3.3 Moles, 3.4 Molar Mass, and 3.5 Percent Composition

Full text

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3.3 Moles,

3.4 Molar Mass,

and

3.5 Percent

Composition

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Collection Terms

A collection term states

a specific number of items.

• 1 dozen donuts

= 12 donuts

• 1 ream of paper

= 500 sheets

• 1 case = 24 cans

Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

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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 12-carbon.

• 6.022 x 1023 atoms of an element (Avogadro’s number).

1 mole element Number of Atoms

1 mole C = 6.022 x 1023 C atoms

1 mole Na = 6.022 x 1023 Na atoms

1 mole Au = 6.022 x 1023 Au atoms

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A mole

• of a covalent compound has Avogadro’s number of molecules.

1 mole CO2 = 6.022 x 1023 CO

2 molecules

1 mole H2O = 6.022 x 1023 H

2O molecules

• of an ionic compound contains Avogadro’s number of formula units.

1 mole NaCl = 6.022 x 1023 NaCl formula units

1 mole K2SO4 = 6.022 x 1023 K

2SO4 formula units

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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 H 2O molecules 1 mole CCl4 = 6.02 x 1023 CCl 4 molecules

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Avogadro’s number 6.022 x 1023 can be written as an equality 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

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Avogadro’s number is used to convert

moles 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 atoms

1 mole Cu = 3.0 x 1023 Cu atoms

Using Avogadro’s Number

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Avogadro’s number is used to convert

particles of a substance to moles.

How many moles of CO2 are in 2.50 x 1024 molecules CO 2? 2.50 x 1024 molecules CO 2 x 1 mole CO2 6.02 x 1023 molecules CO 2 = 4.15 mole CO2

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1. The number of atoms in 2.0 mole Al is A. 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.

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A. The number of atoms in 2.0 mole Al is

2.0 mole Al x 6.02x1023 Al atoms = 1.2 x 1024Al atoms

1 mole Al

Avogadro’s Number

B. The number of moles of S in 1.8 x 1024 atoms S is

1.8 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. Glucose

C6H12O6

In 1 molecule: 6 atoms C 12 atoms H 6 atoms O

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Subscripts State Atoms and Moles

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Factors from Subscripts

The subscripts are used to write conversion factors for moles of each element in 1 mole compound. For aspirin C9H8O4, the following factors can be written:

9 mole C 8 mole H 4 mole O

1 mole C9H8O4 1 mole C9H8O4 1 mole C9H8O4

and

1 mole C9H8O4 1 mole C9H8O4 1 mole C9H8O4

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Learning Check

A. How many mole O are in 0.150 mole

aspirin C

9

H

8

O

4

?

B. How many O atoms are in 0.150 mole

aspirin C

9

H

8

O

4

?

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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 C9H8O4

subscript 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 O

subscript Avogadro’s

factor Number

= 3.61 x 1023 O atoms

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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/mole 35.5 g/mole

40.1 g

Cl 2 71.0 g

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Some One-mole Quantities

32.1 g 55.9 g 58.5 g 294.2 g 342.2 g

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What is the molar mass of each of the following?

A. K2O

B. Al(OH)3

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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

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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/mole 2) 262 g/mole 3) 309 g/mole

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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 CH4

1 mole CH4 16.0 g CH4

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Acetic acid C2H4O2 gives the sour taste to vinegar. Write two molar mass conversion factors for acetic acid.

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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 factors

1 mole acetic acid and 60.0 g acetic acid

60.0 g acetic acid 1 mole acetic acid

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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

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Aluminum is often used to build lightweight bicycle frames. 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 Al 1 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 C6H10S

molar mass factor (inverted)

= 1.97 mole C6H10S

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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 mass number

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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

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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 H 2O molecules 18.0 g H2O 1 mole H2O = 8.03 x 1023 H 2O 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 C 6H10S

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:

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• Thus, the mass of O in a specific mass of CO

2

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 CO

2

.

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QUESTION

Morphine, 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%

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ANSWER

Choice 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)) = 0.716

0.716 × 100 = 71.6 %

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