Monday, April 7 th : “A” Day Tuesday, April 8 th : “B” Day Agenda  Homework Questions/Collect  Go over Sec. 7.1 Quiz Welcome Back !  Section 7.2: “Relative Atomic Mass/Chemical Formulas”  In-Class Assignment: Practice pg. 236: #1-2 Practice pg : #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 Isotopes  Remember: 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 Mass Sample Problem E, pg. 235 The mass of a Cu-63 atom is amu, and that of a Cu-65 atom is 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 Mass Sample Problem E, continued 1. Change each percentage to a decimal (do not round): Cu-63 = 69.17% =.6917 Cu-65 = 30.83% = Multiply each decimal by the atomic mass that goes with it: Cu-63 = (.6917) (62.94 amu) = amu Cu-65 = (.3083) (64.93 amu) = amu 3. Add the atomic masses together: amu amu = amu

Additional Example #1 Chlorine exists as chlorine-35, which has a mass of amu and makes up 75.80% of chlorine atoms. The rest of naturally occurring chlorine is chlorine-37, with a mass of 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% %) 2. Multiply each decimal by the atomic mass that goes with it: Cl-35 = (.7580) ( amu) = amu Cl-37 = (.2420) ( amu) = 8.953amu 3. Add the atomic masses together: amu amu = amu

Additional Example #2 U-234 makes up % of uranium atoms and has a mass of amu. U-235 makes up 0.720% and has a mass of amu. U-238 has a mass of amu and makes up %. What is the average atomic mass of uranium? 1. Change each percentage to a decimal: U-234 = % = U-235 = % = U-238 = % =.99275

Additional Example #2, cont. 2. Multiply each decimal by the atomic mass that goes with it: U-234 = ( ) ( amu) = amu U-235 = ( ) ( ) = 1.69 amu U-238 = (.99275) ( ) = amu 3.Add the atomic masses together: amu amu amu = 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 Composition  Although any sample of compound has many atoms and ions, the formula gives a ratio of those atoms or ions.

Formulas Give Ratios of Polyatomic Ions  Formulas 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 KNO 3 shows a ratio of one K + cation to one NO 3 - anion.

Calculating Molar Mass of Compounds Sample 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. Ba 2+ NO 3 - Two NO 3 - anions are needed to balance the 2 + charge on the barium cation. The simplest formula for barium nitrate is: Ba(NO 3 ) 2

Calculating Molar Mass of Compounds Sample Problem F, continued. To find the molar mass of Ba(NO 3 ) 2, add the atomic masses of each element together. **The 2 outside of the ( ) means that everything inside the ( ) is multiplied by 2** Ba: = g/mol N: 2 (14.01 g/mol) = g/mol O: 6 (16.00 g/mol) = g/mol Molar mass of Ba(NO 3 ) 2 = g/mol

Additional Practice #1 Calculate the molar mass of ammonium sulfite, (NH 4 ) 2 SO 3. Add the atomic masses of each element together: N: 2 (14.01 g/mol) = g/mol H: 8 (1.01 g/mol) = 8.08 g/mol S: g/mol = g/mol O: 3 (16.00 g/mol) = g/mol Molar mass of (NH 4 ) 2 SO 3 = g/mol

Additional Practice #2 Calculate the molar mass of aluminum sulfate, Al 2 (SO 4 ) 3. Add the atomic masses of each element together: Al: 2 (26.98 g/mol) = g/mol S: 3 (32.07 g/mol) = g/mol O: 12 (16.00 g/mol) = g/mol Molar mass of Al 2 (SO 4 ) 3 = g/mol

In-Class Assignment/Homework You Must SHOW WORK!  Practice pg. 236: #1,2  Practice pg : # 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