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Wednesday, April 13 th : “A” Day Thursday, April 14 th : “B” Day Agenda Section 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

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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?

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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!

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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%

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Calculating Average Atomic Mass Sample 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% = Multiply each decimal by it’s atomic mass: Cu-63 = (.6917) (62.94 amu) = amu Cu-65 = (.3083) (64.93) = amu 3. Average atomic mass is found by adding the two atomic masses together amu amu = amu

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Additional Example Calculate the average atomic mass for gallium if 60.00% of its atoms have a mass of amu and 40.00% have a mass of amu. 1. Change each percentage to a decimal: 60.00% / 100% = % / 100% = Multiply each decimal by it’s atomic mass: (.6000) ( amu) = amu (.4000) ( amu) = amu 3.Average atomic mass is found by adding the two atomic masses together: amu amu = amu

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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.

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

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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.

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

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Calculating Molar Mass of Compounds Sample Problem F, continued. 2.To find the molar mass of Ba(NO 3 ) 2, add the atomic masses of each element together. Barium, Ba: g/mol = g/mol Nitrogen, N: 2 (14.01 g/mol) = g/mol Oxygen, O: 6 ( g/mol) = g/mol Molar mass of Ba(NO 3 ) 2 = g/mol

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

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In-Class Assignment/Homework You 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…

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