# AP Chemistry Stoichiometry HW: 1 5 11 13 17 19 25 35 39 49 57 59 63 73 77 89 101.

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AP Chemistry Stoichiometry HW: 1 5 11 13 17 19 25 35 39 49 57 59 63 73 77 89 101

3.1 – Chemical Equations + = Reacts with -> = Yields / Produces Reactants = Substances on the left side of the arrow Products = Substances on the right side of the arrow

Balancing Reactions For now the old method is OK. We’ll learn more about balancing more complicated (redox) reactions later 1.Write correct formulas 2.Balance elements / ions that only appear once on each side 3.Balance others 4.Balance O and H 5.Check

3.1 – Balancing Equations Most can be done by inspection for now! Ex – methane + oxygen -> carbon dioxide + water

Reaction Example CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (g)

3.1 – Balancing Equations Ex – sodium + water -> sodium hydroxide + hydrogen gas

© 2009, Prentice-Hall, Inc. Reaction Types

© 2009, Prentice-Hall, Inc. Combination Reactions (AKA Synthesis) Examples: –2 Mg (s) + O 2 (g)  2 MgO (s) –N 2 (g) + 3 H 2 (g)  2 NH 3 (g) –C 3 H 6 (g) + Br 2 (l)  C 3 H 6 Br 2 (l) In this type of reaction two or more substances react to form one product.

© 2009, Prentice- Hall, Inc. In a decomposition one substance breaks down into two or more substances. Decomposition Reactions Examples: –CaCO 3 (s)  CaO (s) + CO 2 (g) –2 KClO 3 (s)  2 KCl (s) + 3O 2 (g) –2 NaN 3 (s)  2 Na (s) + 3 N 2 (g)

© 2009, Prentice-Hall, Inc. Combustion Reactions Examples: –CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (g) –C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (g) These are generally rapid reactions that produce a flame. Most often involve hydrocarbons reacting with oxygen in the air.

AP Question 4 -Changed for 2008 -Now want BALANCED NET-IONIC -Ask you an additional question -No “choice” in the questions **See additional packet for practice Be sure to know: -Single displacement (metal and nonmetal) -Double displacement -Combustion -Acid / Base (we’ll cover more later)

AP Question 4

3.3 – Formula Weights Atomic mass – Mass of an atom in amu (atomic mass unit) 1 amu = 1/12 of the mass of a Carbon-12 atom Mole = Amount of a substance based on particles Avogadro’s Number = # of particles in 1 mole; Based on 12 grams of Carbon–12 = 6.02 x 10 23 Molar Mass = Mass needed to obtain Avogadro’s Number of particles for any substance.

© 2009, Prentice-Hall, Inc. 3.3 – Atomic Weight / Formula Weight / Molecular Weight A formula weight is the sum of the atomic weights for the atoms in a chemical formula. So, the formula weight of calcium chloride, CaCl 2, would be Ca: 1(40.1 amu) + Cl: 2(35.5 amu) 111.1 amu Formula weights are generally reported for ionic compounds.

© 2009, Prentice-Hall, Inc. Molecular Weight (MW) A molecular weight is the sum of the atomic weights of the atoms in a molecule. For the molecule ethane, C 2 H 6, the molecular weight would be C: 2(12.0 amu) 30.0 amu + H: 6(1.0 amu)

© 2009, Prentice-Hall, Inc. Percent Composition One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation: % element = (number of atoms)(atomic weight) (FW of the compound) x 100

© 2009, Prentice- Hall, Inc. 3.4 - Avogadro’s Number 6.02 x 10 23 particles 1 mole of 12 C has a mass of 12 g.

© 2009, Prentice-Hall, Inc. Molar Mass By definition, a molar mass is the mass of 1 mol of a substance (i.e., g/mol). –The molar mass of an element is the mass number for the element that we find on the periodic table. –The formula weight (in amu’s) will be the same number as the molar mass (in g/mol).

© 2009, Prentice- Hall, Inc. Using Moles Moles provide a bridge from the molecular scale to the real-world scale.

Practice Problems

© 2009, Prentice- Hall, Inc. Mole Relationships One mole of atoms, ions, or molecules contains Avogadro’s number of those particles. One mole of molecules or formula units contains Avogadro’s number times the number of atoms or ions of each element in the compound.

Practice Problems

© 2009, Prentice- Hall, Inc. 3.5 - Calculating Empirical Formulas One can calculate the empirical formula from the percent composition.

© 2009, Prentice- Hall, Inc. Combustion Analysis Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this. –C is determined from the mass of CO 2 produced. –H is determined from the mass of H 2 O produced. –O is determined by difference after the C and H have been determined.

© 2009, Prentice- Hall, Inc. Elemental Analyses Compounds containing other elements are analyzed using methods analogous to those used for C, H and O.

© 2009, Prentice-Hall, Inc. Calculating Empirical Formulas The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA.

© 2009, Prentice-Hall, Inc. Calculating Empirical Formulas C 7 H 7 NO 2

3.5 – Empirical Formulas from Lab Data Determining Molecular Formula -Need to know approximate Molar Mass -Need to know Empirical Formula -Molecular Formula is a multiple of the Empirical Formula

Practice Problem Caproic acid, which is responsible for the foul odor of dirty socks, is composed of C, H and O atoms. Combustion of a 0.225 g sample of this compound produces 0.5112 g carbon dioxide and 0.209 g water. Water is the empirical formula of caproic acid? If the molar mass of the acid is 116 g/mol, what is the molecular formula?

3.6 – Quantitative Information from Balanced Equations -Stoichiometry – Coefficients (numbers in front of substances in a balanced equation) provide relative numbers of moles in the reaction Examples:

© 2009, Prentice- Hall, Inc. Stoichiometric Calculations Starting with the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant).

© 2009, Prentice- Hall, Inc. Stoichiometric Calculations Starting with 1.00 g of C 6 H 12 O 6 … calculate the mass of water that would be produced. C 6 H 12 O 6 + 6 O 2  6 CO 2 + 6 H 2 O

© 2009, Prentice- Hall, Inc. Limiting Reactants You can make cookies until you run out of one of the ingredients. Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat).

© 2009, Prentice- Hall, Inc. How Many Cookies Can I Make? In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make.

© 2009, Prentice- Hall, Inc. Limiting Reactants The limiting reactant is the reactant present in the smallest stoichiometric amount. –In other words, it’s the reactant you’ll run out of first (in this case, the H 2 ).

© 2009, Prentice- Hall, Inc. Limiting Reactants In the example below, the O 2 would be the excess reagent.

Limiting Reactants Best method is to do separate stoichiometry calculations for each reactant. Then compare the answers. The smallest answer is the amount of product that will be produced.

Examples:

© 2009, Prentice-Hall, Inc. Theoretical Yield The theoretical yield is the maximum amount of product that can be made. –In other words it’s the amount of product possible as calculated through the stoichiometry problem. This is different from the actual yield, which is the amount one actually produces and measures.

© 2009, Prentice-Hall, Inc. Percent Yield One finds the percent yield by comparing the amount actually obtained (actual yield) to the amount it was possible to make (theoretical yield). Actual Yield Theoretical Yield Percent Yield =x 100

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