Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Chapter 3 Stoichiometry Chapter 3 Chemical Stoichiometry • Stoichiometry – The study of quantities of materials consumed and produced in chemical reactions. Copyright © Cengage Learning. All rights reserved 2 Section 3.1 Counting by Weighing • • Need average mass of the object. Objects behave as though they were all identical. Return to TOC Copyright © Cengage Learning. All rights reserved 3 Section 3.1 Counting by Weighing Exercise A pile of marbles weigh 394.80 g. 10 marbles weigh 37.60 g. How many marbles are in the pile? Avg. Mass of 1 Marble = 37.60 g = 3.76 g / marble 10 marbles 394.80 g = 105 marbles 3.76 g Return to TOC Copyright © Cengage Learning. All rights reserved 4 Section 3.2 Atomic Masses Counting by Weighing • • Elements occur in nature as mixtures of isotopes. Carbon = 98.89% 12C 1.11% 13C < 0.01% 14C Return to TOC Copyright © Cengage Learning. All rights reserved 5 Section 3.2 Atomic Masses Counting by Weighing Average Atomic Mass for Carbon 98.89% of 12 amu + 1.11% of 13.0034 amu = exact number (0.9889)(12 amu) + (0.0111)(13.0034 amu) = 12.01 amu Return to TOC Copyright © Cengage Learning. All rights reserved 6 Section 3.2 Atomic Masses Counting by Weighing Average Atomic Mass for Carbon • • Even though natural carbon does not contain a single atom with mass 12.01, for stoichiometric purposes, we can consider carbon to be composed of only one type of atom with a mass of 12.01. This enables us to count atoms of natural carbon by weighing a sample of carbon. Return to TOC Copyright © Cengage Learning. All rights reserved 7 Section 3.2 Atomic Masses Counting by Weighing Schematic Diagram of a Mass Spectrometer Return to TOC Copyright © Cengage Learning. All rights reserved 8 Section 3.2 Atomic Masses Counting by Weighing Exercise An element consists of 62.60% of an isotope with mass 186.956 amu and 37.40% of an isotope with mass 184.953 amu. • Calculate the average atomic mass and identify the element. 186.2 amu Rhenium (Re) Return to TOC Copyright © Cengage Learning. All rights reserved 9 Section 3.3 The Mole by Weighing Counting • • • The number equal to the number of carbon atoms in exactly 12 grams of pure 12C. 1 mole of anything = 6.022 x 1023 units of that thing (Avogadro’s number). 1 mole C = 6.022 x 1023 C atoms = 12.01 g C Return to TOC Copyright © Cengage Learning. All rights reserved 10 Section 3.3 The Mole by Weighing Counting Concept Check Calculate the number of iron atoms in a 4.48 mole sample of iron. 2.70×1024 Fe atoms Return to TOC Copyright © Cengage Learning. All rights reserved 11 Section 3.4 Molar Mass • Mass in grams of one mole of the substance: Molar Mass of N = 14.01 g/mol Molar Mass of H2O = 18.02 g/mol (2 × 1.008 g) + 16.00 g Molar Mass of Ba(NO3)2 = 261.35 g/mol 137.33 g + (2 × 14.01 g) + (6 × 16.00 g) Return to TOC Copyright © Cengage Learning. All rights reserved 12 Section 3.4 Molar Mass Concept Check Which of the following is closest to the average mass of one atom of copper? a) b) c) d) e) 63.55 g 52.00 g 58.93 g 65.38 g 1.055 x 10-22 g Return to TOC Copyright © Cengage Learning. All rights reserved 13 Section 3.4 Molar Mass Concept Check Calculate the number of copper atoms in a 63.55 g sample of copper. 6.022×1023 Cu atoms Return to TOC Copyright © Cengage Learning. All rights reserved 14 Section 3.4 Molar Mass Concept Check Which of the following 100.0 g samples contains the greatest number of atoms? a) Magnesium b) Zinc c) Silver Return to TOC Copyright © Cengage Learning. All rights reserved 15 Section 3.4 Molar Mass Exercise Rank the following according to number of atoms (greatest to least): a) 107.9 g of silver b) 70.0 g of zinc c) 21.0 g of magnesium b) a) c) Return to TOC Copyright © Cengage Learning. All rights reserved 16 Section 3.4 Molar Mass Exercise Consider separate 100.0 gram samples of each of the following: H2O, N2O, C3H6O2, CO2 Rank them from greatest to least number of oxygen atoms. H2O, CO2, C3H6O2, N2O Return to TOC Copyright © Cengage Learning. All rights reserved 17 Section 3.5 Learning to Solve Problems Conceptual Problem Solving • Where are we going? • How do we get there? • Read the problem and decide on the final goal. Work backwards from the final goal to decide where to start. Reality check. Does my answer make sense? Is it reasonable? Return to TOC Copyright © Cengage Learning. All rights reserved 18 Section 3.6 Percent Composition of Compounds • Mass percent of an element: mass of element in compound mass % = × 100% mass of compound • For iron in iron(III) oxide, (Fe2O3): 2( 55.85 g) 111.70 g mass % Fe = = × 100% = 69.94% 2( 55.85 g)+ 3( 16.00 g) 159.70 g Return to TOC Copyright © Cengage Learning. All rights reserved 19 Section 3.6 Percent Composition of Compounds Exercise Consider separate 100.0 gram samples of each of the following: H2O, N2O, C3H6O2, CO2 Rank them from highest to lowest percent oxygen by mass. H2O, CO2, C3H6O2, N2O Return to TOC Copyright © Cengage Learning. All rights reserved 20 Section 3.7 Determining the Formula of a Compound Formulas • Empirical formula = CH Simplest whole-number ratio • Molecular formula = (empirical formula)n [n = integer] • Molecular formula = C6H6 = (CH)6 Actual formula of the compound Return to TOC Copyright © Cengage Learning. All rights reserved 21 Section 3.7 Determining the Formula of a Compound Analyzing for Carbon and Hydrogen • Device used to determine the mass percent of each element in a compound. Return to TOC Copyright © Cengage Learning. All rights reserved 22 Section 3.7 Determining the Formula of a Compound Exercise The composition of adipic acid is 49.3% C, 6.9% H, and 43.8% O (by mass). The molar mass of the compound is about 146 g/mol. What is the empirical formula? C3H5O2 What is the molecular formula? C6H10O4 Return to TOC Copyright © Cengage Learning. All rights reserved 23