ChemCh10.3.ppt

(1)

10.3 Percent Composition

and Chemical Formulas

>

Chapter 10 Chemical Quantities

10.1 The Mole: A Measurement of

Matter

10.2 Mole-Mass and Mole-Volume

Relationships

(2)

10.3 Percent Composition

and Chemical Formulas

>

A tag sewn into the

seam of a shirt usually

tells you what fibers

were used to make

the cloth and the

percent of each.

CHEMISTRY

&

YOU

CHEMISTRY

&

YOU

(3)

10.3 Percent Composition

and Chemical Formulas

>

Percent Composition of a Compound

How do you calculate the percent

composition of a compound?

Percent Composition of a

Compound

(4)

10.3 Percent Composition

and Chemical Formulas

>

The relative amounts of the elements in a

compound are expressed as the percent

composition or the percent by mass of

each element in the compound.

Percent Composition of a

Compound

(5)

10.3 Percent Composition

and Chemical Formulas

>

The percent composition of potassium

chromate, K

₂

CrO

₄

, is:

Percent Composition of a

Compound

(6)

10.3 Percent Composition

and Chemical Formulas

>

Percent Composition of a

Compound

K = 40.3%

Cr = 26.8%

+ O = 32.9%

100%

(7)

10.3 Percent Composition

and Chemical Formulas

>

Percent Composition of a

Compound

These percents must total 100%.

(8)

10.3 Percent Composition

and Chemical Formulas

>

If you know the relative masses of each

element in a compound, you can calculate

the percent composition of the compound.

Percent Composition of a

Compound

(9)

10.3 Percent Composition

and Chemical Formulas

>

The percent by mass of an element

in a compound is the number of

grams of the element divided by the

mass in grams of the compound,

multiplied by 100%.

Percent Composition from Mass Data

Percent Composition of a

Compound

(10)

10.3 Percent Composition

and Chemical Formulas

>

When a 13.60-g sample of a

compound containing only

magnesium and oxygen is

decomposed, 5.40 g of oxygen is

obtained. What is the percent

composition of this compound?

Sample

Problem 10.9

Sample

Problem 10.9

Calculating Percent

(11)

10.3 Percent Composition

and Chemical Formulas

>

The percent by mass of an element in a

compound is the mass of that element divided by

the mass of the compound multiplied by 100%.

KNOWNS

mass of compound

= 13.60 g

mass of oxygen

= 5.40 g O

mass of magnesium

= 13.60 g – 5.40 g O = 8.20 g Mg

UNKNOWNS

percent by mass of Mg = ?% Mg

Sample

Problem 10.9

Sample

Problem 10.9

Analyze

List the knowns and the unknowns.

(12)

10.3 Percent Composition

and Chemical Formulas

>

Determine the percent by mass of

Mg in the compound.

Sample

Problem 10.9

Sample

Problem 10.9

Calculate

Solve for the unknowns.

2 % Mg =

mass of Mg

mass of compound

× 100%

= 60.3% Mg

(13)

10.3 Percent Composition

and Chemical Formulas

>

Determine the percent by mass of

O in the compound.

Sample

Problem 10.9

Sample

Problem 10.9

Calculate

Solve for the unknowns.

2 % O =

mass of O

mass of compound

× 100%

5.40 g

=

× 100%

(14)

10.3 Percent Composition

and Chemical Formulas

>

The percents of the elements add up

to 100%.

60.3% + 39.7% = 100%

Sample

Problem 10.9

Sample

Problem 10.9

Evaluate

Does the result make sense?

(15)

10.3 Percent Composition

and Chemical Formulas

>

You can also calculate the percent

composition of a compound using its

chemical formula.

Percent Composition from the

Chemical Formula

Percent Composition of a

Compound

• The subscripts in the formula are used to

calculate the mass of each element in a mole of

that compound.

(16)

10.3 Percent Composition

and Chemical Formulas

>

You can also calculate the percent

composition of a compound using its

chemical formula.

Percent Composition from the

Chemical Formula

Percent Composition of a

Compound

% by mass

of element

mass of element in 1 mol compound

molar mass of compound

× 100%

(17)

10.3 Percent Composition

and Chemical Formulas

>

Propane (C

₃

H

₈

), the fuel

commonly used in gas grills, is

one of the compounds

obtained from petroleum.

Calculate the percent

composition of propane.

Sample

Problem 10.10

Sample

Problem 10.10

(18)

10.3 Percent Composition

and Chemical Formulas

>

Calculate the percent by mass of each element

by dividing the mass of that element in one mole

of the compound by the molar mass of the

compound and multiplying by 100%.

KNOWNS

mass of C in 1 mol C

₃

H

₈

= 3 mol × 12.0 g/mol = 36.0 g

mass of H in 1 mol C

₃

H

₈

= 8 mol × 1.0 g/mol = 8.0 g

molar mass of C

₃

H

₈

= 36.0 g/mol + 8.0 g/mol = 44.0 g/mol

UNKNOWNS

percent by mass of C = ?% C

percent by mass of H = ?% H

Sample

Problem 10.10

Sample

Problem 10.10

Analyze

List the knowns and the unknowns.

(19)

10.3 Percent Composition

and Chemical Formulas

>

Determine the percent by mass of C

in C

₃

H

₈

.

% C =

mass of C in 1 mol C

3 H

8 molar mass of C

₃

H

₈

× 100%

= 81.8% C

=

36.0 g

× 100%

44.0 g

Sample

Problem 10.10

Sample

Problem 10.10

Calculate

Solve for the unknowns.

(20)

10.3 Percent Composition

and Chemical Formulas

>

Determine the percent by mass of H

in C

₃

H

₈

.

% H =

mass of H in 1 mol C

3 H

8 molar mass of C

₃

H

₈

× 100%

= 18% H

=

8.0 g

× 100%

44.0 g

Sample

Problem 10.10

Sample

Problem 10.10

Calculate

Solve for the unknowns.

(21)

10.3 Percent Composition

and Chemical Formulas

>

The percents of the elements add up to

100% when the answers are expressed to

two significant figures (82% + 18% =

100%).

Sample

Problem 10.10

Sample

Problem 10.10

Evaluate

Does the result make sense?

(22)

10.3 Percent Composition

and Chemical Formulas

>

You can use percent composition to

calculate the number of grams of any

element in a specific mass of a compound.

• To do this, multiply the mass of the

compound by a conversion factor based on

the percent composition of the element in

the compound.

Percent Composition as a Conversion

Factor

Percent Composition of a

Compound

(23)

10.3 Percent Composition

and Chemical Formulas

>

Propane is 81.8% carbon and 18%

hydrogen.

• You can use the following conversion factors

to solve for the mass of carbon or hydrogen

contained in a specific amount of propane.

Percent Composition as a Conversion

Factor

Percent Composition of a

Compound

81.8 g C

100 g C

H

and

(24)

10.3 Percent Composition

and Chemical Formulas

>

What information can you get from the

percent composition of a compound?

CHEMISTRY

&

YOU

(25)

10.3 Percent Composition

and Chemical Formulas

>

What information can you get from the

percent composition of a compound?

CHEMISTRY

&

YOU

CHEMISTRY

&

YOU

You can use percent

composition to determine

the mass of an element in

a sample of a compound

of a given size. You can

also determine the

(26)

10.3 Percent Composition

and Chemical Formulas

>

Calculate the mass of carbon and

the mass of hydrogen in 82.0 g of

propane (C

₃

H

₈

).

Sample

Problem 10.11

Sample

Problem 10.11

Calculating the Mass of an

(27)

10.3 Percent Composition

and Chemical Formulas

>

Use the conversion factors based on the percent

composition of propane to make the following

conversions: grams C

₃

H

₈

→ grams C and grams

C

₃

H

₈

→

grams H.

KNOWN

mass of C

₃

H

₈

= 82.0 g

Sample

Problem 10.11

Sample

Problem 10.11

Analyze

List the known and the unknowns.

1 UNKNOWNS

(28)

10.3 Percent Composition

and Chemical Formulas

>

• To calculate the mass of C, first write the

conversion factor to convert from mass of C

₃

H

₈

to mass of C.

81.8 g C

100 g C

₃

H

₈

From Sample

Problem 10.10, the

percent by mass of C

in C

₃

H

₈

is 81.8%.

Sample

Problem 10.11

Sample

Problem 10.11

Calculate

Solve for the unknowns.

2 • Multiply the mass of C

₃

H

₈

by the

conversion factor.

81.8 g C

100 g C

₃

H

₈

(29)

10.3 Percent Composition

and Chemical Formulas

>

• To calculate the mass of H, first write the

conversion factor to convert from mass of C

₃

H

₈

to mass of H.

_{From Sample}

Problem 10.10, the

percent by mass of H

in C

₃

H

₈

is 18%.

Sample

Problem 10.11

Sample

Problem 10.11

Calculate

Solve for the unknowns.

2 • Multiply the mass of C

₃

H

₈

by the

conversion factor.

(30)

10.3 Percent Composition

and Chemical Formulas

>

The sum of the two masses equals 82 g,

the sample size, to two significant figures

(67 g C + 15 g H = 82 g C

₃

H

₈

).

Sample

Problem 10.11

Sample

Problem 10.11

Evaluate

Does the result make sense?

(31)

10.3 Percent Composition

and Chemical Formulas

>

(32)

10.3 Percent Composition

and Chemical Formulas

>

What data can you use to calculate

percent composition?

(33)

10.3 Percent Composition

and Chemical Formulas

>

Empirical Formulas

How can you calculate the empirical

formula of a compound?

Empirical Formulas

(34)

10.3 Percent Composition

and Chemical Formulas

>

The empirical formula of a compound gives

the lowest whole-number ratio of the atoms

or moles of the elements in a compound.

Empirical Formulas

(35)

10.3 Percent Composition

and Chemical Formulas

>

The empirical formula of a compound gives

the lowest whole-number ratio of the atoms

or moles of the elements in a compound.

Empirical Formulas

• An empirical formula may or may not be the

same as a molecular formula.

– For example, the lowest ratio of hydrogen to oxygen

in hydrogen peroxide is 1:1. Thus, the empirical

formula of hydrogen peroxide is HO.

(36)

10.3 Percent Composition

and Chemical Formulas

>

Empirical Formulas

For carbon dioxide, the empirical and

molecular

formulas

are the

same

—

(37)

10.3 Percent Composition

and Chemical Formulas

>

Ethyne (C

₂

H

₂

), also called

The figure below shows two compounds of

carbon and hydrogen having the same

empirical formula (CH) but different

molecular formulas.

Empirical Formulas

(38)

10.3 Percent Composition

and Chemical Formulas

>

The percent composition of a compound

can be used to calculate the empirical

formula of that compound.

Empirical Formulas

(39)

10.3 Percent Composition

and Chemical Formulas

>

The percent composition of a compound

can be used to calculate the empirical

formula of that compound.

• The percent composition tells the ratio of masses of

the elements in a compound.

• The ratio of masses can be changed to ratio of moles

by using conversion factors based on the molar mass

of each element.

• The mole ratio is then reduced to the lowest

whole-Empirical Formulas

(40)

10.3 Percent Composition

and Chemical Formulas

>

A compound is analyzed and

found to contain 25.9% nitrogen

and 74.1% oxygen. What is the

empirical formula of the

compound?

Sample

Problem 10.12

Sample

Problem 10.12

(41)

10.3 Percent Composition

and Chemical Formulas

>

The percent composition gives the ratio of the

mass of nitrogen atoms to the mass of oxygen

atoms in the compound. Change the ratio of

masses to a ratio of moles and reduce this ratio

to the lowest whole-number ratio.

KNOWNS

percent by mass of N

= 25.9% N

percent by mass of O

= 74.1% O

Sample

Problem 10.12

Sample

Problem 10.12

Analyze

List the knowns and the unknown.

(42)

10.3 Percent Composition

and Chemical Formulas

>

Convert the percent by mass of

each element to moles.

25.9 g N ×

1 mol N

14.0 g N

= 1.85 mol N

74.1 g O ×

1 mol

_O

16.0 g O

= 4.63 mol O

The mole ratio of N to O is N

_1.85

O

_4.63

.

Sample

Problem 10.12

Sample

Problem 10.12

Calculate

Solve for the unknown.

2 Percent means

“parts per 100,” so

100.0 g of the

compound

(43)

10.3 Percent Composition

and Chemical Formulas

>

Divide each molar quantity by the smaller

number of moles to get 1 mol for the

element with the smaller number of moles.

4.63 mol O

1.85 = 2.50 mol O

1.85 mol N

1.85 = 1 mol N

Sample

Problem 10.12

Sample

Problem 10.12

Calculate

Solve for the unknown.

(44)

10.3 Percent Composition

and Chemical Formulas

>

Multiply each part of the ratio by the

smallest whole number that will convert

both subscripts to whole numbers.

1 mol N × 2 = 2 mol N

2.5 mol O × 2 = 5 mol O

The empirical formula is N

₂

O

₅

.

Sample

Problem 10.12

Sample

Problem 10.12

Calculate

Solve for the unknown.

(45)

10.3 Percent Composition

and Chemical Formulas

>

The subscripts are whole numbers, and

the percent composition of this empirical

formula equals the percents given in the

original problem.

Sample

Problem 10.12

Sample

Problem 10.12

Evaluate

Does the result make sense?

(46)

10.3 Percent Composition

and Chemical Formulas

>

You are doing an experiment to try to find

the molecular formula of a compound. You

discover the percent composition. Can

(47)

10.3 Percent Composition

and Chemical Formulas

>

You are doing an experiment to try to find

the molecular formula of a compound. You

discover the percent composition. Can

you determine the molecular formula?

(48)

10.3 Percent Composition

and Chemical Formulas

>

Molecular Formulas

How does the molecular formula of a

compound compare with the

empirical formula?

Molecular Formulas

(49)

10.3 Percent Composition

and Chemical Formulas

>

Comparison of Empirical and Molecular Formulas

Formula (name)

Classification of

formula

Molar mass

(g/mol)

CH

Empirical

13 C

₂

H

₂

(ethyne)

Molecular

26 (2 × 13)

C

₆

H

₆

(benzene)

_Molecular

_{78 (6 × 13)}

CH

₂

O (methanol)

Empirical and molecular 30

C

H

O

(ethanoic acid)

Molecular

60 (2 × 30)

Ethyne and benzene have the same

empirical formula—CH.

Interpret

Data

(50)

10.3 Percent Composition

and Chemical Formulas

>

Comparison of Empirical and Molecular Formulas

Formula (name)

Classification of

formula

Molar mass

(g/mol)

CH

Empirical

13 C

₂

H

₂

(ethyne)

Molecular

26 (2 × 13)

C

₆

H

₆

(benzene)

_Molecular

_{78 (6 × 13)}

CH

₂

O (methanal)

Empirical and molecular 30

C

₂

H

₄

O

₂

(ethanoic acid)

Molecular

60 (2 × 30)

C

₆

H

₁₂

O

₆

(glucose)

Molecular

180 (6 × 30)

Methanal, ethanoic acid, and glucose have

the same empirical formula—CH

₂

O. Interpret

Data

(51)

10.3 Percent Composition

and Chemical Formulas

>

Comparison of Empirical and Molecular Formulas

Formula (name)

Classification of

formula

Molar mass

(g/mol)

CH

Empirical

13 C

₂

H

₂

(ethyne)

Molecular

26 (2 × 13)

C

₆

H

₆

(benzene)

_Molecular

_{78 (6 × 13)}

CH

₂

O (methanal)

Empirical and molecular 30

C

H

O

(ethanoic acid)

Molecular

60 (2 × 30)

Notice that the molar masses of the compounds in these

two groups are simple whole-number multiples of the

molar masses of the empirical formulas, CH and CH

₂

O. Interpret

Data

(52)

10.3 Percent Composition

and Chemical Formulas

>

Methanal

(formaldehyde),

ethanoic acid (acetic

acid), and glucose have

the same empirical

formula

—

CH

₂

O. Molecular Formulas

(53)

10.3 Percent Composition

and Chemical Formulas

>

The molecular formula of a compound

is either the same as its experimentally

determined empirical formula, or it is a

simple whole-number multiple of its

empirical formula.

Molecular Formulas

(54)

10.3 Percent Composition

and Chemical Formulas

>

The molecular formula of a compound

is either the same as its experimentally

determined empirical formula, or it is a

simple whole-number multiple of its

empirical formula.

Molecular Formulas

• Once you have determined the empirical

formula of a compound, you can determine

its molecular formula, if you know the

(55)

10.3 Percent Composition

and Chemical Formulas

>

You can calculate the empirical formula

mass (efm) of a compound from its empirical

formula.

Molecular Formulas

(56)

10.3 Percent Composition

and Chemical Formulas

>

You can calculate the empirical formula

mass (efm) of a compound from its empirical

formula.

Molecular Formulas

• Then you can divide the experimentally

determined molar mass by the empirical formula

mass.

• This quotient gives the number of empirical

(57)

10.3 Percent Composition

and Chemical Formulas

>

Calculate the molecular formula of a

compound whose molar mass is 60.0 g/mol

and empirical formula is CH

₄

N. Sample

Problem 10.13

Sample

Problem 10.13

(58)

10.3 Percent Composition

and Chemical Formulas

>

• Divide the molar mass by the empirical

formula mass to obtain a whole number.

• Multiply the empirical formula subscripts by

this value to get the molecular formula.

KNOWNS

empirical formula = CH

₄

N

molar mass

= 60.0 g/mol

Analyze

List the knowns and the unknown.

1 Sample

Problem 10.13

Sample

Problem 10.13

UNKNOWN

(59)

10.3 Percent Composition

and Chemical Formulas

>

First calculate the empirical formula

mass.

efm of CH

₄

N = 12.0 g/mol + 4(1.0 g/mol) + 14.0 g/mol

= 30.0 g/mol

Calculate

Solve for the unknown.

2 Sample

Problem 10.13

(60)

10.3 Percent Composition

and Chemical Formulas

>

• Divide the molar mass by the

empirical formula mass.

Calculate

Solve for the unknown.

2 Sample

Problem 10.13

Sample

Problem 10.13

molar mass

efm

=

60.0 g/mol

30.0 g/mol

= 2

• Multiply the formula subscripts by this

value.

(61)

10.3 Percent Composition

and Chemical Formulas

>

The molecular formula has the molar

mass of the compound.

Evaluate

Does the result make sense?

3 Sample

Problem 10.13

(62)

10.3 Percent Composition

and Chemical Formulas

>

What information, in addition to

empirical formula, is necessary to

(63)

10.3 Percent Composition

and Chemical Formulas

>

What information, in addition to

empirical formula, is necessary to

determine the molecular formula of a

compound?

(64)

10.3 Percent Composition

and Chemical Formulas

>

The percent by mass of an element in a

compound is the number of grams of

the element divided by the mass in

grams of the compound, multiplied by

100%.

The percent composition of a

compound can be used to calculate the

empirical formula of that compound.

Key Concepts

(65)

10.3 Percent Composition

and Chemical Formulas

>

The molecular formula of a compound is

either the same as its experimentally

determined formula, or it is a simple

whole-number multiple of its empirical

formula.

Key Concepts and

Key Equations

mass of element in 1 mol compound

% by mass

of element

mass of element

mass of compound

(66)

10.3 Percent Composition

and Chemical Formulas

>

Glossary Terms

• _{percent composition: the percent by}

mass of each element in a compound

• empirical formula: a formula with the

lowest whole-number ratio of

(67)

10.3 Percent Composition

and Chemical Formulas

>

• The molecular formula of a compound

can be determined by first finding the

percent composition of the compound

and determining the empirical formula.

• Using the empirical formula mass and

the molar mass of the compound, the

molecular formula can be determined.

BIG

IDEA

BIG

IDEA

(68)

10.3 Percent Composition

and Chemical Formulas

>

END OF 10.3