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Chapter 3 Stoichiometry Chemical Stoichiometry • Stoichiometry – The study of quantities of materials consumed and produced in chemical reactions. Copyright © Cengage Learning. All rights reserved 2 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 12C is the standard for atomic mass, with a mass of exactly 12 atomic mass units (u). The masses of all other atoms are given relative to this standard. On this scale 1H = 1.008 amu 16O = 16.00 amu 3 4 Copper, a metal known since ancient times, is used in electrical cables and pennies, among other things. The atomic masses of its two stable isotopes, (69.09 percent) and (30.91 percent), are 62.93 amu and 64.9278 amu, respectively. Calculate the average atomic mass of copper. The relative abundances are given in parentheses. Each isotope contributes to the average atomic mass based on its relative abundance. (0.6909) (62.93 amu) + (0.3091) (64.9278 amu) = 63.55 amu Multiplying the mass of an isotope by its fractional abundance (not percent) will give the contribution to the average atomic mass of that particular isotope. 5 Average Atomic Mass for Carbon 98.89% of 12 u + 1.11% of 13.0034 u = exact number (0.9889)(12 u) + (0.0111)(13.0034 u) = 12.01 u Copyright © Cengage Learning. All rights reserved 6 EXERCISE! An element consists of 62.60% of an isotope with mass 186.956 u and 37.40% of an isotope with mass 184.953 u. • Calculate the average atomic mass and identify the element. 186.2 u Rhenium (Re) Copyright © Cengage Learning. All rights reserved 7 Schematic Diagram of a Mass Spectrometer Copyright © Cengage Learning. All rights reserved 9 10 The Mole (mol): A unit to count numbers of particles Dozen = 12 Pair = 2 The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12.00 grams of 12C 1 mol = NA = 6.0221415 x 1023 Avogadro’s number (NA) 11 Avogadro’s number (NA) 6.022 x 1023 = NA = 1mol 12 eggs Molar mass is the mass of 1 mole of shoes in grams marbles atoms 1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g 1 12C atom = 12.00 amu 1 mole 12C atoms = 12.00 g 12C 1 mole lithium atoms = 6.941 g of Li For any element atomic mass (amu) = molar mass (grams) 13 One Mole of: S C Hg Cu Fe 14 1 12C atom 12.00 g 1.66 x 10-24 g x = 23 12 12.00 amu 6.022 x 10 C atoms 1 amu M = molar mass in g/mol NA = Avogadro’s number 15 • • • The number equal to the number of carbon atoms in exactly 12 grams of pure 12C. 1 mole of something consists of 6.022 × 1023 units of that substance (Avogadro’s number). 1 mole C = 6.022 × 1023 C atoms = 12.01 g C Copyright © Cengage Learning. All rights reserved 16 EXERCISE! Calculate the number of iron atoms in a 4.48 mole sample of iron. 2.70×1024 Fe atoms Copyright © Cengage Learning. All rights reserved 17 • Mass in grams of one mole of the substance: Molar Mass of N = 14.01 g/mol Molar Mass of H2O = 18.02 g/mol (2 × 1.008 g) + 16.00 g Molar Mass of Ba(NO3)2 = 261.35 g/mol 137.33 g + (2 × 14.01 g) + (6 × 16.00 g) Copyright © Cengage Learning. All rights reserved 18 CONCEPT CHECK! Calculate the number of copper atoms in a 63.55 g sample of copper. 6.022×1023 Cu atoms Copyright © Cengage Learning. All rights reserved 19 CONCEPT CHECK! Which of the following 100.0 g samples contains the greatest number of atoms? a) Magnesium b) Zinc c) Silver Copyright © Cengage Learning. All rights reserved 20 EXERCISE! Rank the following according to number of atoms (greatest to least): a) 107.9 g of silver b) 70.0 g of zinc c) 21.0 g of magnesium b) a) Copyright © Cengage Learning. All rights reserved c) 21 EXERCISE! Consider separate 100.0 gram samples of each of the following: H2O, N2O, C3H6O2, CO2 – Rank them from greatest to least number of oxygen atoms. H2O, CO2, C3H6O2, N2O Copyright © Cengage Learning. All rights reserved 22 Conceptual Problem Solving • Where are we going? – Read the problem and decide on the final goal. • How do we get there? – Work backwards from the final goal to decide where to start. • Reality check. – Does my answer make sense? Is it reasonable? Copyright © Cengage Learning. All rights reserved 23 Determining Chemical Formulas mass of " A" in whole mass % " A" 100% mass of the whole • Mass percent of an element: mass of element in compound mass % = × 100% mass of compound • For iron in iron(III) oxide, (Fe2O3): mass % Fe = 2(55.85 g) 111.70 g = n 100% = 69.94% 2(55.85 g)+3(16.00 g) 159.70 g Copyright © Cengage Learning. All rights reserved 24 Mass Percentages from Formulas • Let’s calculate the percent composition of butane, C4H10. First, we need the molecular mass of C4H10. 4 carbons @ 12.0 amu/atom 48.0 amu 10 hydrogens @ 1.00 amu/atom 10.0 amu 1 molecule of C4 H10 58.0 amu Now, we can calculate the percents. amu C % C 5848.0 .0 amu total 100% 82.8%C amu H % H 5810.0 .0 amu total 100% 17.2%H 25 Percent composition of an element in a compound = n x molar mass of element x 100% molar mass of compound n is the number of moles of the element in 1 mole of the compound 2 x (12.01 g) x 100% = 52.14% 46.07 g 6 x (1.008 g) %H = x 100% = 13.13% 46.07 g 1 x (16.00 g) %O = x 100% = 34.73% 46.07 g %C = C2H6O 52.14% + 13.13% + 34.73% = 100.0% 26 EXERCISE! Consider separate 100.0 gram samples of each of the following: H2O, N2O, C3H6O2, CO2 – Rank them from highest to lowest percent oxygen by mass. H2O, CO2, C3H6O2, N2O Copyright © Cengage Learning. All rights reserved 27 Determining Chemical Formulas • Determining the formula of a compound from the percent composition. The percent composition of a compound leads directly to its empirical formula. An empirical formula (or simplest formula) for a compound is the formula of the substance written with the smallest integer (whole number) subscripts. 28 Formulas (Empirical and Molecular) • • Example: Glucose Empirical formula = CH – Simplest whole-number ratio • Molecular formula = (empirical formula)n [n = integer] • Molecular formula = C6H6 = (CH)6 – Actual formula of the compound Copyright © Cengage Learning. All rights reserved 29 Analyzing for Carbon and Hydrogen • Device used to determine the mass percent of each element in a compound. Copyright © Cengage Learning. All rights reserved 30 EXERCISE! The composition of adipic acid is 49.3% C, 6.9% H, and 43.8% O (by mass). The molar mass of the compound is about 146 g/mol. – What is the empirical formula? C3H5O2 – What is the molecular formula? C6H10O4 Copyright © Cengage Learning. All rights reserved 31 Percent Composition and Empirical Formulas Determine the empirical formula of a compound that has the following percent composition by mass: K 24.75, Mn 34.77, O 40.51 percent. 1 mol K nK = 24.75 g K x = 0.6330 mol K 39.10 g K nMn = 34.77 g Mn x 1 mol Mn = 0.6329 mol Mn 54.94 g Mn nO = 40.51 g O x 1 mol O = 2.532 mol O 16.00 g O 32 Percent Composition and Empirical Formulas nK = 0.6330, nMn = 0.6329, nO = 2.532 0.6330 ~ K: ~ 1.0 0.6329 Mn : 0.6329 = 1.0 0.6329 2.532 ~ O: ~ 4.0 0.6329 KMnO4 33 Determining Chemical Formulas • Determining the empirical formula from the percent composition. Benzoic acid is a white, crystalline powder used as a food preservative. The compound contains 68.8% C, 5.0% H, and 26.2% O by mass. What is its empirical formula? In other words, give the smallest wholenumber ratio of the subscripts in the formula Cx HyOz 34 Determining Chemical Formulas Determining the empirical formula from the percent composition. Our 100.0 grams of benzoic acid would contain: 1 mol C 68.8 g C 5.73 mol C 12.0 g 1 mol H 5.0 g H 5.0 mol H 1.0 g 1 mol O 26.2 g O 1.63mol O 16.0 g This isn’t quite a whole number ratio, but if we divide each number by the smallest of the three, a better ratio might 35 emerge. Determining Chemical Formulas • Determining the empirical formula from the percent composition. Our 100.0 grams of benzoic acid would contain: now it’s not too 5.73 mol C 1.63 3.50 difficult to see that the smallest whole 5.0 mol H 1.63 3.0 number ratio is 1.63mol O 1.63 1.00 7:6:2. The empirical formula is C7H6O2 . 36 Chemical Reactions • A representation of a chemical reaction: C2H5OH + 3O2 reactants • 2CO2 + 3H2O products Reactants are only placed on the left side of the arrow, products are only placed on the right side of the arrow. Copyright © Cengage Learning. All rights reserved 37 Balancing Chemical Equations C2H5OH + 3O2 • • • 2CO2 + 3H2O The equation is balanced. All atoms present in the reactants are accounted for in the products. 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water. Copyright © Cengage Learning. All rights reserved 38 • • The balanced equation represents an overall ratio of reactants and products, not what actually “happens” during a reaction. Use the coefficients in the balanced equation to decide the amount of each reactant that is used, and the amount of each product that is formed. Copyright © Cengage Learning. All rights reserved 39 Writing and Balancing the Equation for a Chemical Reaction 1. Determine what reaction is occurring. What are the reactants, the products, and the physical states involved? 2. Write the unbalanced equation that summarizes the reaction described in step 1. 3. Balance the equation by inspection, starting with the most complicated molecule(s). The same number of each type of atom needs to appear on both reactant and product sides. Do NOT change the formulas of any of the reactants or products. Copyright © Cengage Learning. All rights reserved 40 To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved 41 Chemical Reactions and Chemical Equations A process in which one or more substances is changed into one or more new substances is a chemical reaction A chemical equation uses chemical symbols to show what happens during a chemical reaction 3 ways of representing the reaction of H2 with O2 to form H2O reactants products 42 How to “Read” Chemical Equations 2 Mg + O2 2 MgO 2 atoms Mg + 1 molecule O2 makes 2 formula units MgO 2 moles Mg + 1 mole O2 makes 2 moles MgO 48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO IS NOT 2 grams Mg + 1 gram O2 makes 2 g MgO 43 Chemical Reactions: Equations • Writing chemical equations A chemical equation is the symbolic representation of a chemical reaction in terms of chemical formulas. For example, the burning of sodium and chlorine to produce sodium chloride is written 2Na Cl 2 2NaCl The reactants are starting substances in a chemical reaction. The arrow means “yields.” The formulas on the right side of the arrow represent the products. 44 Chemical Reactions: Equations • Writing chemical equations In many cases, it is useful to indicate the states of the substances in the equation. When you use these labels, the previous equation becomes 2Na(s ) Cl 2 (g ) 2NaCl(s ) 45 Chemical Reactions: Equations • Writing chemical equations The law of conservation of mass dictates that the total number of atoms of each element on both sides of a chemical equation must match. The equation is then said to be balanced. CH 4 O 2 CO 2 H 2O Consider the combustion of methane to produce carbon dioxide and water. 46 Chemical Reactions: Equations • Writing chemical equations For this equation to balance, two molecules of oxygen must be consumed for each molecule of methane, producing one molecule of CO2 and two molecules of water. CH 4 2 O 2 CO 2 2 H 2O Now the equation is “balanced.” 47 Balancing Chemical Equations 1. Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation. Ethane reacts with oxygen to form carbon dioxide and water C2H6 + O2 CO2 + H2O 2. Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts. 2C2H6 NOT C4H12 48 Balancing Chemical Equations 3. Start by balancing those elements that appear in only one reactant and one product. C2H6 + O2 2 carbon on left C2H6 + O2 6 hydrogen on left C2H6 + O2 CO2 + H2O start with C or H but not O 1 carbon on right multiply CO2 by 2 2CO2 + H2O 2 hydrogen on right 2CO2 + 3H2O multiply H2O by 3 49 Balancing Chemical Equations 4. Balance those elements that appear in two or more reactants or products. C2H6 + O2 2 oxygen on left 2CO2 + 3H2O multiply O2 by 7 2 4 oxygen + 3 oxygen = 7 oxygen (3x1) on right (2x2) C2H6 + 7 O2 2 2CO2 + 3H2O 2C2H6 + 7O2 4CO2 + 6H2O remove fraction multiply both sides by 2 50 Balancing Chemical Equations 5. Check to make sure that you have the same number of each type of atom on both sides of the equation. 2C2H6 + 7O2 4CO2 + 6H2O 4 C (2 x 2) 4C 12 H (2 x 6) 12 H (6 x 2) 14 O (7 x 2) 14 O (4 x 2 + 6) Reactants 4C 12 H 14 O Products 4C 12 H 14 O 51 Chemical Reactions: Equations • Balance the following equations. O2 P4 PCl 3 N 2O As2S 3 Ca3 (PO4 )2 O2 POCl 3 P4O6 As2O 3 H 3 PO 4 N2 SO 2 Ca(H2 PO 4 )2 52 Chemical Reactions: Equations • Balance the following equations. O 2 2 PCl 3 2 POCl 3 P4 6 N 2O P4O6 6 N 2 2 As2S 3 9 O 2 2 As2O 3 6 SO 2 Ca3 (PO4 )2 4 H 3 PO 4 3Ca(H2 PO 4 )2 53 EXERCISE! Which of the following correctly balances the chemical equation given below? There may be more than one correct balanced equation. If a balanced equation is incorrect, explain what is incorrect about it. CaO + C CaC2 + CO2 I. II. III. IV. CaO2 + 3C 2CaO + 5C CaO + (2.5)C 4CaO + 10C Copyright © Cengage Learning. All rights reserved CaC2 + CO2 2CaC2 + CO2 CaC2 + (0.5)CO2 4CaC2 + 2CO2 Notice • • • • The number of atoms of each type of element must be the same on both sides of a balanced equation. Subscripts must not be changed to balance an equation. A balanced equation tells us the ratio of the number of molecules which react and are produced in a chemical reaction. Coefficients can be fractions, although they are usually given as lowest integer multiples. Copyright © Cengage Learning. All rights reserved 55 Stoichiometric Calculations • Chemical equations can be used to relate the masses of reacting chemicals. Copyright © Cengage Learning. All rights reserved 56 EXERCISE! Consider the following reaction: P4(s) + 5 O2(g) ® 2 P2O5(s) If 6.25 g of phosphorus is burned, what mass of oxygen does it combine with? 8.07 g O2 Copyright © Cengage Learning. All rights reserved 57 EXERCISE! (Part I) Methane (CH4) reacts with the oxygen in the air to produce carbon dioxide and water. Ammonia (NH3) reacts with the oxygen in the air to produce nitrogen monoxide and water. – Write balanced equations for each of these reactions. Copyright © Cengage Learning. All rights reserved 58 EXERCISE! (Part II) Methane (CH4) reacts with the oxygen in the air to produce carbon dioxide and water. Ammonia (NH3) reacts with the oxygen in the air to produce nitrogen monoxide and water. – What mass of ammonia would produce the same amount of water as 1.00 g of methane reacting with excess oxygen? Copyright © Cengage Learning. All rights reserved 59 Stoichiometry: Quantitative Relations in Chemical Reactions • Stoichiometry is the calculation of the quantities of reactants and products involved in a chemical reaction. It is based on the balanced chemical equation and on the relationship between mass and moles. Such calculations are fundamental to most quantitative work in chemistry. Let’s Think About It • Where are we going? – To find the mass of ammonia that would produce the same amount of water as 1.00 g of methane reacting with excess oxygen. • How do we get there? – We need to know: • • How much water is produced from 1.00 g of methane and excess oxygen. How much ammonia is needed to produce the amount of water calculated above. Copyright © Cengage Learning. All rights reserved 61 Calculating Masses of Reactants and Products in Reactions 1. Balance the equation for the reaction. 2. Convert the known mass of the reactant or product to moles of that substance. 3. Use the balanced equation to set up the appropriate mole ratios. 4. Use the appropriate mole ratios to calculate the number of moles of the desired reactant or product. 5. Convert from moles back to grams if required by the problem. Copyright © Cengage Learning. All rights reserved 62 Calculating Masses of Reactants and Products in Reactions Copyright © Cengage Learning. All rights reserved 63 Limiting Reactants • • Limiting reactant – the reactant that runs out first and thus limits the amounts of products that can be formed. Determine which reactant is limiting to calculate correctly the amounts of products that will be formed. Copyright © Cengage Learning. All rights reserved 64 Limiting Reactants To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved 65 A. The Concept of Limiting Reactants Stoichiometric mixture – N2(g) + 3H2(g) 2NH3(g) A. The Concept of Limiting Reactants Limiting reactant mixture N2(g) + 3H2(g) 2NH3(g) A. The Concept of Limiting Reactants • Limiting reactant mixture • N2(g) + 3H2(g) 2NH3(g) • Limiting reactant is the reactant that runs out first. – H2 Limiting Reactants • • • The amount of products that can form is limited by the methane. Methane is the limiting reactant. Water is in excess. Copyright © Cengage Learning. All rights reserved 69 CONCEPT CHECK! Which of the following reaction mixtures could produce the greatest amount of product? Each involves the reaction symbolized by the equation: 2H2 + O2 a) b) c) d) e) 2H2O 2 moles of H2 and 2 moles of O2 2 moles of H2 and 3 moles of O2 2 moles of H2 and 1 mole of O2 3 moles of H2 and 1 mole of O2 Each produce the same amount of product. Copyright © Cengage Learning. All rights reserved 70 Notice • We cannot simply add the total moles of all the reactants to decide which reactant mixture makes the most product. We must always think about how much product can be formed by using what we are given, and the ratio in the balanced equation. Copyright © Cengage Learning. All rights reserved 71 CONCEPT CHECK! • You know that chemical A reacts with chemical B. You react 10.0 g of A with 10.0 g of B. – What information do you need to know in order to determine the mass of product that will be produced? Copyright © Cengage Learning. All rights reserved 72 Let’s Think About It • Where are we going? – To determine the mass of product that will be produced when you react 10.0 g of A with 10.0 g of B. • How do we get there? – We need to know: • • The mole ratio between A, B, and the product they form. In other words, we need to know the balanced reaction equation. The molar masses of A, B, and the product they form. Copyright © Cengage Learning. All rights reserved 73 EXERCISE! You react 10.0 g of A with 10.0 g of B. What mass of product will be produced given that the molar mass of A is 10.0 g/mol, B is 20.0 g/mol, and C is 25.0 g/mol? They react according to the equation: A + 3B 2C Copyright © Cengage Learning. All rights reserved 74 Limiting Reagent • Zinc metal reacts with hydrochloric acid by the following reaction. Zn(s) 2 HCl(aq) ZnCl 2 (aq) H 2 (g ) If 0.30 mol Zn is added to hydrochloric acid containing 0.52 mol HCl, how many moles of H2 are produced? 75 Limiting Reagent Take each reactant in turn and ask how much product would be obtained if each were totally consumed. The reactant that gives the smaller amount is the limiting reagent. 1 mol H 2 0.30 mol Zn 0.30 mol H 2 1 mol Zn 1 mol H 2 0.52 mol HCl 0.26 mol H 2 2 mol HCl Since HCl is the limiting reagent, the amount of H2 produced must be 0.26 mol (0.26*2 = 0.52 g) 76 Reaction Yield Theoretical and Percent Yield • The theoretical yield of product is the maximum amount of product that can be obtained from given amounts of reactants (if all the limiting reagent reacted). • Actual Yield is the amount of product actually obtained from a reaction (experimentally determined). The percentage yield is the actual yield (experimentally determined) expressed as a percentage of the theoretical yield (calculated). actual yield %Yield 100% theoretical yield 77 Theoretical and Percent Yield • To illustrate the calculation of percentage yield, recall that the theoretical yield of H2 in the previous example was 0.26 mol (or 0.52 g) H2. If the actual yield of the reaction had been 0.22 g H2, then 0.22 g H 2 %Yield 100% 42% 0.52 g H 2 • % Yield is an important indicator of the efficiency of a particular laboratory or industrial reaction. 78 EXERCISE! Consider the following reaction: P4(s) + 6F2(g) 4PF3(g) – What mass of P4 is needed to produce 85.0 g of PF3 if the reaction has a 64.9% yield? 46.1 g P4 Copyright © Cengage Learning. All rights reserved 79