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A chemical formula of a compound indicates the number ratio of combining atoms. Chapter 3 Stoichiometry of Chemical Formulas And Equations Subscript: indicates the # of bonded H atoms Leading coefficients: indicate the number of molecules or atoms. We can view covalent and ionic compounds at a atomic level and use special names to describe mass. The Molecular mass is the sum of the atomic masses (in amu) in a single molecule (COVALENT). 1A 2A 3A 4A 5A 6A 7A NN2 O2 O FF2 PP4 SS 8 Cl Cl 2 Se Se 8 Br Br 2 HH 2 2 2 1S 2O SO2 SO2 32.07 amu 4 + 2 x 16.00 amu 64.07 amu NaCl 35.45 amu 58.45 amu Chemistry needs to count #’s of reactants but how? + + 4 amu + 1 O2 32 amu ? grams + ? grams 8 2 2 2 I2I 2 23.00 amu Chemical reactions require specific numbers of molecules or formula units to react. If we can’t see molecules to count them, then how do we do accomplish the counting task? 2H2 2 8 The Formula mass is the sum of atomic masses (in amu) in a single ionic formula unit (IONIC). 1Na 1Cl NaCl 8A Some elements occur in nature as molecules and not as (1) (18) discrete elements as written in the(15) Periodic Table. (2) (13) (14) (16) (17) tetratomic molecules octatomic molecules diatomic molecules Chemistry counts atoms and molecules indirectly by comparing masses of the same number of things using ratio of masses built into the periodic table. 2H2O 36 amu ? grams How do we connect the atomic world to the big world? Masses 12 red marbles 12 yellow marbles 12 purple marbles 12 green marbles 84 g 48 g 40.g 20.g 12 red marble = 84 grams 12 yellow marble = 48 grams 1 red must be 1.75 times heavier than 1 yellow marble (84g/48g =1.75) The scale has 1 mole (6.02 X 1023) of atoms Fe on one side and 1 mole of S atoms on the other. 32.07 g S 55.85 g Fe Think of the mole as a counting unit for a collection of: 6.02 x 1023 things---unlike all other counting numbers it is also linked to masses given in the periodic table!! Special Numbers Defined Unit 12 H2O molecules Fe and S have the same number of atoms counted out but they don’t weigh the same! 55.85 g Fe = 6.022 x 1023 atoms Fe = 1 mol Fe atoms 32.07 g S = 6.022 x 1023 atoms S = 1 mol S atoms KEY RIFF: The periodic table tells us the mass of 1atom of any element in amu, AND it also tells us the mass of 1-mole (6.02 x 1023) of an element in grams! Mass Connection 1 dozen H2O molecules 144 H2O molecules 1 gross H2O molecules 6.02 x 1023 molecules 1 mole H2O molecules none none 18.0 grams of H2O The mole links 6.02 x 1023 entities (atoms, molecules or formula units) to a mass in grams . Atomic Size MOLE molecular mass of water molar mass of water Thinking Small and Thinking Big Covalent Compounds Thinking Small and Thinking Big Ionic Compounds The Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a single molecule. The formula mass (or forumula weight) is the sum of the atomic masses (in amu) in a formula unit. 1S SO2 2O SO2 32.07 amu 1 Na 23.00 amu + 2 x 16.00 amu 64.07 amu 1 Cl 35.45 amu 58.45 amu The Molar mass of a compound is the same number as the molecular mass with units of grams per mole. NaCl The molar mass of an ionic compound is numerically equal to the formula mass, in units of grams/mole. Molar mass of NaCl 1 molecule SO2 = 64.07 amu 1 mole SO2 = 64.07 g SO2 = 58.45 g/mol 1 mole of NaCl(s) = 58.45 grams NaCl(s) 6.022!1023 NaCl units = 58.45 grams NaCl(s) Do You Understand Molecular and Molar Mass? What is the molecular mass of glucose, C6H12O6 Glucose’s molecular formula C6H12O6 Molecular Mass: 6 x 12.011 + 12 x 1.007 + 6 x 15.99 = 180.18 amu Molar Mass: = 180.18 grams C6H12O6 which contains 1 mol C6H12O6 or 6.022 X 1023 molecules C6H12O6 A chemical formula provides lots of information Example: C6H12O6 (MW = 180.16 g/mol) Carbon (C) Hydrogen (H) Oxygen (O) 6 atoms 12 atoms 6 atoms 6(12.01 amu) =72.06 amu 12(1.008 amu) =12.10 amu 6(16.00 amu) =96.00 amu 1023) 1023) Atoms in 1 glucose molecule Mass of 1 glucose molecule Atoms/mole of compound Moles of atoms in 1 mole of compound 1023) 6(6.022 x atoms 12(6.022 x atoms 6(6.022 x atoms 6 moles of atoms 12 moles of atoms 6 moles of atoms 72.06 g 12.10 g 96.00 g Mass/mole of compound Molecular formulas indicates the type and the ratios of combining atoms in a covalent compound. H2O(l) covalent compound There are many conversion factors in a formula. 1 molecule H2O = 1 atom O and 2 atoms of H Molecular mass H2O = (2 x 1.008)+15.99 = 18.00 amu Molar mass = 18.00 g/mol H2O 1 mole H2O = 6.022 x 1023 molecules H2O 1 mole H2O = 2 mol H atoms = 2 x 6.02 x 1023 H atoms 1 mole H2O = 1 mol O atoms = 6.02 x 1023 O atoms We must learn how to recognize conversion factors within a chemical formula, and how to convert grams to moles to molecules to atoms and vis versa. MASS(g) of compound Molar mass (g/mol) MOLES of compound Chemical Formula MOLES of elements in compound Avogadro’s Number MOLECULES or FORMULA UNITS ATOMS in a compound A chemical formula or molecular formula provides lots of information!! Al2(SO4)3 2 atoms of Al and 3 molecules of (SO4)2- = 1 formula unit Al2(SO4)3 2 moles of Al and 3 moles of (SO4)2- = 1 formula unit Al2(SO4)3 1 formula unit Al2(SO4)3 = 342.17 amu Al2(SO4)3 1 mole Al2(SO4)3 = 342.17 g Al2(SO4)3 1 mole Al2(SO4)3 = 6.022 x 1023 formula units Al2(SO4)3 1 mole Al2(SO4)3 = 2 mol Al3+ ions 1 mole Al2(SO4)3 = 3 mol (SO4)2- ions 1 mole Al2(SO4)3 = 12 mol O atoms = 12 x A.N. O atoms 1 mole Al2(SO4)3 = 3 mol S atoms = 3 x A.N. S atoms 1 mole Al2(SO4)3 = 2 mol Al atoms =2 x A.N. Al atoms Calculating the Mass and the Number of Atoms in a Given Number of Moles of an Element (1.) Silver (Ag) is used in jewelry and tableware but no longer in U.S. coins. How many grams of Ag are in 0.0342mol of Ag? g Ag = ? = 0.0342 mol Ag x 107.9 g Ag = 3.69g Ag 1 mol Ag (2). Iron (Fe), the main component of steel, is the most important metal in industrial society. How many Fe atoms are in 95.8g of Fe? 23 Fe atoms = 95.8 g Fe x 1 mol Fe x 6.022x10 atoms Fe 55.85g Fe 1 mol Fe = 1.04x1024 atoms Fe (3). How many molecules of nitrogen dioxide are in 8.92 g of nitrogen dioxide? 1 mol NO2 6.02x1023 molec NO2 molcs NO2=8.92g NO2 46.01g NO2 1 mol NO2 = 1.17x1023 molecules NO2 What’s In A Chemical Formula? Urea, (NH2)2CO, is a nitrogen containing compound used as a fertilizer around the globe? Calculate the following for 25.6 g of urea: a) the molar mass of urea? b) the number of moles of urea in 25.6 g urea? b) # of molecules of urea in 25.6 g of urea? c) # hydrogen atoms present in 25.6 g of urea. Solution To Urea Problem 1. Calculate the molar mass (MM) of urea, (NH2)2CO MM(NH2)2CO = 2 x MN + 4 x MH + MC + MO MM(NH2)2CO = (2 x 14.07g) + 4 (1.007g) + 12.01g + 15.99g MM(NH2)2CO = 60.06 g 2. # moles of (NH2)2CO in 25.6 g M ol (NH2 )2 CO = 25.6 g (NH2 )2 CO × 1 mol (NH2 )2 CO = 0.426 mol (NH2 )2 CO 60.06 g (NH2 )2 CO 3. # molecules of (NH2)2CO in 25.6 g M olec urea = 0.426 mol urea × 6.02 × 1023 molecu urea = 2.57 × 1023 molecu urea 1 mol urea 4. # H atoms in 25.6 g (NH2)2CO #H atoms = 0.426 mol urea × Borax is the common name of a mineral sodium tetraborate, an industrial cleaning adjunct, Na2B4O7. You are given 20.0 g of borax...... (a) what is the formula mass of Na2B4O7 (b) how many moles of borax is 20.0 g? (c) how many moles of boron are present in 20.0 g Na2B4O7? (d) how many grams of boron are present in 20.0 g Na2B4O7? (e) how many atoms of B are present in 20.0g? (f) how many atoms of O are present in 20.0g? (g) how many grams of atomic oxygen are present? 4 mol Hatoms 6.02 × 1023 H atoms × = 1.03 × 1024 atoms 1 mol urea 1 mol urea Borax, Na2B4O7, is the common name of a sodium tetraborate, an industrial cleaning adjunct. Suppose you are given 20.0 g of borax: Solution: (a) The formula mass of Na2B4O7 is (2 ! 23.0) + (4 ! 10.8) + (7 ! 16.0) = 201.2 g. b) # mol borax = (20.0 g borax) / (201.2 g mol–1) = 0.9940 = 0.01 mol of borax, c) # mol B = 20.0 g borax)/(201.2 g mol–1) X 4 mol B/ 1 mol borax = 0.40 mol of B. d) # g B = (0.40 mol) ! (10.8 g B/ mol B) = 4.3 g B e) # atoms B = 0.40 mol B x 6.02 X 1023 atoms B/1 mol B = 2.41 X 1023 atoms B f) # g O = 20.0 g borax)/(201.2 g mol–1) X 7 mol O/ 1 mol borax x 15.99 g O/1 mol O = 11.1 g O A chemical formula determines the % mass of each element in a compound. n x molar mass of element x 100% molar mass of compound n is the number of moles of an element in 1 mole of the compound. Example: C2H4O2 Molar mass = 60.05 g/mole 2 x (12.01 g) x 100% = 40.0% 60.05 g 4 x (1.008 g) %H = x 100% = 6.714% 60.05 g 2 x (16.00 g) %O = x 100% = 53.28% 60.05 g %C = C2H4O2 40.0% + 6.71% + 53.2% = 100.0% An empirical formula shows the simplest most reduced whole-number ratio of the atoms in a compound. A molecular formula shows the whole number ratio of an actual known molecule. molecular H 2O empirical H 2O C6H12O6 CH2O C12H24O12 CH2O C18H36O18 CH2O O3 N 2H 4 O NH2 N8H16 NH2 If we know the chemical formula of a compound, we can determine the % mass of its elements and vis versa. All of the compounds below have the same % by mass and the same empirical formula! All have the same % by mass! %C = a% %H = b% %O = c% 40.0% C Empirical Formula % Mass Element Molar Mass CxHyOz Name formaldehyde acetic acid lactic acid erythrose ribose glucose If we know the % mass of elements in a compound we can determine the empirical formula. With molar mass we get molecular formula Molecular formula GIVEN Molar Mass Assume 100 g sample Molecular Formula CH2O C2H4O2 C3H6O3 C4H8O4 C5H10O5 C6H12O6 6.71% H 53.3% O. Whole-Number M Multiple (g/mol) 1 2 3 4 5 6 Use or Function 30.03 disinfectant; biological preservative 60.05 acetate polymers; vinegar(5% soln) 90.09 sour milk; forms in exercising muscle 120.10 part of sugar metabolism 150.13 component of nucleic acids and B2 180.16 major energy source of the cell Determining the Empirical Formula from Masses of Elements Elemental analysis of a sample of an ionic compound gave the following results: 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. Determine the empirical formula and the name of this compound? Calculate mole ratio Molar Mass Determining the Empirical Formula from Masses of Elements (Elemental Analysis) Elemental analysis of a sample of an ionic compound gave the following results: 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. Determine the empirical formula and the name of this compound? SOLUTION: 2.82 g Na x mol Na = 0.123 mol Na 22.99 g Na 4.35 g Cl x mol Cl 35.45 g Cl = 0.123 mol Cl mol O 7.83 g O x 16.00 g O = 0.489 mol O Na1 Cl1 O3.98 NaClO4 NaClO4 is sodium perchlorate. Determining a Molecular Formula from Elemental Analysis and Molar Mass Dibutyl succinate is an insect repellent used against household ants and roaches. Elemental analysis or analysis indicates that % mass of the composition is 62.58% C, 9.63% H and 27.79% O. Its experimentally determined molecular mass is 230 u. What are the empirical and molecular formulas of dibutyl succinate? Step 1: Determine the mass of each element. Assume a 100 gram sample and use the given data we have: C 62.58 g H 9.63 g O 27.79 g Step 5: Convert to a small whole number ratio. Multiply by 2 to get C5.98H10.98O2 The empirical formula is C6H11O2 Empirical formula mass = 6(12.01) + 1.008(11) + 2 (16.00) = 115 u. Step 2: Convert masses to amounts in moles. Step 6: Now using the empirical formula mass and molecular mass together determine the molecular formula. Empirical formula mass is 115 u. Molecular formula mass is 230 u. Step 3: Write a tentative empirical formula. C5.21H9.55O1.74 Step 4: Convert to small whole numbers. Divide by smallest number of moles C2.99H5.49O Determining a Molecular Formula from Elemental Analysis and Molar Mass During physical activity, lactic acid (M = 90.08 g/mol) forms in muscle tissue and is responsible for muscle soreness at fatigue. Elemental analysis shows that this compound contains 40.0 mass% C, 6.71 mass% H, and 53.3 mass% O. (a) Determine the empirical formula of lactic acid. (b) Determine the molecular formula. Understand what is asked: What is the formula CxHyOz Mass Percent Empirical Formula Molecular Mass Molecular Formula Chemical equations are symbolic representations of “what happens” in a chemical reaction. CH4(g) + O2(g) CO2(g) + H2O(g) unbalanced => No mass conservation CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) The molecular formula is C12H22O4 Determining a Molecular Formula from Elemental Analysis and Molar Mass 1. Assume there are 100. g of lactic acid then use % mass: 40.0 g C 1 mol C 12.01g C 6.71 g H 1 mol H 1.008 g H 3.33 mol C 53.3 g O 6.66 mol H 1 mol O 16.00 g O 3.33 mol O 2. The red numbers are the number of moles of atoms in lactic acid. This is what we use in the formula C3.33 H6.66 O3.33 C3.33 H6.66 O3.33 3.33 3.33 3.33 CH2O empirical formula 3. The molecular formula will be a whole number multiple of the empirical formula determined BY THE MOLAR MASS ratio molar mass of lactate mass of CH2O 90.08 g 30.03 g 3 C3H6O3 is the molecular formula Only balanced chemical equations contain useful stoichiometric conversion factors. NH3 + O2 ===> NO + H2 1 moles NH3 = 1 moles O2 6NH3 + 3O2 ===> 6NO + 9H2 Not Balanced & NOT TRUE! Balance first and ITʼs VALID! Correct Stoichiometric Conversion Factors + Reactants n = Molecular mass/empirical mass = 230 amu/115 amu = 2 “Yields” Products 6 mol NH3 = 3 mol O2 6 mol NH3 = 9 mol H2 3 mol O2 = 9 mol H2 6 mol NH3 = 6 mol NO 3 mol O2 = 6 mol NO 6 mol NO = 9 mol H2 We balance equations using trial and error! __Na3PO4(aq) + __HCl(aq) ==>__H3PO4(aq) + __NaCl(aq) Solutions Na3PO4(aq) + 3HCl(aq) ====> H3PO4(aq) + 3NaCl(aq) __Ba(OH)2(aq) + __HCl(aq) ==> __H2O(l) + __BaCl2(aq) Ba(OH)2(aq) + 2HCl(aq) ====> 2H2O(l) + BaCl2(aq) __C7H14 + __O2 ===> __H2O(l) + __CO2(g) 2C7H14 + 21O2 ====> 14H2O(l) + 14CO2(g) To learn how to balance equations quickly for Exams you have to practice! Try it. Na2SO4 (s) + C(s) C3H8(g) + 5O2(g) Na2S (s) + CO(g) NaOH + H2 (g) Na + H2O A balanced chemical equation contains a lot of very useful chemical information. Learn it and everything gets easy! 3CO2(g) + 4H2O(g) molecules 1 molecule C3H8 + 5 molecules O2 3 molecules CO2 + 4 molecules H2O Mg3N2(s) + H2O(l) Mg(OH)2 (s) + NH3(g) mass of atoms 44.09 amu C3H8 + 160.00 amu O2 132.03 amu CO2 + 72.06 amu H2O H2S (g) + SO2(g) S (s) + H2O(l) amount (mol) 1 mol C3H8 + 5 mol O2 3 mol CO2 + 4 mol H2O CO2 (g) + KOH(s) K2CO3 (g) + H2O(s) mass (g) 44.09 g C3H8 + 160.00 g O2 132.03 g CO2 + 72.06 g H2O total mass (g) 204.09 g Some equations are harder than others..practice! Example: Ethane, C2H6, reacts (is combusted are key words) with O2 to form CO2 and H2O. Write a balanced equation for this reaction. 1. Write the correct formula(s) for reactants and products. C2H6 + O2 CO2 + H2O 2. Start by balancing those elements that appear in only one reactant and one product. C2H6 + O2 2 carbon on left CO2 + H2O 1 carbon on right start with C or H but not O C2H6 + O2 6 hydrogen on left C2H6 + O2 2CO2 + H2O 2 hydrogen on right multiply H2O by 3 2CO2 + 3H2O 4. Balance those elements that appear in two or more reactants or products. C2H6 + O2 2 O on left C 2 H 6 + 7 O2 2 multiply CO2 by 2 204.09 g 2CO2 + 3H2O 4O +3O 2CO2 + 3H2O = 7 oxygen on right multiply O2 by 7 2 remove fraction 2C2H6 + 7O2 4CO2 + 6H2O multiply by 2 4. Balance those elements that appear in two or more reactants or products. C2H6 + O2 2CO2 + 3H2O 2 O on left C 2 H 6 + 7 O2 2 2C2H6 + 7O2 4O + 3 O = 7 oxygen on right 2CO2 + 3H2O 7 multiply O2 by 2 remove fraction multiply both sides by 2 4CO2 + 6H2O 5. Check to make sure that you have the same number of each type of atom on both sides of the equation. 4CO2 + 6H2O 2C2H6 + 7O2 How To Master Stoichiometry! 1. Always write a balanced chemical equation. A balanced chemical equation contains a lot of very useful chemical information. Learn it and everything gets easy! C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) molecules 1 molecule C3H8 + 5 molecules O2 3 molecules CO2 + 4 molecules H2O mass of atoms 44.09 amu C3H8 + 160.00 amu O2 132.03 amu CO2 + 72.06 amu H2O amount (mol) 1 mol C3H8 + 5 mol O2 3 mol CO2 + 4 mol H2O mass (g) 44.09 g C3H8 + 160.00 g O2 132.03 g CO2 + 72.06 g H2O total mass (g) 204.09 g 204.09 g How many CO molecules are required to react with 25 formula units of Fe2O3 as shown below in the balanced equation? 2. Work in moles---not masses....we need to count atoms and molecules using the mole connection to mass. Fe2O3 (s) + 3CO(g) 3. Use dimensional analysis correctly. Grams of Reactant Molar Mass Moles of Reactant 2Fe(s) + 3CO2(g) Grams of Product balanced equation Molar Mass Moles of Product 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? Fe2O3 (s) + 3CO(g) What mass of CO is required to react completely with 146 g of iron (III) oxide? 2Fe(s) + 3CO2(g) Fe2O3 (s) + 3CO(g) # Fe atoms = = 2.50 × 105 f u Fe2 O3 × 2 F e atoms = 5.00 × 105 F e atoms 1 f u Fe2 O3 2Fe(s) + 3CO2(g) How many grams of iron (III) oxide react with excess carbon monoxide if the carbon dioxide produced by the reaction had a mass of 8.65 grams? key riff! It tells you one reactant is in excess and the other is not! Fe2O3 (s) + 3CO(g) What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with excess carbon monoxide? key riff! It tells you one reactant is in excess and the other is not! 2Fe(s) + 3CO2(g) will run out first excess Fe2O3 (s) + 3CO(g) What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with excess carbon monoxide? Fe2O3 (s) + 3CO(g) 2Fe(s) + 3CO2(g) The limiting reagent is the reactant that is runs out (or is consumed) and dictates the amount of product that can be formed. 2Fe(s) + 3CO2(g) Which reactant is the limiting reagent? It’s the same in chemistry. We must “count” the number of each reactant (via moles) and see which is limiting--it’s not how much they weigh! We deal with limiting reagents all the time, we just don’t give it a buzzword like chemists do. Reactants Product Do You Understand Limiting Reagent II? If 25.0 g CH4 is combusted with 40.0 g O2, which reactant is the limiting reactant? Note the absence of “excess” which tells us it is not a limiting reagent problem! There are 2 reactant masses in the problem = limiting reagent! There are two ways you can calculate the answer. Method 1: For both reactants, use balanced equation & stoichiometric factors and compute the amount of any product formed for both (I teach this method!) Number of Copies Possible 87 copies 83 copies 168/2 = 84 328/4 = 82 Method 2: Pick one of the reactant and compute how much of the other reactant you need and compare with the given amount. • If 25.0 g CH 4 is combusted with 40.0 g O2, which reactant is the limiting reactant? • If 25.0 g CH 4 is combusted with 40.0 g O2, which reactant is the limiting reactant? CH4(g) + 2 O2(g) ""# CO2(g) + 2 H2O(g) CH4(g) + 2 O2(g) ""# CO2(g) + 2 H2O(g) Method 1 (Calculating the # moles of product formed by each reactant to determine which reactant makes the least amount) Method 1I (Directly comparing amounts of reactants given in the problem to which is the limiting reagent--less steps) 1. Let’s use the amount of CO2 formed as our “yardstick” of how much product can be made (we could chose H2O). 1 mol CH4 1 mol CO2 × = 1.559 mol CO2 16.04 g CH4 1 mol CH4 1 mol O2 1 mol CO2 mol CO2 = 40.0 g O2 × × = 0.625 mol CO2 32.00 g O2 2 mol O2 mol CO2 = 25.0 g CH4 × O2 must be the limiting reagent as the amount of CO2 produced is the least amount of product! Another stoichiometry example from Silberberg Copper metal is obtained from copper(I) sulfide containing ores in multistep-extractive process. After grinding the ore into fine rocks, it is heated with oxygen gas forming powdered cuprous oxide and gaseous sulfur dioxide. (0) Write a balanced chemical equation for this process (a) How many moles of molecular oxygen are required to fully roast 10.0 mol of copper(I) sulfide? g O2 needed = 25.0 g CH4 × 1 mol CH4 2 mol O2 32.00 g O2 × × = 99.75 g O2 16.04 g CH4 1 mol CH4 1 mol O2 O2 is the limiting reagent as we need 99.75 g of it but we are are only given 40.0 g O2! Thus, the amount of product that can be formed is determined by the amount of O2 not by the amount of methane, CH4. (0) Write a balanced chemical equation for this process Cu2S(s) + O2(g) 2Cu2S(s) + 3O2(g) Cu2O(s) + SO2(g) unbalanced 2Cu2O(s) + 2SO2(g) (a) How many moles of oxygen are required to roast 10.0 mol of copper(I) sulfide? 3 mol O2 = 15.0 mol O mol O2 = ? = 10.0 mol Cu2S x 2 2 mol Cu2S (b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(I) sulfide is roasted? (b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(I) sulfide is roasted? (c) How many kilograms of oxygen are required to form 2.86 kg of copper(I) oxide? g SO2 = 10.0 mol Cu2S x 2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g) (c) How many kilograms of oxygen are required to form 2.86 kg of copper(I) oxide? kg O2 = 2.86 kg Cu2O x 103g Cu2O 1 mol Cu2O x 3mol O2 x x kg Cu2O 143.10g Cu2O 2mol Cu2O kg O2 = 0.959 = 1 kg O2 32.00g O2 x 1000 g O2 1 mol O2 2mol SO2 x 64.07g SO2 = 641g SO2 2mol Cu2S 1 mol SO2 Do You Understand Limiting Reagents? Phosphorus trichloride is a commercially important compound used in the manufacture of pesticides. It is made by the direct combination of phosphorus P4 and gaseous molecular chlorine. Suppose 323 g of chlorine is combined with 125 g P4. Determine the amount of phosphorous trichloride that can be produced when these reactants are combined. NOTE: 2 Reactants With Masses => Limiting Reagent P4 (s) + Cl2 (g) " PCl3 (l) Step 1. Convert words to formulas P4 (s) + 6Cl2 (g) " 4PCl3 (l) Step 2. Balance Step 3. Determine reactant that produces the least product. Do You Understand Limiting Reagents? 124.0 g of solid Al metal is reacted with 601.0 g of iron(III) oxide to produce iron metal and aluminum oxide. Calculate the mass of aluminum oxide formed. mol PCl3 = 323 g Cl2 × 1 mol Cl2 4 mol PCl3 × = 3.04mol PCl3 70.91 g Cl2 6 mol Cl2 1. Write a balanced equation for all problems mol PCl3 = 125 g P4 × 1 mol P4 4 mol PCl3 × = 4.04mol PCl3 123.88 g P4 1 mol P4 2. Two reactant masses and “no excess” = limiting reagent. Step 4. Cl2 is the limiting reagent and determines the amount of PCl3 g PCl3 = 125 g P4 × 1 mol P4 4 mol PCl3 137.32 g PCl3 × × = 417.5 g PCl3 123.88 g P4 6 molCl2 1 mol PCl3 124 g of Al are reacted with 601 g of Fe2O3 2Al + Fe2O3 g Al 124 g Al x Al2O3 + 2Fe mol Al2SO3 produced mol Al 1 mol Al 27.0 g Al g Fe2O3 601 g Fe2O3 x x 1 mol Al2O3 2 mol Al mol Fe2O3 1 mol Fe2O3 160. g Fe2O3 x x 101.96 g Al2O3 = 234 g Al2O3 1 mol Al2O3 mol Al2SO3 produced 1 mol Al2O3 101.96 g Al2O3 x = 383 g Al2O3 1 mol Fe2O3 1 mol Al2O3 Al make the least and is the limiting reagent! 234 g Al2O3 can be produced The % Yield of a chemical reaction is the ratio of product mass obtained in the lab over the theoretical (i.e. calculated) X 100. experimentally determined in lab % Yield = Actual Amount x 100 Theoretical Amount from limiting 3. Work in moles (grams => moles => equation stoichiometry) 4. Determine maximum theoretical amount of product for both reactants. The limiting reagent is the one that produces the least. When we do chemical reactions calculations in the class, they are “ideal and give 100% product. This is the called the “theoretical value”. Real reactions have side-reactions which reduce the amount of product obtained vs the “theoretical amount. We call that experimentally-determined value. A + B C (main product) (reactants) D (side products) Learning Check: Calculating Percent Yield Silicon carbide (SiC) is made by reacting silicon dioxide with powdered carbon (C) under high temperatures. Carbon monoxide is also formed as a by-product. Suppose 100.0 kg of silicon dioxide is processed in the lab and 51.4 kg of SiC is recovered What is the percent yield of SiC from this process? reagent calculation Actual Amount actual amount of product obtained from a reaction in the lab. It’s given in a word problem. Theoretical Amount is the amount of product that is calculated assuming all the limiting reagent reacts. Notice the word excess is missing and the amount of the other reactant. Must assume other reactant is excess! Learning Check: Calculating Percent Yield SiO2(s) + C(s) SiO2(s) + 3C(s) SiC(s) + CO(g) 1. Converts words to formulas SiC(s) + 2CO(g) 2. Balance 3. We are given one reactant, we must assume excess of the other (C) mol SiO2 = 102 kg SiO2 × 1 mol SiO2 1 mol SiC 103 g SiO2 × × = 1664 mol SiC 1 kg SiO2 60.09 g SiO2 1 mol SiO2 kg SiC = 1664 mol SiCl × Forms of Potential Energy Less Stable Physical properties of matter that depend on the Less Stable on the amount of matter are extensive properties. Physical properties that do not change with mass are intensive properties. More Stable More Stable Gravity: PE gained when lifted Spring: stretched vs compressed 40.10 g SiC 1 kg SiC × = 66.73 kg SiC 1 mol SiC 1000 g SiC = 66.73 kg SiC Less Stable Less Stable 4. We get the actual from experiment and theoretical from calculation and plug them into the yield equation (units of mass should be the same) % Yield = kg Actual 51.4 kg × 100 = × 100 = = 77.0% 77.0% kg Theoretical 66.73 kg Putting it all together: Limiting/Percent Yield A student reacts 30.0 g benzene, C6H6, with 65.0 g bromine, Br2, to prepare bromobenzene, C6H5Br in the lab (MW C6H6 = 78.102, C6H5Br 156.99, Br2 = 159.808). After the reaction was complete, the student recovered 56.7 g C6H5Br. Determine the limiting reagent, the theoretical yield, and the overall % yield? More Stable Charge separation: More Stable Fuel Chemical Potential Learning Check: Calculating Percent Yield A student reacts 30.0 g benzene, C6H6, with 65.0 g bromine, Br2, to prepare bromobenzene, C6H5Br. (MW C6H6 = 78.102, C6H5Br 156.99, Br2 = 159.808). After the reaction was complete, the student recovered 56.7 g C6H5Br. Determine the limiting reagent, the theoretical yield, and the overall % yield? Step 1. Converts words to formulas and balance the equation C6H6 + Br2 C6H5Br + HBr Step 2. Determine the reactant that produces the least product. 1. Write a balanced equation 2. Use stoichiometry in balanced equation 3. Find g product predicted limiting reagent actual yield/theoretical yield x 100 4. Calculate percent yield Announcements Event: Exam 1 Coverage: Chapters 1-3 Date: July 18, 2013 Time: Evening 6-7:30PM Room: SOM 111 (Ching Tan Room) Materials Needed: A calculator and a few pens. A periodic table and scrap paper will be provided by the instructor. Multiple-Choice Practice From Blog “Links Tab” http://www.mhhe.com/physsci/chemistry/ silberberg/student/olc/quiz.mhtml http://highered.mcgraw-hill.com/classware/ selfstudy.do?isbn=0072558202 g C6 H5 Br = 30.0 g C6 H6 × g C6 H5 Br = 65.0 g Br2 × 1 mol C6 H6 1 mol C6 H5 Br 156.9 g C6 H5 Br × × = 60.3 g C6 H5 Br 78.1 g C6 H6 1 molC6 H6 1 mol C6 H5 Br 1 mol Br2 1 mol C6 H5 Br 156.9 g C6 H5 Br × × = 63.8 g C6 H5 Br 159.8 g Br2 1 mol Br2 1 mol C6 H5 Br Step 3. C6H6 is the limiting reagent. 60.3 g C6H5Br is theory yield % Yield = g Actual 56.7 g × 100 = × 100 = 94.0% g Theoretical 60.3 g Section 3.5 Fundamentals of Solution Stoichiometry •Distribution of solute in solvent is uniform (mixed) •Components do not separate on standing •Not separable by filtration •Solute/Solvent mix in ratios up to the solubility limit of solute A solution is a homogenous mixture of a solvent plus a solute (focus is aqueous). The solute is the substance(s) dissolved in the solvent. A solvent is the substance present in the larger amount----typically water in aqueous solutions. Solution is the solute + solvent. g solution = g solute + g solvent Section 3.5 Fundamentals of Solution Stoichiometry Seawater is a homogeneous mixture that has many dissolved solutes in a water solvent. 2+ + Mg2+ Ca and K 2- SO4 35 grams of dissolved salts per kilogram of seawater --70+ dissolved components --6 make up >99% the solids Na+ Cl- Chemists express the concentration of a solution by declaring the amount of solute present in a given quantity of solution. Molar Mass M = molarity = moles of solute total liters of solution Solute + Solvent Molar Mass grams solute We can visualize solute particles in a solvent (solute) and solvent molecules not shown. Volume of solution moles solute Molarity 4-Basic Problem Types Using Molarity 1. Simple molarity calculation given mass and total volume (or permutation of this). Add Solvent Increasing Volume Decreases Concentration 2. Mass-Mole-Number of Molecules Involving Solutions 3. Dilution Problems (M1V1 = M2V2) 4. Stoichiometry With Solutions Concentrated Solution More solute particles per unit volume Dilute Solution More solute particles per unit volume Calculate the molarity of a solution that contains 12.5 g of pure sulfuric acid (H2SO4) in 1.75 L of solution (Molar Mass H2SO4 = 98.1g) Preparing Solutions in the Laboratory Weigh the solid needed Add the solvent to final volume in glassware. moles of solute M = molarity = Dissolve the solid total liters of solution Identify the solute here? A •Weigh the solid needed. •Transfer the solid to a volumetric flask that contains about half the final volume of solvent. C Add solvent until the solution reaches its final volume. B Dissolve the solid thoroughly by swirling. How many grams of KI is required to make 500. mL of a 2.80 M KI solution? How many grams of solute are in 1.75 L of 0.460 M sodium monohydrogen phosphate? REWORDED Or suppose I have 500. mL of a 2.80M KI solution. How many grams of KI solute does it contain (it meaning the entire 500. mL volume. g of Na2HPO4 = ? = volume KI M KI moles KI 1.75 L soln X 0.460 moles Na2HPO4 X 141.96 g Na2HPO4 1 mol Na2HPO4 1 L soln M KI grams KI = 114 g Na2HPO4 grams KI = 500. mL x 1L 1000 mL x 2.80 mol KI 1 L soln x 166 g KI 1 mol KI = 232 g KI Calculating the Molarity of a Solution Sucrose has a molar mass of 342.29 g/mol. It is a fine, white, odorless crystalline powder with a pleasing, sweet taste. What is the molarity of an aqueous solution made by placing 75.5 g of 85% pure sugar in 500.00 mL of water? fix this slide for ang Mol sugar = 75.5 g impure × Msugar = 85 g pure sug 1 mol sug × = 0.18748 mol sug 100. g impure 342.29 g sug Preparing Solutions in the Laboratory How you would make up 1.00 L of 0.100M CuSO4? # moles CuSO4 = M x V = 1.0 L soln x 0.100 mol CuSO4 = 0.100 mol x CuSO4.5H2O g CuSO4 = 0.100 mol CuSO4 × 1L mark Dissolve completely mol pure sugar 0.18748 =. = 0.375 total volume solution 0.50000 L 15.96 g CuSO4 159.62 g CuSO4 ·5 H2 O = 15.96 g CuSO4 1 mol CuSO4 Dilute to mark 0.100M CuSO4 What does 3.5 M FeCl3 mean? 3.5 M FeCl3 = 3.5 moles FeCl3 1 Liter solution (1) a homogeneous solution of 3.5 moles of dry 100% pure FeCl3 dissolved in 1.00 Liter total solution volume (not 1 L of liquid!). (3) Note: It does not mean 3.5 moles of FeCl3 is dissolved in 1.00 liter of water! (4) [Fe3+] = 3.5M and [Cl-] = 3 x 3.5 M Given 2 liters of 0.20M Al2(SO4)3, (a) what is the molarity of aluminum sulfate? (b) how many moles of aluminum ions are there in 2L of this solution? (c) what is the molarity of SO42- in this solution? (d) what is the molarity of Al3+ in this solution (5) It can be used as a conversion factor In 2 liters of 0.20M Al2(SO4)3, (a) what is the molarity of aluminum sulfate (b) how many moles of aluminum are there (c) what is the molarity of aluminum ions and sulfate ions? Hydrochloric acid is sold commercially as a 12.0 M solution. How many moles of HCl are in 300.0 mL of 12.0 M solution? (a) [Al2(SO4)3] = 0.20M mol HCl = 300.0 mL 1L 1000 mL 12.0 mol HCl 1 L soln = 3.60 mol HCl (b) mol Al = 2 L X 0.20 mol Al x = 0.4 mol Al (c) [Al2+] = 2 x 0.20M and [SO42-] = 3 x 0.20M Determine the mass of calcium nitrate required to prepare 3.50 L of 0.800 M. Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution. Remember this formula: MiVi Moles of solute in Volume = = MfVf Moles of solute after dilution (f) How would you prepare 500.0 mL of 0.500M HCl from a stock solution of 12.1 M HCl? Vi How many mL of stock are required? Look at the picture carefully and you should see that the number of moles taken from the blue cube on the left (the concentrated stock solution) is the number of moles that ends up in the light white. Mi ! V i Mf ! V f Vf Stock 12.1M HCl 500.00 mL of 0.500M HCl mark Mi n V M= dilute to Mf mol HCl before dilution How would you prepare 60.0 mL of 0.2 M HNO3 from a stock solution of 4.00 M HNO3? Moles of solute in Volume MiVi = Moles of solute after dilution (f) = MfVf How would you prepare 0.80L of isotonic saline from a 6.0M stock solution? Given: Mi = 6.00M 1) Use factor-label and information in problem Vf = 0.80 L Vi = ? L MiVi = MfVf Vi = 2) MiVi = MfVf Mf = 0.15M easy to remember but very mechanical It fails often! MfVf Mi 0.15 M x 0.80M L = 6.00 M HNO3 = 0.020 L 20 mL of stock + 780 mL of water = 800.0 mL of solution How would you prepare 60.0 mL of 0.2 M HNO3 from a stock solution of 4.00 M HNO3? Given: Mi = 4.00 Mf = 0.200 Vf = 0.06 L Vi = ? L MiVi = MfVf Vi = MfVf Mi 0.200 M x 0.060 L = 4.00 M HNO3 = 0.003 L = 3.0 mL 3 mL of acid + 57 mL of water = 60.0 mL of solution If 10.0 mL of 12.0 M HCl is added to enough water to give 100. mL of solution, what is the concentration of the solution? Molarity ties the number of moles of solute to the volume of a solution. grams solute molar mass solute mass of solution Density of solution Volume of Solution Moles of substance Solution Stoichiometry How many milliliters of 0.125 M NaHCO3 are needed to neutralize (react with) 18.0 mL of 0.100 M HCl? HCl(aq) + NaHCO3(aq) => NaCl(aq) + H2O(l) + CO2(g) Volume of HCl Molarity HCl Molarity moles of solute total liters of solution HCl(aq) + NaHCO3(aq) => NaCl(aq) + H2O(l) + CO2(g) # mol HCl = MHCl x L = 0.100M HCl X 0.018L = 1.80 X 10-3 mol HCl # mol NaHCO3 = = 1.80 × 10 1 mol NaHCO3 1 L solution mol HCl × × = .0144 L solution 1 miol HCl 0.125 M NaHCO3 How many liters of “stomach acid” react with a tablet containing 0.10g of magnesium hydroxide? Stoichiometry in Solution Specialized cells in the stomach release HCl to aid in digestion. If too much acid is released, the excess can cause “heartburn” and is sometimes neutralized with “antacids”. A common antacid is magnesium hydroxide, which reacts with the hydrochloric acid to form water and magnesium chloride solution. As a government chemist testing commercial antacids, suppose you use 0.10M HCl to simulate the acid concentration in the stomach. How many liters of “stomach acid” react with a tablet containing 0.10g of magnesium hydroxide? One of the solids present in photographic film is silver bromide. It is prepared by mixing silver nitrate with calcium bromide giving insoluble silver bromide and soluble calcium nitrate . 1. Always translate nomenclature to chemical equation Mg(OH)2(s) + HCl(aq) Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + H2O(l) CaBr2 + 2AgNO3 ==> 2AgBr + Ca(NO3)2 unbalanced MgCl2(aq) + 2H2O(l) balanced 2. Work in moles and use the stoichiometric factors. How many milliliter of 0.125M CaBr2 must be used to react with the solute in 50.0 mL of 0.115M AgNO3? L HCl = ? = L HCl = 0.10 g Mg(OH)2 × 2 mol HCl 1 L HCl 1 mol Mg(OH)2 × × = 3.4 × 10−2 L HCl 58.33 g Mg(OH)2 1 mol Mg(OH)2 0.1 mol HCl = 3.4 x 10-2 L HCl Molarity NaHCO3 Mole to Mole ratio How many milliliters of 0.125 M NaHCO3 solution are needed to neutralize 18.0 mL of 0.100 M HCl? −3 Volume of NaHCO3 One of the solids present in photographic film is silver bromide. It is prepared by mixing silver nitrate with calcium bromide giving insoluble silver bromide and soluble calcium nitrate . CaBr2 + 2AgNO3 ==> 2AgBr + Ca(NO3)2 How many milliliter of 0.125M CaBr2 must be used to react with the solute in 50.0 mL of 0.115M AgNO3? � �� �� �� � � 0.115 mol AgNO3 � � 1 mol CaBr2 � � � � 1000 mL} � 1 L CaBr2 �� �� �� � = 23.0 mLCaBr mL CaBr2 = 0.050 L soln �� 2 � � 2 mol AgNO � � 0.125 mol CaBr � � � 1 L soln 1L 3 2 Suppose 0.050L of 0.010M mercury(II) nitrate reacts with 0.020L of 0.10M sodium sulfide forming mercury(II) sulfide. How many grams of mercury(II) sulfide can form? Hg(NO3)2(aq) + Na2S(aq) g HgS = 0.050 L Hg(NO3 )2 × g HgS = 0.020 L Na2 S × HgS(s) + 2NaNO3(aq) 1 mol HgS 232.7 g HgS 0.010 mol Hg(NO3 )2 × × = 0.12 g HgS 1 L Hg(NO3 )2 1 mol Hg(NO3 )2 1 mol HgS 0.10 mol Na2 S 1 mol HgS 232.7 g HgS × × = 0.47 g HgS 1 L Na2 S 1 mol Na2 S 1 mol HgS Hg(NO is the limiting reagent and 0.12 g HgS can be Hg(NO3)32)(aq) 2(aq) is the limiting reagent and 0.12 g HgS can be produced from produced fromthis thisreaction. reaction. Limiting-Reactant Problems in Solution Mercury and its compounds have many uses, from fillings for teeth (as an alloy with silver, copper, and tin) to the industrial production of chlorine. Suppose 0.050L of 0.010M mercury(II) nitrate reacts with 0.020L of 0.10M sodium sulfide forming mercury(II) sulfide. How many grams of mercury(II) sulfide can form? Step 1: Words to balanced chemical equation Step 2: Is it limiting reagent? Yes or No Step 3: Use given information, balanced equation and good dimensional analysis technique and compute g HgS formed. Step 4: Does it make sense?