2009 Chemistry I: Modern Chemistry Holt, Rinehart, & Winston Chapter 9 – Stoichiometry, pp 299-315 Vocabulary 1 actual yield; 2 composition stoichiometry; 3 excess reactant; 4 limiting reactant; 5 mole ratio; 6 percentage yield; 7 reaction stoichiometry; 8 theoretical yield 1. The measured amount of product obtained from a reaction 2. Quantification dealing with the mass relationships of elements in compounds 3. The substance that is not used up completely in a chemical reaction 4. The reactant limiting the amount of product that can be formed in a chemical reaction 5. Conversion factor relating the amounts n moles of any 2 substances in a chemical reaction 6. The ratio of the actual yield to the theoretical yield, multiplied by 100% 7. Quantification dealing with mass relationships between reactants & products in chemical reactions 8. The maximum amount of product that can be produced from a given amount of reactant Main Ideas I. Reaction Stoichiometry, pp 299-303 A. Use known quantities to determine unknown values for balanced equations 1. Find the actual # of moles of 1 reactant based on given # moles of another reactant a. Use mole ratio of the reactants {coefficients in balanced equation} b. (Reactant you want to find/ Reactant you know)(Given # moles of known reactant)
For example: 4 Al + 3 O2 2 Al2O3 If you have 3 moles of Al, how many moles of O2 are needed?
(3 moles O2 / 4 moles Al)(3 moles Al) = 2.25 moles O2 2. Find the actual # moles of product based on given # moles of one reactant a. Use mole ratio of the reactants to products{coefficients in balanced equation} b. (Product you want to find/ Reactant you know)(Given # moles of known reactant)
For example: 4 Al + 3 O2 2 Al2O3 If you have 3 moles of Al, how many moles of Al2O3 are made?
(2 moles Al2O3 / 4 moles Al)(3 moles Al) = 1.5 moles Al2O3 3. Find the actual # moles of reactant(s) if you are given the moles of a product. a. Use mole ratio of the product to reactant {coefficients in balanced equation} b. (Reactant you want to find/ Product you know)(Given # moles of known product)
For example: 4 Al + 3 O2 2 Al2O3 If you have 3 moles Al2O3, how many moles of O2 are needed?
(3 moles O2/ 2 moles Al2O3)(3 moles Al2O3 ) = 4.5 moles O2 4. Find the # grams if you given the amount in moles a. Calculate the molar mass of the reactants or products in question; use periodic table. DON’T USE COEFFICIENTS FOR THESE CALCULATIONS!!!
For example: Al2O3: 2(27 grams/mole) + 3(16 grams/mole) = 102 grams/mole; Al: 27 g/m;
O2: 2(16g/m) = 32 g/m b. Use mole ratios {What you want over What you know times What is given} as above c. Convert from actual # moles of unknown to grams by multiplying moles by gram/mole For example: Using example #2 above, if I wanted to know # grams of the product, I’d calculate
(1.5 moles Al2O3)(102 g/m) = 153 grams Al2O3 5. Find the # moles if given the mass. a. Convert mass into grams using SI conversions. Milligrams: mg/ 1000 = grams; kilograms: kg x 1000 = grams b. Determine the molar mass of reagents in question as above in 4a. c. Divide the grams given by reagent’s molar mass: grams / grams/mole = moles II. Ideal Stoichiometry Calculations, pp 304-311 A. Conversions of quantities in moles 1. (Amount of given substance in moles)(conversion factor) = unknown quantities a. The conversion factor is again your mole ratio of unknown over known.
For example: Given CO2 + 2LiOH Li2CO3 + H2O; How many moles of LiOH are needed to react
with 20 moles of CO2?
(2 mole LiOH/ 1 mole CO2)(20 mol CO2) = 40 mole LiOH B. Conversions of amounts in moles to mass 1. (# moles of given substance)(mole ratio)( unknown’s molar mass) = grams of unknown a. As always, the mole ratio is what you want to find over what you know already. For example: Using equation above, if I wanted to find the mass of LiOH needed I’d calculate
(20 mol CO2)(2 mole LiOH/ 1 mole CO2) (24 g/m LiOH) = 960 grams LiOH C. Conversions of amounts of mass to moles 1. (# grams given substance/molar mass of given)(mole ratio) = mole of unknown
For example: Given 4NH3 + 5O2 4NO + 6H2O; if I wanted to find the moles of NO given 824 g NH3
with unlimited O2, I’d calculate (824 g NH3/17 grams/mol)(4 mole NO/ 4 mol NH3)= 48.5 m NO D. Mass to mass calculations 1. Only moles are shown in a balanced equation, therefore, convert given grams into moles. 2. Use mole ratios to determine the ACTUAL amount of moles in question. 3. Convert back from actual # moles into grams using molar mass as shown above. III. Limiting Reactants and Percentage Yield A. Usually in real-world applications, one of the reactants is not as plentiful as the others, thus limiting the amount of product that can be generated. Use mole ratios to determine which reactant is limiting and which ones are in excess.
For example: Given SiO2 + 4HF SiF4 + 2H2O in which 6 m of HF is added to 4.5 m SiO2 (6 m HF)(1 m SiO2/ 4 m HF) = 1.5 m SiO2 … so you have 3 m SiO2 in excess. OR… (4.5 m SiO2)(4 m HF/1 m SiO2) = 18 m HF needed … so HF is limiting reactant.
B. Percentage Yield = (Actual Yield/ Theoretical Yield) x 100% 1. Actual Yield is what you experimentally recovered. 2. Theoretical Yield is what the balanced equation said you should get. 3. You can never get more in real life than the theoretical yield; in fact, you won’t even get that! a. In order to get 100% theoretical yield, every molecule must hit and bind and stay bound in the desired formula molecule, all 6.022 x 1023 atoms/mole. No way!