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Copyright©2000 by Houghton Mifflin Company. All rights reserved. 2 Atomic Masses Elements occur in nature as mixtures of isotopes Carbon =98.89% 12 C 1.11% 13 C <0.01% 14 C Carbon atomic mass = 12.01 amu

Atomic Mass Atoms are so small, it is difficult to discuss how much they weigh in grams Use atomic mass units. an atomic mass unit (amu) is one twelfth the mass of a carbon-12 atom This gives us a basis for comparison The decimal numbers on the table are atomic masses in amu

They are not whole numbers Because they are based on averages of atoms and of isotopes. can figure out the average atomic mass from the mass of the isotopes and their relative abundance. add up the percent as decimals times the masses of the isotopes.

The Mole The mole is a number a very large number, but still, just a number 6.022 x 10 23 of anything is a mole a large dozen

Copyright©2000 by Houghton Mifflin Company. All rights reserved. 6 The Mole The number equal to the number of carbon atoms in exactly 12 grams of pure 12 C. 1 mole of anything = 6.022  10 23 units of that thing

Copyright©2000 by Houghton Mifflin Company. All rights reserved. 8 Molar Mass A substance’s molar mass (molecular weight) is the mass in grams of one mole of the compound. CO 2 = 44.01 grams per mole

Copyright©2000 by Houghton Mifflin Company. All rights reserved. 10 Formulas molecular formula = (empirical formula) n [n = integer] molecular formula = C 6 H 6 = (CH) 6 empirical formula = CH

Copyright©2000 by Houghton Mifflin Company. All rights reserved. 11 Empirical Formula Determination 1.Base calculation on 100 grams of compound. 2.Determine moles of each element in 100 grams of compound. 3.Divide each value of moles by the smallest of the values. 4.Multiply each number by an integer to obtain all whole numbers.

Working backwards From percent composition, you can determine the empirical formula. Empirical Formula the lowest ratio of atoms in a molecule Based on mole ratios A sample is 63.68% C, 9.80 %H, 12.38%N, and 14.14%O what is its empirical formula. (question #70; p.119_

Example: Question 79, p.119 Cumene is a compound containing only carbon and hydrogen that is used in the production of acetone and phenol. Combustion of 47.6 mg cumene produces some CO 2 and 42.8 mg water. The molar mass of cumene is between 115 and 125 g/mol. What is the empirical formula and the molecular formula of cumene?

Pure O 2 in CO 2 is absorbed H 2 O is absorbed Sample is burned completely to form CO 2 and H 2 O

Empirical To Molecular Formulas Empirical is lowest ratio Molecular is actual molecule Need Molar mass Ratio of empirical to molar mass will tell you the molecular formula Must be a whole number because...

Chemical Equations Are sentences. Describe what happens in a chemical reaction. Reactants  Products Equations should be balanced Have the same number of each kind of atoms on both sides because...

Copyright©2000 by Houghton Mifflin Company. All rights reserved. 18 Chemical Equation A representation of a chemical reaction: C 2 H 5 OH + 3O 2  2CO 2 + 3H 2 O reactants products

Abbreviations (s),  for product) (g),  for product) (aq) heat  catalyst

Meaning A balanced equation can be used to describe a reaction in molecules and atoms. Not grams. Chemical reactions happen molecules at a time or dozens of molecules at a time or moles of molecules.

Copyright©2000 by Houghton Mifflin Company. All rights reserved. 21 Chemical Equation C 2 H 5 OH + 3O 2  2CO 2 + 3H 2 O The equation is balanced. 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water

Copyright©2000 by Houghton Mifflin Company. All rights reserved. 22 Calculating Masses of Reactants and Products 1.Balance the equation. 2.Convert mass to moles. 3.Set up mole ratios. 4.Use mole ratios to calculate moles of desired substituent. 5.Convert moles to grams, if necessary.

Limiting Reagent Reactant that determines the amount of product formed. The one you run out of first. Makes the least product. Book shows you a ratio method. It works. So does mine

Limiting reagent To determine the limiting reagent requires that you do two stoichiometry problems. Figure out how much product each reactant makes. The one that makes the least is the limiting reagent.

Copyright©2000 by Houghton Mifflin Company. All rights reserved. 26 Solving a Limiting Reactant Problem 1.Balance the equation. 2.Convert masses to moles. 3.Determine which reactant is limiting. 4.Use moles of limiting reactant and mole ratios to find moles of desired product. 5.Convert from moles to grams.

Percent Yield % yield = Actual x 100% Theoretical % yield = what you got x 100% what you could have got Copyright©2000 by Houghton Mifflin Company. All rights reserved. 27