Stoichiometry: Calculations with Chemical Formulas and Equations

Stoichiometry: Calculations with Chemical Formulas and Equations

Lecture Presentation Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations 2012 Pearson Education, Inc. Dr. Subhash C. Goel South GA State College Douglas, GA Chemical Equations

Chemical equations are concise representations of chemical reactions. Stoichiometry 2015 Pearson Education, Inc. What Is in a Chemical Equation? CH4(g) + 2O2(g) CO2(g) + 2H2O(g) Reactants appear on the left side of the equation. Stoichiometry

2015 Pearson Education, Inc. What Is in a Chemical Equation? CH4(g) + 2O2(g) CO2(g) + 2H2O(g) Products appear on the right side of the equation. Stoichiometry 2015 Pearson Education, Inc. What Is in a Chemical Equation? CH4(g) + 2O2(g)

CO2(g) + 2H2O(g) The states of the reactants and products are written in parentheses to the right of each compound. (g) = gas; (l) = liquid; (s) = solid; (aq) = in aqueous solution 2015 Pearson Education, Inc. Stoichiometry What Is in a Chemical Equation? CH4(g) + 2O2(g)

CO2(g) + 2H2O(g) Coefficients are inserted to balance the equation to follow the law of conservation of mass. Stoichiometry 2015 Pearson Education, Inc. Why Do We Add Coefficients Instead of Changing Subscripts to Balance? Hydrogen and oxygen can make water OR hydrogen peroxide: 2 H2(g) + O2(g) 2 H2O(l) H2(g) + O2(g) H2O2(l)

Stoichiometry 2015 Pearson Education, Inc. Balancing Chemical Equation Exercise 1: Balance following equations: Na(s) + H2O(l) NaOH(aq) + H2(g) Stoichiometry 2012 Pearson Education, Inc. Balancing Chemical Equations Exercise 2

Balance these equations by providing the missing coefficients: Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc. Three Types of Reactions Combination reactions Decomposition reactions Combustion reactions Stoichiometry 2015 Pearson Education, Inc.

Combination Reactions In combination reactions two or more substances react to form one product. Examples: 2Mg(s) + O2(g) 2MgO(s) N2(g) + 3H2(g) 2NH3(g) C3H6(g) + Br2(l) C3H6Br2(l) 2012 Pearson Education, Inc.

Stoichiometry Decomposition Reactions In a decomposition reaction one substance breaks down into two or more substances. Examples: CaCO3(s) CaO(s) + CO2(g) 2KClO3(s) 2KCl(s) + O2(g) 2NaN3(s) 2Na(s) + 3N2(g) Sodium azide is used in air-bags of cars. 2012 Pearson Education, Inc.

Stoichiometry Exercise 3 Write a balanced equation for (a) the combination reaction between lithium metal and fluorine gas and (b) the decomposition reaction that occurs when solid barium carbonate is heated (two products form, a solid and a gas). Solution Exercise 4 Write a balanced equation for (a) solid mercury(II) sulfide decomposing into its component elements when heated and (b) aluminum metal combining with

oxygen in the air. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc. Combustion Reactions Combustion reactions are generally rapid reactions that produce a flame. Combustion reactions most often involve

hydrocarbons reacting with oxygen in the air. Examples: CH4(g) + 2O2(g) CO2(g) + 2H2O(g) C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) 2012 Pearson Education, Inc. Stoichiometry Exercise 5: Writing Balanced Equations for Combustion Reactions Write the balanced equation for the reaction that occurs when methanol,CH 3OH(l), is burned in air. Exercise 6

Write the balanced equation for the reaction that occurs when ethanol, C2H5OH(l), burns in air. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc. Formula Weights Stoichiometry 2012 Pearson Education, Inc.

Micro World atoms & molecules Macro World grams Atomic mass is the mass of an atom in atomic mass units (amu) By definition: 1 atom 12C weighs 12 amu On this scale H = 1.008 amu 1

O = 16.00 amu 16 3.1 Formula Weight (FW) A formula weight is the sum of the atomic weights for the atoms in a chemical formula. So, the formula weight of calcium chloride, CaCl2, would be Ca: 1(40.08 amu) + Cl: 2(35.453 amu)

110.99 amu Formula weights are generally reported for ionic compounds. Stoichiometry 2012 Pearson Education, Inc. Molecular Weight (MW) A molecular weight is the sum of the atomic weights of the atoms in a molecule. For the molecule ethane, C2H6, the molecular weight would be C: 2(12.011 amu) + H: 6(1.00794 amu)

30.070 amu Stoichiometry 2012 Pearson Education, Inc. Exercise 7: Calculating Formula Weights Calculate the formula weight of (a) sucrose, C12H22O11 (table sugar), and (b) calcium nitrate, Ca(NO3)2. Exercise 8 Calculate the formula weight of (a) Al(OH)3 and (b) CH3OH. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward

2012 Pearson Education, Inc. Percent Composition One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation: % Element = (number of atoms)(atomic mass) (FW of the compound)

x 100 Stoichiometry 2012 Pearson Education, Inc. Percent Composition So the percentage of carbon in ethane is %C = (2)(12.011 amu) (30.070 amu) 24.022 amu

x 100 = 30.070 amu = 79.887% Stoichiometry 2012 Pearson Education, Inc. Exercise 9: Calculating Percentage Composition Calculate the percentage of carbon, hydrogen, and oxygen (by mass) in C12H22O11. Chemistry, The Central Science, 12th Edition

Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc. Avogadros Number and Moles Stoichiometry 2012 Pearson Education, Inc. Avogadros Number In a lab, we cannot work with individual molecules. They are

too small. 6.02 1023 atoms or molecules is an amount that brings us to lab size. It is ONE MOLE. One mole of 12C has a mass of 12.000 g. 2015 Pearson Education, Inc. Stoichiometry Avogadros Number 6.02 x 1023 particles

1 mole or 1 mole 6.02 x 1023 particles Molar Mass By definition, a molar mass is the mass of 1 mol of a substance (i.e., g/mol). The molar mass of an element is the mass number for the element that we find on the periodic table. The formula weight (in amus) will be the same number as the molar mass (in g/mol).

Stoichiometry 2012 Pearson Education, Inc. A Moles of Particles Contains 6.02 x 1023 particles 1 mole C = 6.02 x 1023 C atoms 1 mole H2O = 6.02 x 1023 H2O molecules 1 mole NaCl

= 6.02 x 1023 Na+ ions and 6.02 x 1023 Cl ions Examples of Moles Moles of elements 1 mole Mg = 6.02 x 1023 Mg atoms 1 mole Au = 6.02 x 1023 Au atoms Moles of compounds 1 mole NH3 = 6.02 x 1023 NH3 molecules 1 mole C9H8O4 = 6.02 x 1023 aspirin molecules

Exercise 10: How many atoms are in 0.551 g of potassium (K) ? 1 mol K = 39.10 g K 1 mol K = 6.022 x 1023 atoms K 1 mol K 6.022 x 1023 atoms K 0.551 g K x x = 1 mol K 39.10 g K 8.49 x 1021 atoms K

Using Moles Moles provide a bridge from the molecular scale to the real-world scale. Stoichiometry 2012 Pearson Education, Inc. Mole Relationships One mole of atoms, ions, or molecules contains Avogadros number of those particles. One mole of molecules or formula units contains Avogadros number times the number of atoms or

ions of each element in the compound. Stoichiometry 2012 Pearson Education, Inc. Exercise 11: Converting Grams to Moles Calculate the number of moles of glucose (C6H12O6) in 5.380 g of C6H12O6. Exercise 12 How many moles of sodium bicarbonate (NaHCO3) are in 508 g of NaHCO3?

Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc. Finding Empirical Formulas Stoichiometry 2012 Pearson Education, Inc. Calculating Empirical Formulas One can calculate the empirical formula from

the percent composition. Stoichiometry 2012 Pearson Education, Inc. Calculating Empirical Formulas Example 13: The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA. Stoichiometry 2012 Pearson Education, Inc.

Calculating Empirical Formulas Assuming 100.00 g of para-aminobenzoic acid, C: H: N: O: 1 mol 12.01 g 1 mol 5.14 g x 1.01 g 1 mol

10.21 g x 14.01 g 1 mol 23.33 g x 16.00 g 61.31 g x = 5.105 mol C = 5.09 mol H = 0.7288 mol N = 1.456 mol O Stoichiometry 2012 Pearson Education, Inc.

Calculating Empirical Formulas Calculate the mole ratio by dividing by the smallest number of moles: C: 7 H: 7 N: O: 5.105 mol 0.7288 mol 5.09 mol

0.7288 mol = 7.005 = 6.984 0.7288 mol 0.7288 mol 1.458 mol 0.7288 mol = 1.000 Stoichiometry

2012 Pearson Education, Inc. = 2.001 Calculating Empirical Formulas These are the subscripts for the empirical formula: C7H7NO2 Stoichiometry 2012 Pearson Education, Inc. Combustion Analysis Compounds containing C, H, and O are routinely analyzed

through combustion in a chamber like the one shown in Figure 3.14. C is determined from the mass of CO2 produced. H is determined from the mass of H2O produced. O is determined by difference after the C and H have been determined. Stoichiometry 2012 Pearson Education, Inc. Quantitative Relationships The coefficients in the balanced equation show relative numbers of molecules of reactants and products. relative numbers of moles of reactants and

products, which can be converted to mass. Stoichiometry 2015 Pearson Education, Inc. Stoichiometric Calculations 2 Na + Cl2 2 NaCl 2 atoms of sodium combines with one molecule of chlorine to produce two molecules of sodium chloride. In terms of mole, 2 moles of sodium combines with one mole of chlorine to produce two mole of sodium chloride.

In grams, 2 moles Na = 2x23 = 46 g 1 mole of Cl2 = 2x 35.45 = 70.9 g 2 moles of NaCl = 2(23 + 35.45) = 116.9 g (Note: Law of Conservation of Mass holds here) The coefficients in the balanced equation give the ratio of moles of reactants and products. Stoichiometry 2012 Pearson Education, Inc. Stoichiometric Calculations Starting with the mass of Substance A, you can use

the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if its a product) or used (if its a reactant). Stoichiometry 2012 Pearson Education, Inc. Stoichiometric Calculations C6H12O6 + 6 O2 6 CO2 + 6 H2O Starting with 1.00 g of C6H12O6

we calculate the moles of C6H12O6 use the coefficients to find the moles of H2O and then turn the moles of water to grams. 2012 Pearson Education, Inc. Stoichiometry Exercise 14 Calculating Amounts of Reactants and Products Solid lithium hydroxide is used in space vehicles to remove the carbon dioxide gas exhaled by astronauts. The hydroxide reacts with the carbon dioxide to form solid lithium carbonate and liquid water. How many grams of carbon dioxide can be absorbed by 1.00 g of lithium hydroxide? 2 LiOH(s) + CO2(g)

Li2CO3(s) + H2O(l) Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc. Limiting Reactants Stoichiometry 2012 Pearson Education, Inc. Limiting Reactants The limiting reactant is the reactant present in

the smallest stoichiometric amount. In other words, its the reactant youll run out of first (in this case, the H2). Stoichiometry 2012 Pearson Education, Inc. Limiting Reactants In the example below, the O2 would be the excess reagent. Stoichiometry 2012 Pearson Education, Inc.

Exercise 15 Calculating the Amount of Product Formed from a Limiting Reactant The most important commercial process for converting N 2 from the air into nitrogen-containing compounds is based on the reaction of N2 and H2 to form ammonia (NH3): N2(g) + 3 H 2(g) 2 NH3(g) How many moles of NH3 can be formed from 3.0 mol of N2 and 6.0 mol of H2? Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc.

Exercise 16 Calculating the Amount of Product Formed from a Limiting Reactant The reaction 2H2 (g) + O 2(g) 2H2 O(g) is used to produce electricity in a hydrogen fuel cell. Suppose a fuel cell contains 150 g of H2(g) and 1500 g of O2(g) (each measured to two significant figures). How many grams of water can form? Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward

2012 Pearson Education, Inc. Theoretical Yield The theoretical yield is the maximum amount of product that can be made. In other words, its the amount of product possible as calculated through the stoichiometry problem. This is different from the actual yield, which is the amount one actually produces and measures. Stoichiometry 2012 Pearson Education, Inc.

Percent Yield One finds the percent yield by comparing the amount actually obtained (actual yield) to the amount it was possible to make (theoretical yield): Percent yield = actual yield theoretical yield x 100 Stoichiometry

2012 Pearson Education, Inc. Exercise 17: Calculating Theoretical Yield and Percent Yield Adipic acid, H2C 6H8O4, used to produce nylon, is made commercially by a reaction between cyclohexane (C 6H 12) and O2: 2 C6H12(l) + 5 O2(g) 2 H2C 6H8O4(l) + 2 H 2O(g) (a) Assume that you carry out this reaction with 25.0 g of cyclohexane and that cyclohexane is the limiting reactant. What is the theoretical yield of adipic acid? (b) If you obtain 33.5 g of adipic acid, what is the percent yield for the reaction? Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward

2012 Pearson Education, Inc. Exercise 17 Continued Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc. ? Calculate the formula weight of the

following compounds from their formulas. Report your answers to three significant figures. calcium hydroxide, Ca(OH)2 methylamine, CH3NH2 Copyright Cengage Learning. All rights reserved. 3 | 55 Ca(OH)2 1 Ca 1(40.08) = 40.08 amu 2O

2(16.00) = 32.00 amu 2H 2(1.008) = 2.016 amu 2 decimal places Total 74.096 74.10 amu CH3NH2 1C 1(12.01) = 12.01 amu 1N 1(14.01) = 14.01 amu 5H 5(1.008) = 5.040 amu

Total 31.060 2 significant figures 31.06 amu Copyright Cengage Learning. All rights reserved. 3 | 56 ? What is the mass in grams of the nitric acid molecule, HNO3? First, find the molar mass of HNO3:

1 H 1(1.008) = 1 N 1(14.01) = 3 O 3(16.00) = Copyright Cengage Learning. All rights reserved. 1.008 14.01 48.00 63.018 (2 decimal places) 63.02 g/mol 3 | 57

? A sample of nitric acid, HNO3, contains 0.253 mol HNO3. How many grams is this? First, find the molar mass of HNO3: 1 H 1(1.008) = 1 N 1(14.01) = 3 O 3(16.00) = (2 decimal places) 63.02 g/mol Copyright Cengage Learning. All rights reserved.

1.008 14.01 48.00 63.018 Note: We need one digit more in the molar mass than in the measured

quantity. 3 | 58 Next, using the molar mass, find the mass of 0.253 mole: 63.02 g 0.253 mole 1mole = 15.94406 g 15.9 g (3 significant figures) Copyright Cengage Learning. All rights reserved.

3 | 59 ? Calcite is a mineral composed of calcium carbonate, CaCO3. 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 CaCO3: 1 Ca 1(40.08) = 1 C 1(12.01) =

3 O 3(16.00) = Copyright Cengage Learning. All rights reserved. 40.08 12.01 48.00 100.09 2 decimal places 100.09 g/mol 3 | 60 Next, find the number of moles in 23.6 g:

1mole 23.6 g 100.09 g 2.35787791 10 1 g 1 2.36 10 g or 0.236 g (3 significant figures) Copyright Cengage Learning. All rights reserved. 3 | 61

? The average daily requirement of the essential amino acid leucine, C6H14O2N, is 2.2 g for an adult. What is the average daily requirement of leucine in moles? First, find the molar mass of leucine: 6C 2O 1N 14 H 6(12.01) =

2(16.00) = 1(14.01) = 14(1.008) = 72.06 32.00 14.01 14.112 132.182 Copyright Cengage Learning. All rights reserved. 2 decimal places 132.18 g/mol

3 | 62 Next, find the number of moles in 2.2 g: 1mole 2.2 g 132.18 g 1.6643 10 2 mol 1.7 10 2 mol or 0.017 mol (2 significant figures) Copyright Cengage Learning. All rights reserved. 3 | 63

? The daily requirement of chromium in the human diet is 1.0 10-6 g. How many atoms of chromium does this represent? Copyright Cengage Learning. All rights reserved. 3 | 64 First, find the molar mass of Cr: 1 Cr 1(51.996) = 51.996

Now, convert 1.0 x 10-6 grams to moles: 23 1mol 6.022 10 atoms 1.0 10 g 51.996 g 1mol 6 =1.158166 1016 atoms 1.2 1016 atoms (2 significant figures)

Copyright Cengage Learning. All rights reserved. 3 | 65 ? Lead(II) chromate, PbCrO4, is used as a paint pigment (chrome yellow). What is the percentage composition of lead(II) chromate? First, find the molar mass of PbCrO4: 1 Pb 1(207.2) = 207.2 1 Cr 1(51.996) = 51.996

4 O 4(16.00) = 64.00 323.196 Copyright Cengage Learning. All rights reserved. (1 decimal place) 323.2 g/mol 3 | 66 Now, convert each to percent composition: Pb : Cr :

O: 207.2 g 100% 64.11% 323.20 g 51.996 g 100% 16.09% 323.20 g 64.00 g 100% 19.80% 323.20 g

Check: 64.11 + 16.09 + 19.80 = 100.00 Copyright Cengage Learning. All rights reserved. 3 | 67 ? The chemical name of table sugar is sucrose, C12H22O11. How many grams of carbon are in 68.1 g of sucrose? First, find the molar mass of C12H22O11: 12 C 12(12.01) = 144.12

11 O 11(16.00) = 176.00 22 H 22(1.008) = 22.176 342.296 Copyright Cengage Learning. All rights reserved. (2 decimal places) 342.30 g/mol 3 | 68 Now, find the mass of carbon in 68.1 g sucrose: 144.12 g carbon 68.1g sucrose 342.30 g sucrose

28.7 g carbon (3 significant figures) Copyright Cengage Learning. All rights reserved. 3 | 69 ? Propane, C3H8, 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: C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

How many grams of O2 are required to burn 20.0 g of propane? Copyright Cengage Learning. All rights reserved. 3 | 70 Molar masses: O2 2(16.00) = 32.00 g C3H8 3(12.01) + 8(1.008) = 44.094 g 1 mol C 3 H8 5 mol O 2 32.00 g O 2 20.0 g C 3 H8

44.094 g C 3 H8 1 mol C 3 H8 1 mol O 2 72.5722320 5 g O 2 72.6 g O2 (3 significant figures) Copyright Cengage Learning. All rights reserved. 3 | 71

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