3.3 Moles, 3.4 Molar Mass, and 3.5 Percent Composition

3.3 Moles,

3.4 Molar Mass,

and

3.5 Percent

Composition

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

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}

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

Samples of One Mole Quantities

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

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

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

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

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.}

A. The number of atoms in 2.0 mole Al is

2.0 mole Al x 6.02x1023 _{Al atoms = 1.2 x 10}24_{Al 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

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

Subscripts State Atoms and Moles

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

Learning Check

A. How many mole O are in 0.150 mole

aspirin C

H

O

?

B. How many O atoms are in 0.150 mole

aspirin C

H

O

?

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}

Molar Mass

The molar mass

• is the mass of one mole of an element or

compound.

• is the atomic mass expressed in grams.

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

Molar Mass from Periodic Table

1 mole Ag 1 mole C 1 mole S = 107.9 g = 12.01 g = 32.07 g

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

Some One-mole Quantities

32.1 g 55.9 g 58.5 g 294.2 g 342.2 g

What is the molar mass of each of the following?

A. K₂O

B. Al(OH)₃

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

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

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

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

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

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

Molar mass factors are used to convert between the grams of a substance and the number of moles.

Calculations Using Molar Mass

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

Learning Check

Allyl sulfide C₆H₁₀S is a

compound that has the odor of garlic. How many moles of C₆H₁₀S are in 225 g?

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

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

Learning Check

How many H₂O molecules are in 24.0 g H₂O? 1) 4.52 x 1023

2) 1.44 x 1025

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

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?

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 10}23 _molecules 114.2 g 1 mole = 1 x 109 _{molecules C} 6H10S

Solution

• 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:

• 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:

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

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

.

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%

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 %