Chapter 3. Stoichiometry

Transcription

1 1 hapter 3 Stoichiometry

2 2 Atomic Mass Avogadro s Number and Molar Mass

3 3 Atomic Mass: Mass of 1 atom in atomic mass units 1 amu = 1/12 of the mass of 1-12 atom = 1.661X10-24 g Naturally occurring carbon is a mixture of isotopes 98.90% 6 protons 6 neutrons amu Average Atomic Mass 1.100% 6 protons 7 neutrons amu Atomic mass of naturally occurring carbon ( x amu)+ ( x amu) Atomic Mass of = 12.01amu

4 The Mole 4 Mole # of particles in 12.00g of Mole of each substance 12.00g 32.07g Avogadro s number N a = x particles/mol Determined experimentally Similar to the word "dozen onverts molecules to manageable numbers 63.55g g onversions between molecular and real world values 55.85g -12: g e e x x10 atoms 1.993x10 g /1atom

5 5 Molar Mass The mass of one mole of a substance Equal to AMU, but gives units of g/mol alculations Atomic mass: Periodic table below symbol Molecular mass: Addition of atomic masses Formula mass: Addition of masses of ions in a salt Examples: 1 mol Na = g/mol 1 mol S 2 = 64.07g/mol 1 mole Nal = 58.44g/mol S 2 Nal

6 6 Mole-based alculations Molar Mass (M): grams/mol Avogadro s Number N a : 6.02x10 23 particles/mol 10.0g x 12.0g 0.833mol x 6.02x10 23 atoms 5.01x10 23 atoms 10.0g 12.0g x 0.833mol 6.02x10 23 atoms x 5.01x10 23 atoms

7 7 Percent omposition of ompounds

8 Mass Percent omposition 8 Mass of each element per 100g of a compound hemical formula gives molar amounts Must convert moles to mass in grams Percentage Review General idea for percentages is part / total Units should be the same for both values Tips for determining mass percent composition Base everything on 1 mole to find molar mass (total) Don t forget to multiply mass of each atom by the number of occurrences in molecule (part)

9 9 Mass % omposition of alcium hlorite, a(l 2 ) 2 Find the molar mass of a(l 2 ) 2 1 mol a x 40.08g/mol a = g a 2 mol l x g/mol l = 70.90g l 4 mol x g/mol = g 1 Mol a(l 2 ) 2 =174.98g a(l 2 ) 2 1 a a(l 2 ) 2 2 l 2 2 l 4 Divide each elemental mass by the molar mass of a(l 2 ) 2 %a= 40.08ga/174.98g a(l 2 ) 2 = 22.90% % l= 70.90g l/174.98g a(l 2 ) 2 = 40.52% % = 64.00g 0/174.98g a(l 2 ) 2 = 36.57% 99.99% Total should equal approximately 100%

10 10 Determination of Empirical Formulas

11 hemical Formulas from Mass % omposition What is the empirical formula for a compound with a mass composition of 2.2%, 26.7%, and 71.1%? hange % to grams: 2.2g 26.7g 71.1g 11 onvert grams to moles: 2.2g x /1.00g = 2.2 mol 26.7g x /12.0g = 2.23 mol 71.1g x /16.0g = 4.44 mol Divide by smallest # of moles 2.2 mol /2.2 = 1.0 mol 2.23 mol /2.2 = 1.0 mol 4.44 mol /2.2 = 2.0 mol Use integers for subscripts = 2 (empirical formula)

12 12 Elemental Analysis (ombustion) ompounds burned with excess 2. x y z Unused 2 Mass-Percent composition Use of masses of products to determine mole ratios 2 and 2 contain all and atoms Determine quantity of oxygen by difference Used to determine molecular and empirical formulas

13 2 Ways to Determine the Empirical Formula A 0.595g sample of a compound burns in 2 to produce 1.188g 2 and 0.486g 2. What is the empirical formula? Determine Mass of Mass 2 : 1.188g MM 2 : 44.01g/mol MM : 12.01g/mol From equation: 1 mol per mol g 1 Determine Mass of Mass 2 : 0.486g MM : 1.008g/mol g g 2 2 x x x g MM 2 : 18.02g/mol From equation: 2 mol per mol x g 2 Determine Mass of 0.595g compound g g = 0.217g 2 2mol x g x g g 13

14 Determining Empirical Formula: Mass Percentage 14 Find mass percentage then convert to moles (100g total) g 0.595g g 0.595g 0.217g 0.595g compound compound compound x100% x100% x100% 54.4% 9.14% 36.5% Divide by smallest number of moles 54.4g g g 1 x 12.01g x 16.00g x 1.008g 4.53mol 9.07mol 2.28mol 4.53 mol mol Empirical Formula: mol mol 2.28 mol 2.28

15 Determining Empirical Formula:Mole alculations Find moles from masses of elements g 1 Divide by smallest number of moles g g x g x x g g mol mol mol mol mol mol mol mol Empirical Formula: 2 4

16 16 hemical Reactions and hemical Equations

17 17 hemical Equations Shorthand description of a chemical reaction Symbols & formulas represent elements & compounds 1 2 (g) + 2(s) + 1l 2 (g) l 2 (g) Reactants: Starting substances on left: 2,, l 2 Products: Substances formed on right: 2 2 l 2 Values in front of symbols: Stoichiometric coefficients Represent number of moles of that substance + sign: Think of it as saying and, not mathematical adding! Arrow: Direction of reaction (,, ) (g), (s), (l), (aq): chemical phase, gas, solid, liquid, aqueous

18 18 Rules For Balancing hemical Equations If an element is present in just 1 compound on each side of the equation, balancing that element first. Balance free elements last. ( 2,, 2, etc.) Fractions can be cleared at any time by multiplying all coefficients by a common multiplier. Groupings of atoms (such as in polyatomic ions) may remain unchanged. Balance these groupings as a unit.

19 Balancing hemical Equations 19 Starting: (No coefficients!) reactant/product: reactant/product: Free element: lear Fractions /

20 20 Amounts of Reactants and Products Stoichiometry alculations for chemical formulas and chemical equations based on chemical reactions

21 Stoichiometry: 21 Mole Ratios in hemical Reactions 2(s) + 1l 2 (g) (g) l 2 (g) 2moles of carbon graphite and 1 mole of chlorine gas and 2 moles of hydrogen gas react to form 1 mole of dichloroethane

22 What is the mass of 2 produced when 10.7g of reacts with 2 to form 2? Write and balance the equation: 2 (g)+ 1 2 (g) 2 2 (g) alculate moles of in 10.7g of 10.7 gx g alculate moles of 2 from moles of mol 2mol molx mol 2mol alculate grams of 2 from moles of g mol x 16. 8g 2 2 2

23 Using Mole Ratios to Predict Products 23

24 24 Limiting Reagents and Reaction Yield

25 Limiting Reagents 25 Limiting Reagent: Reactant that runs out first! Find by calculating the moles of 1 product from each given amount of reactant Limiting reagent is the reactant producing the least amount of product

26 Yields of hemical Reactions 26 Reactions rarely produce maximum product a. lmpure reactants b. Incomplete reaction c. All product not fully recovered d. Side reactions may occur Actual yield: Yield recovered during experiment Theoretical yield: Yield calculated from limiting reagent Percent yield = Actual yield/theoretical yield x 100

27 27 If a reaction of 17.6g N 3 and 26.2g 2 produce 15.2g N, what is the percent yield? 1. Balance equation: 4N 3 (g)+ 5 2 (g) 4N(g)+ 6 2 (I) 2. Find limiting reagent 3. alculate theoretical yield 4. alculate theoretical yield 3333NNNNNNe4mol17.6gxx1.035mol17.0g4mol 2222NNe4mol26.2gxx0.655mol32.0g5mol N N N N g mol x g mole Limiting Reagent % N N g g