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

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  1. Monday, April 7th: “A” DayTuesday, April 8th: “B” DayAgenda • 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”

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

  3. 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…

  4. 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?

  5. Most Elements are a Mixture of Isotopes • Remember: isotopesare 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!

  6. Rules for Calculating Average Atomic Mass • Change each percentage to a decimal by dividing by 100. (move the decimal point 2 places to the left and take away the % sign) • Multiply each decimal by the atomic mass that goes with it. • Add the atomic masses together. That’s it!

  7. 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%

  8. 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 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

  9. 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

  10. 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

  11. 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 • Add the atomic masses together: 0.0117 amu + 1.69 amu + 236.33 amu =238.03 amu

  12. 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.

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

  14. 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 KNO3 shows a ratio of one K+cation to one NO3- anion.

  15. 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

  16. 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/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(NO3)2 = 261.35 g/mol

  17. Additional Practice #1 Calculate 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/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 (NH4)2SO3 = 116.17 g/mol

  18. Additional Practice #2 Calculate 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/mol S: 3 (32.07 g/mol) = 96.21 g/mol O: 12 (16.00 g/mol) = 192.00 g/mol Molar mass of Al2(SO4)3 = 342.17 g/mol

  19. In-Class Assignment/HomeworkYou 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