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Lecture Presentation Chapter 3 Chemical Reactions and Reaction Stoichiometry © 2015 Pearson Education, Inc. James F. Kirby Quinnipiac University Hamden, CT LESSON 1 3-1Equations & 3-2 Types of Rxns Atoms, Molecules, and Ions © 2015 Pearson Education, Inc. Stoichiometry • The study of the mass relationships in chemistry • Based on the Law of Conservation of Mass (Antoine Lavoisier, 1789) “We may lay it down as an incontestable axiom that, in all the operations of art and nature, nothing is created; an equal amount of matter exists both before and after the experiment. Upon this principle, the whole art of performing chemical experiments depends.” —Antoine Lavoisier Chemical Equations Chemical equations are concise representations of chemical reactions. Chemical Reactions & Chemical Equations • Chemical reaction: a process in which a substance or substances change into one or more new substances. Hydrogen gas combusts in the presence of oxygen gas to form water. • Chemical equation: a description of a chemical reaction using chemical symbols to show what happens during the reaction. BALANCE the 2 H2 + O2 2 H2O REACTANTS: Starting compounds; what you start with equation to show conservation of mass PRODUCTS: Resulting compounds; what you end up with 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 What Is in a Chemical Equation? CH4(g) + 2O2(g) Reactants appear on the left side of the equation. CO2(g) + 2H2O(g) What Is in a Chemical Equation? CH4(g) + 2O2(g) CO2(g) + 2H2O(g) Products appear on the right side of the equation. 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 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. Chemical Equations: Additional Information 2HgO (s) 2Hg (l) + O2 (g) (g), (l), (s): shows physical state of reactant or product; i. e., whether reactant or product is a gas (g), liquid (l), or solid (s) STOICHIOMETRIC COEFFICIENTS: show number of moles of reactant or product Chemical Equations: Additional Information KBr (aq) + AgNO3 (aq) KNO3 (aq) + AgBr (aq) (aq): shows that a reactant or product is dissolved in H2O; i. e., the reactant or product is in aqueous solution 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 Interpretation of a Chemical Equation 2 H2O Two molecules + one molecule two molecules Two moles two moles (2)(18.02 g) = 36.04 g 2 H2 + + O2 one mole (2)(2.02 g) = 4.04 g + 32.00 g 36.04 g of reactants 36.04 g of products Balancing Chemical Equations Write chemical equations to describe reactions carried out in the laboratory or other venues. We know the reactants’ identities, but may need to establish identities of products via further research, etc. When identities of reactants and products are established, we can write their formulas. Still, the number of atoms of each element must be the same on both sides of the equation to adhere to the Laws of Conservation of Matter and Mass. Thus, we must BALANCE the chemical equation! 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) 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 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 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 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 12 4 CH (2 6) 14 O(2 (7xx2) 2) 4(6 C 2)6) 1412 OH (4 x 2x + Reactants 4C 12 H 14 O Products 4C 12 H 14 O Try to balance these equations… 1. C6H12O6 C2H5OH + CO2 2. C4H10 + O2 CO2 + H2O 3. P4O10 + H2O H3PO4 4. NaHCO3 + H2SO4 Na2SO4 + CO2 + H2O Homework: Worksheet on Balancing Quiz Types of Reactions 23 Chemical reactions can be classified as • combination reactions • decomposition reactions • single replacement reactions • double replacement reactions • combustion reactions Combination Reactions 24 In a combination reaction, • two or more elements form one product • or simple compounds combine to form one product 2Mg(s) + O2(g) 2MgO(s) 2Na(s) + Cl2(g) 2NaCl(s) SO3(g) + H2O(l) H2SO4(aq) Combination Reaction: MgO 25 Combination Reactions • In combination reactions two or more substances react to form one product. • Examples: – 2 Mg(s) + O2(g) – N2(g) + 3 H2(g) – C3H6(g) + Br2(l) © 2015 Pearson Education, Inc. 2 MgO(s) 2 NH3(g) C3H6Br2(l) Stoichiometry Decomposition Reaction 27 In a decomposition reaction, one substance splits into two or more simpler substances. 2HgO(s) 2KClO3(s) 2Hg(l) + O2(g) 2KCl(s) + 3O2(g) Decomposition Reaction: HgO 28 Decomposition Reactions • In a decomposition reaction one substance breaks down into two or more substances. • Examples: – CaCO3(s) – 2 KClO3(s) – 2 NaN3(s) © 2015 Pearson Education, Inc. CaO(s) + CO2(g) 2 KCl(s) + O2(g) 2 Na(s) + 3 N2(g) Stoichiometry Single Replacement Reaction 30 In a single replacement reaction, one element takes the place of a different element in another reacting a compound. Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g) Fe(s) + CuSO4(aq) FeSO4(aq) + Cu(s) Single Replacement Reaction: ZnCl2 31 The Activity Series for Metals Displacement Reaction M + BC AC + B M is metal BC is acid or H2O B is H2 Ca + 2H2O Ca(OH)2 + H2 Pb + 2H2O Pb(OH)2 + H2 4.4 Double Replacement Reaction 33 In a double replacement, two elements in the reactants exchange places. AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq) ZnS(s) ZnCl2(aq) + H2S(g) + 2HCl(aq) Double Replacement Reaction: BaSO4 34 Combustion Reaction 35 In a combustion reaction, • a carbon-containing compound burns in oxygen gas to form carbon dioxide (CO2) and water (H2O) • energy is released as a product in the form of heat CH4(g) + 2O2(g) CO2(g) + 2H2O(g) + energy Combustion Reaction 36 Insert diagram of following balanced reaction with triangle over the reaction arrow. C3H8(g) + 5O2(g) Δ 3CO2(g) + 4H2O(g) Combustion Reactions • Combustion reactions are generally rapid reactions that produce a flame. • Combustion reactions most often involve oxygen in the air as a reactant. • Examples: – CH4(g) + 2 O2(g) – C3H8(g) + 5 O2(g) © 2015 Pearson Education, Inc. CO2(g) + 2 H2O(g) 3 CO2(g) + 4 H2O(g) Stoichiometry Summary of Reaction Types 38 Learning Check 39 Classify each of the following reactions as combination, decomposition, single replacement, double replacement, or combustion. A. 2Al(s) + 3H2SO4(aq) Al2(SO4)3(s) + 3H2(g) B. Na2SO4(aq) + 2AgNO3(aq) C. 3C(s) + Fe2O3(s) D. C2H4(g) + 2O2(g) Ag2SO4(s) + 2NaNO3(aq) 2Fe(s) + 3CO(g) 2CO2(g) + 2H2O(g) Solution 40 A. 2Al(s) + 3H2SO4(aq) Al2(SO4)3(s) + 3H2(g) Single Replacement B. Na2SO4(aq) + 2AgNO3(aq) C. N2(g) + O2(g) D. C2H4(g) + 2O2(g) 2NO(g) Ag2SO4(s) + 2NaNO3(aq) Double Replacement Combination 2CO2(g) + 2H2O(g) Combustion Learning Check 41 Identify each reaction as combination, decomposition, combustion, single replacement, or double replacement. A. 3Ba(s) + N2(g) Ba3N2(s) B. 2Ag(s) + H2S(aq) Ag2S(s) + H2(g) C. 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(g) D. PbCl2(aq) + K2SO4(aq) E. K2CO3(s) 2KCl(aq) + PbSO4(s) K2O(aq) + CO2(g) Solution 42 Identify each reaction as combination, decomposition, combustion, single replacement, or double replacement. A. 3Ba(s) + N2(g) B. 2Ag(s) + H2S(aq) Ba3N2(s) Combination Ag2S(s) + H2(g) Single Replacement C. C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(g) Combustion Solution 43 Identify each reaction as combination, decomposition, combustion, single replacement, or double replacement. D. PbCl2(aq) + K2SO4(aq) 2KCl(aq) + PbSO4(s) Double Replacement E. K2CO3(s) K2O(aq) + CO2(g) Decomposition Identify the type of redox reaction for each of the following reactions: 2 CuCl Cu + CuCl2 Fe + 2 HCl H2 + FeCl2 S + 3 F2 SF6 2 Ag + PtCl2 2 AgCl + Pt 2 N2O 2 N2 + O2 Homework: Worksheet on Predicting Products QUIZ LESSON 2 3-3 Formula and Molecular Weights, % Comp & 3-4 Avogadro’s # and the Mole Atoms, Molecules, and Ions © 2015 Pearson Education, Inc. Formula Weight (FW) • A formula weight is the sum of the atomic weights for the atoms in a chemical formula. • This is the quantitative significance of a formula. • The formula weight of calcium chloride, CaCl2, would be Ca: 1(40.08 amu) + Cl: 2(35.453 amu) 110.99 amu Stoichiometry © 2015 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 © 2015 Pearson Education, Inc. Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule. 1S SO2 2O SO2 32.07 amu + 2 x 16.00 amu 64.07 amu For any molecule molecular mass (amu) = molar mass (grams) 1 molecule SO2 = 64.07 amu 1 mole SO2 = 64.07 g SO2 3.3 Other Names Related to Mass • Molecular Mass/Molecular Weight: If you have a single molecule, mass is measured in amu’s instead of grams. But, the molecular mass/weight is the same numerical value as 1 mole of molecules. Only the units are different. (This is the beauty of Avogadro’s Number!) • Formula Mass/Formula Weight: Same goes for compounds. But again, the numerical value is the same. Only the units are different. • THE POINT: You may hear all of these terms which mean the SAME NUMBER… just different units Molar Mass: (a.k.a. molecular mass) (a.k.a. molecular weight) The sum of the atomic masses (in a.m.u.) in a molecule. Try this: Calculate the molecular mass of ascorbic acid (i.e., vitamin C), the chemical formula of which is C6H8O6. Molar Mass • A molar mass is the mass of 1 mol of a substance (i.e., g/mol). • The molar mass of an element is the atomic weight for the element from the periodic table. If it is diatomic, it is twice that atomic weight. • The formula weight (in amu’s) will be the same number as the molar mass (in g/mol). © 2015 Pearson Education, Inc. Stoichiometry Ionic Compounds and Formulas • Remember, ionic compounds exist with a three-dimensional order of ions. There is no simple group of atoms to call a molecule. • As such, ionic compounds use empirical formulas and formula weights (not molecular weights). Stoichiometry © 2015 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: (number of atoms)(atomic weight) % Element = (FW of the compound) × 100 Stoichiometry © 2015 Pearson Education, Inc. Percent Composition of a Chemical Compound % composition of an element = (n)(molar mass of element) ------------------------------------ x 100 molar mass of compound • Useful for verification of purity of compounds for research, etc. 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% 3.5 Example: Percent Composition of the Elements in Beryl Aquamarine (Beryl, Be3Al2Si6O18) Emerald (Beryl, Be3Al2Si6O18) • Moles of atoms: • Molar masses of: Beryllium (Be) Aluminum (Al) Silicon (Si) Oxygen (O) Beryllium (Be) Aluminum (Al) Silicon (Si) Oxygen (O) Beryl (Be3Al2Si6O18) What are the percents of the elements involved? NEXT SLIDE Try this one! Determine the % of each element in the mineral desmine (CaAl2Si7O18.7H2O). Find the percentage composition of a compound that contains 1.94 g of carbon, O.48 g of hydrogen, and 2.58 g of sulfur in 5.00-g sample of the compound. Percent Composition So the percentage of carbon in ethane is (2)(12.011 amu) %C = = (30.070 amu) 24.022 amu 30.070 amu × 100 = 79.887% Stoichiometry © 2015 Pearson Education, Inc. We deal with quantities involving more than one object all the time…. Pair (2 items) Dozen (12 items) Gross (144 items) What would a mole of any substance look like? WHAT IS A MOLE ? Avogadro’s 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 The mole, as defined by the SI system of units: A mole is the amount of a substance that contains as many elementary entities (i.e., atoms , molecules, or other particles) as there are atoms in exactly 12 grams of the carbon-12 isotope. 6.022 x NA 23 10 atoms, molecules, etc. = Avogadro’s number Amadeo Avogadro The Mole Q: how long would it take to spend a mole of $1 coins if they were being spent at a rate of 1 billion per second? Mollionaire Q: how long would it take to spend a mole of $1 coins if they were being spent at a rate of 1 billion per second? A: $ 6.02 x 1023 / $1 000 000 000 = 6.02 x 1014 payments = 6.02 x 1014 seconds 6.02 x 1014 seconds / 60 = 1.003 x 1013 minutes 1.003 x 1013 minutes / 60 = 1.672 x 1011 hours 1.672 x 1011 hours / 24 = 6.968 x 109 days 6.968 x 109 days / 365.25 = 1.908 x 107 years A: It would take 19 million years 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) One Mole of: S C Hg Cu Fe 3.2 Mole Relationships • One mole of atoms, ions, or molecules contains Avogadro’s number of those particles. • One mole of molecules or formula units contains Avogadro’s number times the number of atoms or ions of each element in the compound. Stoichiometry © 2015 Pearson Education, Inc. 1 mole of 12C atoms = 12.00 g of 12C atoms 1 mole of 12C atoms = 6.022 x 1023 atoms of 12C 6.022 x 1023 atoms of 12C = 12.00 g of 12C What is the mass(in grams) of one ATOM of 12C? Using Moles Moles provide a bridge from the molecular scale to the real-world scale. Stoichiometry © 2015 Pearson Education, Inc. 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 1 amu = 1.66 x 10-24 g or 1 g = 6.022 x 1023 amu M = molar mass in g/mol NA = Avogadro’s number Molar Mass: The mass, usually in grams (g), of 1 mole of atoms or molecules of a substance. 1 mole of X = molar mass of X (in grams) 1 mole of X = 6.022 x 1023 atoms or molecules of X Do You Understand Molar Mass? How many atoms are in 0.551 g of potassium (K) ? Do You Understand Molar Mass? 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 3.2 Practice Problem: (a) How many moles of magnesium (Mg) are in 88.8 g of Mg? (b) How many atoms of Mg are contained in this many moles of Mg? How many H2O molecules are in a 9.00-g sample of water? a. 0.500 b. 3.01 × 1023 c. 2.71 × 1024 d. 1.08 × 1023 © 2015 Pearson Education, Inc. How many H2O molecules are in a 9.00-g sample of water? a. 0.500 b. 3.01 × 1023 c. 2.71 × 1024 d. 1.08 × 1023 © 2015 Pearson Education, Inc. What number would you use to convert (a) moles of CH4 to grams of CH4 grams and (b) number of molecules of CH4 to moles of CH4? (a) a. b. c. d. Avogadro’s number, 6.02 × 1023 particles/mol Inverse of molar mass of CH4, 1 mol CH4/16.0 g CH4 Molar mass of CH4, 16.0 g CH4/1 mol CH4 Formula weight of CH4, 16.0 amu © 2015 Pearson Education, Inc. What number would you use to convert (a) moles of CH4 to grams of CH4 grams and (b) number of molecules of CH4 to moles of CH4? (a) a. b. c. d. Avogadro’s number, 6.02 × 1023 particles/mol Inverse of molar mass of CH4, 1 mol CH4/16.0 g CH4 Molar mass of CH4, 16.0 g CH4/1 mol CH4 Formula weight of CH4, 16.0 amu © 2015 Pearson Education, Inc. What number would you use to convert (a) moles of CH4 to grams of CH4 grams and (b) number of molecules of CH4 to moles of CH4? (b) a. Inverse of Avogadro’s number, 1 mol/6.02 × 1023 molecules b. Inverse of molar mass of CH4, 1 mol CH4/16.0 g CH4 c. Molar mass of CH4, 16.0 g CH4/1 mol CH4 d. Formula weight of CH4, 16.0 amu © 2015 Pearson Education, Inc. What number would you use to convert (a) moles of CH4 to grams of CH4 grams and (b) number of molecules of CH4 to moles of CH4? (b) a. Inverse of Avogadro’s number, 1 mol/6.02 × 1023 molecules b. Inverse of molar mass of CH4, 1 mol CH4/16.0 g CH4 c. Molar mass of CH4, 16.0 g CH4/1 mol CH4 d. Formula weight of CH4, 16.0 amu © 2015 Pearson Education, Inc. Hwk: page 112-120: 23, 27, 29, 35, 37, 39, 44 Quiz to follow Atoms, Molecules, and Ions © 2015 Pearson Education, Inc. LESSON 3 3-5 Empirical Formulas, Molecular Formulas, Combustion Analysis Atoms, Molecules, and Ions © 2015 Pearson Education, Inc. Determining Empirical Formulas One can determine the empirical formula from the percent composition by following these three steps. Stoichiometry © 2015 Pearson Education, Inc. Determining Empirical Formulas— an Example 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 © 2015 Pearson Education, Inc. Determining Empirical Formulas— an Example Assuming 100.00 g of para-aminobenzoic acid, C: H: N: O: 1 mol 12.01 g 1 mol 5.14 g × 1.01 g 1 mol 10.21 g × 14.01 g 1 mol 23.33 g × 16.00 g 61.31 g × = 5.105 mol C = 5.09 mol H = 0.7288 mol N = 1.456 mol O Stoichiometry © 2015 Pearson Education, Inc. Determining Empirical Formulas— an Example Calculate the mole ratio by dividing by the smallest number of moles: © 2015 Pearson Education, Inc. C: 5.105 mol 0.7288 mol = 7.005 ≈ 7 H: 5.09 mol 0.7288 mol = 6.984 ≈ 7 N: 0.7288 mol 0.7288 mol = 1.000 O: 1.458 mol 0.7288 mol = 2.001 ≈ 2 Stoichiometry Determining Empirical Formulas— an Example These are the subscripts for the empirical formula: C7H7NO2 Stoichiometry © 2015 Pearson Education, Inc. Using percent composition in reverse: • Can determine the empirical formula of a compound Example: Ascorbic acid (a.k.a. vitamin C) Elemental Composition (mass %): C 40.92 % H 4.58 % O 54.50 % Goal: Determine the empirical formula of ascorbic acid. Example: Ascorbic acid (a.k.a. vitamin C) Step 1: Convert mass percentages to grams of each element. If a mass of the compound is given, multiply that mass by the fraction (mass % / 100) of each element to get grams of each element. If no mass of compound is given, assume its mass to be 100 g. Mass of C = 40.92 g C Mass of H = 4.58 g H Mass of O = 54.50 g O Example: Ascorbic acid (a.k.a. vitamin C) Step 2: Calculate the number of moles of each element in the compound. How to do this? Use the molar mass of each element as follows: 1 mole C Moles of C = 40.92 g C x ---------------- = 3.407 mole C 12.01 g C 1 mole H Moles of H = 4.58 g H x ---------------- = 4.54 mole H 1.008 g H 1 mole O Moles of O = 54.50g O x ---------------- = 3.406 mole O 16.00 g O Example: Ascorbic acid (a.k.a. vitamin C) Step 3: Calculate the mole ratio of each element in the compound to the element with the smallest (or smaller, if 2) number of moles. Which element has the smallest number of moles? oxygen Thus we obtain the empirical formula 3.407 C: ------------- = 1.000 1 3.406 4.54 H: ----------- = 1.333 1.33 3.406 3.406 O: ------------- = 1.000 1 3.406 CH1.33O Is this formula our final result? Example: Ascorbic acid (a.k.a. vitamin C) Step 4: By trial and error, convert fractional subscripts in the formula into whole numbers (integers). This means changing 1.33 into an integer: 1.33 x 2 = 2.66 1.33 x 3 = 3.99 4 Multiply each subscript by 3: Thus the empirical formula for ascorbic acid is: C3H4O3 Determining a Molecular Formula • Remember, the number of atoms in a molecular formula is a multiple of the number of atoms in an empirical formula. • If we find the empirical formula and know a molar mass (molecular weight) for the compound, we can find the molecular formula. Stoichiometry © 2015 Pearson Education, Inc. Determining a Molecular Formula— an Example • The empirical formula of a compound was found to be CH. It has a molar mass of 78 g/mol. What is its molecular formula? • Solution: Whole-number multiple = 78/13 = 6 The molecular formula is C6H6. Stoichiometry © 2015 Pearson Education, Inc. Molecular Formula Calculation of the Molecular Formula A compound has an empirical formula of NO2. The colourless liquid, used in rocket engines has a molar mass of 92.0 g/mole. What is the molecular formula of this substance? empirical formula mass: 14.01+2 (16.00) = 46.01 g/mol n = molar mass = 92.0 g/mol emp. f. mass 46.01 g/mol n = 2 2(NO2) = N2O4 Molecular Formulas from Empirical Formulas: To get the actual, molecular formula of the compound, we must know: The approximate molar mass of the compound. The empirical formula of the compound. Molecular Formulas from Empirical Formulas: Example: A sample of a compound containing boron (B) and hydrogen (H) was analyzed and found to contain 6.4442 g B and 1.8031 g H. The molar mass of the compound is about 30 g. What is the molecular formula of this compound? 1 mole B Moles of B = 6.4442 g B x ---------------- = 0.59608 mole B 10.811 g B 0.59608 B: ------------- = 1.000 1 0.59608 1 mole H Moles of H = 1.8031 g H x ---------------- = 1.7888 mole H 1.008 g H 1.7888 H: ----------- = 3.0009 3 0.59608 Molecular Formulas from Empirical Formulas: Example: A sample of a compound containing boron (B) and hydrogen (H) was analyzed and found to contain 6.4442 g B and 1.8031 g H. The molar mass of the compound is about 30 g. What is the molecular formula of this compound? The empirical molar mass is: The empirical formula is: BH3 molar mass 30 g ---------------------------- = ------------ = 2.2 2 empirical molar mass 13.835 g (10.811 g) + (3)(1.008 g) = 13.835 g The molecular formula is: (2)(BH3) B2H6 Cyclohexane, a commonly used organic solvent is 85.6% C and 14.4% H by mass with a molar mass of 84.2 g/mol? What is the molecular formula? Stoichiometry © 2015 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 the difference after C and H have been determined. Stoichiometry © 2015 Pearson Education, Inc. Experimental Determination of Empirical Formulas: O2 Unused O2 Ethanol Heat Elemental analyzer for determination of carbon and hydrogen H2O absorber CO2 absorber (From Fig. 3.5, p. 81, Chang, 7th ed.) • Combustion of sample in presence of O2 (Ethanol + O2 CO2 + H2O) • H2O and CO2 absorber tubes weighed prior to combustion, then weighed again after combustion is complete. • Masses of H2O and CO2 obtained by difference, with masses of H and C in original sample obtained by calculation (see next slide). Experimental Determination of Empirical Formulas: O2 Unused O2 Ethanol Heat Elemental analyzer for determination of carbon and hydrogen H2O absorber CO2 absorber (From Fig. 3.5, p. 81, Chang, 7th ed.) Consider the following elemental analysis: An 11.5 g sample of absolute ethanol is combusted in an apparatus such as the one shown above. Combustion of the 11.5 g ethanol sample produced 22.0 g of CO2 and 13.5 g of H2O. Find: (a) The masses of C, H, and O in the sample (b) The empirical formula for ethanol Experimental Determination of Empirical Formulas: O2 Unused O2 Ethanol Heat Elemental analyzer for determination of carbon and hydrogen H2O absorber CO2 absorber (From Fig. 3.5, p. 81, Chang, 7th ed.) 1 mole CO2 1 mole C 12.01 g C mass of C = 22.0 g CO2 x ------------------- x ----------------- x ---------------- = 6.00 g C 44.01 g CO2 1 mole CO2 1 mole C 1 mole H2O 2 moles H 1.008 g H mass of H = 13.5 g H2O x ------------------- x ----------------- x ---------------- = 1.51 g H 18.02 g H2O 1 mole H2O 1 mole H Experimental Determination of Empirical Formulas: O2 Unused O2 Ethanol Heat Elemental analyzer for determination of carbon and hydrogen H2O absorber CO2 absorber (From Fig. 3.5, p. 81, Chang, 7th ed.) mass of O = 11.5 g ethanol - (6.00 g C + 1.51 g H) = 4.0 g O (obtained by difference) 6.00 g C 1.51 g H 4.0 g O Calculate moles of C, H, O… Combust 11.5 g ethanol Collect 22.0 g CO2 and 13.5 g H2O g CO2 mol CO2 mol C gC 6.0 g C = 0.5 mol C g H2O mol H2O mol H gH 1.5 g H = 1.5 mol H g of O = g of sample – (g of C + g of H) 4.0 g O = 0.25 mol O Empirical formula C0.5H1.5O0.25 Divide by smallest subscript (0.25) Empirical formula C2H6O Isopropyl alcohol, sold as rubbing alcohol, is composed of C, H, and O. Combustion of 0.255g of isopropyl alcohol produces 0.561g of CO2 and 0.306g of H2O. Determine the empirical formula. Stoichiometry © 2015 Pearson Education, Inc. Caproic acid, responsible for the odor of dirty socks, is composed of C, H, and O atoms. Combustion of a 0.255g sample of this compound produces 0.512g CO2 and 0.209g of H2O. What is the empirical formula? If caproic acid has a molar mass of 116 g/mol, what is the molecular formula? Stoichiometry © 2015 Pearson Education, Inc. Hwk: page 112-120: 45, 49, 51, 53, 55, 59 QUIZ Atoms, Molecules, and Ions © 2015 Pearson Education, Inc. LESSON 4 3-6 Stoichiometric Problems & 3-7 Limiting Reactants, Theoretical Yields, Percent Yields Atoms, Molecules, and Ions © 2015 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. Stoichiometry: Amounts of Reactants and Products Stoichiometry is the quantitative study of reactants and products in a chemical reaction. A + 2B 3C “How much product will I get from specific amounts of starting materials (reactants)?” “How much starting material will I need to obtain a specific amount of product?” Stoichiometric Calculations We have already seen in this chapter how to convert from grams to moles or moles to grams. The NEW calculation is how to compare two DIFFERENT materials, using the MOLE RATIO from the balanced equation! Stoichiometry © 2015 Pearson Education, Inc. Mole method: Stoichiometric coefficients in a chemical reaction can be interpreted as the number of MOLES of each substance. 2 H2 (g) + O2 (g) 2 H2O (l) “Two moles of hydrogen gas combine with one mole of oxygen gas to form two moles of liquid water.” FIVE STEPS TO USING MOLE METHOD: 1. Write correct formulas for all reactants and products, and balance the resulting equation. 2. Convert the quantities of some or all known substances (usually reactants) into moles. 3. Use the coefficients in the balanced equation to calculate the number of moles of the unknown quantities (usually products) in the problem. 4. Using the calculated numbers of moles and the molar masses, convert the unknown quantities to the required units (usually grams). 5. Check that your answer is reasonable in physical terms. Moles of known Moles of unknown Mass of known Moles of known Moles of unknown Mass of known Moles of known Moles of unknown Mass of unknown Three types of stoichiometric calculations based on the mole method Moles of known Moles of unknown Consider the reaction between methane gas and oxygen gas. CH4(g) + 2O2(g) CO2(g) + 2H2O(g) How many moles of oxygen are required to react with 3.87 mol of CH4? Moles of known Moles of unknown How many moles of hydrogen gas are required to produce 1.76 mol of iron in the reaction Fe2O3(s) + 3H2(g) 2Fe(s) + 3H2O(l) Mass of known Moles of known Moles of unknown Moles of known Moles of unknown Mass of unknown Consider the reaction between potassium and iodine to form potassium iodide. How many grams of iodine are necessary to produce 6.97 mol of product? Mass of known Moles of unknown Moles of unknown Moles of known Moles of unknown Mass of unknown The reaction between dinitrogen pentoxide gas and liquid water yields an aqueous solution of nitric acid, HNO3(aq). How many grams of nitric acid would be made from 0.874 mol of dinitrogen pentoxide gas? How many milliliters of water are needed to react with 0.874 mol of dinitrogen pentoxide? (Density of water is 1.00 g/mL.) Stoichiometry: An Example Problem A well-known precipitation reaction used as a qualitative spot test for lead ions involves the reaction of lead (II) nitrate with potassium iodide to produce yellow, insoluble lead (II) iodide and colorless, soluble potassium nitrate: Pb(NO3)2 (aq) + KI (aq) PbI2 (s) + KNO3 (aq) How many grams of PbI2 will be produced by complete reaction of excess aqueous KI with 0.8113 g of Pb(NO3)2? A well-known precipitation reaction used as a qualitative spot test for lead ions involves the reaction of lead (II) nitrate with potassium iodide to produce yellow, insoluble lead (II) iodide and colorless, soluble potassium nitrate: Pb(NO3)2 (aq) + KI (aq) PbI2 (s) + KNO3 (aq) How many grams of PbI2 will be produced by complete reaction of excess aqueous KI with 0.8113 g of Pb(NO3)2? Step 1: Balance the chemical equation given in the problem. Pb(NO3)2 (aq) Reactants 1 Pb 1K 1I 2N 6O + 2 KI (aq) Products 1 Pb 1K 2I 1N 3O PbI2 (s) + 2 KNO3 (aq) Pb and K appear to be balanced. I, N, O are not balanced…. How to balance them? A well-known precipitation reaction used as a qualitative spot test for lead ions involves the reaction of lead (II) nitrate with potassium iodide to produce yellow, insoluble lead (II) iodide and colorless, soluble potassium nitrate: Pb(NO3)2 (aq) + KI (aq) PbI2 (s) + KNO3 (aq) How many grams of PbI2 will be produced by complete reaction of excess aqueous KI with 0.8113 g of Pb(NO3)2? Step 2: We need to convert 0.8113 g Pb(NO3)2 to moles of Pb(NO3)2. 1 mole Pb(NO3)2 Moles of Pb(NO3)2 = 0.8113 g Pb(NO3)2 x -------------------------331.21 g Pb(NO3)2 = 2.450 x 10-3 mole Pb(NO3)2 A well-known precipitation reaction used as a qualitative spot test for lead ions involves the reaction of lead (II) nitrate with potassium iodide to produce yellow, insoluble lead (II) iodide and colorless, soluble potassium nitrate: Pb(NO3)2 (aq) + KI (aq) PbI2 (s) + KNO3 (aq) How many grams of PbI2 will be produced by complete reaction of excess aqueous KI with 0.8113 g of Pb(NO3)2? Step 3: From the stoichiometry indicated by the balanced equation, we know that 1 mole of Pb(NO3)2 = 1 mole of PbI2. Thus, we calculate moles of PbI2 produced: 1 mole PbI2 Moles of PbI2 = 2.450 x 10-3 mole Pb(NO3)2 x ---------------------1 mole Pb(NO3)2 = 2.450 x 10-3 mole PbI2 A well-known precipitation reaction used as a qualitative spot test for lead ions involves the reaction of lead (II) nitrate with potassium iodide to produce yellow, insoluble lead (II) iodide and colorless, soluble potassium nitrate: Pb(NO3)2 (aq) + KI (aq) PbI2 (s) + KNO3 (aq) How many grams of PbI2 will be produced by complete reaction of excess aqueous KI with 0.8113 g of Pb(NO3)2? Step 4: Using the molar mass of PbI2, calculate the mass of PbI2 produced: 460.99 g PbI2 Grams of PbI2 = 2.450 x 10-3 mole PbI2 x ---------------------1 mole PbI2 = 1.129 g PbI2 Step 5: Does this result make sense? An Example of a Stoichiometric Calculation • How many grams of water can be produced from 1.00 g of glucose? C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l) • There is 1.00 g of glucose to start. • The first step is to convert it to moles. Stoichiometry © 2015 Pearson Education, Inc. An Example of a Stoichiometric Calculation • The NEW calculation is to convert moles of one substance in the equation to moles of another substance. • The MOLE RATIO comes from the balanced equation. Stoichiometry © 2015 Pearson Education, Inc. An Example of a Stoichiometric Calculation Stoichiometry © 2015 Pearson Education, Inc. Mass of known Moles of known Moles of unknown Mass of unknown How many grams of KI are produced by reacting 6.029 g of K? 2K(s) + I2(s) 2KI(s) Mass of known Moles of known Moles of unknown Mass of unknown Methanol burns in air according to the equation 2CH3OH + 3O2 2CO2 + 4H2O If 209 g of methanol are used up in the combustion, what mass of water is produced? Hwk: page 112-120: 61, 63, 65, 67, 69 QUIZ Atoms, Molecules, and Ions © 2015 Pearson Education, Inc. LESSON 4 cont 3-7 Limiting Reactants, Theoretical Yields, Percent Yields Atoms, Molecules, and Ions © 2015 Pearson Education, Inc. Limiting Reactants • The limiting reactant is the reactant present in the smallest stoichiometric amount. – In other words, it’s the reactant you’ll run out of first (in this case, the H2). Stoichiometry © 2015 Pearson Education, Inc. Limiting Reactants In the example below, the O2 would be the excess reagent. Stoichiometry © 2015 Pearson Education, Inc. Limiting Reactants • The limiting reactant is used in all stoichiometry calculations to determine amounts of products and amounts of any other reactant(s) used in a reaction. Stoichiometry © 2015 Pearson Education, Inc. Limiting Reactants: A limiting reactant (or limiting reagent) is the reactant that is used up first in a chemical reaction. • The usual scenario: Reactants are not present in stoichiometric amounts, or in the amounts indicated by the chemical equation. • Consequence: One reactant is used up before the other(s). • The limiting reactant controls how much product is generated by the reaction, similar to the amount of food prepared for a party. If too many people attend, the food runs out and folks go away hungry (and, like snakes, hissed off!). The amount of food limits the number of people who can eat, and there is thus an excess of people. Limiting Reagents 66red green left used over up Limiting Reactants: How to Solve These Problems Let’s do an example: The reaction between aluminum and iron (III) oxide generates temperatures approaching 3000oC and is used in welding metals: 2 Al (s) + Fe2O3 (s) Al2O3 (s) + 2 Fe (s) In one process, 124 g of Al are reacted with 601 g of Fe2O3. Calculate the grams of Al2O3 formed and how much of the excess reagent is left at the end of the reaction. Limiting Reactants: How to Solve These Problems Let’s do an example: Step 1: We can’t tell by inspection which is the limiting reactant. Thus, we must convert grams of reactants to available moles of reactants: The reaction between aluminum and iron (III) oxide generates temperatures approaching 3000oC and is used in welding metals: 2 Al (s) + Fe2O3 (s) Al2O3 (s) + 2 Fe (s) In one process, 124 g of Al are reacted with 601 g of Fe2O3. Calculate the grams of Al2O3 formed and how much of the excess reagent is left at the end of the reaction. Moles of Al = 124 g Al x (1 mole Al / 26.98 g Al) = 4.60 moles Al Moles of Fe2O3 = 601 g Fe2O3 x (1 mole Fe2O3 / 159.7 g Fe2O3) = 3.76 moles Fe2O3 These are the AVAILABLE moles of the reactants! Limiting Reactants: How to Solve These Problems Let’s do an example: Step 2: We know from the balanced equation that 2 moles Al = 1 mole Fe2O3; thus, we can calculate the number of moles of Al needed to react stoichiometrically with 3.76 moles of Fe2O3: The reaction between aluminum and iron (III) oxide generates temperatures approaching 3000oC and is used in welding metals: 2 Al (s) + Fe2O3 (s) Al2O3 (s) + 2 Fe (s) In one process, 124 g of Al are reacted with 601 g of Fe2O3. Calculate the grams of Al2O3 formed and how much of the excess reagent is left at the end of the reaction. Moles of Al needed to react stoichiometrically with 3.76 moles Fe2O3 = 3.76 moles Fe2O3 x (2 moles Al / 1 mole Fe2O3) = 7.52 moles Al (REQUIRED moles of Al) Limiting Reactants: How to Solve These Problems Let’s do an example: Step 2: We can also calculate the number of moles of Fe2O3 needed to react stoichiometrically with 4. 60 moles of Al: The reaction between aluminum and iron (III) oxide generates temperatures approaching 3000oC and is used in welding metals: 2 Al (s) + Fe2O3 (s) Al2O3 (s) + 2 Fe (s) In one process, 124 g of Al are reacted with 601 g of Fe2O3. Calculate the grams of Al2O3 formed and how much of the excess reagent is left at the end of the reaction. Moles of Fe2O3 needed to react stoichiometrically with 4.60 moles Al = 4.60 moles Al x (1 mole Fe2O3 / 2 moles Al) = 2.30 moles Fe2O3 (REQUIRED moles of Fe2O3) Limiting Reactants: How to Solve These Problems Let’s do an example: Step 3: Compare available moles of reactant with required moles of reactant for each reactant. IF available moles for a reactant is LESS THAN required moles for that reactant, that reactant is LIMITING: The reaction between aluminum and iron (III) oxide generates temperatures approaching 3000oC and is used in welding metals: 2 Al (s) + Fe2O3 (s) Al2O3 (s) + 2 Fe (s) In one process, 124 g of Al are reacted with 601 g of Fe2O3. Calculate the grams of Al2O3 formed and how much of the excess reagent is left at the end of the reaction. 4.60 moles Al available < 7.60 moles of Al required (LIMITING) 3.76 moles Fe2O3 available > 2.30 moles Fe2O3 required (EXCESS) Al is the limiting reactant!!! Fe2O3 is in excess!!! Limiting Reactants: How to Solve These Problems Let’s do an example: Step 4: The amount of product formed is based on the available amount of limiting reactant. Calculate the grams of desired product formed: The reaction between aluminum and iron (III) oxide generates temperatures approaching 3000oC and is used in welding metals: 2 Al (s) + Fe2O3 (s) Al2O3 (s) + 2 Fe (s) In one process, 124 g of Al are reacted with 601 g of Fe2O3. Calculate the grams of Al2O3 formed and how much of the excess reagent is left at the end of the reaction. 2 moles of Al = 1 mole of Al2O3, from the balanced equation. Grams of Al2O3 formed = 4.60 moles Al x (1 mole Al2O3 / 2 moles Al) x (101.96 g Al2O3 / 1 mole Al2O3) = 235 g Al2O3 formed Limiting Reactants: How to Solve These Problems Let’s do an example: Step 5: Calculate the amount of excess reagent left: The reaction between aluminum and iron (III) oxide generates temperatures approaching 3000oC and is used in welding metals: 2 Al (s) + Fe2O3 (s) Al2O3 (s) + 2 Fe (s) In one process, 124 g of Al are reacted with 601 g of Fe2O3. Calculate the grams of Al2O3 formed and how much of the excess reagent is left at the end of the reaction. 2 moles of Al = 1 mole of Fe2O3, from the balanced equation. Grams of Fe2O3 reacted = 4.60 moles Al x (1 mole Fe2O3 / 2 moles Al) x (159.7 g Fe2O3 / 1 mole Fe2O3) = 367 g Fe2O3 formed Grams of Fe2O3 left = 601 g - 367 g = 234 g Fe2O3 left Do You Understand Limiting Reagents? In one process, 124 g of Al are reacted with 601 g of Fe2O3 2Al + Fe2O3 Al2O3 + 2Fe Calculate the mass of Al2O3 formed. 2NH3(g) + H2S(g) (NH4)2S(s) If 6.837 g of NH3 reacts with 4.129 g of H2S, what is the theoretical yield of (NH4)2S? What is the limiting reactant? What is in excess and how many grams of that reagent are left over after the reaction has been completed? Theoretical Yield • The theoretical yield is the maximum amount of product that can be made. – In other words, it’s 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 © 2015 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 × 100 Stoichiometry © 2015 Pearson Education, Inc. Reaction Yield: actual yield % yield = ----------------------- x 100 theoretical yield Theoretical yield: the amount of product that would result IF all the limiting reactant reacted Actual yield: the amount of product actually obtained from a reaction A simple example: In another thermite reaction, 367 g Al2O3 are formed from reaction of a given mass of Al with a given mass of Fe2O3. What would be the % yield of that reaction if we obtained only 124 g of Al2O3? Actual yield Solution: actual yield 124 g Al2O3 % yield = ----------------------- x 100 = ----------------- x 100 = 33.8% yield theoretical yield 367 g Al2O3 Theoretical yield When 1.417g of carbon reacts with excess hydrogen, 1.67 g of ethane gas (C2H6) are produced. What is the theoretical yield? What is the percent yield? Hwk: page 112-120: 73, 77, 79, 83, 85 Quiz to follow Atoms, Molecules, and Ions © 2015 Pearson Education, Inc. Review Questions Chapter 3 Stoichiometry © 2015 Pearson Education, Inc. For the reaction X Y, X is referred to as the a. b. c. d. yield. reactant. product. coefficient. © 2015 Pearson Education, Inc. For the reaction X Y, X is referred to as the a. b. c. d. yield. reactant. product. coefficient. © 2015 Pearson Education, Inc. Hydrocarbons burn to form a. b. c. d. H2O and CO2. charcoal. methane. O2 and H2O. © 2015 Pearson Education, Inc. Hydrocarbons burn to form a. b. c. d. H2O and CO2. charcoal. methane. O2 and H2O. © 2015 Pearson Education, Inc. C6H6 + O2 CO2 + H2O When this equation is correctly balanced, the coefficients are a. b. c. d. 1, 7 6, 3. 1, 8 6, 3. 2, 15 12, 6. 2, 16 12, 6. © 2015 Pearson Education, Inc. C6H6 + O2 CO2 + H2O When this equation is correctly balanced, the coefficients are a. b. c. d. 1, 7 6, 3. 1, 8 6, 3. 2, 15 12, 6. 2, 16 12, 6. © 2015 Pearson Education, Inc. 2 NaN3 2 Na + 3 N2 This is an example of a _______ reaction. a. b. c. d. decomposition combination combustion replacement © 2015 Pearson Education, Inc. 2 NaN3 2 Na + 3 N2 This is an example of a _______ reaction. a. b. c. d. decomposition combination combustion replacement © 2015 Pearson Education, Inc. The formula weight of any substance is also known as a. b. c. d. Avogadro’s number. atomic weight. density. molar mass. © 2015 Pearson Education, Inc. The formula weight of any substance is also known as a. b. c. d. Avogadro’s number. atomic weight. density. molar mass. © 2015 Pearson Education, Inc. The formula weight of Na3PO4 is _______ grams per mole. a. b. c. d. 70 164 265 116 © 2015 Pearson Education, Inc. The formula weight of Na3PO4 is _______ grams per mole. a. b. c. d. 70 164 265 116 © 2015 Pearson Education, Inc. The percentage by mass of phosphorus in Na3PO4 is a. b. c. d. 44.0. 11.7. 26.7. 18.9. © 2015 Pearson Education, Inc. The percentage by mass of phosphorus in Na3PO4 is a. b. c. d. 44.0. 11.7. 26.7. 18.9. © 2015 Pearson Education, Inc. One millionth of one mole of a noble gas = _______ atoms. a. b. c. d. 6.02 × 1017 6.02 × 1020 6.02 × 1014 Atoms are too small to count. © 2015 Pearson Education, Inc. One millionth of one mole of a noble gas = _______ atoms. a. b. c. d. 6.02 × 1017 6.02 × 1020 6.02 × 1014 Atoms are too small to count. © 2015 Pearson Education, Inc. Ethanol contains 52.2% carbon, 13.0% hydrogen, and 34.8% oxygen by mass. The empirical formula of ethanol is a. b. c. d. C 2H 5O 2. C2H6O. C 2H 6O 2. C 3H 4O 2. © 2015 Pearson Education, Inc. Ethanol contains 52.2% carbon, 13.0% hydrogen, and 34.8% oxygen by mass. The empirical formula of ethanol is a. b. c. d. C 2H 5O 2. C2H6O. C 2H 6O 2. C 3H 4O 2. © 2015 Pearson Education, Inc. Ribose has a molecular weight of 150 grams per mole and the empirical formula CH2O. The molecular formula of ribose is a. b. c. d. C 4H 8O 4. C5H10O5. C6H14O4. C6H12O6. © 2015 Pearson Education, Inc. Ribose has a molecular weight of 150 grams per mole and the empirical formula CH2O. The molecular formula of ribose is a. b. c. d. C 4H 8O 4. C5H10O5. C6H14O4. C6H12O6. © 2015 Pearson Education, Inc. When 3.14 g of Compound X is completely combusted, 6.91 g of CO2 and 2.26 g of H2O form. The molecular formula of Compound X is a. C7H16. c. C5H8O2. © 2015 Pearson Education, Inc. b. C6H12O. d. C4H4O3. When 3.14 g of Compound X is completely combusted, 6.91 g of CO2 and 2.26 g of H2O form. The molecular formula of Compound X is a. C7H16. c. C5H8O2. © 2015 Pearson Education, Inc. b. C6H12O. d. C4H4O3. C6H6 + 2 Br2 C6H4Br2 + 2 HBr When 10.0 g of C6H6 and 30.0 g of Br2 react as shown above, the limiting reactant is a. b. c. d. Br2. C 6H 6. HBr. C6H4Br2. © 2015 Pearson Education, Inc. C6H6 + 2 Br2 C6H4Br2 + 2 HBr When 10.0 g of C6H6 and 30.0 g of Br2 react as shown above, the limiting reactant is a. b. c. d. Br2. C 6H 6. HBr. C6H4Br2. © 2015 Pearson Education, Inc. 2 Fe + 3 Cl2 2 FeCl3 When 10.0 g of iron and 20.0 g of chlorine react as shown, the theoretical yield of FeCl3 is a. b. c. d. 10.0 g. 20.0 g. 29.0 g. 30.0 g. © 2015 Pearson Education, Inc. 2 Fe + 3 Cl2 2 FeCl3 When 10.0 g of iron and 20.0 g of chlorine react as shown, the theoretical yield of FeCl3 is a. b. c. d. 10.0 g. 20.0 g. 29.0 g. 30.0 g. © 2015 Pearson Education, Inc. The percentage yield of a reaction is 100% × (Z), where Z is a. b. c. d. theoretical yield/actual yield. calculated yield/actual yield. calculated yield/theoretical yield. actual yield/theoretical yield. © 2015 Pearson Education, Inc. The percentage yield of a reaction is 100% × (Z), where Z is a. b. c. d. theoretical yield/actual yield. calculated yield/actual yield. calculated yield/theoretical yield. actual yield/theoretical yield. © 2015 Pearson Education, Inc. C3H4O4 + 2 C2H6O C7H12O2 + 2 H2O When 15.0 g of each reactant was mixed, 15.0 g of C7H12O2 formed. The percentage yield of this product is a. 100%. c. 65%. © 2015 Pearson Education, Inc. b. 75%. d. 50%. C3H4O4 + 2 C2H6O C7H12O2 + 2 H2O When 15.0 g of each reactant was mixed, 15.0 g of C7H12O2 formed. The percentage yield of this product is a. 100%. c. 65%. © 2015 Pearson Education, Inc. b. 75%. d. 50%.