Wednesday, April 13th: “A” DayThursday, April 14th: “B” Day

AgendaSection 7.2: “Relative Atomic Mass and Chemical

Formulas”In-Class Assignments:

Sec. 7.2 review, pg. 240: 1-12 (10-12 abc ONLY)Homework:

Concept Review: “Relative Atomic Mass and Chemical Formulas”

Quiz next time over section 7.2

Section 7.2: “Relative Atomic Mass and Chemical Formulas”

You have learned to use the periodic table to find the atomic mass of an element.

However, most atomic masses are written to at least 3 decimal places.

Why?

Most Elements are a Mixture of IsotopesRemember: isotopes are atoms of the same

element that have different numbers of neutrons.

Because they have different numbers of neutrons, isotopes have different atomic masses.

The periodic table reports average atomic mass, which is a weighted average of the atomic mass of an element’s isotopes.

If you know the abundance of each isotope, you can calculate the average atomic mass of an element – and that’s what we’re going to do!

Calculating Average Atomic MassSample Problem E, pg. 235

The mass of a Cu-63 atom is 62.94 amu, and that of a Cu-65 atom is 64.93 amu.

(amu = atomic mass unit)Using the data below, find the average atomic

mass of copper. abundance of Cu-63 = 69.17% abundance of Cu-65 = 30.83%

Calculating Average Atomic MassSample Problem E, continued

1. Change each percentage to a decimal by dividing it by 100. (move the decimal point 2 places to the LEFT) Cu-63 = 69.17% ⁄ 100% = .6917 Cu-65 = 30.83% ⁄ 100% = .3083

2. Multiply each decimal by it’s atomic mass:Cu-63 = (.6917) (62.94 amu) = 43.54 amuCu-65 = (.3083) (64.93) = 20.01 amu

3. Average atomic mass is found by adding the two atomic masses together.

43.54 amu + 20.01 amu = 63.55 amu

Additional ExampleCalculate the average atomic mass for gallium if 60.00%

of its atoms have a mass of 68.926 amu and 40.00% have a mass of 70.925 amu.

1. Change each percentage to a decimal:60.00% / 100% = .600040.00% / 100% = .4000

2. Multiply each decimal by it’s atomic mass:(.6000) (68.926 amu) = 41.36 amu(.4000) (70.925 amu) = 28.37 amu

3. Average atomic mass is found by adding the two atomic masses together:41.36 amu + 28.37 amu = 69.73 amu

Chemical Formulas and Moles

A compound’s chemical formula tells you which elements, as well as how much of each, are present in a compound.

Formulas for covalent compounds show the elements and the number of atoms of each element in a molecule.

Formulas for ionic compounds show the simplest ratio of cations and anions in any pure sample.

Formulas Express CompositionAlthough any sample of compound has many atoms and

ions, the formula gives a ratio of those atoms or ions.

Formulas Give Ratios of Polyatomic IonsFormulas for polyatomic ions show the

simplest ratio of cations and anions.

They also show the elements and the number of atoms of each element in each ion.

For example, the formula KNO3 shows a ratio of one K+ cation to one NO3

- anion.

Calculating Molar Mass of CompoundsSample Problem F, pg. 239

Find the molar mass of barium nitrate.

1. Barium is a 2+ cation and nitrate is a 1- anion. Ba 2+ NO3

-

2. Two NO3- anions are needed to balance the 2+

charge on the barium cation.

3. The simplest formula for barium nitrate is: Ba(NO3)2

Calculating Molar Mass of CompoundsSample Problem F, continued.

2. To find the molar mass of Ba(NO3)2, add the atomic masses of each element together.

Barium, Ba: 137.33 g/mol = 137.33 g/mol Nitrogen, N: 2 (14.01 g/mol) = 28.02 g/mol Oxygen, O: 6 ( 16.00 g/mol) = 96.00 g/molMolar mass of Ba(NO3)2 = 261.35 g/mol

Additional PracticeCalculate the molar mass of (NH4)2SO3.

Add the atomic masses of each element together:

Nitrogen, N: 2 (14.01 g/mol) = 28.02 g/molHydrogen, H: 8 (1.01 g/mol) = 8.08 g/molSulfur, S: 32.07 g/mol = 32.07 g/molOxygen, O: 3 (16.00 g/mol) = 48.00 g/molMolar mass of (NH4)2SO3 = 116.17 g/mol

In-Class Assignment/HomeworkYou Must SHOW WORK!

Section 7.2 review, pg. 240: 1-12 (10-12 a,b,c ONLY)

Homework:Concept Review: “Relative Atomic Mass and

Chemical Formulas”

We’ll have a quiz over this section next time…