Chapter 10 Relationships in Chemical Reactions Section 10.1 Conversion Factors from a Chemical Equation Goal 1 The coefficients in a chemical equation give us the conversion factors to get from the number of particles of one substance, grouped into moles, to the number of particles of another substance in a chemical change. a chemical equation, or a reaction for which the equation is known, and the number of moles of one species in the reaction, calculate the number of moles of any other species. A 2 + 3 B 2 2 AB 3 How many moles of oxygen are needed to completely react with 2.34 moles of methane (CH 4 ) in a combustion reaction? Carbon dioxide and steam are the products. CH 4 + 2 O 2 CO 2 + 2 H 2 O PER relationship from equation: PER 2 mol O 2 1
GIVEN: 2.34 mol CH 4 WANTED: mol O 2 1 mol CH /2 mol O PER/PATH: mol CH 4 2 4 mol O 2 Section 10.2 2.34 mol CH 4 2 mol O 2 = 4.68 mol O 2 Mass-Mass Stoichiometry Goal 2 Stoichiometry: The quantitative relationships between the substances involved in a chemical reaction, established by the equation for the reaction a chemical equation, or a reaction for which the equation can be written, and the number of grams or moles of one species in the reaction, find the number of grams or moles of any other species. A stoichiometry problem asks, How much or how many? Prerequisite Skills for Stoichiometry Fundamental Stoichiometry Pattern: Write chemical formulas Ch 6 Calculate molar masses from chemical formulas Use molar masses to change mass to moles and moles to mass Sect 7.4 Sect 7.5 Macroscopic Measurable Macroscopic Measurable Write and balance chemical equations Ch 8 Use the equation to change from moles of one species to moles of another Sect 10.1 2
Mass Mass Stoichiometry Pattern: How to Solve a Stoichiometry Problem: The Stoichiometry Path Mass Molar Mass g PER mol Equation mol PER mol Molar Mass g PER mol Mass Step 1: Change the mass of the given species to moles. Step 2: Change the moles of the given species to moles of the wanted species. Step 3: Change the moles of the wanted species to mass. How many grams of carbon dioxide are produced when 10.0 g of methane, CH 4, is burned? GIVEN: 10.0 g CH 4 WANTED: g CO 2 CH 4 + 2 O 2 CO 2 + 2 H 2 O PER/PATH: g CH 4 /mol CH 4 mol CH 4 / /mol CO 2 g CO2 PER/PATH: g CH 4 /mol CH 4 10.0 g CH 4 mol CH 4 / mol CO2 /mol CO 2 g CO2 mol CO 2 = 27.4 g CO 2 Goal 3 Section 9.3 Percent Yield two of the following, or information from which two of the following may be determined, calculate the third: theoretical yield, actual yield, percent yield. 3
The actual yield of a chemical reaction is usually less than the theoretical yield predicted by a stoichiometry calculation because: reactants may be impure the reaction may not go to completion other reactions may occur Percentage yield expresses the ratio of actual yield to theoretical yield: % yield = actual yield 100 theoretical yield Determine the percentage yield if 6.97 grams of ammonia is produced from the reaction of 6.22 grams of nitrogen with excess hydrogen. STEP 1: Calculate the theoretical yield STEP 2: Use the given actual yield and the calculated theoretical yield to find the percentage yield GIVEN: 6.22 g N 2 N + H 2 3 NH 2 2 3 PER/PATH: WANTED: g NH 3 g N 2 /mol N 2 mol N 2 /1 mol N 2 mol NH3 /mol NH 3 g NH 3 PER/PATH: g N 2 /mol N 2 mol N 2 /1 mol N 2 mol NH3 % yield = actual yield 100 theoretical yield 6.22 g N 2 /mol NH 3 g NH 3 1 mol N 2 mol NH 3 1 mol N 2 = 7.56 g NH 3 = 6.97 g NH 3 (act) 100 7.56 g NH 3 = 92.2% 4
Once a percentage yield has been determined for a reaction, it can be used in stoichiometry calculations For example, a 92% yield means 92 g (act) PER 100 g 5