3. 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. 239-240: #1-4 (1 & 2: a,b,c,d only)  Homework: Concept Review: “Relative Atomic Mass and Chemical Formulas”

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

5. 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…

6. 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?

7. 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!

8. 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!

9. Calculating Average Atomic Mass Sample 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%

10. 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% =.3083 2. Multiply each decimal by the atomic mass that goes with it: Cu-63 = (.6917) (62.94 amu) = 43.54 amu Cu-65 = (.3083) (64.93 amu) = 20.01 amu 3. Add the atomic masses together: 43.54 amu + 20.01 amu = 63.55 amu

11. Additional Example #1 Chlorine 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 amu Cl-37 = (.2420) (36.996 amu) = 8.953amu 3. Add the atomic masses together: 26.51 amu + 8.953 amu = 35.46 amu

12. Additional Example #2 U-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 % = 0.00720 U-238 = 99.275% =.99275

13. Additional Example #2, cont. 2. Multiply each decimal by the atomic mass that goes with it: U-234 = (0.0000500) (234.041amu) = 0.0117 amu U-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

14. 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.

15. Formulas Express Composition  Although any sample of compound has many atoms and ions, the formula gives a ratio of those atoms or ions.

16. 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.

17. 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

18. 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: = 137.33 g/mol N: 2 (14.01 g/mol) = 28.02 g/mol O: 6 (16.00 g/mol) = 96.00 g/mol Molar mass of Ba(NO 3 ) 2 = 261.35 g/mol

19. 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) = 28.02 g/mol H: 8 (1.01 g/mol) = 8.08 g/mol S: 32.07 g/mol = 32.07 g/mol O: 3 (16.00 g/mol) = 48.00 g/mol Molar mass of (NH 4 ) 2 SO 3 = 116.17 g/mol

20. 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) = 53.96 g/mol S: 3 (32.07 g/mol) = 96.21 g/mol O: 12 (16.00 g/mol) = 192.00 g/mol Molar mass of Al 2 (SO 4 ) 3 = 342.17 g/mol

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