Chapter 3 Calculations with Chemical Formulas and Equations

Contents and Concepts Mass and Moles of Substances Here we will establish a critical relationship between the mass of a chemical substance and the quantity of that substance (in moles). 1. Molecular Mass and Formula Mass 2. The Mole Concept Copyright Cengage Learning. All rights reserved. 3 2

Determining Chemical Formulas Explore how the percentage composition and mass percentage of the elements in a chemical substance can be used to determine the chemical formula. 3. Mass Percentages from the Formula 4. Elemental Analysis: Percentages of C, H, and O 5. Determining Formulas Copyright Cengage Learning. All rights reserved. 3 3

Stoichiometry: Quantitative Relations in Chemical Reactions Develop a molar interpretation of chemical equations, which then allows for calculation of the quantities of reactants and products. 6. Molar Interpretation of a Chemical Equation 7. Amounts of Substances in a Chemical Equation 8. Limiting Reactant: Theoretical and Percentage Yield Copyright Cengage Learning. All rights reserved. 3 4

Learning Objectives Mass and Moles of Substances 1. Molecular Mass and Formula Mass a. Define the terms molecularand formula mass of a substance. b. Calculate the formula mass from a formula. c. Calculate the formula mass from molecular models. Copyright Cengage Learning. All rights reserved. 3 5

2. The Mole Concept a. Define the quantity called the mole. b. Learn Avogadro s number. c. Understand how the molar mass is related to the formula mass of a substance. d. Calculate the mass of atoms and molecules. e. Perform calculations using the mole. f. Convert from moles of substance to grams of substance. g. Convert from grams of substance to moles of substance. h. Calculate the number of molecules in a given mass of a substance. Copyright Cengage Learning. All rights reserved. 3 6

Determining Chemical Formulas 3. Mass Percentages from the Formula a. Define mass percentage. b. Calculate the percentage composition of the elements in a compound. c. Calculate the mass of an element in a given mass of compound. Copyright Cengage Learning. All rights reserved. 3 7

4. Elemental Analysis: Percentages of C, H, and O a. Describe how C, H, and O combustion analysis is performed. b. Calculate the percentage of C, H, and O from combustion data. Copyright Cengage Learning. All rights reserved. 3 8

5. Determining Formulas a. Define empirical formula. b. Determine the empirical formula of a binary compound from the masses of its elements. c. Determine the empirical formula from the percentage composition. d. Understand the relationship between the molecular mass of a substance and its empirical formula mass. e. Determine the molecular formula from the percentage composition and molecular mass. Copyright Cengage Learning. All rights reserved. 3 9

Stoichiometry: Quantitative Relations in Chemical Reactions 6. Molar Interpretation of a Chemical Equation a. Relate the coefficients in a balanced chemical equation to the number of molecules or moles (molar interpretation). Copyright Cengage Learning. All rights reserved. 3 10

7. Amounts of Substances in a Chemical Equation a. Use the coefficients in a balanced chemical equation to perform calculations. b. Relate the quantities of reactant to the quantity of product. c. Relate the quantities of two reactants or two products. Copyright Cengage Learning. All rights reserved. 3 11

8. Limiting Reactant: Theoretical and Percentage Yield a. Understand how a limiting reactant determines how many moles of product are formed during a chemical reaction and how much excess reactant remains. b. Calculate with a limiting reactant involving moles. c. Calculate with a limiting reactant involving masses. d. Define and calculate the theoretical yield of chemical reactions. e. Determine the percentage yield of a chemical reaction. Copyright Cengage Learning. All rights reserved. 3 12

Molecular Mass The sum of the atomic masses of all the atoms in a molecule of the substance. Formula Mass The sum of the atomic masses of all atoms in a formula unit of the compound, whether molecular or not. Copyright Cengage Learning. All rights reserved. 3 13

Calculate the formula weight of the following compounds from their formulas. Report your answers to three significant figures. calcium hydroxide, Ca(OH) 2 methylamine, CH 3 NH 2 Copyright Cengage Learning. All rights reserved. 3 14

Ca(OH) 2 1 Ca 1(40.08) = 40.08 amu 2 O 2(16.00) = 32.00 amu 2 H 2(1.008) = 2.016 amu Total 74.095 3 significant figures 74.1 amu CH 3 NH 2 1 C 1(12.01) = 12.01 amu 1 N 1(14.01) = 14.01 amu 5 H 5(1.008) = 5.040 amu Total 31.060 3 significant figures 31.1 amu Copyright Cengage Learning. All rights reserved. 3 15

What is the mass in grams of the nitric acid molecule, HNO 3? First, find the molar mass of HNO 3 : 1 H 1(1.008) = 1.008 1 N 1(14.01) = 14.01 3 O 3(16.00) = 48.00 63.018 (2 decimal places) 63.02 g/mol Copyright Cengage Learning. All rights reserved. 3 16

Mole, mol The quantity of a given amount of substance that contains as many molecules or formula units as the number of atoms in exactly 12 g of carbon-12. Avogadro s Number, N A The number of atoms in exactly 12 g of carbon-12 N A = 6.02 10 23 (to three significant figures). Copyright Cengage Learning. All rights reserved. 3 17

Next, convert this mass of one mole to one molecule using Avogadro s number: 63.02 g mol x 6.02 x10 1mol 23 molecules 1.046843854 x10 22 g Themassof one HNO 22 3 molecule is 1.05 x10 g. (3 significant figures) Copyright Cengage Learning. All rights reserved. 3 18

A sample of nitric acid, HNO 3, contains 0.253 mol HNO 3. How many grams is this? First, find the molar mass of HNO 3 : 1 H 1(1.008) = 1.008 1 N 1(14.01) = 14.01 3 O 3(16.00) = 48.00 63.018 (2 decimal places) 63.02 g/mol Copyright Cengage Learning. All rights reserved. 3 20

Calcite is a mineral composed of calcium carbonate, CaCO 3. A sample of calcite composed of pure calcium carbonate weighs 23.6 g. How many moles of calcium carbonate is this? First, find the molar mass of CaCO 3 : 1 Ca 1(40.08) = 40.08 1 C 1(12.01) = 12.01 3 O 3(16.00) = 48.00 100.09 2 decimal places 100.09 g/mol Copyright Cengage Learning. All rights reserved. 3 22

The average daily requirement of the essential amino acid leucine, C 6 H 14 O 2 N, is 2.2 g for an adult. What is the average daily requirement of leucine in moles? First, find the molar mass of leucine: 6 C 6(12.01) = 72.06 2 O 2(16.00) = 32.00 1 N 1(14.01) = 14.01 14 H 14(1.008) = 14.112 132.182 2 decimal places 132.18 g/mol Copyright Cengage Learning. All rights reserved. 3 24

First, find the molar mass of Cr: 1 Cr 1(51.996) = 51.996 Now, convert 1.0 x 10-6 grams to moles: 1.0 x10 6 g x 1mol 51.996 g x 6.02 x10 23 1mol atoms =1.157781368 x 10 16 atoms 1.2 x 10 16 atoms (2 significant figures) Copyright Cengage Learning. All rights reserved. 3 27

Lead(II) chromate, PbCrO 4, is used as a paint pigment (chrome yellow). What is the percentage composition of lead(ii) chromate? First, find the molar mass of PbCrO 4 : 1 Pb 1(207.2) = 207.2 1 Cr 1(51.996) = 51.996 4 O 4(16.00) = 64.00 323.196 (1 decimal place) 323.2 g/mol Copyright Cengage Learning. All rights reserved. 3 28

Now, convert each to percent composition: Pb: Cr: 207.2 g x100% 64.11% 323.20 g 51.996 g x100% 16.09% 323.20 g O : 64.00 g 323.20 g x100% 19.80% Check: 64.11 + 16.09 + 19.80 = 100.00 Copyright Cengage Learning. All rights reserved. 3 29

The chemical name of table sugar is sucrose, C 12 H 22 O 11. How many grams of carbon are in 68.1 g of sucrose. First, find the molar mass of C 12 H 22 O 11 : 12 C 12(12.01) = 144.12 11 O 11(16.00) = 176.00 22 H 22(1.008) = 22.176 342.296 (2 decimal places) 342.30 g/mol Copyright Cengage Learning. All rights reserved. 3 30

Percentage Composition The mass percentage of each element in the compound. The composition is determined by experiment, often by combustion. When a compound is burned, its component elements form oxides for example, CO 2 and H 2 O. The CO 2 and H 2 O are captured and weighed to determine the amount of C and H in the original compound. Copyright Cengage Learning. All rights reserved. 3 32

Benzene is a liquid compound composed of carbon and hydrogen; it is used in the preparation of polystyrene plastic. A sample of benzene weighing 342 mg is burned in oxygen and forms 1156 mg of carbon dioxide. What is the percentage composition of benzene? Copyright Cengage Learning. All rights reserved. 3 33

Strategy 1. Use the mass of CO 2 to find the mass of carbon from the benzene. 2. Use the mass of benzene and the mass of carbon to find the mass of hydrogen. 3. Use these two masses to find the percent composition. Copyright Cengage Learning. All rights reserved. 3 34

First, find the mass of C in 1156 mg of CO 2 : 1156 x 10 3 g CO 2 1mol CO 2 44.01 g CO 2 x 1mol C 1mol CO 2 x 12.01 g C 1mol CO 2 3.15463758 1 1mg 2 x10 gc 10 3 g = 315.5 mg C Copyright Cengage Learning. All rights reserved. 3 35

Next, find the mass of H in the benzene sample: 342 mg benzene -315.5 mg C 26.5 mg H (the decimal is not significant) Now, we can find the percentage composition: 315.5 mg x100% 92.3% C 342 mg 26.5 mg x100% 7.7% H 342 mg Copyright Cengage Learning. All rights reserved. 3 36

Empirical Formula (Simplest Formula) The formula of a substance written with the smallest integer subscripts. For example: The empirical formula for N 2 O 4 is NO 2. The empirical formula for H 2 O 2 is HO Copyright Cengage Learning. All rights reserved. 3 37

Determining the Empirical Formula Beginning with percent composition: 1. Assume exactly 100 g so percentages convert directly to grams. 2. Convert grams to moles for each element. 3. Manipulate the resulting mole ratios to obtain whole numbers. Copyright Cengage Learning. All rights reserved. 3 38

Manipulating the ratios: Divide each mole amount by the smallest mole amount. If the result is not a whole number: Multiply each mole amount by a factor. For example: If the decimal portion is 0.5, multiply by 2. If the decimal portion is 0.33 or 0.67, multiply by 3. If the decimal portion is 0.25 or 0.75, multiply by 4. Copyright Cengage Learning. All rights reserved. 3 39

Benzene is composed of 92.3% carbon and 7.7% hydrogen. What is the empirical formula of benzene? 1mol C 92.3 g C 7.685 mol 12.01 g C 1mol H 7.7 g H 7.64 mol H 1.008 g H C 7.685 1 7.64 7.64 1 7.64 Empirical formula: CH Copyright Cengage Learning. All rights reserved. 3 40

To determine the molecular formula: 1. Compute the empirical formula weight. 2. Find the ratio of the molecular weight to the empirical formula weight. molecular weight n empirical formula weight 3. Multiply each subscript of the empirical formula by n. Copyright Cengage Learning. All rights reserved. 3 42

Benzene has the empirical formula CH. Its molecular weight is 78.1 amu. What is its molecular formula? Empirical formula weight 13.02 amu 78.1 13.02 6 Molecular formula C 6 H 6 Copyright Cengage Learning. All rights reserved. 3 43

Sodium pyrophosphate is used in detergent preparations. It is composed of 34.5% Na, 23.3% P, and 42.1% O. What is its empirical formula? 1mol Na 34.5 g Na 22.99 g Na 1mol P 23.3 g P 30.97 g P 1.501 mol Na 0.7523 mol P 1mol O 42.1 g O 2.631 mol O 16.00 g O 1.501 0.7523 0.7523 0.7523 2.631 0.7523 2.00 1.00 3.50 x x x 2 2 2 Empirical formula Na 4 P 2 O 7 4 2 7 Copyright Cengage Learning. All rights reserved. 3 44

Hexamethylene is one of the materials used to produce a type of nylon. It is composed of 62.1% C, 13.8% H, and 24.1% N. Its molecular weight is 116 amu. What is its molecular formula? 1mol C 62.1 g C 12.01 g C 13.8 1mol H g H 1.008 g H 1mol N 24.1 g N 14.01 g N 5.171 mol 13.69 mol 1.720 mol Copyright Cengage Learning. All rights reserved. 3 45 C H H 5.171 1.720 13.69 1.720 1.720 1.720 3 8 1 Empirical formula C 3 H 8 N

The empirical formula is C 3 H 8 N. Find the empirical formula weight: 3(12.01) + 8(1.008) + 1(14.01) = 58.104 amu n 116 58.10 2 Molecular formula: C 6 H 16 N 2 Copyright Cengage Learning. All rights reserved. 3 46

Stoichiometry The calculation of the quantities of reactants and products involved in a chemical reaction. Interpreting a Chemical Equation The coefficients of the balanced chemical equation may be interpreted in terms of either (1) numbers of molecules (or ions or formula units) or (2) numbers of moles, depending on your needs. Copyright Cengage Learning. All rights reserved. 3 47

To find the amount of B (one reactant or product) given the amount of A (another reactant or product): 1. Convert grams of A to moles of A Using the molar mass of A 2. Convert moles of A to moles of B Using the coefficients of the balanced chemical equation 3. Convert moles of B to grams of B Using the molar mass of B Copyright Cengage Learning. All rights reserved. 3 48

Propane, C 3 H 8, is normally a gas, but it is sold as a fuel compressed as a liquid in steel cylinders. The gas burns according to the following equation: C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(g) How many grams of CO 2 are produced when 20.0 g of propane is burned? Copyright Cengage Learning. All rights reserved. 3 49

Molar masses C 3 H 8 : 3(12.01) + 8(1.008) = 44.094 g CO 2 : 1(12.01) + 2(16.00) = 44.01 g 20.0 g C 3 H 8 1mol 44.094 C 3 H g C 3 8 H 8 3 mol 1mol CO C 3 H 2 8 44.01 g CO 1mol CO 2 2 59.88569873 g CO 2 59.9 g CO 2 (3 significant figures) Copyright Cengage Learning. All rights reserved. 3 50

Propane, C 3 H 8, is normally a gas, but it is sold as a fuel compressed as a liquid in steel cylinders. The gas burns according to the following equation: C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(g) How many grams of O 2 are required to burn 20.0 g of propane? Copyright Cengage Learning. All rights reserved. 3 51

Molar masses: O 2 2(16.00) = 32.00 g C 3 H 8 3(12.01) + 8(1.008) = 44.094 g 20.0 g C 3 H 8 1mol 44.094 C 3 H g C 3 8 H 8 5 mol 1mol C O 3 2 H 8 32.00 1mol g O O 2 2 72.5722320 5 g O 2 72.6 g O 2 (3 significant figures) Copyright Cengage Learning. All rights reserved. 3 52

Limiting Reactant The reactant that is entirely consumed when a reaction goes to completion. Once one reactant has been completely consumed, the reaction stops. Any problem giving the starting amount for more than one reactant is a limiting reactant problem. Copyright Cengage Learning. All rights reserved. 3 53

All amounts produced and reacted are determined by the limiting reactant. How can we determine the limiting reactant? 1. Use each given amount to calculate the amount of product produced. 2. The limiting reactant will produce the lesser or least amount of product. Copyright Cengage Learning. All rights reserved. 3 54

Magnesium metal is used to prepare zirconium metal, which is used to make the container for nuclear fuel (the nuclear fuel rods): ZrCl 4 (g) + 2Mg(s) 2MgCl 2 (s) + Zr(s) How many moles of zirconium metal can be produced from a reaction mixture containing 0.20 mol ZrCl 4 and 0.50 mol Mg? Copyright Cengage Learning. All rights reserved. 3 55

0.20 mol ZrCl 1mol Zr 4 1mol ZrCl4 0.20 mol Zr 1mol Zr 0.50 molmg 2molMg 0.25 mol Zr ZrCl 4 is the limiting reactant. 0.20 mol Zr will be produced. Copyright Cengage Learning. All rights reserved. 3 56

Urea, CH 4 N 2 O, is used as a nitrogen fertilizer. It is manufactured from ammonia and carbon dioxide at high pressure and high temperature: 2NH 3 + CO 2 (g) CH 4 N 2 O + H 2 O In a laboratory experiment, 10.0 g NH 3 and 10.0 g CO 2 were added to a reaction vessel. What is the maximum quantity (in grams) of urea that can be obtained? How many grams of the excess reactant are left at the end of the reactions? Copyright Cengage Learning. All rights reserved. 3 57

Molar masses NH 3 1(14.01) + 3(1.008) = 17.02 g CO 2 1(12.01) + 2(16.00) = 44.01 g CH 4 N 2 O 1(12.01) + 4(1.008) + 2(14.01) + 1(16.00) = 60.06 g 10.0 10.0 gnh gco 3 2 1mol 17.024 1mol NH CO 2 44.01gCO 3 gnh 2 3 1mol 1mol 1mol CH CH 2mol CO 60.06 1mol 13.6 gch CH CO 2 is the limiting reactant. 13.6 g CH 4 N 2 O will be produced. gch Copyright Cengage Learning. All rights reserved. 3 58 4 N 4 2 2 N NH O 2 3 O 60.06 1mol gch CH 17.6 gch 4 4 N N 2 4 N 2 O 4 4 N 2 O N 4 2 2 O N O O 2 O

To find the excess NH 3, we find how much NH 3 reacted: 1mol CO2 2molNH3 17.02 gnh 10.0 gco2 44.01gCO2 1mol CO2 1molNH3 7.73460577 1g NH 3 7.73 gnh3 reacted Now subtract the amount reacted from the starting amount: 10.0 at start -7.73 reacted 2.27 g remains 2.3 g NH 3 is left unreacted. (1 decimal place) 3 Copyright Cengage Learning. All rights reserved. 3 59

Actual Yield The amount of product that is actually obtained. This is a measured amount. Percentage Yield percentage yield actual yield theoretica l yield x100% Copyright Cengage Learning. All rights reserved. 3 61

2NH 3 + CO 2 (g) CH 4 N 2 O + H 2 O When 10.0 g NH 3 and 10.0 g CO 2 are added to a reaction vessel, the limiting reactant is CO 2. The theoretical yield is 13.6 of urea. When this reaction was carried out, 9.3 g of urea was obtained. What is the percent yield? Theoretical yield = 13.6 g Actual yield = 9.3 g 9.3g 13.6g x 100% = 68% yield (2 significant figures) Copyright Cengage Learning. All rights reserved. 3 62