2 Atomic MassAtoms are so small, it is difficult to discuss how much they weigh in grams.Use atomic mass units.an atomic mass unit (amu) is 1/12th the mass of a carbon-12 atom.This gives us a basis for comparison.The decimal numbers on the Periodic Table are atomic masses in amu.
3 They are not whole numbers based on averages of atoms & isotopes.figure average atomic mass from mass of the isotopes & relative abundance.add up % as decimals x isotope masses.
4 ExampleThere are two isotopes of carbon 12C with a mass of amu(98.892%), and 13C with a mass of amu (1.108%)12C: x =13C : x =Total atomic mass = amu for C
5 The Mole The mole is a very large number. 6.02 x 1023 of anything is a mole.The number of atoms in exactly 12 grams of carbon-12.Makes the numbers on the table the mass of the average atom.
6 Molar mass mass of 1 mole of a substance. often called molecular weight or gram formula mass.To determine the molar mass of an element, look on the table.To determine the molar mass of a compound, add up the molar masses of the elements that make it up.
7 Find the molar mass of CH4 12.01 + 1.008(4) = 16.04 g Mg3P (3) (2) = gCa(NO3) (2) (6) = gAl2(Cr2O7) (2)+52.00(6)+16.00(21) = gCaSO4 · 2H2O (4) (4) (2) = g
8 Percent Composition Percent of each element in a compound. Find the mass of each element, divide by the total mass, multiply by a 100.Easiest if you use a mole of the compound.
9 Percent Composition Ex: Find the % composition of CH4 Molar mass of = CH4 = gramsC = grams x 100 = 74.88%gramsH = grams x 100 = 25.12%
10 Working backwardsFrom percent composition, you can determine the empirical formula.Empirical Formula the lowest ratio of atoms in a molecule.Based on mole ratios.
11 Empirical FormulaA sample is 71.65% Cl, 24.27%C, and 4.07%H. Find empirical & molecular formulas? (molar mass = g/mol)Step 1: convert mass % to mass (g).Cl: g Cl x 1 mol Cl = mol Cl.35.45 g ClC: g C x 1 mol C = mol C12.01 g C
12 Empirical Formula H: 4.07 g H x 1 mol H = 4.04 mol H 1.008 g H Step 2: Divide each mole value by the smallest number of moles present.
13 Empirical Formula 2.021 mol Cl = 1 mol Cl 2.021 mol 2.021 mol C = 1 mol C4.04 mol H = 1.99 mol H = 2 mol HEmpirical formula = ClCH2
14 Molecular Formula Empirical Formula = ClCH2 To find molecular formula, compare empirical formula with molar mass:Molar mass = g/mol = 2Empirical mass g/molMolecular formula = (ClCH2)2 = Cl2C2H4
15 Chemical Equations Are sentences. Describe what happens in a chemical reaction.Reactants ® ProductsEquations should be balanced.Have the same number of each kind of atoms on both sides.
16 Balancing equations CH4 + O2 ® CO2 + H2O 1 C 1 4 H 2 2 O 3 Reactants Products1C14H22O3
17 Balancing equations CH4 + O2 ® CO2 + 2 H2O 1 C 1 4 H 2 4 2 O 3 ReactantsProducts1C14H242O3
18 Balancing equations CH4 + O2 ® CO2 + 2 H2O 1 C 1 4 H 2 4 4 2 O 3 ReactantsProducts1C14H2442O3
19 Balancing equations CH4 + 2O2 ® CO2 + 2 H2O 1 C 1 4 H 2 4 4 4 2 O 3 ReactantsProducts1C14H24442O3
20 Abbreviations (s) , ¯ (for product) (g) , (for product) (aq) heat D catalyst
21 MeaningA 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.
22 StoichiometryGiven an amount of either starting material or product, determining the other quantities.use conversion factors from:molar mass (g - mole)balanced equation (mole - mole)
23 Limiting ReagentReactant that determines the amount of product formed.The one you run out of first.Makes the least product.Book shows you a ratio method.
24 Limiting reagentTo 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.
25 Excess Reagent The reactant you don’t run out of. The amount of stuff you make is the yield.The theoretical yield is the amount you would make if everything went perfect.The actual yield is what you make in the lab.
26 Percent Yield % yield = Actual x 100% Theoretical % yield = what you got x 100% what you could have got