Chapter 07 - Quantitative Composition of Compounds

Foundations of College Chemistry, 14th Ed.

Morris Hein and Susan Arena

Black pearls are composed of calcium carbonate, CaCO₃. The pearls can be measured by either weighing or counting.

7 Quantitative Composition of

Compounds

2

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}

Basic Chemistry Copyright © 2011 Pearson Education, Inc.

3

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 10}23 _{atoms of an element (Avogadro’s}

number).

1 mol of Element Number of Atoms

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

Moles can be used to describe elements, particles or compounds.

1 mol of atoms = 6.022 x 1023 atoms

1 mol of molecules = 6.022 x 1023 molecules

1 mol of electrons = 6.022 x 1023 electrons

Mole is often abbreviated as mol.

Avogadro’s number can be used as a conversion factor.

1 mol

6.022 x 1023 objects 1 mol

6.022 x 1023 objects

The Mole

5

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 of Cu?

0.50 mol Cu x 6.022 x 1023 _{Cu atoms}

1 mol Cu

= 3.0 x 1023 _{Cu atoms}

6

1. The number of atoms in 2.0 mol of Al is

2. The number of moles of S in 1.8 x 10

_atoms

of S is

Learning Check

7

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

1 molecule: 6 atoms of C 12 atoms of H 6 atoms of O 1 mol: 6 mol of C 12 mol of H 6 mol of O

8

Factors from Subscripts

Subscripts used for conversion factors

• _{relate moles of each element in 1 mol of 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

Example – Molar Ratios

•

_{How many moles of oxygen atoms are in 2.7}

moles of aspirin, C

H

O

?

10

Molar Mass

The

molar mass

•

_{is the mass of one mol of an element}

or compound

•

_{is the atomic mass expressed in grams}

207.2 g Pb 1 mol Pb

We can use both the mol and molar mass as conversion factors.

How many moles of lead does 15.0 g of Pb represent?

Calculate

= 7.24 x 10-2 mol Pb

15.0 g Pb ×

Solution Map g Pb mol Pb

The conversion factor relates g of Pb to moles of Pb.

1 mol Pb

207.2 g Pb or

207.2 g Pb 1 mol Pb

(Obtain molar mass from the periodic table.)

Using the Mole and

Molar Mass Concepts

Example – Molar Mass

•

_{How many moles of carbon are in 12.4 g}

sample of coal, which is composed of carbon?

•

_{If there are 1.72 x 10}

_{atoms of Ne in a neon}

14

Molar Mass of a Compound

For a compound, the molar mass 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

Calculating molecular (formula) masses

•

_KCl

–

_{39.09 + 35.45 = 74.54 amu}

•

C

H

–

_{2(12.01) + 6(1.008) = 30.08 amu}

•

Ca

(PO

)

The molar mass of a compound contains Avogadro’s number of formula units/molecules.

H₂ O H₂O

2 x (6.022 x 1023)

H atoms 6.022 x 10

23

O atoms 6.022 x 10

23

H₂O molecules

2 mol H atoms 1 mol O atoms 1 mol H₂O molecules

2 x 1.008 g = 2.016 g H 16.00 g O 18.016 g H₂O

Reminder: Pay close attention to whether the desired unit involves atoms or formula units/molecules.

Example Cl₂

Contains 2 mol of Cl atoms but only 1 mol of Cl₂ molecules.

Molar Mass of Compounds

Example – Molar Mass

•

What is the molar mass of aspirin, C

H

O

?

18

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

19

Acetic acid HC

H

O

gives the sour taste to vinegar. Write

an equality and two molar mass conversion factors for

acetic acid.

Write the molar conversion factors for the moles of

carbon in acetic acid.

Do the same for hydrogen.

Learning Check

20 Aluminum is used to build lightweight bicycle

frames. How many grams of Al are in 3.00 mol of Al?

The artificial sweetener aspartame (NutraSweet), C₁₄H₁₈N₂O₅, is used to sweeten diet foods, coffee, and soft drinks. How many moles of aspartame are present in 225 g of aspartame?

Moles to Grams

21

Learning Check

How many molecules of H₂O are in 24.0 g of H₂O?

Example – Mole Conversions

•

_{Hydrogen cyanide (HCN) is a volatile, colorless}

liquid with the odor of certain fruit pits (peach

and cherry). The compound is highly toxic.

What is the mass of 3.14 x 10

_{molecules of}

Percent = parts per 100 parts

Percent composition: mass percent of each element in a compound

Molar mass: total mass (100%) of a compound

% composition is independent of sample size

% composition can be determined by: 1. Knowing the compound’s formula or

2. Using experimental data

Percent Composition of Compounds

Two Step Strategy

1. Calculate the molar mass of the compound. 2. Divide the total mass of each element by the compound’s molar mass and multiply by 100.

Total element mass Compound molar mass

% of the element = × 100

Percent Composition from the

Compound’s Formula

Calculate the percent composition of K₂S.

Step 1 Calculate compound molar mass

MM_K2S = 2(39.10) g + 32.07 g = 110.3 g

Step 2 Calculate % composition of each element.

2(39.10) g K 110.3 g

% K = 100

Notice the sum of the percentages must equal 100%. This provides another way of determining the %

composition of a specific element, if the other %s are known.

×

32.07 g S 110.3 g

% S = × 100

= 70.90 % K

= 29.10 % S

Percent Composition of Compounds

Example

•

Calculate the % composition of

K

CrO

.

Two Step Strategy

1. Calculate the mass of the compound formed.

2. Divide the mass of each element by the total mass and multiply by 100.

Total element mass Total compound mass

% of the element = _{× 100}

Percent Composition from

Experimental Data

When heated in air, 1.63 g of Zn reacts with

0.40 g of oxygen to give ZnO. Calculate the percent composition of the compound formed.

Step 1 Calculate the mass of the compound formed.

Mass compound = Mass_Zn + Mass_O = 1.63 g + 0.40 g

Step 2 Calculate % composition

Total should be +/0.5% of 100100.3%

= 2.03 g compound

1.63 g Zn 2.03 g

% Zn = × 100 = 80.3 % Zn

0.40 g O 2.03 g

% O = × 100 = 20. % O

Percent Composition from

Experimental Data

Example

•

Aluminum chloride forms by reaction

of 13.43 g of Al with 53.18 g of

chlorine.

•

What is the percent composition of Cl

Empirical Formula: smallest whole number ratio of atoms in a compound

Molecular Formula: actual formula of a compound.

Represents the total number of atoms in one formula unit of the compound.

Whole number multiple of the empirical formula

Example Acetylene (C₂H₂) and Benzene (C₆H₆)

Both have the same empirical formula CH. Each compound is a multiple of CH.

Acetylene C₂H₂ = (CH)₂ Benzene C₆H₆ = (CH)₆

Empirical and Molecular Formula

Each compound has very different chemical and physical properties even though they share

the same empirical formula.

Compounds with the same empirical formula have the same percent composition.

Formula Composition

% C %H Molar Mass (g/mol) CH (empirical formula) 92.3 7.7 13.02

C₂H₂ (acetylene) 92.3 7.7 26.04 (2 x 13.02) C₆H₆(benzene) 92.3 7.7 78.16 (6 x 13.02)

Molar mass = molar mass of the empirical unit multiple of the unit

×

Empirical and Molecular Formula

32

A. What is the empirical formula for C

H

?

B. What is the empirical formula for C

H

?

1) C

H

2) C

H

3) C

H

C. Which is a possible molecular formula for CH

O?

1) C

H

O

2) C

H

O

3) C

H

O

Learning Check

33

A compound has an empirical formula

SN.

If there are four atoms of N in one

molecule, what is the molecular formula?

1) SN

2) SN

3) S

N

Learning Check

To calculate an empirical formula, you need to know:

1. The elements present in the compound

2. The atomic masses of each element (from the Periodic Table)

3. The ratio (by mass or %) of the combined elements

Calculating Empirical Formulas

Strategy to Calculate an Empirical Formula:

1. Assume a starting mass of the compound (usually

100.0 g) and express the mass of each element in grams. 2. Convert g of each element to mol using molar mass. (These numbers may or may not be whole numbers.) 3. Divide each of the mole amounts from Step 2 by the smallest mole amount. The new numbers are the

subscripts in the empirical formula.

Special Case:

If fractions are encountered, multiply by a common factor to provide whole numbers for each subscript.

Calculating Empirical Formulas

Calculate the empirical formula for a compound that contains 11.19% H and 88.79% O.

Step 1 Find amounts of each element

In a 100.0 g sample, there are

11.19 g H and 88.79 g O

Step 2 Convert g to moles using element molar masses

= 11.10 mol H 11.19 g H _× 1 mol H

1.008 g H

= 5.549 mol O 88.79 g O × 1 mol O

16.00 g O

Calculating Empirical Formulas

Step 3 Convert to whole numbers by dividing by the smallest mole amount.

= 2.000 11.10 mol H

5.549 mol O

Empirical formula is H₂O

= 1.000 5.549 mol O

5.549 mol O

Calculating Empirical Formulas

Examples

•

Calculate the empirical formula for a

compound that contains 56.68% K, 8.68% C

and 34.73% O.

•

The major air pollutant in coal burning countries is a

colorless, pungent gaseous compound containing only

sulfur and oxygen. Chemical analysis of a 1.078 g

sample of this gas showed that it contained 0.540 g of S

and 0.538 g of O. What is the empirical formula of this

Example – Empirical Formula

•

_{TNT, trinitrotoluene, is composed of 37.01%}

40

A

molecular formula

•

_{is equal or a multiple of its empirical formula}

•

_{has a molar mass that is the product of the}

empirical formula mass multiplied by a small

integer

molar mass = a small integer

empirical mass

•

_{is obtained by multiplying the subscripts in the}

empirical formula by the same small integer

Relating Molecular and Empirical Formulas

If molar mass is known, the molecular formula can be calculated from the empirical formula.

Molecular formula is a multiple of the empirical formula. Need to determine the value of n.

Solving for n

Molar mass

Mass of empirical formula

n = = number of empirical units in the molecular formula

Calculating the Molecular Formula from

the Empirical Formula

A compound with the empirical formula NH₂ was found to have a molar mass of 32.05 g.

What is the molecular formula?

Molecular formula = (NH₂)₂ = N₂H₄

Molar mass

Mass of empirical formula

n = = number of empirical units in the molecular formula

32.05

14.01 + 2(1.008)

n = = 2

Calculating the Molecular Formula from

the Empirical Formula

Example

•

A compound with the empirical formula NO

was found to have a molar mass of 92.00 g.

What is the molecular formula?

•

_{Nicotine is an alkaloid found in the nightshade family of}

plants. It is present in small amounts in eggplant and

tomato plants, but makes up 0.6-3.0% of the dry weight

of tobacco. Nicotine has an empirical formula of C

H

N

and a molar mass of about 160 g/mol. What is the

Propylene contains 14.3 % H and 85.7 % C and has a molar mass of 42.08 g. What is its molecular formula?

Plan Calculate empirical formula and then determine the

molecular formula

Step 1 Find compound masses

Step 2 Convert g to moles

= 14.2 mol H 14.3 g H × 1 mol H

1.008 g H

In 100.0 g of compound, 14.3 g H and 85.7 g C

= 7.14 mol C 85.7 g C × 1 mol C

12.01 g C

Calculating the Molecular Formula from

the Empirical Formula

Step 3 Convert to whole numbers by dividing by the smallest mole amount.

= 1.99 14.2 mol H

7.14 mol

= 1.00 7.14 mol C

7.14 mol

Empirical formula = CH₂

With EF, calculate the molecular formula

Molecular formula = (CH₂)₃ = C₃H₆

42.08 g

12.01 + 2(1.008) g

n = = 3

Calculating the Molecular Formula from

the Empirical Formula

Calculate the molecular formula for a compound that

contains 80.0% C and 20.0% H with a molar mass of 30.00 g.

Plan Calculate empirical and then

molecular formula

Step 1 Find compound masses

Step 2 Convert g to moles

= 19.8 mol H 20.0 g H _× 1 mol H

1.008 g H

= 6.66 mol C 80.0 g C _× 1 mol C

12.01 g C

In 100.0 g of compound, 20.0 g H and 80.0 g C a. CH₃

b. CH₂

c. C₂H₆

d. C₂H₄

Calculating the Molecular Formula from

the Empirical Formula

Example

•

Calculate the molecular formula for a

compound that contains 80.0% C

and 20.0% H with a molar mass of

Apply the concept of the mole, molar mass,

and Avogadro’s number to solve chemistry problems.

7.1 The Mole

Calculate the molar mass of a compound.

7.2 Molar Mass of Compounds

Calculate the percent composition of a compound from its chemical composition and from experimental data.

7.3 Percent Composition of Compounds

Learning Objectives

Determine the empirical formula for a compound from its percent composition.

7.4 Calculating Empirical Formulas

Compare an empirical formula to a molecular formula and calculate a molecular formula from an

empirical formula, using the molar mass.

7.5 Calculating the Molecular Formula from the Empirical Formula

Learning Objectives