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Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu How to Use This Presentation To View the presentation as a slideshow with effects select “View” on the menu bar and click on “Slide Show.” To advance through the presentation, click the right-arrow key or the space bar. From the resources slide, click on any resource to see a presentation for that resource. From the Chapter menu screen click on any lesson to go directly to that lesson’s presentation. You may exit the slide show at any time by pressing the Esc key.

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 1 Avogadro’s Number and Molar Conversions Bellringer List as many common counting units as you can. Determine how many groups of each unit in your list are present in each of the following amounts: 500 goldfish 150 unicycles 50 jet planes Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Objectives Identify the mole as the unit used to count particles, whether atoms, ions, or molecules. Use Avogadro’s number to convert between amount in moles and number of particles. Solve problems converting between mass, amount in moles, and number of particles using Avogadro’s number and molar mass. Section 1 Avogadro’s Number and Molar Conversions Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Avogadro’s Number and the Mole The SI unit for amount is called the mole (mol). A mole is the number of atoms in exactly 12 grams of carbon-12. Scientists use the mole to make counting large numbers of particles easier. The number of particles in a mole is called Avogadro’s Number. Avogadro’s number is 6.02214199  10 23 units/mole. Section 1 Avogadro’s Number and Molar Conversions Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Avogadro’s Number and the Mole, continued The Mole Is a Counting Unit Section 1 Avogadro’s Number and Molar Conversions The mole is used to count out a given number of particles, whether they are atoms, molecules, formula units, ions, or electrons. The mole is just one kind of counting unit: 1 dozen = 12 objects 1 hour = 3600 seconds 1 mole = 6.022  10 23 particles Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Avogadro’s Number and the Mole, continued Amount in Moles Can Be Converted to Number of Particles Counting units are used to make conversion factors. The definition of one mole is 6.022  10 23 particles = 1 mol The conversion factor is Section 1 Avogadro’s Number and Molar Conversions Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Avogadro’s Number and the Mole, continued Choose the Conversion Factor That Cancels the Given Units All conversion factors are equal to 1, so you can use them to convert among different units. You can tell which conversion factor to use, because the needed conversion factor should cancel the units of the given quantity to give you the units of the answer or the unknown quantity. Section 1 Avogadro’s Number and Molar Conversions Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Converting Amount in Moles to Number of Particles Sample Problem A Find the number of molecules in 2.5 mol of sulfur dioxide. Section 1 Avogadro’s Number and Molar Conversions Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Converting Amount in Moles to Number of Particles Sample Problem A Solution 2.5 mol SO 2  ? = ? molecules SO 2 You are converting from the unit mol to the unit molecules. The conversion factor must have the units of molecules/mol. You use 6.022  10 23 molecules/1 mol. Section 1 Avogadro’s Number and Molar Conversions Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Converting Amount in Moles to Number of Particles Sample Problem A Solution, continued 2.5 mol SO 2  ? = ? molecules SO 2 Section 1 Avogadro’s Number and Molar Conversions 2.5 mol SO 2 = 1.5  10 24 molecules SO 2 Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Avogadro’s Number and the Mole, continued Number of Particles Can Be Converted to Amount in Moles The reverse calculation is similar to that in Sample Problem A but the conversion factor is inverted to get the correct units in the answer. example: How many moles are 2.54  10 22 iron(III) ions? Section 1 Avogadro’s Number and Molar Conversions Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Converting Number of Particles to Amount in Moles Sample Problem B A sample contains 3.01  10 23 molecules of sulfur dioxide, SO 2. Determine the amount in moles. Section 1 Avogadro’s Number and Molar Conversions Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Sample Problem B Solution 3.01 × 10 23 molecules SO 2  ? = ? mol SO 2 You are converting from the unit molecules to the unit mol. The conversion factor must have the units of mol/molecules. You use 1 mol/6.022  10 23 molecules. Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Number of Particles to Amount in Moles

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Sample Problem B Solution, continued 3.01  10 23 molecules SO 2  ? = ? mol SO 2 Section 1 Avogadro’s Number and Molar Conversions 3.01  10 23 molecules SO 2 = 0.500 mol SO 2 Chapter 7 Converting Number of Particles to Amount in Moles

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Molar Mass Relates Moles to Grams Amount in Moles Can Be Converted to Mass The molar mass is the mass in grams of one mole of an element or compound. Molar mass is numerically equal to the atomic mass of monatomic elements and the formula mass of compounds and diatomic elements. The units for molar mass are g/mol. Molar mass can be used as a conversion factor in problems converting between mass and amount. Section 1 Avogadro’s Number and Molar Conversions Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Molar Mass Relates Moles to Grams, continued The Mole Plays a Central Part in Chemical Conversions To convert from number of particles to mass, you must use a two-part process: First, convert number of particles to amount in moles. Second, convert amount in moles to mass in grams. One step common to many problems in chemistry is converting to amount in moles. Section 1 Avogadro’s Number and Molar Conversions Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Sample Problem C Find the mass in grams of 2.44  10 24 atoms of carbon, whose molar mass is 12.01 g/mol. Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Number of Particles to Mass

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Converting Number of Particles to Mass Sample Problem C Solution First part: 2.44  10 24 atoms  ? = ? mol Select the conversion factor that will take you from number of atoms to amount in moles. You use 1 mol/6.022  10 23 atoms. Section 1 Avogadro’s Number and Molar Conversions Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Sample Problem C Solution, continued Second part: ? mol  ? = ? g Select the conversion factor that will take you from amount in moles to mass in grams. You use the molar mass of carbon, 12.01 g C/1 mol. Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Number of Particles to Mass

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Molar Mass Relates Moles to Grams, continued Mass Can Be Converted to Amount in Moles Converting from mass to number of particles is the reverse of the operation in the previous problem. To convert from mass to number of particles, you must use a two-part process: First, convert mass in grams to amount in moles. Second, convert amount in moles to number of particles. Section 1 Avogadro’s Number and Molar Conversions Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Converting Mass to Number of Particles Sample Problem D Find the number of molecules present in 47.5 g of glycerol, C 3 H 8 O 3. The molar mass of glycerol is 92.11 g/mol. Section 1 Avogadro’s Number and Molar Conversions Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Sample Problem D Solution First part: 47.5 g  ? = ? mol Select the conversion factor that will take you from mass in grams to amount in moles. You use the inverse of the molar mass of glycerol: Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Mass to Number of Particles

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Sample Problem D Solution, continued Second part: ? mol  ? = ? molecules Select the conversion factor that will take you from amount in moles to number of particles. You use. Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Mass to Number of Particles

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Remember that an answer must never be given to more significant figures than is appropriate. Round molar masses from the periodic table to two significant figures to the right of the decimal point. Section 1 Avogadro’s Number and Molar Conversions Molar Mass Relates Moles to Grams, continued Remember to Round Consistently Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 2 Relative Atomic Mass and Chemical Formulas Bellringer Compare the masses of a roll of pennies and a roll of dimes. Both contain 50 coins. Why are the masses of the rolls different when both rolls contain the same number of coins? If given a roll of mixed coins, what information would you need to determine the mass of the roll? Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Objectives Use a periodic table or isotopic composition data to determine the average atomic masses of elements. Infer information about a compound from its chemical formula. Determine the molar mass of a compound from its formula. Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Average Atomic Mass and the Periodic Table Most Elements are a Mixture of Isotopes Isotopes are atoms that have different numbers of neutrons than other atoms of the same element do. Average atomic mass is a weighted average of the atomic mass of an element’s isotopes. If you know the abundance of each isotope, you can calculate the average atomic mass of an element. Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Calculating Average Atomic Mass Sample Problem E The mass of a Cu-63 atom is 62.94 amu, and that of a Cu-65 atom is 64.93 amu. Using the data below, find the average atomic mass of copper. abundance of Cu-63 = 69.17% abundance of Cu-65 = 30.83% Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Sample Problem E Solution The contribution of each isotope is equal to its atomic mass multiplied by the fraction of that isotope. contribution of Cu-63:62.94 amu × 0.6917 contribution of Cu-65:64.93 amu × 0.3083 Average atomic mass is the sum of the individual contributions: (62.94 amu × 0.6917) + (64.93 amu × 0.3083) = 63.55 amu Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7 Calculating Average Atomic Mass

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chemical Formulas and Moles Formulas Express Composition A compound’s chemical formula tells you which elements, as well as how much of each, are present in a compound. Formulas for covalent compounds show the elements and the number of atoms of each element in a molecule. Formulas for ionic compounds show the simplest ratio of cations and anions in any pure sample. Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Any sample of compound has many atoms and ions, and the formula gives a ratio of those atoms or ions. Section 2 Relative Atomic Mass and Chemical Formulas Chemical Formulas and Moles, continued Formulas Express Composition, continued Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chemical Formulas and Moles, continued Formulas Give Ratios of Polyatomic Ions Formulas for polyatomic ions show the simplest ratio of cations and anions. They also show the elements and the number of atoms of each element in each ion. For example, the formula KNO 3 indicates a ratio of one K + cation to one anion. Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chemical Formulas and Moles, continued Formulas Are Used to Calculate Molar Masses The molar mass of a molecular compound is the sum of the masses of all the atoms in the formula expressed in g/mol. The molar mass of an ionic compound is the sum of the masses of all the atoms in the formula expressed in g/mol. Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Calculating Molar Mass of Compounds Sample Problem F Find the molar mass of barium nitrate, Ba(NO 3 ) 2. Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Sample Problem F Solution Find the number of moles of each element in 1 mol Ba(NO 3 ) 2 : Each mole has 1 mol Ba, 2 mol N, and 6 mol O. Use the periodic table to find the molar mass of each element in the formula: molar mass of Ba: 137.33 g/mol molar mass of N: 14.01 g/mol molar mass of O: 16.00 g/mol Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7 Calculating Molar Mass of Compounds

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Sample Problem F Solution, continued Multiply the molar mass of each element by the number of moles of each element. Add these masses to get the total molar mass of Ba(NO 3 ) 2. mass of 1 mol Ba = 1  137.33 g/mol = 137.33 g/mol mass of 2 mol N = 2  14.01 g/mol = 28.02 g/mol + mass of 6 mol O = 6  16.00 g/mol = 96.00 g/mol Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7 Calculating Molar Mass of Compounds molar mass of Ba(NO 3 ) 2 = 261.35 g/mol

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Objectives Determine a compound’s empirical formula from its percentage composition. Determine the molecular formula or formula unit of a compound from its empirical formula and its formula mass. Calculate percentage composition of a compound from its molecular formula or formula unit. Section 3 Formulas and Percentage Composition Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Using Analytical Data The percentage composition is the percentage by mass of each element in a compound. Percentage composition helps verify a substance’s identity. Percentage composition also can be used to compare the ratio of masses contributed by the elements in two different substances. Section 3 Formulas and Percentage Composition Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Using Analytical Data, continued Determining Empirical Formulas Section 3 Formulas and Percentage Composition An empirical formula is a chemical formula that shows the simplest ratio for the relative numbers and kinds of atoms in a compound. An actual formula shows the actual ratio of elements or ions in a single unit of a compound. For example, the empirical formula for ammonium nitrate is NH 2 O, while the actual formula is NH 4 NO 2. Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Using Analytical Data, continued Determining Empirical Formulas, continued Section 3 Formulas and Percentage Composition You can use the percentage composition for a compound to determine its empirical formula. 1.Convert the percentage of each element to g. 2.Convert from g to mol using the molar mass of each element as a conversion factor. 3.Compare these amounts in mol to find the simplest whole-number ratio among the elements. Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Determining an Empirical Formula from Percentage Composition Sample Problem G Chemical analysis of a liquid shows that it is 60.0% C, 13.4% H, and 26.6% O by mass. Calculate the empirical formula of this substance. Section 3 Formulas and Percentage Composition Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Determining an Empirical Formula from Percentage Composition Sample Problem G Solution Assume that you have a 100.0 g sample, and convert the percentages to grams. for C:60.0%  100.0 g = 60.0 g C for H:13.4%  100.0 g = 13.4 g H for O:26.6%  100.0 g = 26.6 g O Section 3 Formulas and Percentage Composition Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Determining an Empirical Formula from Percentage Composition Sample Problem G Solution, continued Convert the mass of each element into the amount in moles, using the reciprocal of the molar mass. Section 3 Formulas and Percentage Composition Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Determining an Empirical Formula from Percentage Composition Sample Problem G Solution, continued The formula can be written as C 5 H 13.3 O 1.66, but you divide by the smallest subscript to get whole numbers. Section 3 Formulas and Percentage Composition The empirical formula is C 3 H 8 O. Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Using Analytical Data, continued Molecular Formulas Are Multiples of Empirical Formulas Section 3 Formulas and Percentage Composition The formula for an ionic compound shows the simplest whole-number ratio of the large numbers of ions in a crystal of the compound. A molecular formula is a whole-number multiple of the empirical formula. The molar mass of any compound is equal to the molar mass of the empirical formula times a whole number, n. Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Determining a Molecular Formula from an Empirical Formula Sample Problem H The empirical formula for a compound is P 2 O 5. Its experimental molar mass is 284 g/mol. Determine the molecular formula of the compound. Section 3 Formulas and Percentage Composition Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Determining a Molecular Formula from an Empirical Formula Sample Problem H Solution Find the molar mass of the empirical formula P 2 O 5. 2  molar mass of P = 61.94 g/mol + 5  molar mass of O = 80.00 g/mol molar mass of P 2 O 5 = 141.94 g/mol Section 3 Formulas and Percentage Composition Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Determining a Molecular Formula from an Empirical Formula Sample Problem H Solution, continued Section 3 Formulas and Percentage Composition Chapter 7 n (empirical formula) = 2 (P 2 O 5 ) = P 4 O 10

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Using Analytical Data, continued Chemical Formulas Can Give Percentage Composition Section 3 Formulas and Percentage Composition If you know the chemical formula of any compound, then you can calculate the percentage composition. From the subscripts, determine the mass contributed by each element and add these to get molar mass. Divide the mass of each element by the molar mass. Multiply by 100 to find the percentage composition of that element. Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu CO and CO 2 are both made up of C and O, but they have different percentage compositions. Using Analytical Data, continued Chemical Formulas Can Give Percentage Composition Section 3 Formulas and Percentage Composition Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Using a Chemical Formula to Determine Percentage Composition Sample Problem I Calculate the percentage composition of copper(I) sulfide, Cu 2 S, a copper ore called chalcocite. Section 3 Formulas and Percentage Composition Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Using a Chemical Formula to Determine Percentage Composition Sample Problem I Solution Find the molar mass of Cu 2 S. 2 mol  63.55 g Cu/mol = 127.10 g Cu + 1 mol  32.07 g S/mol = 32.07 g S molar mass of Cu 2 S = 159.17 g/mol Section 3 Formulas and Percentage Composition Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Using a Chemical Formula to Determine Percentage Composition Sample Problem I Solution, continued Calculate the fraction that each element contributes to the total mass by substituting the masses into the equations below and rounding correctly. Section 3 Formulas and Percentage Composition 79.852% Cu Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Using a Chemical Formula to Determine Percentage Composition Sample Problem I Solution, continued Section 3 Formulas and Percentage Composition 20.15% S Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Standardized Test Preparation Understanding Concepts Chapter 7 1.Element A has two isotopes. One has an atomic mass of 120 and constitutes 60%; the other has an atomic mass of 122 and constitutes 40%. Which range below includes the average atomic mass of Element A? A.less than 120 B.between 120 and 121 C.between 121 and 122 D.greater than 122

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Standardized Test Preparation Understanding Concepts Chapter 7 1.Element A has two isotopes. One has an atomic mass of 120 and constitutes 60%; the other has an atomic mass of 122 and constitutes 40%. Which range below includes the average atomic mass of Element A? A.less than 120 B.between 120 and 121 C.between 121 and 122 D.greater than 122

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 2.Which of the following can be determined from the empirical formula of a compound alone? F.the true formula of the compound G.the molecular mass of the compound H.the percentage composition of the compound I.the arrangement of atoms within a molecule of the compound Standardized Test Preparation Understanding Concepts Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 2.Which of the following can be determined from the empirical formula of a compound alone? F.the true formula of the compound G.the molecular mass of the compound H.the percentage composition of the compound I.the arrangement of atoms within a molecule of the compound Standardized Test Preparation Understanding Concepts Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 3.How many ions are in 0.5 moles of NaCl? A.1.204  10 23 B.3.011  10 23 C.6.022  10 23 D.9.033  10 23 Standardized Test Preparation Understanding Concepts Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 3.How many ions are in 0.5 moles of NaCl? A.1.204  10 23 B.3.011  10 23 C.6.022  10 23 D.9.033  10 23 Standardized Test Preparation Understanding Concepts Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 4.How many moles of calcium (mass = 40.1) are in a serving of milk containing 290 mg of calcium? Standardized Test Preparation Understanding Concepts Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 4.How many moles of calcium (mass = 40.1) are in a serving of milk containing 290 mg of calcium? Answer: 7.23  10 –3 mol Standardized Test Preparation Understanding Concepts Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 6.Antimony has two isotopes. One, amounting to 57.3% of the atoms, has a mass of 120.9 amu. The other, 42.7% of the atoms, has a mass of 122.9 amu. What is the average atomic mass of antimony? Standardized Test Preparation Understanding Concepts Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 6.Antimony has two isotopes. One, amounting to 57.3% of the atoms, has a mass of 120.9 amu. The other, 42.7% of the atoms, has a mass of 122.9 amu. What is the average atomic mass of antimony? Answer: 121.8 amu Standardized Test Preparation Understanding Concepts Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Read the passage below. Then answer the questions. In 1800 two English chemists, Nicholson and Carlisle, discovered that when an electric current is passed through water, hydrogen and oxygen were produced in a 2:1 volume ratio and a 1:8 mass ratio. This evidence helped to support John Dalton's theory that matter consisted of atoms, demonstrating that water consists of the two elements in a constant proportion. If the same number of moles of each gas occupy the same volume, then each molecule of water must consist of twice as much hydrogen as oxygen, even though the mass of hydrogen is only one-eighth that of oxygen. Standardized Test Preparation Reading Skills Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 7.Based on this experiment, what is the empirical formula of water? F.HO G.H 2 O H.H 2 O 8 I.O 8 H Standardized Test Preparation Reading Skills Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 7.Based on this experiment, what is the empirical formula of water? F.HO G.H 2 O H.H 2 O 8 I.O 8 H Standardized Test Preparation Reading Skills Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 8.How would the experimental result have been different if hydrogen gas existed as individual atoms while oxygen formed molecules with two atoms bound by a covalent bond? A.The result would be the same. B.The ratio of hydrogen to oxygen would be 1:1. C.The ratio of hydrogen to oxygen would be 1:4. D.The ratio of hydrogen to oxygen would be 4:1. Standardized Test Preparation Reading Skills Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 8.How would the experimental result have been different if hydrogen gas existed as individual atoms while oxygen formed molecules with two atoms bound by a covalent bond? A.The result would be the same. B.The ratio of hydrogen to oxygen would be 1:1. C.The ratio of hydrogen to oxygen would be 1:4. D.The ratio of hydrogen to oxygen would be 4:1. Standardized Test Preparation Reading Skills Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 10.How many moles of oxygen atoms are there in 100.0 moles of carbon dioxide? F.66.7 G.72.7 H.100.0 I.200.0 Standardized Test Preparation Interpreting Graphics Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 10.How many moles of oxygen atoms are there in 100.0 moles of carbon dioxide? F.66.7 G.72.7 H.100.0 I.200.0 Standardized Test Preparation Interpreting Graphics Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 11.Explain why the percentage of oxygen in carbon dioxide is not twice the percentage of oxygen in carbon monoxide, if there are twice as many oxygen atoms. Standardized Test Preparation Interpreting Graphics Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 11.Explain why the percentage of oxygen in carbon dioxide is not twice the percentage of oxygen in carbon monoxide, if there are twice as many oxygen atoms. Answer: Although there are twice as many oxygen atoms in carbon dioxide, the percentage composition is based on the mass, not the number of atoms. Standardized Test Preparation Interpreting Graphics Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 12.If you did not know the true formulas for carbon monoxide and carbon dioxide, what information would you need beyond what is provided in the illustration in order to calculate them? A.the percentage compositions B.the atomic masses of carbon and oxygen C.the melting and boiling points of each compound D.the number of atoms of each element in the compound Standardized Test Preparation Interpreting Graphics Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 12.If you did not know the true formulas for carbon monoxide and carbon dioxide, what information would you need beyond what is provided in the illustration in order to calculate them? A.the percentage compositions B.the atomic masses of carbon and oxygen C.the melting and boiling points of each compound D.the number of atoms of each element in the compound Standardized Test Preparation Interpreting Graphics Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 13.How many grams of carbon are contained in 200.0 grams of carbon dioxide? F.27.29 G.42.88 H.54.58 I.85.76 Standardized Test Preparation Interpreting Graphics Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 13.How many grams of carbon are contained in 200.0 grams of carbon dioxide? F.27.29 G.42.88 H.54.58 I.85.76 Standardized Test Preparation Interpreting Graphics Chapter 7