3.3 moles, 3.4 molar mass, and 3.5 percent...
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Collection Terms
A collection term statesa specific number of items.
• 1 dozen donuts
= 12 donuts
• 1 ream of paper
= 500 sheets
• 1 case = 24 cans
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A mole is a collection that contains
• the same number of particles as there are pure 12-carbon atoms in 12.0 g of carbon.
• 6.022 x 1023 atoms of an element (Avogadro’s number).
1 mole element Number of Atoms1 mole C = 6.022 x 1023 C atoms1 mole Na = 6.022 x 1023 Na atoms1 mole Au = 6.022 x 1023 Au atoms
A Mole of Atoms
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A mole • of a covalent compound has Avogadro’s number of
molecules.1 mole CO2 = 6.022 x 1023 CO2 molecules1 mole H2O = 6.022 x 1023 H2O molecules
• of an ionic compound contains Avogadro’s number of formula units.
1 mole NaCl = 6.022 x 1023 NaCl formula units1 mole K2SO4 = 6.022 x 1023 K2SO4 formula units
A Mole of a Compound
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Samples of One Mole Quantities
1 mole C = 6.02 x 1023 C atoms
1 mole Al = 6.02 x 1023 Al atoms
1 mole S = 6.02 x 1023 S atoms
1 mole H2O = 6.02 x 1023 H2O molecules
1 mole CCl4 = 6.02 x 1023 CCl4 molecules
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Avogadro’s number 6.022 x 1023 can be written as anequality and two conversion factors.
Equality:1 mole = 6.022 x 1023 particles
Conversion Factors:6.022 x 1023 particles and 1 mole
1 mole 6.022 x 1023 particles
Avogadro’s Number
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Avogadro’s number is used to convertmoles of a substance to particles.
How many Cu atoms are in 0.50 mole Cu?
0.50 mole Cu x 6.022 x 1023 Cu atoms1 mole Cu
= 3.0 x 1023 Cu atoms
Using Avogadro’s Number
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Avogadro’s number is used to convertparticles of a substance to moles.
How many moles of CO2 are in 2.50 x 1024 molecules CO2?
2.50 x 1024 molecules CO2 x 1 mole CO2
6.02 x 1023 molecules CO2
= 4.15 mole CO2
Using Avogadro’s Number
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1. The number of atoms in 2.0 mole Al isA. 2.0 Al atoms.B. 3.0 x 1023 Al atoms.C. 1.2 x 1024 Al atoms.
2. The number of moles of S in 1.8 x 1024 atoms S is A. 1.0 mole S atoms.B. 3.0 mole S atoms.C. 1.1 x 1048 mole S atoms.
Learning Check
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A. The number of atoms in 2.0 mole Al is2.0 mole Al x 6.02x1023 Al atoms = 1.2 x 1024Al atoms
1 mole AlAvogadro’s Number
B. The number of moles of S in 1.8 x 1024 atoms S is1.8 x 1024 atoms S x 1 mole S
6.02 x 1023 S atoms Avogadro’s
Number= 3.0 moles of S atoms
Solution
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Subscripts and Moles
The subscripts in a formula show• the relationship of atoms in the formula.• the moles of each element in 1 mole of compound.
GlucoseC6H12O6
In 1 molecule: 6 atoms C 12 atoms H 6 atoms OIn 1 mole: 6 mole C 12 mole H 6 mole O
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Subscripts State Atoms and Moles
9 mole C 8 mole H 4 mole O
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Factors from Subscripts
The subscripts are used to write conversion factors formoles of each element in 1 mole compound. For aspirinC9H8O4, the following factors can be written:
9 mole C 8 mole H 4 mole O 1 mole C9H8O4 1 mole C9H8O4 1 mole C9H8O4and 1 mole C9H8O4 1 mole C9H8O4 1 mole C9H8O4
9 mole C 8 mole H 4 mole O
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Learning Check
A. How many mole O are in 0.150 mole aspirin C9H8O4?
B. How many O atoms are in 0.150 mole aspirin C9H8O4?
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A. How many mole O are in 0.150 mole aspirin C9H8O4?0.150 mole C9H8O4 x 4 mole O = 0.600 mole O
1 mole C9H8O4subscript factor
B. How many O atoms are in 0.150 mole aspirin C9H8O4?0.150 mole C9H8O4 x 4 mole O x 6.02 x 1023 O atoms
1 mole C9H8O4 1 mole Osubscript Avogadro’sfactor Number
= 3.61 x 1023 O atoms
Solution
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Molar Mass
The molar mass
• is the mass of one mole of an element or compound.
• is the atomic mass expressed in grams.
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Molar Mass from Periodic Table
Molar mass is the atomic mass expressed in grams.
1 mole Ag 1 mole C 1 mole S= 107.9 g = 12.01 g = 32.07 g
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Molar Mass of a Compound
The molar mass of a compound is the sum of the molar masses of the elements in the formula.
Example: Calculate the molar mass of CaCl2 to the tenths decimal place.
Element Number of Moles
Atomic Mass Total Mass
Ca 1 40.1 g/mole35.5 g/mole
40.1 gCl 2 71.0 gCaCl2 111.1 g
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Some One-mole Quantities
32.1 g 55.9 g 58.5 g 294.2 g 342.2 g
One-Mole Quantities
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What is the molar mass of each of the following?
A. K2O
B. Al(OH)3
Learning Check
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A. K2O 94.2 g/mole
2 mole K (39.1 g/mole) + 1 mole O (16.0 g/mole)78.2 g + 16.0 g = 94.2 g
B. Al(OH)3 78.0 g/mole
1 mole Al (27.0 g/mole) + 3 mol O (16.0 g/mole)+ 3 mol H (1.01 g/mole) 27.0 g + 48.0 g + 3.03 g = 78.0 g
Solution
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Prozac, C17H18F3NO, is an antidepressant that inhibits the uptake of serotonin by the brain. What is the molar mass of Prozac?
1) 40.1 g/mole2) 262 g/mole 3) 309 g/mole
Learning Check
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Prozac, C17H18F3NO, is an antidepressant that inhibits the uptake of serotonin by the brain. What is the molar mass of Prozac?
3) 309 g/mole
17 C (12.0) + 18 H (1.01) + 3 F (19.0)
+ 1 N (14.0) + 1 O (16.0) =
204 + 18.2 + 57.0 + 14.0 + 16.0 = 309 g/mole
Solution
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Molar mass conversion factors• are written from molar mass.• relate grams and moles of an element or compound.
Example: Write molar mass factors for methane, CH4, used in gas stoves and gas heaters.Molar mass:
1 mol CH4 = 16.0 g
Conversion factors:16.0 g CH4 and 1 mole CH41 mole CH4 16.0 g CH4
Molar Mass Factors
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Acetic acid C2H4O2 gives the sour taste to vinegar. Write two molar mass conversion factors for acetic acid.
Learning Check
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Acetic acid C2H4O2 gives the sour taste to vinegar.Write two molar mass factors for acetic acid.
Calculate molar mass:24.0 + 4.04 + 32.0 = 60.0 g/mole
1 mole of acetic acid = 60.0 g acetic acid
Molar mass factors1 mole acetic acid and 60.0 g acetic acid 60.0 g acetic acid 1 mole acetic acid
Solution
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Molar mass factors are used to convert between the grams of a substance and the number of moles.
Calculations Using Molar Mass
Grams Molar mass factor Moles
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Aluminum is often used to build lightweight bicycleframes. How many grams of Al are in 3.00 mole Al?
Molar mass equality: 1 mole Al = 27.0 g Al
Setup with molar mass as a factor:
3.00 mole Al x 27.0 g Al = 81.0 g Al1 mole Al
molar mass factor for Al
Moles to Grams
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Learning Check
Allyl sulfide C6H10S is a compound that has the odor of garlic. How many moles of C6H10S are in 225 g?
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Calculate the molar mass of C6H10S. (6 x 12.0) + (10 x 1.01) + (1 x 32.1) = 114.2 g/mole
Set up the calculation using a mole factor. 225 g C6H10S x 1 mole C6H10S
114.2 g C6H10Smolar mass factor (inverted)
= 1.97 mole C6H10S
Solution
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Grams, Moles, and Particles
A molar mass factor and Avogadro’s number convert• grams to particles.
molar mass Avogadro’s number
g mole particles
• particles to grams.
Avogadro’s molar massnumber
particles mole g
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Learning Check
How many H2O molecules are in 24.0 g H2O?
1) 4.52 x 1023
2) 1.44 x 1025
3) 8.03 x 1023
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How many H2O molecules are in 24.0 g H2O?
3) 8.02 x 1023
24.0 g H2O x 1 mole H2O x 6.02 x 1023 H2O molecules 18.0 g H2O 1 mole H2O
= 8.03 x 1023 H2O molecules
Solution
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Learning Check
If the odor of C6H10S can be detected from 2 x 10-13
g in one liter of air, how many molecules of C6H10S are present?
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If the odor of C6H10S can be detected from 2 x 10-13 g in one liter of air, how many molecules of C6H10S are present?
2 x 10-13 g x 1 mole x 6.02 x 1023 molecules 114.2 g 1 mole
= 1 x 109 molecules C6H10S
Solution
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• Percent Composition of Compounds: What is the mass percentage of C in CO2?
• The mass percentage is calculated using the following equation:
• If a sample consisting of 1 mole of CO2 is used, the mole-based relationships given earlier show that 1 mole CO2=44.01 g CO2 (12.01 g C + 32.00 g O). Thus, the mass of C in a specific mass of CO2 is known, and the problem is solved as follows:
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• What is the mass percentage of oxygen in CO2?
• The mass percentage is calculated using the following equation:
• Once again, a sample consisting of 1 mole of CO2 is used to take advantage of the mole-based relationships given earlier where:
1 mole CO2 = 44.01g CO2 = 12.01 g C + 32.00g O
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• Thus, the mass of O in a specific mass of CO2 is known, and the problem is solved as follows:
• We see that the % C + % O = 100% , which should be the case because C and O are the only elements present in CO2.
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QUESTIONMorphine, derived from opium plants, has the potential for use and abuse. It’s formula is C17H19NO3. What percent, by mass, is the carbon in this compound?
1. 42.5%2. 27.9%3. 71.6%4. This cannot be solved until the mass of the sample is given.
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ANSWERChoice 3 is correct. First determine the molar mass of the compound, then divide that into the total mass of carbon present, and finally, multiply that by 100.
(17 × 12.01) / ((17 × 12.01) + (19 × 1.008) + (1 × 14.01) + 3 ×16.00)) = 0.7160.716 × 100 = 71.6 %
Section 3.5: Percent Composition of Compounds