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

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!

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 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

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

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. 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

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

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

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…