Quantity Relationships in Chemical Reactions Chapter 10 Quantity Relationships in Chemical Reactions Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Section 10.1 Conversion Factors from a Chemical Equation Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Goal 1 Given 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. Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

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. A2 + 3 B2 2 AB3 1 mol A2 3 mol B2 1 mol A2 2 mol AB3 3 mol B2 2 mol AB3 Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

How many moles of oxygen are needed Example: How many moles of oxygen are needed to completely react with 2.34 moles of methane (CH4) in a combustion reaction? Carbon dioxide and steam are the products. Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

CH4 + O2 CO2 + H2O 2 2 PER relationship from equation: 1 mol CH4 PER 2 mol O2 Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

GIVEN: 2.34 mol CH4 WANTED: mol O2 1 mol CH4/2 mol O2 PER/PATH: mol CH4 mol O2 X 2 mol O2 1 mol CH4 2.34 mol CH4 = 4.68 mol O2 Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Section 10.2 Mass-Mass Stoichiometry Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Goal 2 Given 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. Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

The quantitative relationships between the Stoichiometry: The quantitative relationships between the substances involved in a chemical reaction, established by the equation for the reaction A stoichiometry problem asks, “How much or how many?” Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Prerequisite Skills for Stoichiometry Write chemical formulas Ch 6 Calculate molar masses from chemical formulas Sect 7.4 Use molar masses to change mass to moles and moles to mass Sect 7.5 Write and balance chemical equations Ch 8 Use the equation to change from moles of one species to moles of another Sect 10.1 Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Fundamental Stoichiometry Pattern: Given Macroscopic Measurable Quantity Moles of Wanted Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Mass–Mass Stoichiometry Pattern: Moles of Given Quantity Moles of Wanted Quantity Given Mass Wanted Mass Molar Mass g PER mol Equation mol PER mol Molar Mass g PER mol Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

How to Solve a Stoichiometry Problem: The Stoichiometry Path 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. Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Example: How many grams of carbon dioxide are produced when 10.0 g of methane, CH4, is burned? GIVEN: 10.0 g CH4 WANTED: g CO2 CH4 + O2 2 CO2 + H2O 2 16.04 g CH4/mol CH4 PER/PATH: g CH4 1 mol CO2/1 mol CH4 mol CH4 mol CO2 44.01 g CO2/mol CO2 g CO2 Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

16.04 g CH4/mol CH4 PER/PATH: g CH4 1 mol CO2/1 mol CH4 mol CH4 44.01 g CO2/mol CO2 g CO2 X 1 mol CH4 16.04 g CH4 10.0 g CH4 X 1 mol CO2 1 mol CH4 X 44.01 g CO2 mol CO2 = 27.4 g CO2 Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Section 9.3 Percent Yield Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Goal 3 Given two of the following, or information from which two of the following may be determined, calculate the third: theoretical yield, actual yield, percent yield. Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

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 Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Percentage yield expresses the ratio of actual yield to theoretical yield: % yield = actual yield X 100 theoretical yield Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

STEP 1: Calculate the theoretical yield Example: 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 Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

GIVEN: 6.22 g N2 WANTED: g NH3 (theo) N2 + H2 3 2 NH3 28.02 g N2/mol N2 PER/PATH: g N2 2 mol NH3/1 mol N2 mol N2 mol NH3 17.03 g NH3/mol NH3 g NH3 Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

PER/PATH: g N2 28.02 g N2/mol N2 mol N2 2 mol NH3/1 mol N2 mol NH3 17.03 g NH3/mol NH3 g NH3 X 1 mol N2 28.02 g N2 X 2 mol NH3 1 mol N2 6.22 g N2 X 17.03 g NH3 mol NH3 = 7.56 g NH3 (theo) Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

% yield = actual yield X 100 theoretical yield = 6.97 g NH3 (act) X 100 7.56 g NH3 (theo) = 92.2% Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

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 (theo) Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.