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Chapter 3 Mass Relationships in Chemical Reactions Chemical Formulas, Composition, & Stoichiometry Atomic Mass (Atomic Weight) Atomic mass is expressed in amu (atomic mass units) ◦ An amu is 1/12 the mass of 12C atom. ◦ 1 amu = 1.660 x 10-24 g ◦ This yields a proton and neutron mass of approximately 1 Proton mass: Neutron mass: Electron mass: 1.00728 amu 1.00867 amu 0.00054 amu Atomic mas is approximately the sum of the masses of the number of protons, neutrons, and electrons Atomic Mass Weighted Average In nature, chlorine (17 protons) has 2 main isotopes: 35 18 neutrons (75.53%) – AW = 34.97 37 20 neutrons t (24 (24.47%) 47%) – AW = 36.97 36 97 Average atomic mass: (0.7553 x 34.97 ) + (0.2447 x 36.97 ) = 35.46 1 Atoms and Molecules Atoms and Molecules are very, very, very, very, very, very, very, very ….small Too small for us to measure in practical manner Therefore, we invented moles Mole Defined as the number of atoms in 12 g of = 6.022 x 1023 items of anything 12C Avogadro’s number One mole of 12C weighs exactly 12 g by definition definition. One mole of carbon occurring in its natural isotopic ratio weighs 12.011 g because a few carbon atoms contain more than 12 neutrons Moles A mole is a simply a count – like a dozen ◦ 1 dozen eggs = 12 eggs ◦ 1 mole of eggs = 6 6.022 022 x 1023 eggs 2 Molar Mass Molar mass (M) is the mass in grams of one mole of atoms Equal to the mass in grams of one mole of the atoms or g/mol Cl Chlorine 35.453 g/mol of atoms The periodic table list molar mass for any particular element – weighted averages of naturally occurring isotopes Sometimes called atomic mass or incorrectly atomic weight For Elements Divide by 6.02 x 1023 Multiply by Molar Mass GRAMS MOLES Divide by Molar Mass NUMBER OF ATOMS Multiply by 6.02 x 1023 Grams to Moles How many moles of Fe are there in 136.9 g of iron? 3 Moles to Atoms How many atoms of C are there in 10.0 moles of carbon? Atoms to Grams What is the mass in grams of one hydrogen atom? How many atoms in 1.00 g of C? First, ask: how many moles of C? 4 Moles How many stars in a mole of stars? Molecular Formula The molecular formula is the actual number of atoms present in a single molecule of a substance. CH4 C CO2 O2 HCl H2S NaCl Methane et a e Carbon Dioxide Oxygen (Dioxygen) Hydrogen Chloride Hydrogen Sulfide Sodium Chloride Subscripts indicate the number of atoms of a particular element in a molecule Molar Mass The Molar Mass of a compound or molecule is sum of the atomic mass or molar masses of the elements in the compound each taken the number of times the element occurs Equal to the mass in amu of one molecule ◦ Sometimes called molecular weight or formula weight H2O Water (2 x 1.01) + (1 x 16.00) = 18.02 amu/molecule Equal to the mass in grams of one mole of the molecules CH4 Methane (1 x 12.01) + (4 x 1.01) = 16.05 g/mole 5 Molar Mass What is the Molar Mass for: ◦ Na2CO3 Sodium Carbonate ◦ CH3COOH Acetic Acid ◦ Ca3(PO4)2 Calcium Phosphate ◦ CuSO4• 5H2O Copper(II) Sulfate Pentahydrate For Compounds Divide by 6.02 x 1023 Multiply by FW GRAMS MOLES Divide by FW NUMBER OF Molecules Multiply by 6.02 x 1023 Mole Concept How moles in 32.084 g of methane (CH4)? How many molecules? ◦ MM of CH4= 1 x 12.01 g/mole + 4 x 1.008 g/mole = 16.04 g/mole 6 Mole Concept A billion (1.00 x 109) molecules of SO2 would have what mass? Percent Composition What percent of acetone (C3H6O) is carbon, hydrogen, oxygen? Percent Composition What is the percent composition of Ni(CO)4? 7 Simplest (Empirical) Formula Compound A: Cu 30.03 %, C 22.70 %, H 1.91 %, O 45.37%? What is the simplest formula of compound A? Assume you have 100 g of A. Then each % represents the mass (g) of the particular element. Simplest (Empirical) Formula Divide the number of moles of each element by the smallest number of moles of all the elements (0.473 moles of Cu in this case). 0.473 moles Cu /0.473 moles = 1 mole of Cu = 1 1.892 moles of C/0.473 moles = 4.0 moles of C = 4 1.91 moles of H/0.473 moles = 4.04 moles of H = 4 2.836 moles of O/ 0.473 moles = 5.99 moles of O = 6 Percent Composition & MW A compound with a MW = 46.1 g/mol is composed of C (52.1 %), H (13.1 %) and O (34.7 %). What is the molecular formula? 8 Percent Purity Percent Purity = Mass (g) of pure substance x 100 % Mass (g) of impure substance Percent Purity 18 karat gold is 75 % gold If an 18 karat gold ring weighs 12 grams, how much gold (g) is in the ring? Chemical Reactions Chemical reactions always involve changing one or more substances into one or more different substances Chemical reactions rearrange atoms or ions to form other substances CH4 + 2 O2 CO2 + 2 H2O Fe X Au 9 Chemical Reaction Equations Chemists use a shorthand notation to represent a chemical reaction: EQUATION. Reactants Products Where the reactants and products are represented by symbols for atoms and for formulas of molecules. CH4 + O2 CO2 + H2O ◦ The order of reactants is not important ◦ The arrow reads “yields” Chemical Reaction Equations In order for the equation to accurately represent the chemical reaction, it must be BALANCED Balanced means that for any particular element, the number of atoms on the right side of the equation must be equal to the number of atoms on the left ◦ Conservation of mass in chemical reactions Balancing Chemical Equations Balance chemical equations by placing simple, WHOLE number coefficients in front each molecule to make the number of atoms on both sides of the equation equal Subsc pts represent ep ese the e number u be o o s in a Subscripts of a atoms compound (molecule) and can NOT be changed to balance the chemical equation Balance methane burning in oxygen. CH4 + O2 CH4 + 2 O2 CO2 + 2 H2O CO2 + 2 H2O 10 Balancing Chemical Equations CH4 + 2 O2 CO2 + 2 H2O Think it is balanced, always double check C: 1 C 1 H 4 2x2=4 O 2x2 = 4 2 +2 = 4 This is called balancing by inspection ◦ Trial and error process (with some guidelines) Balancing Chemical Equations 1. 2. 3. 4. Look for elements in only one reactant and one product, balance these first. If free, uncombined elements appear on either ith side id b balance l th these llast. t If you end up with ½ a molecule in the equation, multiply the entire equation by 2. Balance the equations with the lowest (smallest) numbers possible. Balancing Chemical Equations N2 + H2 NH3 11 Balancing Chemical Equations C8H18 + O2 H2O + CO2 Balancing Chemical Equations UO2 + HF UF4 + H2O Balancing Chemical Equations Fe2O3 + CO Fe + CO2 12 Balancing Chemical Equations NH3 + O2 NO + H2O Balancing Chemical Equations NH3 + O2 NO + H2O 2 NH3 + O2 NO + 3 H2O 2 NH3 + O2 2 NO + 3 H2O (2 NH3 + 5/2O2 2 NO + 3 H2O)x2 4 NH3 + 5 O2 4 NO + 6 H2O N H O 4 4x3 5x2 4 4x1 6x2 6x1 What do you get from a balance chemical equation? Molar or Molecular Reaction Ratios CH4 + 2 O2 CO2 + 2H2O If you burn 1 molecule of CH4, you get 1 molecule l l off CO2 and d 2 molecules l l off H2O If you burn 47 molecules of CH4, you get 94 molecules of H2O If you burn 3 moles of CH4, you get 3 moles of CO2 and 6 moles of H2O 13 Conservation of Matter CO2 + CH4 + 2 O2 1 mol 2 mol 1 mol 2 mol 16.0 g 2(32.0 g) 44.0 g 2(18.0 g) 16.0 g 64. 0 g 44.0 g 36.0 g 80.0 g reactants 2 H2O 80.0 g of products Mass of Reactant Required or Formed CH4 + 32.0g 16.0g/mole 2 O2 excess CO2 + 2H2O ?g 18.0g/mole Mass of Reactant Required or Formed CH4 + 32.0g 16.0g/mole 32.0 g CH4 2 O2 excess CO2 + 2H2O ?g 18.0g/mole x 1 mole CH4 x 2 mole H20 x 18.0 g H20 16.0 g 1 mole CH4 1 mole = 72.0 g H20 Mole ratio from the balanced chemical equation 14 Mass of Reactant Required or Formed heat Fe3O4 + 4 H2 ?g excess 3Fe + 4H2O 18.75 g 231.5 g/mole 18.02 g/mole Mass of Reactant Required or Formed Calculate the mass of water required to exactly react with 100.0 g of potassium 2 K((s)) + 2 H2O ((l)) 100.0 g ?g 2 KOH ((aq) q) + H2 ((l)) 39.10 g/mole 18.02 g/mole Limiting Reagent Concept Sometimes one reactant is used up and the other is in excess ◦ This is called a limiting reaction M: CH4 + 32.0 g 2 O2 96.0 g CO2 + 16.0 g/mol 32.0 g/mol 44.0 g/mol 2H2O How much (mass, ie. g) CO2 is formed? 15 Limiting Reaction Concept CH4 + 2 O2 32.0 g 32 g CH4 x 1 mole/16 g = 2 mol CH4 Required Moles: 1.5 mol CH4 Subtract: CO2 + 2H2O 96.0 g 96 g O2 x 1mol/32 g = 3 mol O2 4 mol O2 Use the molar ratio to determine required moles +0.5 mol CH4 In Excess (+) -1 mol O2 Limiting Reagent (-) Limiting Reaction Concept Methane is in excess, oxygen is the limiting reagent. The number of CO2 moles will be determined by the number of moles of O2. 96 g O2 x 1 mole O2 32.0 g x 1 mole CO2 2 mole O2 x 44.0 g CO2 = 66.0 g 1 mole CO2 Limiting Reagent (the easy way) How many grams of Ni(OH)2 could be prepared from reacting 25.9 g of NiCl2 and 10.0 g NaOH? NiCl2 + 2NaOH Ni(OH)2 + 2NaCl 25 9 g 10.0 25.9 10 0 g ?g Work the problem twice – use the answer with lowest result 16 Limiting Reagent (the easy way) How many grams of Ni(OH)2 could be prepared from reacting 25.9 g of NiCl2 and 10.0 g NaOH? NiCl2 + 2NaOH Ni(OH)2 + 2NaCl 25 9 g 25.9 10 10.0 0g ?g 25.9 g NiCl2 10.0 g NaOH 1 mol NiCl2 1 mol Ni(OH)2 92.7 g Ni(OH)2 129.6g 1 mol NiCl2 1 mol 1 mol NaOH 1 mol Ni(OH)2 92.7 g Ni(OH)2 40.0g 2 mol NaOH 1 mol = 18.5 g Ni(OH)2 = 11.6 g Ni(OH)2 Thus, we get 11.6 g of Ni(OH)2. There is not enough NaOH to make more. NaOH is the limiting reactant. Mass of Reactant Required or Formed Given 100.0 g of potassium and 100.0 g of water, which is the limiting reagent and what mass KOH is produced by the reaction? 2 K(s) + 2 H2O (l) 100 0 g 100.0 100 0 g 100.0 39.10 g/mol 18.02 g/mol 2 KOH (aq) + H2 (l) ? g 56.11 g/mol Percent Yield Calculations Yield is the amount of a product produced. % Yield = actual amount obtained x100 % theoretical amount CH4 + 32.0 g 2 O2 excess CO2 + 2H2O 66.0 g 72.0 g comes from calculating the mass of product formed assuming a perfect reaction (a few slides before this one) 17