Introductory Chemistry Essentials Nivaldo J. Tro Fourth Edition

Introductory Chemistry Essentials Nivaldo J. Tro Fourth Edition

Pearson Education Limited Edinburgh Gate Harlow Essex CM20 2JE England and Associated Companies throughout the world Visit us on the World Wide Web at: www.pearsoned.co.uk Pearson Education Limited 2014 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without either the prior written permission of the publisher or a licence permitting restricted copying in the United Kingdom issued by the Copyright Licensing Agency Ltd, Saffron House, 6 10 Kirby Street, London EC1N 8TS. All trademarks used herein are the property of their respective owners. The use of any trademark in this text does not vest in the author or publisher any trademark ownership rights in such trademarks, nor does the use of such trademarks imply any affiliation with or endorsement of this book by such owners. ISBN 10: 1-292-02072-5 ISBN 13: 978-1-292-02072-3 British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library Printed in the United States of America

CHAPTER IN REVIEW CHEMICAL PRINCIPLES The Mole Concept: The mole is a specific number (6.022 * 10 23 ) that allows us to easily count atoms or molecules by weighing them. One mole of any element has a mass equivalent to its atomic mass in grams, and a mole of any compound has a mass equivalent to its formula mass in grams. The mass of 1 mol of an element or compound is its molar mass. Chemical Formulas and Chemical Composition: Chemical formulas indicate the relative number of each kind of element in a compound. These numbers are based on atoms or moles. By using molar masses, we can use the information in a chemical formula to determine the relative masses of each kind of element in a compound. We can then relate the mass of a sample of a compound to the masses of the elements contained in the compound. Empirical and Molecular Formulas from Laboratory Data: We can refer to the relative masses of the elements within a compound to determine the empirical formula of the compound. If the chemist also knows the molar mass of the compound, he or she can also determine its molecular formula. RELEVANCE The Mole Concept: The mole concept allows us to determine the number of atoms or molecules in a sample from its mass. Just as a hardware store customer wants to know the number of nails in a certain weight of nails, so we want to know the number of atoms in a certain mass of atoms. Since atoms are too small to count, we use their mass. Chemical Formulas and Chemical Composition: The chemical composition of compounds is important because it lets us determine how much of a particular element is contained within a particular compound. For example, an assessment of the threat to the Earth s ozone layer from chlorofluorocarbons (CFCs) requires knowing how much chlorine is in a particular CFC. Empirical and Molecular Formulas from Laboratory Data: The first thing a chemist wants to know about an unknown compound is its chemical formula, because the formula reveals the compound s composition. Chemists often arrive at formulas by analyzing compounds in the laboratory either by decomposing them or by synthesizing them to determine the relative masses of the elements they contain. CHEMICAL SKILLS Converting between Moles and Number of Atoms (Section 3) You are given moles of copper and asked to find the number of copper atoms. EXAMPLES EXAMPLE 13 Converting between Moles and Number of Atoms Calculate the number of atoms in 4.8 mol of copper. FIND: 4.8 mol Cu Cu atoms To convert between moles and number of atoms, use Avogadro s number, 6.022 * 10 23 atoms = 1 mol, as a conversion factor. mol Cu Cu atoms 6.022 10 23 Cu atoms 1 mol Cu 1 mol Cu = 6.022 * 10 23 Cu atoms (Avogadro s number, from inside back cover) 218

4.8 mol Cu * 6.022 * 1023 Cu atoms 1 mol Cu 2.9 * 10 24 Cu atoms The units, Cu atoms, are correct. The answer makes physical sense because the number is very large, as you would expect for nearly 5 moles of atoms. = Converting between Grams and Moles (Section 3) You are given the number of moles of aluminum and asked to find the mass of aluminum in grams. EXAMPLE 14 Converting between Grams and Moles Calculate the mass of aluminum (in grams) of 6.73 moles of aluminum. 6.73 mol Al FIND: g Al Use the molar mass of aluminum to convert between moles and grams. mol Al g Al 26.98 g Al 1 mol Al 26.98 g Al = 1 mol Al (molar mass of Al from periodic table) 6.73 mol Al * 26.98 g Al = 182 g Al 1 mol Al The units, g Al, are correct. The answer makes physical sense because each mole has a mass of about 27 g; therefore, nearly 7 moles should have a mass of nearly 190 g. 219

Converting between Grams and Number of Atoms or Molecules (Section 3) You are given the mass of a zinc sample and asked to find the number of Zn atoms that it contains. First use the molar mass of the element to convert from grams to moles, and then use Avogadro s number to convert moles to number of atoms. EXAMPLE 15 Converting between Grams and Number of Atoms or Molecules Determine the number of atoms in a 48.3-g sample of zinc. 48.3 g Zn FIND: Zn atoms g Zn mol Zn 1 mol Zn 6.022 10 23 Zn atoms 65.39 g Zn 1 mol Zn number of Zn atoms 65.39 g Zn = 1 mol Zn (molar mass of Zn from periodic table) 1 mol = 6.022 * 10 23 atoms (Avogadro s number, from inside back cover) 48.3 g Zn * 1 mol Zn 65.39 g Zn * 6.022 * 1023 Zn atoms 1 mol Zn = 4.45 * 10 23 Zn atoms The units, Zn atoms, are correct. The answer makes physical sense because the number of atoms in any macroscopic-sized sample should be very large. Converting between Moles of a Compound and Moles of a Constituent Element (Section 5) You are given the number of moles of sulfuric acid and asked to find the number of moles of oxygen. To convert between moles of a compound and moles of a constituent element, use the chemical formula of the compound to determine a ratio between the moles of the element and the moles of the compound. EXAMPLE 16 Converting between Moles of a Compound and Moles of a Constituent Element Determine the number of moles of oxygen in 7.20 mol of H 2 SO 4. FIND: 7.20 mol H 2 SO 4 mol O mol H 2 SO 4 4 mol O 1 mol H 2 SO 4 mol O 4 mol O : 1 mol H 2 SO 4 220 7.20 mol H 2 SO 4 * 4 mol O 1 mol H 2 SO 4 = 28.8 mol O The units, mol O, are correct. The answer makes physical sense because the number of moles of an element in a compound is equal to or greater than the number of moles of the compound itself.

Converting between Grams of a Compound and Grams of a Constituent Element (Section 5) EXAMPLE 17 Converting between Grams of a Compound and Grams of a Constituent Element Find the grams of iron in 79.2 g of Fe 2 O 3. You are given the mass of iron (III) oxide and asked to find the mass of iron contained within it. Use the molar mass of the compound to convert from grams of the compound to moles of the compound. Then use the chemical formula to obtain a conversion factor to convert from moles of the compound to moles of the constituent element. Finally, use the molar mass of the constituent element to convert from moles of the element to grams of the element. FIND: 79.2 g Fe 2 O 3 g Fe g Fe 2 O 3 mol Fe 2 O 3 1 mol Fe 2 O 3 2 mol Fe 159.70 g Fe 2 O 3 1 mol Fe 2 O 3 mol Fe 55.85 g Fe 1 mol Fe g Fe Molar mass Fe 2 O 3 = 2(55.85) + 3(16.00) = 159.70 g/mol 2 mol Fe : 1 mol Fe 2 O 3 (from given chemical formula) 79.2 g Fe 2 O 3 * 1 mol Fe 2O 3 159.70 g Fe 2 O 3 * 2 mol Fe 1 mol Fe 2 O 3 * 55.85 g Fe 1 mol Fe = 55.4 g Fe The units, g Fe, are correct. The answer makes physical sense because the mass of a constituent element within a compound should be less than the mass of the compound itself. 221

Using Mass Percent Composition as a Conversion Factor (Section 6) You are given the mass of titanium(iv) oxide and the mass percent titanium in the oxide. You are asked to find the mass of titanium in the sample. EXAMPLE 18 Using Mass Percent Composition as a Conversion Factor Determine the mass of titanium in 57.2 g of titanium(iv) oxide. The mass percent of titanium in titanium(iv) oxide is 59.9%. FIND: g Ti 57.2 g TiO 2 59 g Ti 100 g TiO 2 Use the percent composition as a conversion factor between grams of titanium(iv) oxide and grams of titanium. g TiO 2 59.9 g Ti 100 g TiO 2 g Ti 59.9 g Ti : 100 g TiO 2 57.2 g TiO 2 * 59.9 g Ti 100 g TiO 2 = 34.3 g Ti The units, g Ti, are correct. The answer makes physical sense because the mass of an element within a compound should be less than the mass of the compound itself. Determining Mass Percent Composition from a Chemical Formula (Section 7) You are given the formula of potassium oxide and asked to determine the mass percent of potassium within it. The solution map shows how the information derived from the chemical formula can be substituted into the mass percent equation to yield the mass percent of the element. EXAMPLE 19 Determining Mass Percent Composition from a Chemical Formula Calculate the mass percent composition of potassium in potassium oxide (K 2 O). K 2 O FIND: Mass % K Chemical formula Mass % K 2 Molar mass K Mass % K 100% Molar mass K 2 O Mass percent of element X Mass of element X in 1 mol of compound = Mass of 1 mol of compound (mass percent equation, from Section 6) * 100% 222