chapter 7 chemical quantities the mole atomic mass and formula mass copyright © 2008 by pearson...

Chapter 7 Chemical Quantities

The MoleAtomic Mass and

Formula Mass

Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings

Chapter 7 Slide 2 of 75

The Mole



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 © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings


A mole (mol) is a collection that contains

• The same number of particles as there are carbon atoms in 12.01 g of carbon.

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

1 mol C = 6.022 x 1023 C atoms

1 mol Na = 6.022 x 1023 Na atoms

1 mol Au = 6.022 x 1023 Au atoms

A Mole of Atoms


A mole

• Of a covalent compound has Avogadro’s number of _________________.

1 mol CO2 = 6.022 x 1023 CO2 molecules

1 mol H2O = 6.022 x 1023 H2O molecules

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

1 mol NaCl = 6.022 x 1023 NaCl formula units

1 mol K2SO4 = 6.022 x 1023 K2SO4 formula units

A Mole of A Compound


Samples of One Mole Quantities



Avogadro’s number 6.022 x 1023 can be written asan equality and two conversion factors.

As an equality:

1 mol particles = 6.022 x 1023 particles

As conversion Factors:

6.022 x 1023 particles and 1 mol particles 1 mol particles 6.022 x 1023 particles

Avogadro’s Number


Using Avogadro’s Number

Avogadro’s number• Converts moles of a substance to

the number of particles.

How many Cu atoms are in 0.50 mol

Cu?

0.50 mol Cu x 6.022 x 1023 Cu atoms

1 mol Cu

= 3.0 x 1023 Cu atoms Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings


Using Avogadro’s NumberAvogadro’s number• Converts particles of a substance to

moles.

How many moles of CO2 are

2.50 x 1024 CO2 molecules?

2.50 x 1024 CO2 x 1 mol CO2

6.022 x 1023 CO2

= 4.15 mol CO2


The number of atoms in 2.0 mol Al is

2.0 mol Al x 6.022 x 1023 Al atoms =

1 mol Al

1.2 x 1024 Al atoms

Learning Check


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

1.8 x 1024 S atoms x 1 mol S =

6.022 x 1023 S atoms

3.0 mol S atoms

Learning Check


Subscripts and Moles

The subscripts in a formula state• The relationship of atoms in the formula.• The moles of each element in 1 mol of compound.

Glucose

C6H12O6

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

In 1 mol: 6 mol C 12 mol H 6 mol O


Subscripts State Atoms and Moles

1 mol C9H8O4 = 9 mol C 8 mol H 4 mol O



Factors from Subscripts

Subscripts used for conversion factors• Relate moles of each element in 1 mol compound. • For aspirin C9H8O4 can be written as:

9 mol C 8 mol H 4 mol O

1 mol C9H8O4 1 mol C9H8O4 1 mol C9H8O4

and

1 mol C9H8O4 1 mol C9H8O4 1 mol C9H8O4

9 mol C 8 mol H 4 mol O


Learning Check

How many moles O are in 0.150 mol aspirin C9H8O4?

0.150 mol C9H8O4 x 4 mol O = 0.600 mol O

1 mol C9H8O4

subscript factor


Learning Check

How many O atoms are in 0.150 mol aspirin C9H8O4?

0.150 mol C9H8O4 x 4 mol O x 6.022 x 1023 O atoms

1 mol C9H8O4 1 mol O

subscript Avogadro’s

factor number

= 3.61 x 1023 O atoms


Atomic Mass

Atomic mass is the• Mass of a single atom in

atomic mass units (amu).• Mass of an atom

compared to a 12C atom.• Number below the

symbol of an element. (the average atomic mass)


Periodic Table and Atomic Mass

Ag has atomic mass = 107.9 amu

C has atomic mass = 12.01 amu

S has atomic mass = 32.07 amu


Atomic Mass Factors

The atomic mass • Can be written as an equality.

Example: 1 P atom = 30.97 amu

• Can be written as two ___________ ________.

Example: 1 P atom and 30.97 amu

30.97 amu 1 P atom


Uses of Atomic Mass Factors

The atomic mass factors are used to convert • A specific number of atoms to mass (amu).• An amount in amu to ___________ of atoms.


Using Atomic Mass Factors

1. What is the mass in amu of 75 P atoms?

75 P atoms x 30.97 amu = 2323 amu (2.323 x103 amu)

1 P atom

2. How many Cu atoms have a mass of 4.500 x 105 amu?

4.500 x 105 amu x 1 Cu atom = 7081 Cu atoms

63.55 amu


Learning Check

What is the mass in amu of 75 silver atoms?

75 Ag atoms x 107.9 amu = 8093 amu

1 Ag atom


Learning Check

How many gold atoms have a mass of

1.85 x 105 amu?

1.85 x 105 amu x 1 Au atom = 939 Au atoms

197.0 amu


Formula MassThe formula mass is• The mass in amu of a compound.• The _____ of the atomic masses of the

elements in a formula.



Calculating Formula Mass

To calculate the formula mass of Na2SO4

• Multiply the atomic mass of each element by its subscript, then total the masses of the atoms.

2 Na x 22.99 amu = 45.98 amu 1 Na

1 S x 32.07 amu = 32.07 amu 142.05 amu 1 S

4 O x 16.00 amu = 64.00 amu1 O


Learning Check

Using the periodic table, calculate the formula mass of aluminum sulfide Al2S3.

2 Al x 26.98 amu = 53.96 amu

1 Al

3 S x 32.07 amu = 96.21 amu

1 S 150.17 amu


Molar Mass



Molar Mass

The molar mass • Is the mass of one

mol of an element or compound.

• Is the atomic mass expressed in grams.



Molar Mass from the Periodic Table

Molar mass • Is the atomic

mass expressed in grams.



Give the molar mass for:

A. 1 mol K atoms =

B. 1 mol Sn atoms =

Learning Check


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.

Element Number of Moles

Atomic Mass Total Mass

Ca 1 40.08 g/mol 40.08 g

Cl 2 35.45 g/mol 70.90 g

CaCl2 110.98 g


Molar Mass of K3PO4

Calculate the molar mass of K3PO4.

Element Number of Moles

Atomic Mass Total Mass in K3PO4

K 3 39.10 g/mol 117.3 g

P 1 30.97 g/mol 30.97 g

O 4 16.00 g/mol 64.00 g

K3PO4 212.3 g


Some One-Mol Quantities

32.07 g 55.85 g 58.44 g 294.20 g 342.30 g



A. K2O 94.20 g/mol

2 mol K (39.10 g/mol) + 1 mol O (16.00 g/mol)

78.20 g + 16.00 g

= 94.20 g

B. Al(OH)3 78.00 g/mol

1 mol Al (26.98 g/mol) + 3 mol O (16.00 g/mol) + 3 mol H (1.008 g/mol) 26.98 g + 48.00 g + 3.024 g = 78.00 g

Learning Check


Calculations Using Molar Mass



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 cook tops and gas heaters.

Molar mass:

1 mol CH4 = 16.04 g

Conversion factors:

16.04 g CH4 and 1 mol CH4

1 mol CH4 16.04 g CH4

Molar Mass Factors


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

Calculate molar mass:

24.02 + 4.032 + 32.00 = 60.05 g/mol

1 mol of acetic acid = 60.05 g acetic acid

Molar mass factors

1 mol acetic acid and 60.05 g acetic acid 60.05 g acetic acid 1 mol acetic acid

Learning Check


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


Aluminum is used to build lightweight bicycleframes. How many grams of Al are 3.00 mol Al?

Molar mass equality: 1 mol Al = 26.98 g Al

Setup with molar mass as a factor:

3.00 mol Al x 26.98 g Al = 80.9 g Al1 mol Al

molar mass factor for Al

Moles to Grams


Learning CheckAllyl sulfide C6H10S is a compound that has the odor of garlic. How many moles of C6H10S are in 225 g C6H10S?

Action Plan: Calculate the molar mass, and convert 225 g to moles.


Grams Molar mass factor Moles


Calculate the molar mass of C6H10S.

(6 x 12.01) + (10 x 1.008) + (1 x 32.07)

= 114.21 g/mol

Set up the calculation using a mole factor.

225 g C6H10S x 1 mol C6H10S

114.21 g C6H10S

molar mass factor(inverted)

= 1.97 mol C6H10S

Solution


Grams, Moles, and Particles

A molar mass factor and Avogadro’s number

convert• Grams to particles molar mass Avogadro’s

number

(g mol particles)• Particles to grams

Avogadro’s molar mass number

(particles mol g)


Learning Check

How many H2O molecules are in 24.0 g H2O?

A) 4.52 x 1023

B) 1.44 x 1025

C) 8.02 x 1023


Learning Check

How many H2O molecules are in 24.0 g H2O?

24.0 g H2O x 1 mol H2O x 6.022 x 1023 H2O molecules

18.02 g H2O 1 mol H2O

= 8.02 x 1023 H2O molecules


Learning CheckIf 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 mol x 6.022 x 1023 molecules

114.21 g 1 mol

= 1 x 109 molecules C6H10S


Percent Composition

and

Empirical Formulas



Percent Composition

Percent composition

• Is the percent by mass of each element in a formula.

Example: Calculate the percent composition of CO2.

CO2 = 1 C(12.01g) + 2 O(16.00 g) = 44.01 g/mol)

12.01 g C x 100 = 27.29 % C 44.01 g CO2

32.00 g O x 100 = 72.71 % O 44.01 g CO2


What is the percent composition of lactic acid, C3H6O3, a compound that appears in the blood after vigorous activity?

→

Learning Check


STEP 1

3C(12.01) + 6H(1.008) + 3O(16.00) = 90.08 g/mol

36.03 g C + 6.048 g H + 48.00 g O = 90.08 g/mol

STEP 2

%C = 36.03 g C x 100 = 40.00% C 90.08 g cpd

%H = 6.048 g H x 100 = 6.714% H 90.08 g cpd

%O = 48.00 g O x 100 = 53.29% O 90.08 g cpd

Solution

C3H6O3


Learning Check

The chemical isoamyl acetate C7H14O2 contributes to the odor of pears. What is the percent carbon in isoamyl acetate?



Molar mass C7H14O2 = 7C(12.01) + 14H(1.008)

+ 2O(16.00) = 130.18 g/mol

Total C = 7C(12.01) = g

% C = total g C x 100 total g cpd

% C = 84.07 g C x 100 = 64.58 % C 130.18 g cpd

Solution

C7H14O2


The empirical formula

• Is the simplest whole number ratio of the atoms.

• Is calculated by dividing the subscripts in the actual (molecular) formula by a whole number to give the lowest ratio.

C5H10O5 5 = C1H2O1 = CH2Oactual (molecular) empirical formula

formula

Empirical Formulas


Some Molecular and Empirical Formulas

• The molecular formula is the same or a multiple of the empirical.



1. What is the empirical formula for _________?

A) C2H4 B) CH2 C) CH

2. What is the empirical formula for _________?

A) C4H7 B) C6H12 C) C8H14

3. Which is a possible molecular formula for ______?

A) C4H4O4 B) C2H4O4 C) C3H6O3

Learning Check P-1


A compound contains 7.31 g Ni and 20.0 g Br. Calculate its empirical (simplest) formula.

Lets ask, what does the empirical formula tell us?

The ratio of atoms of each element,

ratio of moles of each element.

Action Plan: Convert mass (g) to moles.

Learning Check


Convert 7.31 g Ni and 20.0 g Br to moles.

7.31 g Ni x 1 mol Ni = 0.125 mol Ni

58.69 g Ni

20.0 g Br x 1 mol Br = 0.250 mol Br

79.90 g Br

Divide by smallest:

0.125 mol Ni = 1 Ni 0.250 mol Br = 2 Br

0.125 0.125

Write ratio as subscripts: NiBr2

Solution


Converting Decimals to Whole Numbers

When the number of moles for an element is adecimal, all the moles are multiplied by a small

integer to obtain whole number.



Aspirin is 60.0% C, 4.5 % H and 35.5 % O. Calculate its empirical (simplest) formula.

We have percentages, not grams.

However, the percentages are really an expression of grams of an element in 100. (exact) grams of compound.

Learning Check


STEP 1. Calculate the moles of each element in 100. g of the compound.

100. g aspirin contains 60.0% C or 60.0 g C, 4.5% H or 4.5 g H, and 35.5% O or 35.5 g O.

60.0 g C x 1 mol C = 5.00 mol C 12.01 g C

4.5 g H x 1 mol H = 4.5 mol H 1.008 g H

35.5 g O x 1mol O = 2.22 mol O

16.00 g O

Solution


Solution (continued)

STEP 2. Divide by the smallest number of mol.

5.00 mol C = 2.25 mol C (decimal)

2.22 4.5 mol H = 2.0 mol H2.222.22 mol O = 1.00 mole O2.22



3. Use the lowest whole number ratio as subscripts

When the moles are not whole numbers, multiply by a factor to give whole numbers, in this case x 4.

C: 2.25 mol C x 4 = 9 mol CH: 2.0 mol H x 4 = 8 mol HO: 1.00 mol O x 4 = 4 mol O

Using these whole numbers as subscripts the simplest formula is

C9H8O4


Molecular Formulas



A molecular formula• Is a multiple (or equal) of its empirical formula.

• Has a molar mass that is the empirical formula mass multiplied by a whole number.

molar mass = a whole number empirical mass

• Is obtained by multiplying the empirical formula by a whole number.

Relating Molecular and Empirical Formulas


Diagram of Molecular and Empirical Formulas

A small integer links• A molecular formula and its empirical

formula.• A molar mass and its empirical formula

mass.



Determine the molecular formula of compound thathas a molar mass of 78.11 g and an empiricalformula of CH.

STEP 1. Empirical formula mass of CH = 13.02 g

STEP 2. Divide the molar mass by the empirical mass.

78.11 g = 5.999 ~ 6 13.02 g

STEP 3. Multiply each subscript in C1H1 by 6.

molecular formula = C1x 6 H1 x 6 = C6H6

Finding the Molecular Formula


Some Compounds with Empirical Formula CH2O


formaldehyde


A compound has a molar mass of 176.1g and an empirical formula of C3H4O3. What is the molecular formula?

What do we need to do?

From the empirical formula, determine the

empirical mass.

Then determine the whole number integer.

Learning Check


A compound has a formula mass of 176.1 and an empirical formula of C3H4O3. What is the molecular formula?

C3H4O3 = 88.06 g/EF

176.1 g (molar mass) = 2.00 88.06 g (empirical mass)

Molecular formula = 2 x empirical formulaC3 x 2H4 x 2O3 x 2 = C6H8O6

Solution


A compound contains C 24.27%, H 4.07%, and Cl 71.65%. The molar mass is about 99 g. What are the empirical and molecular formulas?

STEP 1. Calculate the empirical formula.

Write the mass percents as the grams in a 100.00-g sample of the compound.

C 24.27 g H 4.07 g Cl 71.65 g

Molecular Formula


Finding the Molecular Formula (Continued)Calculate the number of moles of each element.

24.27 g C x 1 mol C = 2.021 mol C 12.01 g C

4.07 g H x 1 mol H = 4.04 mol H 1.008 g H

71.65 g Cl x 1 mol Cl = 2.021 mol Cl 35.45 g Cl


Finding the Molecular Formula (Continued)Divide by the smallest number of moles:

2.021 mol C = 1 mol C

2.021

4.04 mol H = 2 mol H

2.021

2.02 mol Cl = 1 mol Cl

2.021

Empirical formula = C1H2Cl1 = CH2Cl

Calculate empirical mass (EM) CH2Cl = 49.48 g


Finding the Molecular Formula (Continued)

STEP 2. Divide molar mass by empirical mass.Molar mass = 99 g = 2

Empirical mass 49.48 g

STEP 3. Multiply the empirical formula by the small integer to determine the molecular formula.

2 x (CH2Cl)

C1 x2 H2 x 2 Cl1x 2 = C2H4Cl2


A compound is 27.4% S, 12.0% N and 60.6 % Cl. If the compound has a molar mass of 351 g, what is the molecular formula?

Learning Check


In 100. g, there are 27.4 g S, 12.0 g N, and 60.6 g Cl.

27.4 g S x 1 mol S = 0.854 mol S 32.07 g S

12.0 g N x 1 mol N = 0.857 mol N 14.01 g N

60.6 g Cl x 1mol Cl = 1.71 mol Cl 35.45 g Cl

Solution


Divide by the smallest number of moles

0.854 mol S /0.854 = 1.00 mol S

0.857 mol N/0.854 = 1.00 mol N

1.71 mol Cl/0.854 = 2.00 mol Cl

empirical formula = SNCl2 = 116.98 g

Molar Mass/ Empirical mass

351 g = 3

116.98 g molecular formula = (SNCl2)3 = S3N3Cl6


chapter 7 chemical quantities the mole atomic mass and formula mass copyright © 2008 by pearson...

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