Download presentation

Presentation is loading. Please wait.

Published byMagdalen Williamson Modified over 2 years ago

1
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro Set up the problem in the normal way. You are given the number of silver atoms and asked to find the number of moles. The conversion factor is Avogadro’s number. EXAMPLE 6.1 Converting between Moles and Number of Atoms Given: 1.1 × 10 22 Find : mol Ag Conversion Factor : 1 mol Ag = 6.022 × 10 23 Ag atoms A silver ring contains 1.1 × 10 22 silver atoms. How many moles of silver are in the ring? Build the solution map, beginning with silver atoms and ending at moles. How many gold atoms are in a pure gold ring containing 8.83 × 10 –2 mol Au? SKILLBUILDER 6.1 Converting between Moles and Number of Atoms Follow the solution map to solve the problem. Beginning with 1.1 × 10 22 Ag atoms, use the conversion factor to get to moles of Ag. FOR MORE PRACTICE Example 6.13, Problems 17, 18, 19, 20. Solution Map : Solution :

2
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro Set up the problem in the normal way. You are given the mass of sulfur and asked to find the number of moles. The conversion factor is the molar mass of sulfur. EXAMPLE 6.2 The Mole Concept—Converting between Grams and Moles Given: 57.8 g S Find : mol S Conversion Factor : 32.07 g S = 1 mol S Calculate the number of moles of sulfur in 57.8 g of sulfur. Draw a solution map showing the conversion from g S to mol S. The conversion factor is the molar mass of sulfur. Calculate the number of grams of sulfur in 2.78 mol of sulfur. SKILLBUILDER 6.2 The Mole Concept—Converting between Grams and Moles Follow the solution map to solve the problem. Begin with 57.8 g S and use the conversion factor to get to mol S. FOR MORE PRACTICE Example 6.14; Problems 25, 26, 27, 28, 29, 30. Solution : Solution Map :

3
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro Set up the problem in the normal way. You are given the mass of aluminum and asked to find the number of aluminum atoms. The required conversion factors are the molar mass of aluminum and the number of atoms in a mole. EXAMPLE 6.3 The Mole Concept—Converting between Grams and Number of Atoms Given: 16.2 g Al Find : Al atoms Conversion Factor : 26.98 g Al = 1 mol Al (molar mass of Al) 6.022 × 10 23 = 1 mol How many aluminum atoms are there in an aluminum can with a mass of 16.2 g? The solution map has two steps. In the first step, convert from g Al to mol Al. In the second step, convert from mol Al to the number of Al atoms. Calculate the mass of 1.23 × 10 24 helium atoms. SKILLBUILDER 6.3 The Mole Concept—Converting between Grams and Number of Atoms Follow the solution map to solve the problem, beginning with 16.2 g Al and multiplying by the appropriate conversion factors to arrive at Al atoms. Solution Map : Solution :

4
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 6.3 The Mole Concept—Converting between Grams and Number of Atoms Continued FOR MORE PRACTICE Example 6.15; Problems 35, 36, 37, 38, 39, 40, 41, 42.

5
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro Set up the problem in the normal way. You are given moles of water and asked to find the mass. The conversion factor is the molar mass of water. EXAMPLE 6.4 The Mole Concept—Converting between Grams and Moles for Compounds Given: 1.75 H 2 O Find : g H 2 O Conversion Factor : H 2 O molar mass = 2(Atomic mass H) + 1(Atomic mass O) = 2(1.01) + 1(16.00) = 18.02 g/mol Calculate the mass (in grams) of 1.75 mol of water. Draw a solution map showing the conversion from mol H 2 O to g H 2 O. Calculate the number of moles of NO 2 in 1.18 g of NO 2. SKILLBUILDER 6.4 The Mole Concept—Converting between Grams and Moles for Compounds Follow the solution map to solve the problem. Begin with 1.75 mol of water and use the molar mass to convert to grams of water. Solution Map : Solution :

6
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 6.4 The Mole Concept—Converting between Grams and Moles for Compounds Continued FOR MORE PRACTICE Problems 45, 46, 47, 48.

7
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro Set up the problem in the normal way. You are given the number of NO 2 molecules and asked to find the mass. The required conversion factors are the molar mass of NO 2 and the number of molecules in a mole. EXAMPLE 6.5 The Mole Concept—Converting between Mass of a Compound and Number of Molecules Given: 4.78 × 10 24 NO 2 molecules Find : g NO 2 Conversion Factor : 6.022 × 10 23 = 1 mol NO 2 molar mass = 1(Atomic mass N) + 2(Atomic mass O) = 14.01 + 2(16.00) = 46.01 g/mol Find the mass of 4.78 X 10 24 NO 2 molecules. The solution map has two steps. In the first step, convert from molecules of NO 2 to moles of NO 2 In the second step, convert from moles of NO 2 to mass of NO 2. How many H 2 O molecules are in a sample of water with a mass of 3.64 g? SKILLBUILDER 6.5 The Mole Concept—Converting between Mass of a Compound and Number of Molecules Using the solution map as a guide, begin with molecules of NO 2 and multiply by the appropriate conversion factors to arrive at g NO 2. Solution Map : Solution :

8
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 6.5 The Mole Concept—Converting between Mass of a Compound and Number of Molecules Continued FOR MORE PRACTICE Problems 49, 50, 51, 52.

9
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro Set up the problem in the normal way. You are given the number of moles of CaCO 3 and asked to find the number of moles of O. The conversion factor is obtained from the chemical formula, which indicates three O atoms for every CaCO 3 unit. EXAMPLE 6.6 Chemical Formulas as Conversion Factors—Converting between Moles of a Compound and Moles of a Constituent Element Determine the number of moles of O in 1.7 mol of CaCO 3. The solution map shows how the conversion factor from the chemical formula converts moles of CaCO 3 into moles of O. Determine the number of moles of O in 1.4 mol of H 2 SO 4. SKILLBUILDER 6.6 Chemical Formulas as Conversion Factors—Converting between Moles of a Compound and Moles of a Constituent Element Follow the solution map to solve the problem. The subscripts in a chemical formula are exact, so they never limit significant figures. FOR MORE PRACTICE Example 6.16; Problems 61, 62. Solution Map : Solution : Given: 1.7 mol CaCO 3 Find : mol O Conversion Factor : 3 mol O 1 mol CaCO 3

10
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro Extract the important information from the problem in the normal way. You are given the mass of carvone and asked to find the mass of one of its constituent elements. You need three conversion factors. The first is the molar mass of carvone. The second conversion factor is the relationship between moles of C and moles of carvone from the molecular formula. The third conversion factor is the molar mass of carbon. EXAMPLE 6.7 Chemical Formulas as Conversion Factors—Converting between Grams of a Compound and Grams of a Constituent Element Carvone (C 10 H 14 O) is the main component of spearmint oil. It has a pleasant aroma and mint flavor. Carvone is often added to chewing gum, liqueurs, soaps, and perfumes. Find the mass of carbon in 55.4 g of carvone. Given: 55.4 g C 10 H 14 O Find : g C Conversion Factors : Molar mass = 10(12.01) + 14(1.01) + 1(16.00) = 120.1 + 14.14 + 16.00 = 150.2 g/mol 10 mol C 1 mol C 10 H 14 O 1 mol C = 12.01 g C

11
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 6.7 Chemical Formulas as Conversion Factors—Converting between Grams of a Compound and Grams of a Constituent Element Continued The solution map is based on Grams Mole Mole Grams Remember, the conversion factor obtained from the chemical formula (10 mol C = 1 mol C 10 H 14 O) applies only to moles; since we are given grams of carvone, we must first convert from g to mol. Solution Map :

12
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 6.7 Chemical Formulas as Conversion Factors—Converting between Grams of a Compound and Grams of a Constituent Element Continued FOR MORE PRACTICE Example 6.17; Problems 65, 66, 67, 68. Determine the mass of oxygen in a 5.8-g sample of sodium bicarbonate (NaHCO 3 ). SKILLBUILDER 6.7 Chemical Formulas as Conversion Factors—Converting between Grams of a Compound and Grams of a Constituent Element Determine the mass of oxygen in a 7.20-g sample of Al 2 (SO 4 ) 3. SKILLBUILDER PLUS Follow the solution map to solve the problem, beginning with g C 10 H 14 O and multiplying by the appropriate conversion factors to arrive at g C. Solution :

13
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro Set up the problem in the normal way. The conversion factor is the mass percent of sodium in sodium chloride. EXAMPLE 6.8 Using Mass Percent Composition as a Conversion Factor The FDA recommends that you consume less than 2.4 g of sodium per day. How many grams of sodium chloride can you consume and still be within the FDA guidelines? Sodium chloride is 39% sodium by mass. The solution map starts with g Na and uses the mass percent as a conversion factor to get to g NaCl. If someone consumes 22 g of sodium chloride, how much sodium does that person consume? Sodium chloride is 39% sodium by mass. SKILLBUILDER 6.8 Using Mass Percent Composition as a Conversion Factor Follow the solution map to solve the problem, beginning with grams Na and ending with grams of NaCl. The answer is 6.2 g NaCl, the amount of salt you can consume and still be within the FDA guideline. Solution Map : Solution : ► Twelve and a half salt packets contain 6.2 g NaCl. Given: 2.4 g Na Find : g NaCl Conversion Factor : 39 g Na 100 g NaCl

14
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 6.8 Using Mass Percent Composition as a Conversion Factor Continued FOR MORE PRACTICE Example 6.19; Problems 73, 74, 75, 76.

15
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro You are asked to find the mass percent of Cl given the molecular formula of freon-114. The required equation, given earlier, shows how to compute the mass percent of an element in a compound based on the formula for the compound. EXAMPLE 6.9 Mass Percent Composition Calculate the mass percent of Cl in freon-114 (C 2 Cl 4 F 2 ). The solution map simply shows how the mass of Cl in 1 mol of C 2 Cl 4 F 2 and the molar mass of C 2 Cl 4 F 2 when substituted into the mass percent equation, yield the mass percent of Cl. Calculate the necessary parts of the equation and substitute the values into the equation to find mass percent Cl. Given: C 2 Cl 4 F 2 Find : Mass % Cl Conversion Factor : Solution Map : Solution :

16
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 6.9 Mass Percent Composition Continued FOR MORE PRACTICE Example 6.20; Problems 77, 78, 79, 80, 81, 82. Acetic acid (HC 2 H 3 O 2 ) is the active ingredient in vinegar. Calculate the mass percent composition of O in acetic acid. SKILLBUILDER 6.9 Mass Percent Composition

17
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro 2. Convert each of the masses in Step 1 to moles by using the appropriate molar mass for each element as a conversion factor. EXAMPLE 6.11 Given: 24.5 g N 70.0 g O Find : empirical formula Obtaining an Empirical Formula from Experimental Data 1. Write down (or compute) as given the masses of each element present in a sample of the compound. If you are given mass percent composition, assume a 100-g sample and compute the masses of each element from the given percentages EXAMPLE 6.10 Given: In a 100-g sample: 60.00 g C 4.48 g H 35.53 g O Find : empirical formula A compound containing Nitrogen and oxygen is decomposed in the laboratory and produces 24.5 g of nitrogen and 70.0 g of oxygen. Calculate the empirical formula of the compound. A laboratory analysis of aspirin determined the following mass percent composition: C 60.00% H 4.48% O 35.53% Find the empirical formula. Solution:

18
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 6.11 N 1.75 O 4.38 Obtaining an Empirical Formula from Experimental Data Continued 3. Write down a pseudoformula for the compound, using the moles of each element (from Step 2) as subscripts. EXAMPLE 6.10 C 4.996 H 4.44 O 2.221 4. Divide all the subscripts in the formula by the smallest subscript 5. If the subscripts are not whole numbers, multiply all the subscripts by a small whole number (see the following table) to get whole-number subscripts

19
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 6.11 Obtaining an Empirical Formula from Experimental Data Continued EXAMPLE 6.10 SKILLBUILDER 6.10SKILLBUILDER 6.11 A sample of a compound is decomposed in the laboratory and produces 165 g of carbon, 27.8 g of hydrogen, and 220.2 g O. Calculate the empirical formula of the compound. Ibuprofen, an aspirin substitute, has the following mass percent composition: C 75.69%; H 8.80%; O 15.51%. Calculate the empirical formula of the compound. FOR MORE PRACTICE Problems 83, 84, 85, 86.Example 6.21; Problems 87, 88, 89, 90.

20
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro Begin by setting up the problem in the normal way. You must recognize this problem as one requiring a special procedure EXAMPLE 6.12 Calculating an Empirical Formula from Reaction Data Given: 3.24 g Ti 5.40 g oxide Find : empirical formula A3.24-g sample of titanium reacts with oxygen to form 5.40 g of the metal oxide. What is the formula of the oxide? 1. Write down (or compute) the masses of each element present in a sample of the compound. In this case, we are given the mass of the initial Ti sample and the mass of its oxide after the sample reacts with oxygen. The mass of oxygen is the difference between the mass of the oxide and the mass of titanium. 2. Convert each of the masses in Step 1 to moles by using the appropriate molar mass for each element as a conversion factor. Solution : 3.24 g Ti Mass O = Mass oxide – Mass titanium = 5.40 g – 3.24 g = 2.16 g O 3. Write down a pseudoformula for the compound, using the moles of each element obtained in Step 2 as subscripts. Ti 0.0677 O 0.135

21
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro 4. Divide all the subscripts in the formula by the smallest subscript. EXAMPLE 6.12 Calculating an Empirical Formula from Reaction Data Continued 5. If the subscripts are not whole numbers, multiply all the subscripts by a small whole number to get whole number subscripts. this last step is unnecessary. The correct empirical formula is TiO 2. A 1.56-g sample of copper reacts with oxygen to form 1.95 g of the metal oxide. What is the formula of the oxide? SKILLBUILDER 6.12 Calculating an Empirical Formula from Reaction Data FOR MORE PRACTICE Problems 91, 92, 93, 94.

22
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro Set up the problem in the normal way. You are given the empirical formula and the molar mass of a compound and asked to find its molecular formula. EXAMPLE 6.13 Calculating Molecular Formula from Empirical Formula and Molar Mass Given: empirical formula = C 5 H 4 molar mass = 128.16 g/mol Find :molecular formula Naphthalene is a compound containing carbon and hydrogen that is often used in mothballs. Its empirical formula is C 5 H 4 and its molar mass is 128.16 g/mol Find its molecular formula. The molecular formula is n times the empirical formula. To find n, divide the molar mass by the empirical formula molar mass. Therefore, the molecular formula is 2 times the empirical formula. Solution : Molecular formula = C 5 H 4 × 2 = C 10 H 8 Butane is a compound containing carbon and hydrogen that is used as a fuel in butane lighters. Its empirical formula is C 2 H 5 and its molar mass is 58.12 g/mol. Find its molecular formula. SKILLBUILDER 6.13 Calculating Molecular Formula from Empirical Formula and Molar Mass Solution :

23
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 6.13 Calculating Molecular Formula from Empirical Formula and Molar Mass Continued A compound with the following mass percent composition has a molar mass of 60.10 g/mol. Find its molecular formula. C 39.97% H 13.41% N 46.62% SKILLBUILDER PLUS FOR MORE PRACTICE Example 6.22; Problems 95, 96, 97, 98.

24
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 6.14 Converting between Moles and Number of Atoms Calculate the number of atoms in 4.8 mol of copper. Given: 4.8 mol Cu Find : Cu atoms Conversion Factor : 1 mol Cu = 6.022 × 10 23 Cu atoms Solution Map : Solution :

25
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 6.15 Converting between Grams and Moles Calculate the mass of aluminum (in grams) of 6.73 moles of aluminum. Given: 6.73 mol Al Find : g Al Conversion Factor : 26.98 g Al = 1 mol Al Solution Map : Solution :

26
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 6.16 Converting between Grams and Number of Atoms or Molecules Determine the number of atoms in a 48.3-g sample of zinc. Given: 48.3 g Zn Find : Zn atoms Conversion Factors : 65.39 g Zn = 1 mol Zn 1 mol = 6.022 × 10 23 atoms Solution Map : Solution :

27
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 6.17 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. Given: 7.20 mol H 2 SO 4 Find : mol O Conversion Factor : 4 mol O 1 mol H 2 SO 4 Solution Map : Solution :

28
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 6.18 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. Given: 79.2 g Fe 2 O 3 Find : g Fe Conversion Factors : 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 Solution Map : Solution :

29
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 6.19 Calculating Mass Percent Composition from Experimental Data A 3.52-g sample of chromium reacts with fluorine to produce 7.38 g of the metal fluoride. What is the mass percent composition of chromium in the fluoride? Given: 3.52 g Cr 7.38 g metal fluoride Find : Cr mass percent Solution :

30
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 6.20 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%. Given: 57.2 g TiO 2 Find : g Ti Conversion Factor : 59.9 g Ti 100 g TiO 2 Solution Map : Solution :

31
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 6.21 Determining Mass Percent Composition from a Chemical Formula Calculate the mass percent composition of potassium in potassium oxide (K 2 O). Given: K 2 O Find : mass % K Equation : Solution Map : Solution :

32
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 6.22 Determining an Empirical Formula from Experimental Data A laboratory analysis of vanillin, the flavoring agent in vanilla, determined the following mass percent composition: C, 63.15%; H, 5.30%; O, 31.55%. Determine the empirical formula of vanillin. Given: In a 100-g sample, we have 63.15 g C, 5.30 g H, and 31.55 g O. Find : empirical formula Solution :

33
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 6.23 Calculating a Molecular Formula from an Empirical Formula and Molar Mass Acetylene, a gas often used in welding torches, has the empirical formula CH and a molar mass of 26.04 g/mol. Find its molecular formula. Given: empirical formula = CH molar mass = 26.04 g/mol Find : molecular formula Solution :

Similar presentations

Presentation is loading. Please wait....

OK

Molar Mass & Percent Composition

Molar Mass & Percent Composition

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on dairy farm management Ppt on coalition government big Ppt on verbs for grade 3 Ppt on properties of rational numbers class 8 Disaster management ppt on uttarakhand india Ppt on pin diode circuit 7 segment led display ppt online Interactive ppt on classification Ppt on ready to serve beverages in spanish Export pdf to ppt online converter