Stoichiometry Chapter 12

12.1: Arithmetic of Equations A balanced equation is the recipe for a reaction. Chemists use balanced equations as a basis to calculate how much reactant is needed or product is formed The calculation of these quantities in chemical reactions is called STOICHIOMETRY.

Interpreting Chemical Equations A balanced chemical equation can be interpreted in terms of many different quantities Number of atoms Molecules Moles Mass Volume

Atoms/ Molecules # of Atoms # of Molecules on the atomic level, a balanced equation indicates the number of and type of atoms that make up both the reactants and products The number of atoms is not changed in the reaction. # of Molecules A balanced equation also establishes a ratio of how molecules interact Look at the formation of ammonia The ratio established is 1:3:2 1 molecule of atmospheric nitrogen reacts with 3 molecules of atmospheric hydrogen to form 2 molecules of ammonia

Moles A balanced equations gives you the number of moles in both the reactants and products We use the coefficients to establish this This is the most important information that an equation can give us. Using this we can calculate exactly how much of each reactant is needed to give a desired product Due to recombination of elements, however, the # of moles in reactants does not always equal the # of moles in the products.

Mass/ Volume Mass Volume Balanced equations always obey conservation of mass laws This means the mass of reactants is always equal to mass of the products no matter is the # of moles changes Volume If standard pressure is assumed, then a balanced equation can also tell you the volume of products produced Remember 1 moles = 22.4 Liters

Mass Conservation Mass and Atoms are always conserved inside of a reaction Volume, Molecules, and # of moles can change This depends on the reaction

Homework Pg. 358 Questions 5-10 Due tomorrow

12.2 Chemical Calculations Writing and Using Mole Ratios Mole ratio – a conversion factor derived from the coefficients of a balanced chemical equation, interpreted in terms of moles. Mole ratios can be used to calculate conversions between moles of products and reactants, just reactants, or just products

Mole to Mole Calculations How many moles of ammonia are produced from 0.60 moles of nitrogen?? 0.60 mol N x 2 mol NH3/1 mole N = ??

Mass to Mass Calculations No balance/scale can measure directly in moles; instead we work in the world of grams so being able to convert from moles to grams or grams to moles is very important The process for converting from grams of reactant to grams of product follows this sequence Grams to moles; moles to moles (using mole ratio from balanced equation); then finally moles to grams of product

Mass to Mass cont. Steps in Solving Mass to Mass problems 1. change mass of given to moles by using molar mass 2. change moles of given to moles of wanted using mole ratio 3. change moles of wanted to grams of wanted using molar mass of wanted.

Mass to Mass cont. How many grams of NH3 are produced when 5.40 grams of hydrogen reacts with an excess of nitrogen Remember our balanced equation: N2 + 3H2  2NH3 Take a shot at solving this follow the steps from the previous slide.

Practice Problems Page 361 in textbook #13 and #14 Try to solve these quickly. 13/14 equation: CaC2 + 2H2O  C2H2 + Ca(OH)2

Other Stoichiometric Calculations We can also interconvert moles and grams into representative particles (molecules/atoms) and into volumes (assuming standard pressure) Think back……. What number did we use to convert to molecules/particles? What number did we use to convert to volume @ STP?

Other Calculations cont. Representative Particles The conversion factor we use is AVOGADRO’S NUMBER 6.02 X 1023 = 1 MOLE Practice #15 Volume The conversion factor used for volume always assumes standard temperature and pressure. 1 mole = 22.4 Liters Practice #17 and #19

Assignment 12. 2 Section Review Pg. 366 Questions 22-24 Will grade tomorrow

12.3 Limiting Reagent In any equation an insufficient quantity of any reactant will limit the amount of product produced Limiting reagent – the reactant that is completely used up in a chemical reaction Excess reagent – reactant that is not completely used up in a chemical reaction There is a little bit of this left over.

Calculations of Limiting Reagent 2Cu + S  Cu2S Given quantities: 80.0 grams Copper 25.0 grams Sulfur Which is limiting? How do we calculate this?

Solution 2Cu + S  Cu2S 80.0 g. Cu x 1mol/63.5gCu = 1.26 mol Cu 25.0 g. S x 1mol/32.1g.S = 0.779 mol S 1.26 mol Cu x 1mol S/ 2mol Cu = 0.630 mol S Copper (Cu) is the limiting reagent Practice Problems #25 and #26 on page 370

Limiting Reagent to find Qty of Product produced Using same equation and quantities from previous problem: 2Cu + S  Cu2S L.R. = 1.26 mol Cu Solution: 1.26 mol Cu x 1mol Cu2S/ 2mol Cu x 159.1 g. Cu2S/ 1mol Cu2S = 100grams Cu2S Try Practice Problems #27 and #28 on pg. 371

Percent Yield Percent yield measures the efficiency of a reaction, when carried out under laboratory conditions. Theoretical vs Actual Yield Theoretical yield – maximum amount of product that can be produced from given amounts of reactants Actual yield – The amount of product produced when the reaction is performed

Percent Yield Percent Yield = Actual yield/ Theoretical yield x 100 Never is larger than 100% Side reactions, impure reactants, and inaccurate measurement cause few, if any, reactions to run to 100% completion.

Example: Pg. 374 Sample 12.9 Practice Problems pg. 374 (29 and 30) Given: mass of CaCO3 = 24.8 g 1 mol of CaCO3 = 100.1 g 1 mol CaO = 56.1 g Unknown: theoretical yield of CaO in grams Equation: CaCO3  CaO + CO2 Practice Problems pg. 374 (29 and 30)

Example 2 Sample 12.10 on pg. 375 Givens: actual yield of CaO = 13.1 grams Theoretical yield of CaO = 13.9 grams Unknown: Percent Yield of CaO Practice Problems: pg. 375 (31 and 32)

Assignment 12.3 Review Pg. 375 Questions 33-35