Chemical Calculations 7.1-7.7 CHAPTER 7 Chemical Calculations 7.1-7.7 1

The Mole A number of atoms, ions, or molecules that is large enough to see and handle. A mole = number of things Just like a dozen = 12 things One mole = 6.022 x 1023 things Avogadro’s number = 6.022 x 1023 Symbol for Avogadro’s number is NA. 16

The Mole 17

The Mole How do we know when we have a mole? count it out weigh it out Molar mass - mass in grams numerically equal to the atomic weight of the element in grams. H has an atomic weight of 1.00794 g 1.00794 g of H atoms = 6.022 x 1023 H atoms Mg has an atomic weight of 24.3050 g 24.3050 g of Mg atoms = 6.022 x 1023 Mg atoms 17

The Mole 17

The Mole Example 2-1: Calculate the mass of a single Mg atom in grams to 3 significant figures. 18

The Mole Example 2-1: Calculate the mass of a single Mg atom in grams to 3 significant figures. 19

The Mole Example 2-1: Calculate the mass of a single Mg atom in grams to 3 significant figures. 20

The Mole Example 2-1: Calculate the mass of a single Mg atom, in grams, to 3 significant figures. 21

The Mole Example 2-2: Calculate the number of atoms in one-millionth of a gram of Mg to 3 significant figures. 22

The Mole Example 2-2: Calculate the number of atoms in one-millionth of a gram of Mg to 3 significant figures. 23

The Mole Example 2-2: Calculate the number of atoms in one-millionth of a gram of Mg to 3 significant figures. 24

The Mole Example 2-2: Calculate the number of atoms in one-millionth of a gram of Mg to 3 significant figures. 25

The Mole Example 2-3. How many atoms are contained in 1.67 moles of Mg? 28

The Mole Example 2-3. How many atoms are contained in 1.67 moles of Mg? 28

The Mole Example 2-3. How many atoms are contained in 1.67 moles of Mg? 28

The Mole Example 2-3. How many atoms are contained in 1.67 moles of Mg? 28

The Mole Example 2-3. How many atoms are contained in 1.67 moles of Mg? 28

The Mole Example 2-4: How many moles of Mg atoms are present in 73.4 g of Mg? You do it! 29

The Mole Example 2-4: How many moles of Mg atoms are present in 73.4 g of Mg? 31

The Mole Example 2-4: How many moles of Mg atoms are present in 73.4 g of Mg? 31

IT IS IMPERATIVE THAT YOU KNOW HOW TO DO THESE PROBLEMS The Mole Example 2-4: How many moles of Mg atoms are present in 73.4 g of Mg? IT IS IMPERATIVE THAT YOU KNOW HOW TO DO THESE PROBLEMS 31

Formula Weights, Molecular Weights, and Moles How do we calculate the molar mass of a compound? add atomic weights of each atom The molar mass of propane, C3H8, is: 32

Formula Weights, Molecular Weights, and Moles The molar mass of calcium nitrate, Ca(NO3)2 , is: You do it! 32

Formula Weights, Molecular Weights, and Moles 33

Formula Weights, Molecular Weights, and Moles One Mole of Contains Cl2 or 70.90g 6.022 x 1023 Cl2 molecules 2(6.022 x 1023 ) Cl atoms C3H8 You do it! 34

Formula Weights, Molecular Weights, and Moles One Mole of Contains Cl2 or 70.90g 6.022 x 1023 Cl2 molecules 2(6.022 x 1023 ) Cl atoms C3H8 or 44.11 g 6.022 x 1023 C3H8 molecules 3 (6.022 x 1023 ) C atoms 8 (6.022 x 1023 ) H atoms 34

Formula Weights, Molecular Weights, and Moles Example 2-5: Calculate the number of C3H8 molecules in 74.6 g of propane. 38

Formula Weights, Molecular Weights, and Moles Example 2-5: Calculate the number of C3H8 molecules in 74.6 g of propane. 38

Formula Weights, Molecular Weights, and Moles Example 2-5: Calculate the number of C3H8 molecules in 74.6 g of propane. 38

Formula Weights, Molecular Weights, and Moles Example 2-5: Calculate the number of C3H8 molecules in 74.6 g of propane. 38

Formula Weights, Molecular Weights, and Moles Example 2-8. Calculate the number of O atoms in 26.5 g of Li2CO3. You do it! 40

Formula Weights, Molecular Weights, and Moles Example 2-8. Calculate the number of O atoms in 26.5 g of Li2CO3. 41

Percent Composition % composition = mass of an individual element in a compound divided by the total mass of the compound x 100% Determine the percent composition of C in C3H8. 45

Percent Composition and Formulas of Compounds What is the percent composition of H in C3H8? You do it! 46

Percent Composition What is the percent composition of H in C3H8? 47

Percent Composition Example 2-10: Calculate the percent composition of Fe2(SO4)3 to 3 significant figures. You do it! 48

Percent Composition Example 2-10: Calculate the percent composition of Fe2(SO4)3 to 3 sig. fig. 49

Percent Composition 49

Empirical and Molecular Formulas Empirical Formula - smallest whole-number ratio of atoms present in a compound Molecular Formula - actual numbers of atoms of each element present in a molecule of the compound We determine the empirical and molecular formulas of a compound from the percent composition of the compound. 50

Empirical And Molecular Formulas 51

Empirical Formulas Example 2-11: A compound contains 24.74% K, 34.76% Mn, and 40.50% O by mass. What is its empirical formula? Make the simplifying assumption that we have 100.0 g of compound. In 100.0 g of compound there are: 24.74 g of K 34.76 g of Mn 40.50 g of O 51

Empirical Formulas 52

Empirical Formulas 53

Empirical Formulas 54

Empirical Formulas 55

Empirical Formulas 56

Empirical Formulas Example 2-12: A sample of a compound contains 6.541g of Co and 2.368g of O. What is empirical formula for this compound? You do it! 57

Empirical Formulas Example 2-12: A sample of a compound contains 6.541g of Co and 2.368g of O. What is empirical formula for this compound? 58

Empirical Formulas Example 2-12: A sample of a compound contains 6.541g of Co and 2.368g of O. What is empirical formula for this compound? 59

Molecular Formulas Example 2-13: A compound is found to contain 85.63% C and 14.37% H by mass. In another experiment its molar mass is found to be 56.1 g/mol. What is its molecular formula? short cut method 61

Determination of Molecular Formulas 62

Calculations Based on Chemical Equations

Calculations Based on Chemical Equations Example 3-1: How many CO molecules are required to react with 25 formula units of Fe2O3?

Calculations Based on Chemical Equations Example 3-2: How many iron atoms can be produced by the reaction of 2.50 x 105 formula units of iron (III) oxide with excess carbon monoxide?

Calculations Based on Chemical Equations Example 3-2: How many iron atoms can be produced by the reaction of 2.50 x 105 formula units of iron (III) oxide with excess carbon monoxide?

Calculations Based on Chemical Equations Example 3-2: How many iron atoms can be produced by the reaction of 2.50 x 105 formula units of iron (III) oxide with excess carbon monoxide?

Calculations Based on Chemical Equations Example 3-3: What mass of CO is required to react with 146 g of iron (III) oxide?

Calculations Based on Chemical Equations Example 3-3: What mass of CO is required to react with 146 g of iron (III) oxide?

Calculations Based on Chemical Equations Example 3-3: What mass of CO is required to react with 146 g of iron (III) oxide?

Calculations Based on Chemical Equations Example 3-4: What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with carbon monoxide?

Calculations Based on Chemical Equations Example 3-4: What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with carbon monoxide?

Calculations Based on Chemical Equations Example 3-4: What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with carbon monoxide?

Calculations Based on Chemical Equations Example 3-5: What mass of iron (III) oxide reacted with carbon monoxide if the carbon dioxide produced by the reaction had a mass of 8.65 grams? You do it!

Calculations Based on Chemical Equations Example 3-5: What mass of iron (III) oxide reacted with carbon monoxide if the carbon dioxide produced by the reaction had a mass of 8.65 grams?

Calculations Based on Chemical Equations Example 3-6: How many pounds of carbon monoxide would react with 125 pounds of iron (III) oxide? You do it!

Calculations Based on Chemical Equations YOU MUST BE PROFICIENT WITH THESE TYPES OF PROBLEMS!!!

Limiting Reactant Concept Kitchen example of limiting reactant concept. 1 packet of muffin mix + 2 eggs + 1 cup of milk 12 muffins How many muffins can we make with the following amounts of mix, eggs, and milk?

Limiting Reactant Concept Mix Packets Eggs Milk 1 1 dozen 1 gallon limiting reactant is the muffin mix 2 1 dozen 1 gallon 3 1 dozen 1 gallon 4 1 dozen 1 gallon 5 1 dozen 1 gallon 6 1 dozen 1 gallon 7 1 dozen 1 gallon limiting reactant is the dozen eggs

Limiting Reactant Concept Example 3-7: Suppose a box contains 87 bolts, 110 washers, and 99 nuts. How many sets, each consisting of one bolt, two washers, and one nut, can you construct from the contents of one box?

Limiting Reactant Concept Example 3-8: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110. g of oxygen?

Limiting Reactant Concept Example 3-8: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110. g of oxygen?

Limiting Reactant Concept Example 3-8: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110. g of oxygen?

Limiting Reactant Concept Example 3-8: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110. g of oxygen? Determine which mass makes the most product

Limiting Reactant Concept

Limiting Reactant Concept Which is limiting reactant? Limiting reactant is O2. What is maximum mass of sulfur dioxide? Maximum mass is 147 g.

Percent Yields from Reactions Theoretical yield is calculated by assuming that the reaction goes to completion. Determined from the limiting reactant calculation. Actual yield is the amount of a specified pure product made in a given reaction. In the laboratory, this is the amount of product that is formed in your beaker, after it is purified and dried. Percent yield indicates how much of the product is obtained from a reaction.

Percent Yields from Reactions Example 3-9: A 10.0 g sample of ethanol, C2H5OH, was boiled with excess acetic acid, CH3COOH, to produce 14.8 g of ethyl acetate, CH3COOC2H5. What is the percent yield?

Percent Yields from Reactions

Percent Yields from Reactions

Percent Yields from Reactions