# Stoichiometry Calculations with Chemical Formulas and Equations Chapter

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Stoichiometry: Calculations with Chemical Formulas and Equations Chapter 3 1

Chemical Equations 2 H 2(g) + O 2(g) 2 H 2 O(g) • The materials you start with are called Reactants. • The materials you make are called Products. • The numbers in front of the compounds (H 2 and H 2 O) are called coefficients. – Coefficients are multipliers, in this equation 2 in front of the H 2 indicates that there are 2 molecules of H 2 in the equation. Chapter 3 2

Chemical Equations 2 H 2(g) + O 2(g) 2 H 2 O(g) • Notice that the number of hydrogen atoms and oxygen atoms on the reactant side and the product side is equal. Law of Conservation of Mass Matter cannot be created or lost in any chemical reaction. Chapter 3 3

Chemical Equations Balancing Chemical Reactions ___NH 4 NO 3(s) ___N 2 O(g) + ___H 2 O(g) Chapter 3 4

Chemical Equations Balancing Chemical Reactions ___NH 4 NO 3(s) ___N 2 O(g) + ___H 2 O(g) Reactants Products N 2 H 4 H 2 O 3 O 2 Chapter 3 5

Chemical Equations Balancing Chemical Reactions ___NH 4 NO 3(s) ___N 2 O(g) + _2_H 2 O(g) Reactants Products N 2 H 4 H 24 O 3 O 23 Chapter 3 6

Chemical Equations Balancing Chemical Reactions ___Mg 3 N 2(s) + ___H 2 O(l) ___Mg(OH)2(s) + ___NH 3(aq) Chapter 3 7

Chemical Equations Balancing Chemical Reactions ___Mg 3 N 2(s) + ___H 2 O(l) ___Mg(OH)2(s) + ___NH 3(aq) Reactants Products Mg 3 Mg 1 N 2 N 1 H 2 H 5 O 1 O 2 Chapter 3 8

Chemical Equations Balancing Chemical Reactions ___Mg 3 N 2(s) + ___H 2 O(l) _3_Mg(OH)2(s) + ___NH 3(aq) Reactants Products Mg 3 Mg 13 N 2 N 1 H 2 H 59 O 1 O 26 Chapter 3 9

Chemical Equations Balancing Chemical Reactions ___Mg 3 N 2(s) + ___H 2 O(l) _3_Mg(OH)2(s) + _2_NH 3(aq) Reactants Products Mg 3 Mg 13 N 2 N 12 H 5 9 12 O 1 O 26 Chapter 3 10

Chemical Equations Balancing Chemical Reactions ___Mg 3 N 2(s) + _6_H 2 O(l) _3_Mg(OH)2(s) + _2_NH 3(aq) Reactants Products Mg 3 Mg 13 N 2 N 12 H 2 12 H 5 9 12 O 16 O 26 Chapter 3 11

Patterns of Chemical Reactivity Combustion Reactions This is the combination of a substance with oxygen: C 3 H 8(g) + 5 O 2(g) 3 CO 2(g) + 4 H 2 O(l) • Combustion of hydrocarbons yield CO 2 and H 2 O 4 Fe(s) + 3 O 2(g) 2 Fe 2 O 3(s) Chapter 3 12

Patterns of Chemical Reactivity Combustion Reactions This is the combination of a substance with oxygen: C 3 H 8(g) + 5 O 2(g) 3 CO 2(g) + 4 H 2 O(l) • Less oxygen is required if the reactant has oxygen. C 3 H 7 OH(l) + 4 ½ O 2(g) 3 CO 2(g) + 4 H 2 O(l) Chapter 3 13

Patterns of Chemical Reactivity Combination and Decomposition Reactions Combination Reactions Two or more substances react to form one product. Ca. O(s) + H 2 O(l) Ca(OH)2(s) Chapter 3 14

Patterns of Chemical Reactivity Combination and Decomposition Reactions One substance reacts to form two or more substances. Ca. CO 3(s) Ca. O(s) + CO 2(g) Chapter 3 15

Atomic and Molecular Weights The Atomic Mass Scale The mass of individual atoms is measured in atomic mass units (amu). 1 amu = 1. 66054 x 10 -24 g Chapter 3 16

Atomic and Molecular Weights The Atomic Mass Scale The amu is derived from carbon-12 (12 C). The mass of one 12 C is exactly 12 amu. Chapter 3 17

Atomic and Molecular Weights Average Atomic Mass • The elements are actually a mixture of isotopes. • The atomic mass shown in the periodic table is a weighted average of the various isotopes. Naturally occurring C: 98. 892 % 12 C, 1. 108 % 13 C. Average mass of C: (0. 98892)(12 amu) + (0. 01108)(13 amu) = 12. 011 amu Chapter 3 18

Atomic and Molecular Weights Average Atomic Mass Atomic weight (AW) The average atomic mass (atomic weight) of an element. Atomic weights are listed on the periodic table. Chapter 3 19

Atomic and Molecular Weights Formula weight (FW) The sum of the atomic weights of each atom in the chemical formula. Example: CO 2 Chapter 3 20

Atomic and Molecular Weights Formula weight (FW) The sum of the atomic weights of each atom in the chemical formula. Example: CO 2 Formula Weight = 1(AW, carbon) + 2(AW, oxygen) Formula Weight = 1(12. 011 amu) + 2(16. 0 amu) Formula Weight = 44. 0 amu Chapter 3 21

Atomic and Molecular Weights Formula and Molecular Weights Molecular weight The sum of the atomic weights of each atom in the molecular formula. • Formula weight is the general term, molecule weight refers specifically to molecular compounds. Chapter 3 22

Atomic and Molecular Weights Percentage Composition from Formulas Chapter 3 23

Atomic and Molecular Weights Percentage Composition from Formulas Example: Calculate the percent oxygen in CH 3 CH 2 OH. Formula weight ethanol: 2(12. 01 amu) + 6(1. 01 amu) + 1(16. 00 amu) = 46. 08 amu Mass of oxygen: 1(16. 00 amu) = 16. 00 amu Chapter 3 24

Atomic and Molecular Weights Percentage Composition from Formulas Example: Calculate the percent oxygen in CH 3 CH 2 OH. Formula weight ethanol: 2(12. 01 amu) + 6(1. 01 amu) + 1(16. 00 amu) = 46. 08 amu Mass of oxygen: 1(16. 00 amu) = 16. 00 amu % oxygen Chapter 3 25

Atomic and Molecular Weights Percentage Composition from Formulas Example: Calculate the percent oxygen in CH 3 CH 2 OH. Formula weight ethanol: 2(12. 01 amu) + 6(1. 01 amu) + 1(16. 00 amu) = 46. 08 amu Mass of oxygen: 1(16. 00 amu) = 16. 00 amu % oxygen Chapter 3 26

The Mole That amount of matter that contains as many objects as exactly 12 g of 12 C. 12 g 12 C gives 6. 02 x 1023 atoms of 12 C 6. 02 x 1023 is Avogadro’s Number Chapter 3 27

Molar Mass in grams of 1 mole of substance (g/mol). This is determined by the sum of atomic weights (in grams) of the atoms in the formula of the molecule. Example: CO 2 1(12. 011 g/mol)+2(16. 00 g/mol) = Chapter 3 28

Molar Mass in grams of 1 mole of substance (g/mol). This is determined by the sum of atomic weights (in grams) of the atoms in the formula of the molecule. Example: CO 2 1(12. 011 g/mol)+2(16. 00 g/mol) = 44. 01 g/mol Chapter 3 29

The Mole Interconverting Masses, Moles, and Numbers of Particles Chapter 3 30

The Moles « Numbers of Particles Chapter 3 31

The Mole Mass Moles Chapter 3 32

The Moles Mass Chapter 3 33

The Mole A sample of female sex hormone, estradiol, C 18 H 24 O 2, contains 3. 0 x 1020 atoms of hydrogen. How many atoms of carbon does it contain? Chapter 3 34

The Mole A sample of female sex hormone, estradiol, C 18 H 24 O 2, contains 3. 0 x 1020 atoms of hydrogen. How many atoms of carbon does it contain? Chapter 3 35

The Mole A sample of female sex hormone, estradiol, C 18 H 24 O 2, contains 3. 0 x 1020 atoms of hydrogen. How many atoms of carbon does it contain? Chapter 3 36

The Mole A sample of female sex hormone, estradiol, C 18 H 24 O 2, contains 3. 0 x 1020 atoms of hydrogen. How many atoms of carbon does it contain? Chapter 3 37

The Mole A sample of female sex hormone, estradiol, C 18 H 24 O 2, contains 3. 0 x 1020 atoms of hydrogen. How many atoms of carbon does it contain? Chapter 3 38

The Mole A sample of female sex hormone, estradiol, C 18 H 24 O 2, contains 3. 0 x 1020 atoms of hydrogen. How many molecules of estradiol does it contain? Chapter 3 39

The Mole A sample of female sex hormone, estradiol, C 18 H 24 O 2, contains 3. 0 x 1020 atoms of hydrogen. How many molecules of estradiol does it contain? Chapter 3 40

The Mole A sample of female sex hormone, estradiol, C 18 H 24 O 2, contains 3. 0 x 1020 atoms of hydrogen. How many molecules of estradiol does it contain? Chapter 3 41

The Mole A sample of female sex hormone, estradiol, C 18 H 24 O 2, contains 3. 0 x 1020 atoms of hydrogen. How many molecules of estradiol does it contain? Chapter 3 42

The Mole A sample of female sex hormone, estradiol, C 18 H 24 O 2, contains 3. 0 x 1020 atoms of hydrogen. How many molecules of estradiol does it contain? Chapter 3 43

The Mole A sample of female sex hormone, estradiol, C 18 H 24 O 2, contains 3. 0 x 1020 atoms of hydrogen. How many moles of estradiol does it contain? – Since we know the number of molecules, we can calculate the number of moles Chapter 3 44

The Mole A sample of female sex hormone, estradiol, C 18 H 24 O 2, contains 3. 0 x 1020 atoms of hydrogen. How many moles of estradiol does it contain? – Since we know the number of molecules, we can calculate the number of moles Chapter 3 45

The Mole A sample of female sex hormone, estradiol, C 18 H 24 O 2, contains 3. 0 x 1020 atoms of hydrogen. How many moles of estradiol does it contain? – Since we know the number of molecules, we can calculate the number of moles Chapter 3 46

The Mole A sample of female sex hormone, estradiol, C 18 H 24 O 2, contains 3. 0 x 1020 atoms of hydrogen. How many moles of estradiol does it contain? – Since we know the number of molecules, we can calculate the number of moles Chapter 3 47

The Mole A sample of female sex hormone, estradiol, C 18 H 24 O 2, contains 3. 0 x 1020 atoms of hydrogen. What is the mass of this sample in grams? – Now we will convert the moles to grams Chapter 3 48

The Mole A sample of female sex hormone, estradiol, C 18 H 24 O 2, contains 3. 0 x 1020 atoms of hydrogen. What is the mass of this sample in grams? – Now we will convert the moles to grams Chapter 3 49

The Mole A sample of female sex hormone, estradiol, C 18 H 24 O 2, contains 3. 0 x 1020 atoms of hydrogen. What is the mass of this sample in grams? – Now we will convert the moles to grams Chapter 3 50

The Mole A sample of female sex hormone, estradiol, C 18 H 24 O 2, contains 3. 0 x 1020 atoms of hydrogen. What is the mass of this sample in grams? – Now we will convert the moles to grams Chapter 3 51

The Mole A sample of female sex hormone, estradiol, C 18 H 24 O 2, contains 3. 0 x 1020 atoms of hydrogen. What is the mass of this sample in grams? – Now we will convert the moles to grams Chapter 3 52

The Mole A sample of female sex hormone, estradiol, C 18 H 24 O 2, contains 3. 0 x 1020 atoms of hydrogen. What is the mass of this sample in grams? – Now we will convert the moles to grams Chapter 3 53

Empirical Formulas from Analyses Chapter 3 54

Empirical Formulas from Analyses Analysis Hg 73. 9% Cl 26. 1% - assume 100 g sample Hg 73. 9 g Cl 26. 1 g -convert grams to moles Hg 73. 9 g / 200. 59 g/mol = 0. 367 mol Cl 26. 1 g/ 35. 45 g/mol = 0. 736 mol - determine the empirical formula by using the moles of the elements to get the smallest whole number ratio of the elements. Chapter 3 55

Empirical Formulas from Analyses Analysis Hg 73. 9% Cl 26. 1% - assume 100 g sample Hg 73. 9 g Cl 26. 1 g -convert grams to moles Hg 73. 9 g / 200. 59 g/mol = 0. 367 mol Cl 26. 1 g/ 35. 45 g/mol = 0. 736 mol - determine the empirical formula by using the moles of the elements to get the smallest whole number ratio of the elements. Chapter 3 56

Empirical Formulas from Analyses Analysis Hg 73. 9% Cl 26. 1% - assume 100 g sample Hg 73. 9 g Cl 26. 1 g - convert grams to moles Hg 73. 9 g / 200. 59 g/mol = 0. 367 mol Cl 26. 1 g/ 35. 45 g/mol = 0. 736 mol - determine the empirical formula by using the moles of the elements to get the smallest whole number ratio of the elements. Chapter 3 57

Empirical Formulas from Analyses Analysis Hg 73. 9% Cl 26. 1% - assume 100 g sample Hg 73. 9 g Cl 26. 1 g - convert grams to moles Hg 73. 9 g / 200. 59 g/mol = 0. 367 mol Cl 26. 1 g/ 35. 45 g/mol = 0. 736 mol - determine the empirical formula by using the moles of the elements to get the smallest whole number ratio of the elements. Chapter 3 58

Empirical Formulas from Analyses Molecular Formula from Empirical Formula To determine the molecular formula from an empirical formula, you must have the molecular weight of the substance. Chapter 3 59

Empirical Formulas from Analyses Molecular Formula from Empirical Formula To determine the molecular formula from an empirical formula, you must have the molecular weight of the substance. Chapter 3 60

Empirical Formulas from Analyses Molecular Formula from Empirical Formula To determine the molecular formula from an empirical formula, you must have the molecular weight of the substance. Chapter 3 61

Empirical Formulas from Analyses Molecular Formula from Empirical Formula Empirical formula: CH Empirical formula weight: 13. 019 g/mol Molecular weight: 78. 114 g/mol Chapter 3 62

Empirical Formulas from Analyses Molecular Formula from Empirical Formula Empirical formula: CH Empirical formula weight: 13. 019 g/mol Molecular weight: 78. 114 g/mol Chapter 3 63

Empirical Formulas from Analyses Molecular Formula from Empirical Formula Empirical formula: CH Empirical formula weight: 13. 019 g/mol Molecular weight: 78. 114 g/mol Chapter 3 64

Empirical Formulas from Analyses Molecular Formula from Empirical Formula Empirical formula: CH Empirical formula weight: 13. 019 g/mol Molecular weight: 78. 114 g/mol Chapter 3 65

Empirical Formulas from Analyses Molecular Formula from Empirical Formula Empirical formula: CH Empirical formula weight: 13. 019 g/mol Molecular weight: 78. 114 g/mol Chapter 3 66

Empirical Formulas from Analyses Determine the empirical formula of the compound with the following compositions by mass: C, 10. 4%; S, 27. 8%, Cl, 61. 7%. Assume a 100 g sample, so C 10. 4% 10. 4 g S 27. 8% 27. 8 g Cl 61. 7% 61. 7 g Moles of each element C 10. 4 g/12. 011 g/mol = 0. 866 mol S 27. 8 g/32. 066 g/mol = 0. 867 mol Cl 61. 7 g/35. 453 g/mol = 1. 74 mol Chapter 3 67

Empirical Formulas from Analyses Determine the empirical formula of the compound with the following compositions by mass: C, 10. 4%; S, 27. 8%, Cl, 61. 7%. Moles of each element C 10. 4 g/12. 011 g/mol = 0. 866 mol S 27. 8 g/32. 066 g/mol = 0. 867 mol Cl 61. 7 g/35. 453 g/mol = 1. 74 mol Chapter 3 68

Empirical Formulas from Analyses Determine the empirical formula of the compound with the following compositions by mass: C, 10. 4%; S, 27. 8%, Cl, 61. 7%. Moles of each element C 10. 4 g/12. 011 g/mol = 0. 866 mol S 27. 8 g/32. 066 g/mol = 0. 867 mol Cl 61. 7 g/35. 453 g/mol = 1. 74 mol Chapter 3 69

Empirical Formulas from Analyses Determine the empirical formula of the compound with the following compositions by mass: C, 10. 4%; S, 27. 8%, Cl, 61. 7%. Moles of each element C 10. 4 g/12. 011 g/mol = 0. 866 mol S 27. 8 g/32. 066 g/mol = 0. 867 mol Cl 61. 7 g/35. 453 g/mol = 1. 74 mol Chapter 3 70

Empirical Formulas from Analyses Combustion Analysis Chapter 3 71

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0. 1005 g sample of menthol is combusted, producing 0. 2829 g of CO 2 and 0. 1159 g of H 2 O. What is the empirical formula for menthol? Mass of Carbon mass CO 2 moles C grams C Chapter 3 72

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0. 1005 g sample of menthol is combusted, producing 0. 2829 g of CO 2 and 0. 1159 g of H 2 O. What is the empirical formula for menthol? Mass of Carbon mass CO 2 moles C grams C Chapter 3 73

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0. 1005 g sample of menthol is combusted, producing 0. 2829 g of CO 2 and 0. 1159 g of H 2 O. What is the empirical formula for menthol? Mass of Carbon mass CO 2 moles C grams C Chapter 3 74

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0. 1005 g sample of menthol is combusted, producing 0. 2829 g of CO 2 and 0. 1159 g of H 2 O. What is the empirical formula for menthol? Mass of Carbon mass CO 2 moles C grams C Chapter 3 75

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0. 1005 g sample of menthol is combusted, producing 0. 2829 g of CO 2 and 0. 1159 g of H 2 O. What is the empirical formula for menthol? Mass of Carbon mass CO 2 moles C grams C Chapter 3 76

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0. 1005 g sample of menthol is combusted, producing 0. 2829 g of CO 2 and 0. 1159 g of H 2 O. What is the empirical formula for menthol? Mass of Hydrogen mass H 2 O moles H grams H Chapter 3 77

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0. 1005 g sample of menthol is combusted, producing 0. 2829 g of CO 2 and 0. 1159 g of H 2 O. What is the empirical formula for menthol? Mass of Hydrogen mass H 2 O moles H grams H Chapter 3 78

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0. 1005 g sample of menthol is combusted, producing 0. 2829 g of CO 2 and 0. 1159 g of H 2 O. What is the empirical formula for menthol? Mass of Hydrogen mass H 2 O moles H grams H Chapter 3 79

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0. 1005 g sample of menthol is combusted, producing 0. 2829 g of CO 2 and 0. 1159 g of H 2 O. What is the empirical formula for menthol? Mass of Hydrogen mass H 2 O moles H grams H Chapter 3 80

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0. 1005 g sample of menthol is combusted, producing 0. 2829 g of CO 2 and 0. 1159 g of H 2 O. What is the empirical formula for menthol? Mass of Hydrogen mass H 2 O moles H grams H Chapter 3 81

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0. 1005 g sample of menthol is combusted, producing 0. 2829 g of CO 2 and 0. 1159 g of H 2 O. What is the empirical formula for menthol? Mass of Oxygen mass O = mass of sample – (mass C +mass H) Chapter 3 82

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0. 1005 g sample of menthol is combusted, producing 0. 2829 g of CO 2 and 0. 1159 g of H 2 O. What is the empirical formula for menthol? Mass of Oxygen mass O = mass of sample – (mass C +mass H) Chapter 3 83

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0. 1005 g sample of menthol is combusted, producing 0. 2829 g of CO 2 and 0. 1159 g of H 2 O. What is the empirical formula for menthol? Now we can determine the empirical formula Mass of elements: C 0. 07721 g H 0. 01299 g O 0. 01030 g Chapter 3 84

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0. 1005 g sample of menthol is combusted, producing 0. 2829 g of CO 2 and 0. 1159 g of H 2 O. What is the empirical formula for menthol? Now we can determine the empirical formula Moles of elements: C 0. 07721 g/12. 011 g/mol = H 0. 01299 g/1. 01 g/mol = O 0. 01030 g/16. 00 g/mol = Chapter 3 85

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0. 1005 g sample of menthol is combusted, producing 0. 2829 g of CO 2 and 0. 1159 g of H 2 O. What is the empirical formula for menthol? Now we can determine the empirical formula Moles of elements: C 0. 07721 g/12. 011 g/mol = 0. 006428 mol H 0. 01299 g/1. 01 g/mol = 0. 01286 mol O 0. 01030 g/16. 00 g/mol = 0. 0006438 mol Chapter 3 86

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0. 1005 g sample of menthol is combusted, producing 0. 2829 g of CO 2 and 0. 1159 g of H 2 O. What is the empirical formula for menthol? Now we can determine the empirical formula Moles of elements: C 0. 07721 g/12. 011 g/mol = 0. 006428 mol H 0. 01299 g/1. 01 g/mol = 0. 01286 mol O 0. 01030 g/16. 00 g/mol = 0. 0006438 mol Chapter 3 87

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0. 1005 g sample of menthol is combusted, producing 0. 2829 g of CO 2 and 0. 1159 g of H 2 O. What is the empirical formula for menthol? Now we can determine the empirical formula Moles of elements: C 0. 07721 g/12. 011 g/mol = 0. 006428 mol H 0. 01299 g/1. 01 g/mol = 0. 01286 mol O 0. 01030 g/16. 00 g/mol = 0. 0006438 mol Chapter 3 88

Empirical Formulas from Analyses Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0. 1005 g sample of menthol is combusted, producing 0. 2829 g of CO 2 and 0. 1159 g of H 2 O. What is the empirical formula for menthol? Now we can determine the empirical formula Moles of elements: C 0. 07721 g/12. 011 g/mol = 0. 006428 mol H 0. 01299 g/1. 01 g/mol = 0. 01286 mol O 0. 01030 g/16. 00 g/mol = 0. 0006438 mol Chapter 3 89

Quantitative Information • The coefficients in a balanced equation represent both the number of molecules and the number of moles in a reaction. • The coefficients can also be used to derive ratios between any two substances in the chemical reaction. 2 H 2 : 1 O 2 2 H 2 : 2 H 2 O 1 O 2 : 2 H 2 O • The ratios can be used to predict - The amount of product formed - The amount of reactant needed Chapter 3 90

Quantitative Information Chapter 3 91

Quantitative Information 2 C 4 H 10(l) + 13 O 2(g) 8 CO 2(g) + 10 H 2 O(g) How many grams of CO 2 are formed if 1. 00 g of butane (C 4 H 10) is allowed to react with excess oxygen? Chapter 3 92

Quantitative Information 2 C 4 H 10(l) + 13 O 2(g) 8 CO 2(g) + 10 H 2 O(g) How many grams of CO 2 are formed if 1. 00 g of butane (C 4 H 10) is allowed to react with excess oxygen? 1. Moles of C 4 H 10 F. W. 58. 124 g/mol Chapter 3 93

Quantitative Information 2 C 4 H 10(l) + 13 O 2(g) 8 CO 2(g) + 10 H 2 O(g) How many grams of CO 2 are formed if 1. 00 g of butane (C 4 H 10) is allowed to react with excess oxygen? 1. Moles of C 4 H 10 F. W. 58. 124 g/mol Chapter 3 94

Quantitative Information 2 C 4 H 10(l) + 13 O 2(g) 8 CO 2(g) + 10 H 2 O(g) How many grams of CO 2 are formed if 1. 00 g of butane (C 4 H 10) is allowed to react with excess oxygen? 2. Ratio of C 4 H 10: CO 2 3. 2 C 4 H 10 : 8 CO 2 4. or Chapter 3 95

Quantitative Information 2 C 4 H 10(l) + 13 O 2(g) 8 CO 2(g) + 10 H 2 O(g) How many grams of CO 2 are formed if 1. 00 g of butane (C 4 H 10) is allowed to react with excess oxygen? 3. Set-up ratio and proportion between known and unknown quantities Chapter 3 96

Quantitative Information 2 C 4 H 10(l) + 13 O 2(g) 8 CO 2(g) + 10 H 2 O(g) How many grams of CO 2 are formed if 1. 00 g of butane (C 4 H 10) is allowed to react with excess oxygen? 3. Set-up ratio and proportion between known and unknown quantities Chapter 3 97

Quantitative Information 2 C 4 H 10(l) + 13 O 2(g) 8 CO 2(g) + 10 H 2 O(g) How many grams of CO 2 are formed if 1. 00 g of butane (C 4 H 10) is allowed to react with excess oxygen? 3. Set-up ratio and proportion between known and unknown quantities Chapter 3 98

Quantitative Information 2 C 4 H 10(l) + 13 O 2(g) 8 CO 2(g) + 10 H 2 O(g) How many grams of CO 2 are formed if 1. 00 g of butane (C 4 H 10) is allowed to react with excess oxygen? 4. Convert the moles of unknown substance into the desired units Chapter 3 99

Quantitative Information 2 C 4 H 10(l) + 13 O 2(g) 8 CO 2(g) + 10 H 2 O(g) How many grams of CO 2 are formed if 1. 00 g of butane (C 4 H 10) is allowed to react with excess oxygen? 4. Convert the moles of unknown substance into the desired units 5. FW of CO 2: 44. 011 g/mol Chapter 3 100

Quantitative Information 2 C 4 H 10(l) + 13 O 2(g) 8 CO 2(g) + 10 H 2 O(g) How many grams of CO 2 are formed if 1. 00 g of butane (C 4 H 10) is allowed to react with excess oxygen? 4. Convert the moles of unknown substance into the desired units 5. FW of CO 2: 44. 011 g/mol Chapter 3 101

Quantitative Information 2 C 4 H 10(l) + 13 O 2(g) 8 CO 2(g) + 10 H 2 O(g) How many grams of CO 2 are formed if 1. 00 g of butane (C 4 H 10) is allowed to react with excess oxygen? 4. Convert the moles of unknown substance into the desired units 5. FW of CO 2: 44. 011 g/mol Chapter 3 102

Limiting Reactants “What runs out first” 2 C 8 H 18 + 25 O 2 16 CO 2 + 18 H 2 O Chapter 3 103

Limiting Reactants “What runs out first” 2 C 8 H 18 + 25 O 2 16 CO 2 + 18 H 2 O • If you have 2 moles of C 8 H 18 and 20 moles of O 2 all the O 2 will be used and the reaction will stop Chapter 3 104

Limiting Reactants “What runs out first” 2 C 8 H 18 + 25 O 2 16 CO 2 + 18 H 2 O • If you have 2 moles of C 8 H 18 and 20 moles of O 2 all the O 2 will be used and the reaction will stop • O 2 is call the limiting reagent (reactant) Chapter 3 105

Limiting Reactants “What runs out first” 2 C 8 H 18 + 25 O 2 16 CO 2 + 18 H 2 O • If you have 2 moles of C 8 H 18 and 20 moles of O 2 all the O 2 will be used and the reaction will stop • O 2 is call the limiting reagent (reactant) Limiting Reagent – The reagent present in the smallest stoichiometric quantity in a mixture of reactants. Chapter 3 106

Limiting Reactants Example 2 C 8 H 18 + 25 O 2 16 CO 2 + 18 H 2 O Determine the limiting reagent of this reaction if 10. 0 grams of C 8 H 18 and 25. 0 grams of O 2 are allowed to react. Chapter 3 107

Limiting Reactants Example 2 C 8 H 18 + 25 O 2 16 CO 2 + 18 H 2 O Determine the limiting reagent of this reaction if 10. 0 grams of C 8 H 18 and 25. 0 grams of O 2 are allowed to react. 1. Convert grams to moles Chapter 3 108

Limiting Reactants Example 2 C 8 H 18 + 25 O 2 16 CO 2 + 18 H 2 O Determine the limiting reagent of this reaction if 10. 0 grams of C 8 H 18 and 25. 0 grams of O 2 are allowed to react. 1. Convert grams to moles FW(C 8 H 18) 114. 268 g/mol FW(O 2) = 32. 00 g/mol Chapter 3 109

Limiting Reactants Example 2 C 8 H 18 + 25 O 2 16 CO 2 + 18 H 2 O Determine the limiting reagent of this reaction if 10. 0 grams of C 8 H 18 and 25. 0 grams of O 2 are allowed to react. 1. Convert grams to moles FW(C 8 H 18) 114. 268 g/mol FW(O 2) = 32. 00 g/mol Chapter 3 110

Limiting Reactants Example 2 C 8 H 18 + 25 O 2 16 CO 2 + 18 H 2 O Determine the limiting reagent of this reaction if 10. 0 grams of C 8 H 18 and 25. 0 grams of O 2 are allowed to react. 1. Convert grams to moles FW(C 8 H 18) 114. 268 g/mol FW(O 2) = 32. 00 g/mol Chapter 3 111

Limiting Reactants Example 2 C 8 H 18 + 25 O 2 16 CO 2 + 18 H 2 O Determine the limiting reagent of this reaction if 10. 0 grams of C 8 H 18 and 25. 0 grams of O 2 are allowed to react. 2. Divide each reagent by its own coefficient Chapter 3 112

Limiting Reactants Example 2 C 8 H 18 + 25 O 2 16 CO 2 + 18 H 2 O Determine the limiting reagent of this reaction if 10. 0 grams of C 8 H 18 and 25. 0 grams of O 2 are allowed to react. 2. Divide each reagent by its own coefficient Chapter 3 113

Limiting Reactants Example 2 C 8 H 18 + 25 O 2 16 CO 2 + 18 H 2 O Determine the limiting reagent of this reaction if 10. 0 grams of C 8 H 18 and 25. 0 grams of O 2 are allowed to react. 3. The substance with the smallest calculated value will be the limiting reagent. In this case, O 2 is the limiting reagent. Chapter 3 114

Limiting Reactants Theoretical Yields - The calculated amount of product based on the limiting reagent (Theoretical yield). Chapter 3 115

Limiting Reactants Theoretical Yields - The calculated amount of product based on the limiting reagent (Theoretical yield). Chapter 3 116

Practice Problems 3. 6, 3. 18, 3. 32, 3. 40, 3. 48, 3. 60, 3. 72 Chapter 3 117