Chapter 9 Stoichiometry Section 9.1 Introduction to Stoichiometry Standard.e.: Students know how to calculate the masses of reactant and products in a chemical reaction from the mass of one of the reactants or products and the relevant atomic masses. Language bjective: We will calculate the moles of a reactant or product using the moles of the reactant or product. Content bjective: I will show through writing how to calculate the moles of a reactant or product using the moles of the reactant or product. Types of Stoichiometry Problems Given is in moles and unknown is in moles. o moles of given moles of unknown Given is in moles and unknown is in mass (grams). o moles of given moles of unknown mass of unknown Given is in mass and unknown is in moles. o mass of given moles of given moles of unknown Given is in mass and unknown is in mass. o mass of given moles of given moles of unknown mass of unknown Writing and Using Mole Ratios Mole Ratio: a conversion factor derived from the coefficients of a balanced chemical equation interpreted in terms of moles. o In chemical calculations, mole ratios are used to convert between: Moles of reactants Moles of reactants and moles of products Moles of products N (g) + H (g) NH (g) o Three mole ratios can be derived from the balanced equation above: o 1 mol N mol NH mol H mol H 1 mol N mol NH Example: 4 Li + Li 4 mol Li 4 mol Li 1 mol 1 mol mol Li mol Li **continued on next page**
Section 9. Ideal Stoichiometric Calculations Mole-Mole Calculations How many moles of ammonia (NH ) are produced when 0.60 mol of N reacts with H o Step 1: Write out equation and balance it o Step : Determine mole ratio of known (N ) to unknown (NH ). o Step : Multiply given amount of known by the correct mole ratio that will cancel out the known units, leaving you the unknown units N + H NH 1 mol N mol NH 0.60 mol N mol NH 1. mol NH 1 1 mol N Example: How many moles of Chlorine are produced when. moles sodium chloride decomposes? NaCl Na + Cl mol NaCl 1 mol Cl. mol NaCl 1 1 mol Cl 1.6 mol Cl mol NaCl Homework: page 01 # and page 11 #1 a, b (balance, and then a and b each have answers)
Section 9. Ideal Stoichiometric Calculations Standard.e.: Students know how to calculate the masses of reactant and products in a chemical reaction from the mass of one of the reactants or products and the relevant atomic masses. Language bjective: We will calculate the moles or mass of a reactant or product using the moles or mass of the reactant or product. Content bjective: I will show through writing how to calculate the moles or mass of a reactant or product using the moles or mass of the reactant or product. Mole-Mass Calculations How many grams of aluminum oxide (Al ) are produced when 0.50 mol of Al reacts with o Step 1: Write out equation and balance it o Step : Multiply by the mole ratio to get from known to unknown moles. o Step : Multiply the moles of unknown by the molar mass of the unknown to get grams. 4Al + Al 4 mol Al Molar Mass Al = 101.96 g/mol mol Al 0.50 mol Al 1 mol Al 4 mol Al 101.96 g Al 1 mol Al 5 g Al Example: How many grams of Copper are needed to react with Sulfur to form 5.70 moles of copper sulfide? Cu + S Cu S 5.70 mol Cu 1 S mol Cu 6.55 g Cu 74 g Cu 1 mol CuS 1 mol Cu Mass-Mole Calculations Calculate the number of moles of H produced by the reaction of 5.40 g of with an excess of H. *excess means that there is enough to react. o Step 1: Write out equation and balance it. o Step : Multiply the given amount by the molar mass of the known to get moles. o Step : Multiply by the mole ratio to get from known to unknown moles. H + H 5.40 g 1 1 mol.00 g mol H 0.8 mol 1 mol H
Example: How many moles of hydrogen are needed to react with nitrogen to form 10.5 grams of ammonia (NH )? N + H NH 10.5 g NH 1 1 mol NH 17.04 g NH mol H mol NH 0.94 mol H Example: How many moles of Lithium are formed by the decomposition of.40 grams of Lithium xide? 4Li + Li.40 g Li 1 1 mol Li 4 mol Li 0.8 mol 9.88 g Li mol Li Li Homework: page 11 #-
Section 9. Ideal Stoichiometric Calculations Standard.e.: Students know how to calculate the masses of reactant and products in a chemical reaction from the mass of one of the reactants or products and the relevant atomic masses. Language bjective: We will calculate the mass of a reactant or product using the mass of the reactant or product. Content bjective: I will show through writing how to calculate the mass of a reactant or product using the mass of the reactant or product. Mass-Mass Calculations Calculate the number of grams of C produced by the reaction of.5 g of C with an excess of. o Step 1: Write out equation and balance it. o Step : Multiply the given amount by the molar mass of the known to get moles. o Step : Multiply by the mole ratio to get from known to unknown moles. o Step 4: Multiply the moles of unknown by the molar mass of the unknown to get grams. C + C.5 g C 1 1 mol C mol C 8.01 g C 7.58 g C 1.01 g C mol C 1 mol C Example: How many grams of xygen are needed to make 0.0 grams of N? N + N 0.0 g N 1 1 mol N 46.01 g N 1 mol mol N.00 g 1 mol 6.96 g Example: How many grams of S are produced when 55. grams of S reacts with an excess of. S + S 55. g 1 S 1 mol S 64.07 g S mol S mol S 80.07 g S 1 mol S 69.1 g S Homework: page 11 #4 and practice problem #1 on page 11
Section 9. Limiting Reactants and Percentage Yield Standard..: Students know how to calculate percent yield in a chemical reaction. Language bjective: We will find the limiting reactant in a chemical reaction. Content bjective: I will show through writing how to find the limiting reactant in a chemical reaction. Limiting reactant: the substance that determines the amount of product that can be formed by a reaction. Excess reactant: the substance that is not completely used up in a reaction. Keep in mind that the reactant that is present in the smaller amount by mass or volume is not necessarily the limiting reactant. Determining Limiting Reactant What is the limiting reactant when 80.0g Cu reacts with 5.0g S? Cu(s) + S(s) Cu S(s) o Using the grams of reactant given, determine the moles of reactant. o Using the balanced chemical equation, determine the mole ratio. o Multiply the moles of one reactant by the mole ratio to get the needed amount of the other reactant and compare to the amount you have. 80.0 g Cu 1 5.0 g S 1 1.6 mol Cu 1 0.780 mol S 1 1 mol Cu 1.6 mol Cu 6.55 g Cu 1 mol S 0.780 mol S.07 g S 1 mol S 0.60 mol S mol Cu mol Cu 1.56 mol Cu 1 mol S Limiting reactant is Copper *When you need more than you have, that is your limiting reactant. *When you have more than you need, that is your excess reactant. Example: If 4.00 mol C H 4 is reacted with 9.50 mol, identify the limiting reactant and calculate the moles of excess reactant remaining. C H 4 (g) + (g) C (g) + H (g) *Since we already have moles, we know what we have, we just have to find what we need. 4.00 mol C 1 H 4 mol 1 mol CH Limiting Reactant is 9.50 mol 1 1 mol CH mol 4 4 1.0 mol.17 mol C H 4 Have-Need= Remaining 4.00-.17 = 0.8 moles
Using a Limiting Reactant to Find Quantity of a Product What is the maximum number of grams of Cu S that can be formed when 45.0g Cu reacts with 4.0g S? Cu(s) + S(s) Cu S(s) o Find the limiting reactant and then use the moles of limiting reactant to calculate moles and then grams of Cu S. 45.0 g Cu 1 4.0 g S 1 0.708 mol Cu 1 0.748 mol S 1 1 mol Cu 0.708 mol Cu 6.55 g Cu 1 mol S 0.748 mol S.07 g S 1 mol S 0.54 mol S mol Cu mol Cu 1.50 mol Cu 1 mol S Limiting Reactant is Copper, so we use the given amount of Copper to find grams of Cu S. We already have converted to moles in the first part, so start with moles. 0.708 mol Cu 1 1 mol CuS 159.17 g CuS 56. g Cu mol Cu 1 mol CuS Example Calculate the moles of I produced when 80.0g I 5 reacts with 8.0g C. I 5 (g) + 5C(g) 5C (g) + I (g) 80.0 g I 1 8.0 g C 1 5 0.40 mol I 1 1 mol I 5.80 g I 5 0.40 mol I 1 mol C 1.00 mol C 8.01 g C 1.00 mol C 1 5 5 mol C 1 mol I 5 1. mol C 1 mol I 5 0.00 mol I 5 mol C 5 5 S Limiting Reactant is C, so we use the given amount of C to find moles of I. We already have converted to moles in the first part, so start with moles. 1.00 mol C 1 Homework: page 1 #- 1 mol I 0.00 mol I 5 mol C
Section 9. Limiting Reactants and Percentage Yield Standard..: Students know how to calculate percent yield in a chemical reaction. Language bjective: We will calculate the percent yield in a chemical reaction. Content bjective: I will show through writing how to calculate the percent yield in a chemical reaction. Percent Yield Theoretical Yield: the maximum amount of product that could be formed from given amounts of reactants. Actual Yield: the amount of product that actually forms when the reaction is carried out in the laboratory. Percent Yield: the ratio of the actual yield to the theoretical yield expressed as a percent. Percent Yield = actual yield x 100% theoretical yield The percent yield is a measure of the efficiency of a reaction carried out in the laboratory. Factors that can cause the actual yield to be less than the theoretical yield: Reactions do not always go to completion. Side reactions may occur. Poor laboratory technique. Calculating Theoretical Yield What is the theoretical yield in grams of Ca if 4.8g CaC is heated? CaC (s) Ca(s) + C (g) o Calculate the theoretical yield by converting the mass of the reactant to the mass of the product, using molar mass and mole ratios. 4.8 g CaC 1 1 mol CaC 100.09 g CaC 1 mol Ca 1 mol CaC 56.08 g Ca 1.9 g Ca 1 mol Ca Example What is the theoretical yield of AlCl in moles when.00 mol Al reacts with an excess of Cl? Al + Cl AlCl.00 mol Al 1 mol AlCl mol Al.00 mol AlCl
Calculating Percent Yield What is the percent yield if 1.1g Ca is actually produced when 4.8g CaC is heated? CaC (s) Ca(s) + C (g) o Find the theoretical yield (1.9g Ca) and plug both the theoretical and actual yields into the percent yield equation. 1.1 g Ca Percent Yield 100% 94.% 1.9 g Ca Example What is the percent yield if.85mol AlCl is actually produced when.00mol Al reacts with an excess of Cl? Al + Cl AlCl.85 mol AlCl Percent Yield 1.00 mol AlCl 100% 95.0% Another Way To Look At It Think of percent yield as a grade on a test. The Theoretical Yield is the number of points possible: 70 pts The Actual Yield is the number of points you earned: 64 pts Percent yield = 64/70 x 100% = 91% Homework: page 18 #1 &