10.3 Percent Composition and Chemical Formulas. Chapter 10 Chemical Quantities Percent Composition and Chemical Formulas
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1 Chapter 10 Chemical Quantities 101 The Mole: A Measurement of Matter 102 Mole-Mass and Mole-Volume Relationships 103 Percent Composition and Chemical Formulas 1
2 CHEMISTRY & YOU What does the percent composition of a compound tell you? A tag sewn into the seam of a shirt usually tells you what fibers were used to make the cloth and the percent of each 2
3 Percent Composition of a Compound Percent Composition of a Compound How do you calculate the percent composition of a compound? 3
4 Percent Composition of a Compound The relative amounts of the elements in a compound are expressed as the percent composition or the percent by mass of each element in the compound 4
5 Percent Composition of a Compound The percent composition of potassium chromate, K 2 CrO 4, is: K = 403% Cr = 268% O = 329% 5
6 Percent Composition of a Compound These percents must total 100% K = 403% Cr = 268% + O = 329% 100% 6
7 Percent Composition of a Compound These percents must total 100% The percent composition of a compound is always the same 7
8 Percent Composition of a Compound Percent Composition from Mass Data If you know the relative masses of each element in a compound, you can calculate the percent composition of the compound 8
9 Percent Composition of a Compound Percent Composition from Mass Data The percent by mass of an element in a compound is the number of grams of the element divided by the mass in grams of the compound, multiplied by 100% % by mass of element = mass of element mass of compound 100% 9
10 Sample Problem 109 Calculating Percent Composition from Mass Data When a 1360-g sample of a compound containing only magnesium and oxygen is decomposed, 540 g of oxygen is obtained What is the percent composition of this compound? 10
11 Sample Problem Analyze List the knowns and the unknowns The percent by mass of an element in a compound is the mass of that element divided by the mass of the compound multiplied by 100% KNOWNS mass of compound = 1360 g mass of oxygen = 540 g O mass of magnesium = 1360 g 540 g O = 820 g Mg UNKNOWNS percent by mass of Mg =?% Mg percent by mass of O =?% O 11
12 Sample Problem Calculate Solve for the unknowns Determine the percent by mass of Mg in the compound % Mg = mass of Mg mass of compound 100% = 100% 820 g 1360 g = 603% Mg 12
13 Sample Problem Calculate Solve for the unknowns Determine the percent by mass of O in the compound % O = mass of O mass of compound 100% 540 g = 100% 1360 g = 397% O 13
14 Sample Problem Evaluate Does the result make sense? The percents of the elements add up to 100% 603% + 397% = 100% 14
15 15 Percent Composition from the Chemical Formula You can also calculate the percent composition of a compound using its chemical formula Percent Composition of a Compound The subscripts in the formula are used to calculate the mass of each element in a mole of that compound Using the individual masses of the elements and the molar mass, you can calculate the percent by mass of each element
16 Percent Composition from the Chemical Formula You can also calculate the percent composition of a compound using its chemical formula Percent Composition of a Compound % by mass of element mass of element in 1 mol compound = molar mass of compound 100% 16
17 Sample Problem 1010 Calculating Percent Composition from a Formula Propane (C 3 H 8 ), the fuel commonly used in gas grills, is one of the compounds obtained from petroleum Calculate the percent composition of propane 17
18 Sample Problem Analyze List the knowns and the unknowns Calculate the percent by mass of each element by dividing the mass of that element in one mole of the compound by the molar mass of the compound and multiplying by 100% KNOWNS mass of C in 1 mol C 3 H 8 = 3 mol 120 g/mol = 360 g mass of H in 1 mol C 3 H 8 = 8 mol 10 g/mol = 80 g molar mass of C 3 H 8 = 360 g/mol + 80 g/mol = 440 g/mol 18 UNKNOWNS percent by mass of C =?% C percent by mass of H =?% H
19 Sample Problem Calculate Solve for the unknowns Determine the percent by mass of C in C 3 H 8 % C = mass of C in 1 mol C 3 H 8 molar mass of C 3 H g = 100% 440 g 100% = 818% C 19
20 Sample Problem Calculate Solve for the unknowns Determine the percent by mass of H in C 3 H 8 % H = mass of H in 1 mol C 3 H 8 molar mass of C 3 H 8 80 g = 100% 440 g 100% = 18% H 20
21 Sample Problem Evaluate Does the result make sense? The percents of the elements add up to 100% when the answers are expressed to two significant figures (82% + 18% = 100%) 21
22 Percent Composition of a Compound Percent Composition as a Conversion Factor You can use percent composition to calculate the number of grams of any element in a specific mass of a compound To do this, multiply the mass of the compound by a conversion factor based on the percent composition of the element in the compound 22
23 Percent Composition as a Conversion Factor Propane is 818% carbon and 18% hydrogen You can use the following conversion factors to solve for the mass of carbon or hydrogen contained in a specific amount of propane 818 g C 100 g C 3 H 8 and 18 g H 100 g C 3 H 8 Percent Composition of a Compound 23
24 CHEMISTRY & YOU What information can you get from the percent composition of a compound? 24
25 CHEMISTRY & YOU What information can you get from the percent composition of a compound? You can use percent composition to determine the mass of an element in a sample of a compound of a given size You can also determine the empirical formula of the compound 25
26 Sample Problem 1011 Calculating the Mass of an Element in a Compound Using Percent Composition Calculate the mass of carbon and the mass of hydrogen in 820 g of propane (C 3 H 8 ) 26
27 Sample Problem Analyze List the known and the unknowns Use the conversion factors based on the percent composition of propane to make the following conversions: grams C 3 H 8 grams C and grams C 3 H 8 grams H KNOWN mass of C 3 H 8 = 820 g UNKNOWNS mass of carbon =? g C mass of hydrogen =? g H 27
28 Sample Problem Calculate Solve for the unknowns To calculate the mass of C, first write the conversion factor to convert from mass of C 3 H 8 to mass of C 818 g C 100 g C 3 H 8 From Sample Problem 1010, the percent by mass of C in C 3 H 8 is 818% 28 Multiply the mass of C 3 H 8 by the conversion factor 818 g C 820 g C 3 H 8 = 671 g C 100 g C 3 H 8
29 Sample Problem Calculate Solve for the unknowns To calculate the mass of H, first write the conversion factor to convert from mass of C 3 H 8 to mass of H From Sample 18 g H 100 g C 3 H 8 Problem 1010, the percent by mass of H in C 3 H 8 is 18% 29 Multiply the mass of C 3 H 8 by the conversion factor 18 g H 820 g C 3 H 8 = 15 g H 100 g C 3 H 8
30 Sample Problem Evaluate Does the result make sense? The sum of the two masses equals 82 g, the sample size, to two significant figures (67 g C + 15 g H = 82 g C 3 H 8 ) 30
31 What data can you use to calculate percent composition? 31
32 What data can you use to calculate percent composition? You can calculate percent composition if you know the mass of a compound and the masses of the elements contained in the compound, or if you know the chemical formula, the molar mass of the compound, and the atomic weights of the elements contained in the compound 32
33 Empirical Formulas Empirical Formulas How can you calculate the empirical formula of a compound? 33
34 Empirical Formulas The empirical formula of a compound gives the lowest whole-number ratio of the atoms or moles of the elements in a compound An empirical formula may or may not be the same as a molecular formula 34
35 Empirical Formulas 35 The empirical formula of a compound gives the lowest whole-number ratio of the atoms or moles of the elements in a compound An empirical formula may or may not be the same as a molecular formula For example, the lowest ratio of hydrogen to oxygen in hydrogen peroxide is 1:1 Thus, the empirical formula of hydrogen peroxide is HO The molecular formula, H 2 O 2, has twice the number of atoms as the empirical formula Notice that the ratio of hydrogen to oxygen is still the same, 1:1
36 Empirical Formulas For carbon dioxide, the empirical and molecular formulas are the same CO 2 36
37 Empirical Formulas The figure below shows two compounds of carbon and hydrogen having the same empirical formula (CH) but different molecular formulas 37 Ethyne (C 2 H 2 ), also called acetylene, is a gas used in welders torches Styrene (C 8 H 8 ) is used in making polystyrene
38 Empirical Formulas The percent composition of a compound can be used to calculate the empirical formula of that compound 38
39 Empirical Formulas The percent composition of a compound can be used to calculate the empirical formula of that compound The percent composition tells the ratio of masses of the elements in a compound The ratio of masses can be changed to ratio of moles by using conversion factors based on the molar mass of each element The mole ratio is then reduced to the lowest wholenumber ratio to obtain the empirical formula of the compound 39
40 Sample Problem 1012 Determining the Empirical Formula of a Compound A compound is analyzed and found to contain 259% nitrogen and 741% oxygen What is the empirical formula of the compound? 40
41 Sample Problem Analyze List the knowns and the unknown The percent composition gives the ratio of the mass of nitrogen atoms to the mass of oxygen atoms in the compound Change the ratio of masses to a ratio of moles and reduce this ratio to the lowest whole-number ratio KNOWNS percent by mass of N = 259% N percent by mass of O = 741% O 41 UNKNOWN empirical formula = N? O?
42 Sample Problem Calculate Solve for the unknown Convert the percent by mass of each element to moles 259 g N 1 mol N 140 g N 741 g O 1 mol O 160 g O = 185 mol N = 463 mol O Percent means parts per 100, so 1000 g of the compound contains 259 g N and 741 g O The mole ratio of N to O is N 185 O
43 Sample Problem Calculate Solve for the unknown Divide each molar quantity by the smaller number of moles to get 1 mol for the element with the smaller number of moles 185 mol N mol O 185 = 1 mol N = 250 mol O The mole ratio of N to O is N 1 O 25 43
44 Sample Problem Calculate Solve for the unknown Multiply each part of the ratio by the smallest whole number that will convert both subscripts to whole numbers 1 mol N 2 = 2 mol N 25 mol O 2 = 5 mol O The empirical formula is N 2 O 5 44
45 Sample Problem Evaluate Does the result make sense? The subscripts are whole numbers, and the percent composition of this empirical formula equals the percents given in the original problem 45
46 You are doing an experiment to try to find the molecular formula of a compound You discover the percent composition Can you determine the molecular formula? 46
47 You are doing an experiment to try to find the molecular formula of a compound You discover the percent composition Can you determine the molecular formula? You can determine the empirical formula This might be the same as the molecular formula, or it might not You would need more data to be sure of the molecular formula 47
48 Molecular Formulas Molecular Formulas How does the molecular formula of a compound compare with the empirical formula? 48
49 Interpret Data Ethyne and benzene have the same empirical formula CH Formula (name) Comparison of Empirical and Molecular Formulas Classification of formula CH Empirical 13 Molar mass (g/mol) C 2 H 2 (ethyne) Molecular 26 (2 13) C 6 H 6 (benzene) Molecular 78 (6 13) CH 2 O (methanol) Empirical and molecular 30 C 2 H 4 O 2 (ethanoic acid) Molecular 60 (2 30) C 6 H 12 O 6 (glucose) Molecular 180 (6 30) 49
50 Interpret Data Methanal, ethanoic acid, and glucose have the same empirical formula CH 2 O Formula (name) Comparison of Empirical and Molecular Formulas Classification of formula CH Empirical 13 Molar mass (g/mol) C 2 H 2 (ethyne) Molecular 26 (2 13) C 6 H 6 (benzene) Molecular 78 (6 13) CH 2 O (methanal) Empirical and molecular 30 C 2 H 4 O 2 (ethanoic acid) Molecular 60 (2 30) C 6 H 12 O 6 (glucose) Molecular 180 (6 30) 50
51 Interpret Data Notice that the molar masses of the compounds in these two groups are simple whole-number multiples of the molar masses of the empirical formulas, CH and CH 2 O Formula (name) Comparison of Empirical and Molecular Formulas Classification of formula CH Empirical 13 Molar mass (g/mol) C 2 H 2 (ethyne) Molecular 26 (2 13) C 6 H 6 (benzene) Molecular 78 (6 13) CH 2 O (methanal) Empirical and molecular 30 C 2 H 4 O 2 (ethanoic acid) Molecular 60 (2 30) C 6 H 12 O 6 (glucose) Molecular 180 (6 30) 51
52 Molecular Formulas Methanal (formaldehyde), ethanoic acid (acetic acid), and glucose have the same empirical formula CH 2 O 52
53 Molecular Formulas The molecular formula of a compound is either the same as its experimentally determined empirical formula, or it is a simple whole-number multiple of its empirical formula 53
54 Molecular Formulas The molecular formula of a compound is either the same as its experimentally determined empirical formula, or it is a simple whole-number multiple of its empirical formula Once you have determined the empirical formula of a compound, you can determine its molecular formula, if you know the compound s molar mass 54
55 Molecular Formulas You can calculate the empirical formula mass (efm) of a compound from its empirical formula This is simply the molar mass of the empirical formula 55
56 Molecular Formulas 56 You can calculate the empirical formula mass (efm) of a compound from its empirical formula Then you can divide the experimentally determined molar mass by the empirical formula mass This quotient gives the number of empirical formula units in a molecule of the compound and is the multiplier to convert the empirical formula to the molecular formula
57 Sample Problem 1013 Finding the Molecular Formula of a Compound Calculate the molecular formula of a compound whose molar mass is 600 g/mol and empirical formula is CH 4 N 57
58 Sample Problem Analyze List the knowns and the unknown Divide the molar mass by the empirical formula mass to obtain a whole number Multiply the empirical formula subscripts by this value to get the molecular formula KNOWNS empirical formula = CH 4 N molar mass = 600 g/mol UNKNOWN 58 molecular formula = C? H? N?
59 Sample Problem Calculate Solve for the unknown First calculate the empirical formula mass efm of CH 4 N = 120 g/mol + 4(10 g/mol) g/mol = 300 g/mol 59
60 Sample Problem Calculate Solve for the unknown Divide the molar mass by the empirical formula mass molar mass efm = 600 g/mol 300 g/mol = 2 Multiply the formula subscripts by this value (CH 4 N) 2 = C 2 H 8 N 2 60
61 Sample Problem Evaluate Does the result make sense? The molecular formula has the molar mass of the compound 61
62 What information, in addition to empirical formula, is necessary to determine the molecular formula of a compound? 62
63 What information, in addition to empirical formula, is necessary to determine the molecular formula of a compound? Molecular formula can be determined if the empirical formula and the molecular mass of a compound are known 63
64 Key Concepts The percent by mass of an element in a compound is the number of grams of the element divided by the mass in grams of the compound, multiplied by 100% The percent composition of a compound can be used to calculate the empirical formula of that compound 64
65 Key Concepts and Key Equations The molecular formula of a compound is either the same as its experimentally determined formula, or it is a simple whole-number multiple of its empirical formula % by mass of element mass of element = 100% mass of compound % by mass of element = mass of element in 1 mol compound molar mass of compound 100% 65
66 Glossary Terms percent composition: the percent by mass of each element in a compound empirical formula: a formula with the lowest whole-number ratio of elements in a compound; the empirical formula of hydrogen peroxide (H 2 O 2 ) is HO 66
67 BIG IDEA The Mole and Quantifying Matter The molecular formula of a compound can be determined by first finding the percent composition of the compound and determining the empirical formula Using the empirical formula mass and the molar mass of the compound, the molecular formula can be determined 67
68 END OF
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