, Chapter 6
58
• Collect Needed Conversion Factors:
Molar Mass C10H14O = 10(atomic mass C) + 14(atomic mass H) + 1(atomic mass O)
= 10(12.01) + 14(1.01) + (16.00)
= 150.2 g/mol
1 mole C10H14O = 150.2 g C10H14O
1 mole C10H14O 10 mol C
1 mole C = 12.01 g C
Information
Given: 55.4 g C10H14O
Find: g C
Example:Find the mass of carbon in 55.4 g of carvone, (C10H14O).
Tro's Introductory Chemistry, Chapter 6
59
• Write a Solution Map for converting the units :
gC10H14O
C mol 1
C g 0112.
OHC g 50.21
OHC mol 1
1410
1410
Information
Given: 55.4 g C10H14O
Find: g C
CF: 1 mol C10H14O = 150.2 g
1 mol C10H14O 10 mol C
1 mol C = 12.01 g
Example:Find the mass of carbon in 55.4 g of carvone, (C10H14O).
molC10H14O
molC
gC
OHC mol 1
C mol 10
1410
Tro's Introductory Chemistry, Chapter 6
60
C mole 1
C g 2.011
OHC mol 1
C mol 10
OHC g 50.21
OHC mole 1OHC g 5.45
14101410
14101410
• Apply the Solution Map:
= 44.2979 g C
= 44.3 g C • Sig. Figs. & Round:
Information
Given: 55.4 g C10H14OFind: g C
CF: 1 mol C10H14O = 150.2 g
1 mol C10H14O 10 mol C 1 mol C = 12.01 g
SM: g C10H14O mol C10H14O mol C g C
Example:Find the mass of carbon in 55.4 g of carvone, (C10H14O).
Tro's Introductory Chemistry, Chapter 6
61
• Check the Solution:
55.4 g C10H14O = 44.3 g C
The units of the answer, g C, are correct.The magnitude of the answer makes sense since
the amount of C is less than the amount of C10H14O.
Information
Given: 55.4 g C10H14OFind: g C
CF: 1 mol C10H14O = 150.2 g
1 mol C10H14O 10 mol C 1 mol C = 12.01 g
SM: g C10H14O mol C10H14O mol C g C
Example:Find the mass of carbon in 55.4 g of carvone, (C10H14O).
Tro's Introductory Chemistry, Chapter 6
62
Percent Composition• Percentage of each element in a compound
By mass
• Can be determined from 1. the formula of the compound2. the experimental mass analysis of the
compound• The percentages may not always total to 100%
due to rounding
100%wholepart
Percentage
Tro's Introductory Chemistry, Chapter 6
63
Mass Percent as a Conversion Factor
• the mass percent tells you the mass of a constituent element in 100 g of the compoundthe fact that NaCl is 39% Na by mass means that
100 g of NaCl contains 39 g Na
• this can be used as a conversion factor100 g NaCl 39 g Na
Na g NaCl g 100
Na g 39 NaCl g NaCl g
Na g 39
NaCl g 100 Na g
Tro's Introductory Chemistry, Chapter 6
64
Example - Percent Composition from the Formula C2H5OH
1. Determine the mass of each element in 1 mole of the compound
2 moles C = 2(12.01 g) = 24.02 g6 moles H = 6(1.008 g) = 6.048 g1 mol O = 1(16.00 g) = 16.00 g
2. Determine the molar mass of the compound by adding the masses of the elements
1 mole C2H5OH = 46.07 g
Tro's Introductory Chemistry, Chapter 6
65
Sample - Percent Composition from the Formula C2H5OH
3. Divide the mass of each element by the molar mass of the compound and multiply by 100%
52.14%C100%46.07g24.02g
13.13%H100%46.07g6.048g
34.73%O100%46.07g16.00g
Tro's Introductory Chemistry, Chapter 6
66
Example – Percent Composition of Carvone if a 30.0 g sample contains 24.0 g of C, 3.2 g
O and the rest H?
1. Determine the masses of all the elements in the sample
C = 24.0 g, O = 3.2 g
H = 30.0 g – (24.0 g + 3.2 g) = 2.8 g
67
Example – Percent Composition of Carvone if a 30.0 g sample contains 24.0 g of C, 3.2 g
O and the rest H?2. Divide the mass of each element by the
total mass of the sample then multiply by 100% to give its percentage
80.0%C100%30.0g24.0g
11%O100%30.0g3.2g
9.3%H100%30.0g2.8g
Tro's Introductory Chemistry, Chapter 6
68
Empirical Formulas• The simplest, whole-number ratio of atoms in a
molecule is called the Empirical Formulacan be determined from percent composition or
combining masses
• The Molecular Formula is a multiple of the Empirical Formula
% A mass A (g) moles A100g MMA
% B mass B (g) moles B100g MMB
moles Amoles B
Tro's Introductory Chemistry, Chapter 6
69
Empirical FormulasHydrogen PeroxideMolecular Formula = H2O2
Empirical Formula = HOBenzeneMolecular Formula = C6H6
Empirical Formula = CHGlucoseMolecular Formula = C6H12O6
Empirical Formula = CH2O
Tro's Introductory Chemistry, Chapter 6
70
Finding an Empirical Formula1) convert the percentages to grams
a) skip if already grams
2) convert grams to molesa) use molar mass of each element
3) write a pseudoformula using moles as subscripts4) divide all by smallest number of moles5) multiply all mole ratios by number to make all whole
numbersa) if ratio ?.5, multiply all by 2; if ratio ?.33 or ?.67, multiply
all by 3, etc. b) skip if already whole numbers
Example 6.11Finding an Empirical Formula
from Experimental Data
Tro's Introductory Chemistry, Chapter 6
72
Example:• A laboratory analysis of aspirin determined the following
mass percent composition. Find the empirical formula.
C = 60.00%
H = 4.48%
O = 35.53%
Tro's Introductory Chemistry, Chapter 6
73
Example:Find the empirical formula of aspirin with the given mass percent composition.
• Write down the given quantity and its units.
Given: C = 60.00%
H = 4.48%
O = 35.53%
Therefore in 100 g of aspirin there are 60.00 g C,
4.48 g H and 35.53 g O
Tro's Introductory Chemistry, Chapter 6
74
• Write down the quantity to find and/or its units.
Find: empirical formula, CxHyOz
Information
Given: 60.00 g C, 4.48 g H, 35.53 g O
Example:Find the empirical formula of aspirin with the given mass percent composition.
Tro's Introductory Chemistry, Chapter 6
75
• Collect Needed Conversion Factors:
1 mole C = 12.01 g C
1 mole H = 1.01 g H
1 mole O = 16.00 g O
Information
Given: 60.00 g C, 4.48 g H, 35.53 g O
Find: empirical formula, CxHyOz
Example:Find the empirical formula of aspirin with the given mass percent composition.
Tro's Introductory Chemistry, Chapter 6
76
• Write a Solution Map:
gC, H, O
molC, H, O
molratio
empiricalformula
Information
Given: 60.00 g C, 4.48 g H, 35.53 g O
Find: empirical formula, CxHyOz
CF: 1 mol C = 12.01 g; 1 mol H = 1.01 g; 1 mol O = 16.00 g
Example:Find the empirical formula of aspirin with the given mass percent composition.
77
• Apply the Solution Map:calculate the moles of each element
Information
Given: 60.00 g C, 4.48 g H, 35.53 g O
Find: empirical formula, CxHyOz
CF: 1 mol C = 12.01 g; 1 mol H = 1.01 g; 1 mol O = 16.00 gSM: g C,H,O mol C,H,O
mol ratio empirical formula
Example:Find the empirical formula of aspirin with the given mass percent composition.
C mol 9964C g 12.01
C mol 1C g 0060 ..
H mol 444H g 1.01
H mol 1H g .484 .
O mol 2212O g 16.00
O mol 1O g 5.533 .
Tro's Introductory Chemistry, Chapter 6
78
• Apply the Solution Map:write a pseudoformula
Information
Given: 4.996 mol C, 4.44 mol H,
2.221 mol O
Find: empirical formula, CxHyOz
CF: 1 mol C = 12.01 g; 1 mol H = 1.01 g; 1 mol O = 16.00 gSM: g C,H,O mol C,H,O
mol ratio empirical formula
Example:Find the empirical formula of aspirin with the given mass percent composition.
C4.996H4.44O2.221
Tro's Introductory Chemistry, Chapter 6
79
• Apply the Solution Map: find the mole ratio by dividing by the smallest number of moles
Information
Given: C4.996H4.44O2.221
Find: empirical formula, CxHyOz
CF: 1 mol C = 12.01 g; 1 mol H = 1.01 g; 1 mol O = 16.00 gSM: g C,H,O mol C,H,O
mol ratio empirical formula
Example:Find the empirical formula of aspirin with the given mass percent composition.
122.25
2.221
2.221
2.221
4.44
2.221
4.996
OHC
OHC
Tro's Introductory Chemistry, Chapter 6
80
• Apply the Solution Map:multiply subscripts by factor to give whole number
Information
Given: C2.25H2O1
Find: empirical formula, CxHyOz
CF: 1 mol C = 12.01 g; 1 mol H = 1.01 g; 1 mol O = 16.00 gSM: g C,H,O mol C,H,O
mol ratio empirical formula
Example:Find the empirical formula of aspirin with the given mass percent composition.
122.25 OHC{ } x 4
C9H8O4
Tro's Introductory Chemistry, Chapter 6
81
All these molecules have the same Empirical Formula. How are the
molecules different?
Name Molecular
Formula
Empirical
Formula
glyceraldehyde C3H6O3 CH2O
erythrose C4H8O4 CH2O
arabinose C5H10O5 CH2O
glucose C6H12O6 CH2O
Tro's Introductory Chemistry, Chapter 6
82
All these molecules have the same Empirical Formula. How are the
molecules different?
Name Molecular
Formula
Empirical
Formula
Molar
Mass, g
glyceraldehyde C3H6O3 CH2O 90
erythrose C4H8O3 CH2O 120
arabinose C5H10O5 CH2O 150
glucose C6H12O6 CH2O 180
Tro's Introductory Chemistry, Chapter 6
83
Molecular Formulas
• The molecular formula is a multiple of the empirical formula
• To determine the molecular formula you need to know the empirical formula and the molar mass of the compound
Molar Massreal formula
Molar Massempirical formula
= factor used to multiply subscripts
Tro's Introductory Chemistry, Chapter 6
84
Example – Determine the Molecular Formula of Cadinene if it has a molar mass of
204 g and an empirical formula of C5H8
1. Determine the empirical formula• May need to calculate it as previous
C5H8
2. Determine the molar mass of the empirical formula
5 C = 60.05 g, 8 H = 8.064 g
C5H8 = 68.11 g
Tro's Introductory Chemistry, Chapter 6
85
3. Divide the given molar mass of the compound by the molar mass of the empirical formula
Round to the nearest whole number
3g 11.68
g 204
Example – Determine the Molecular Formula of Cadinene if it has a molar mass of
204 g and an empirical formula of C5H8
Tro's Introductory Chemistry, Chapter 6
86
4. Multiply the empirical formula by the factor above to give the molecular formula
(C5H8)3 = C15H24
Example – Determine the Molecular Formula of Cadinene if it has a molar mass of
204 g and an empirical formula of C5H8