Monday, April 7th: “A” DayTuesday, April 8th: “B” Day

AgendaHomework Questions/CollectGo over Sec. 7.1 Quiz Welcome Back!Section 7.2: “Relative Atomic Mass/Chemical

Formulas”In-Class Assignment:

Practice pg. 236: #1-2Practice pg. 239-240: #1-4 (1 & 2: a,b,c,d only)

Homework: Concept Review: “Relative Atomic Mass and Chemical Formulas”

Homework Questions/Problems?Sec. 7.1 review, pg. 233…

Sec 7.1 Quiz:“Avogadro’s Number and Molar

Conversions”This quiz seemed to give some of you problems,

so I wanted to take some time to go over it…

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!

Rules for Calculating Average Atomic Mass

1. Change each percentage to a decimal by dividing by 100. (move the decimal point 2 places to the left and take away the % sign)

2. Multiply each decimal by the atomic mass that goes with it.

3. Add the atomic masses together.

That’s it!

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 (do not round): Cu-63 = 69.17% = .6917 Cu-65 = 30.83% = .3083

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

3. Add the atomic masses together:

43.54 amu + 20.01 amu = 63.55 amu

Additional Example #1Chlorine exists as chlorine-35, which has a mass of 34.969

amu and makes up 75.80% of chlorine atoms. The rest of naturally occurring chlorine is chlorine-37, with a mass of 36.996 amu. What is the average atomic mass of chlorine?

1. Change each percentage to a decimal: Cl-35 = 75.80% = .7580 Cl-37 = 24.20% = .2420 (100% - 75.80%)

2. Multiply each decimal by the atomic mass that goes with it:Cl-35 = (.7580) (34.969 amu) = 26.51 amuCl-37 = (.2420) (36.996 amu) = 8.953amu

3. Add the atomic masses together:

26.51 amu + 8.953 amu = 35.46 amu

Additional Example #2U-234 makes up 0.00500% of uranium atoms and has a mass

of 234.041 amu. U-235 makes up 0.720% and has a mass of 235.044 amu. U-238 has a mass of 238.051 amu and makes up 99.275%. What is the average atomic mass of uranium?

1. Change each percentage to a decimal: U-234 = 0.00500% = 0.0000500 U-235 = 0.720 % = 0.00720U-238 = 99.275% = .99275

Additional Example #2, cont.2. Multiply each decimal by the atomic mass that goes with

it:U-234 = (0.0000500) (234.041amu) = 0.0117 amuU-235 = (0.00720) (235.044) = 1.69 amu U-238 = (.99275) (238.051) = 236.33 amu

3. Add the atomic masses together:

0.0117 amu + 1.69 amu + 236.33 amu =238.03 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.**Before you can find the molar mass, you need to write

the chemical formula**

Barium is a 2+ cation and nitrate is a 1- anion. Ba2+ NO3

-

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

on the barium cation.

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

Calculating Molar Mass of CompoundsSample Problem F, continued.

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

**The 2 outside of the ( ) means that everything inside the ( ) is multiplied by 2**

Ba: = 137.33 g/molN: 2 (14.01 g/mol) = 28.02 g/mol

O: 6 (16.00 g/mol) = 96.00 g/molMolar mass of Ba(NO3)2 = 261.35 g/mol

Additional Practice #1Calculate the molar mass of ammonium sulfite,

(NH4)2SO3.

Add the atomic masses of each element together:N: 2 (14.01 g/mol) = 28.02 g/molH: 8 (1.01 g/mol) = 8.08 g/molS: 32.07 g/mol = 32.07 g/molO: 3 (16.00 g/mol) = 48.00 g/molMolar mass of (NH4)2SO3 = 116.17 g/mol

Additional Practice #2Calculate the molar mass of aluminum sulfate,

Al2(SO4)3.

Add the atomic masses of each element together:Al: 2 (26.98 g/mol) = 53.96 g/molS: 3 (32.07 g/mol) = 96.21 g/molO: 12 (16.00 g/mol) = 192.00 g/mol

Molar mass of Al2(SO4)3 = 342.17 g/mol

In-Class Assignment/HomeworkYou Must SHOW WORK!

Practice pg. 236: #1,2Practice pg. 239-240: # 1-4 (1 & 2: a,b,c,d only)Homework:

Concept Review: “Relative Atomic Mass and Chemical Formulas”

Next time:Sec. 7.2 work day/quiz