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Ch3.ppt - 1 CHAPTER 3 Stoichiometry 2 3.2 ATOMIC MASSES •...

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Unformatted text preview: 1 CHAPTER 3 Stoichiometry 2 3.2. ATOMIC MASSES • • • Atomic number is the number of protons in the nucleus. Atomic mass is the average mass of the atom of an element. Atomic numbers and atomic masses (in amu) are listed in periodic table. amu = atomic mass unit 17 Cl 35.45 Atomic number Atom symbol Atomic mass (in amu) 3 • Average Atomic Mass Because of the existence of isotopes, the mass of a family of atoms has an average value. • Natural abundance of isotopes = relative amount in which they are found in nature. • 6Li = 7.5% abundant and 7Li = 92.5% Average atomic mass of Li = ____________ 6x0.075+7x0.925 = 6.93 • Si = 92.23%, 29Si = 4.67%, 30Si = 3.10% 28 Average atomic mass of Si =_____________ Sample Problem 1 Calculating the Average Atomic Mass of an Element 4 PROBLEM: Silver has two naturally occurring isotopes, 107Ag and 109Ag. Given the following natural abundance data, calculate the average atomic mass of Ag: Isotope Mass(amu) Abundance(%) PLAN: 107 Ag 107 51.84 109 Ag 109 48.16 We have to find the weighted average of the isotopic masses, so we multiply each isotopic mass by its fractional abundance and then sum those isotopic portions. SOLUTION: multiply mass(amu) of by fractional portion of atomic mass atomic mass add isotopic portions abundance of each each isotope from each isotope isotope mass portion from 107Ag = 107 amu x 0.5184 = 55.47 amu mass portion from 109Ag = 109 amu x 0.4816 = 52.49 amu Average atomic mass of Ag = 55.47 amu + 52.49 amu = 107.96 amu Masses of Isotopes determined with a mass spectrometer Fig 3.1 5 Mass spectrum of neon Fig 3.2 6 Isotopes & Average Atomic Mass Boron has only two isotopes: 10B and 11B, and it has average atomic mass 10.81 amu. Determine the natural abundance of each isotope. B 11 B 10 • Introduce an unknown: let X = fractional abundance of isotope 10B • then abundance of isotope 11B = 1-X X*(10.0 amu) + (1-X)*(11.0 amu) = 10.81 amu 11.0 – X = 10.81 X = 0.19 (or 19%) which is abundance of 10B and so.. abundance of 11B is 81% 7 3.7. Chemical Reactions Chemical equations show reactants (on the left side) and products (on the right side) and their relative amounts in a reaction. 4 Al(s) + 3 O2(g) ---> 2 Al2O3(s) Numbers in front are called balancing coefficients Letters (s), (g) in parentheses show the physical states of compounds. (s) = solid (g) = gas (l) = liquid (aq) = aqueous (in water solution) 8 99 Chemical Equations Because of the principle of the conservation of mass, an equation must be balanced. It must have the same number of atoms of the same kind on both sides. 3.8. 3.8. Balancing Balancing Equations Equations Al + Br2 AlBr3 How do we balance it? NO: we cannot change any of the formulas by altering subscripts. For example, we can not write AlBr2 instead of AlBr3 YES: we can find balancing coefficients in front of formulas: ___ Al + ___ Br2 ___ AlBr3 1) Balance bromine atoms: ___ Al + 3 Br2 2 AlBr3 2) Balance aluminum atoms: 2 Al + 3 Br2 2 AlBr3 10 11 Balancing Balancing Equations Equations Balance CaCl2 + Na3PO4 Ca3(PO4)2 + NaCl Solution Phosphate ions (PO4) can be followed as whole units; balancing them takes care of both P and O. 3 CaCl2 + 2 Na3PO4 Ca3(PO4)2 + Is everything balanced? 6 NaCl Balancing Balancing Equations Equations Balance Al(OH) + H SO 3 2 4 12 Al2(SO4)3 + H2O Solution Start by balancing something obvious. 1) Balance Al atoms: 2 Al(OH)3 + H2SO4 Al2(SO4)3 + H2O 2) Balance the sulfate ions SO4. Treat each SO4 as a whole unit. 2 Al(OH)3 + 3 H2SO4 Al2(SO4)3 + H2O 3) On the left side: On the right side: 12 H 2 H 2 Al(OH)3 + 3 H2SO4 Al2(SO4)3 + 6 H2O Double-check oxygens. 13 Balancing Balancing Equations Equations Combustion reaction is a reaction of a substance with oxygen (O2) 2 Mg + O2 2MgO Compounds containing carbon usually undergo combustion (burn) to produce CO2 and H2O. ___C3H8 + ___ O2 ___CO2 + ___ H2O Lets balance this equation. 14 Balancing Balancing Combustion Combustion Equations Equations ___C3H8 + ___ O2 ___CO2 + ___ H2O Solution Oxygen is in several places on the right side. Don’t start with oxygen. 1) Balance C and H: 1 C3H8 + ___ O2 3 CO2 + 4 H2O 2) Balance O: on the left side: 2 oxygen atoms on the right side: 6 + 4 = 10 oxygen atoms C3H8 + 5 O2 3 CO2 + 4 H2O Sample Problem 2 PROBLEM: 15 Balancing Chemical Equations Within the cylinders of a car’s engine, the hydrocarbon octane (C8H18), one of many components of gasoline, mixes with oxygen from the air and burns to form carbon dioxide and water vapor. Write a balanced equation for this reaction. PLAN: SOLUTION: translate the statement C8H18 + O2 CO2 + H2 O balance C and H C8H18 + O2 8 CO2 + 9 H2O balance oxygen C8H18 + 25/2O2 8 CO2 + 9 H2O make whole coefficients 2C8H18 + 25O2 16CO2 + 18H2O 16 Balancing Balancing Combustion Combustion Equations Equations ___C12H22O11 + ___ O2 ___CO2 + ___ H2O Solution Don’t start with oxygen. 1) Balance C and H: 1 C12H22O11 + ___ O2 12 CO2 + 11 H2O 2) Balance O: On the left side: 11 + 2 = 13 oxygen atoms On the right side: 24 + 11 = 35 oxygen atoms and the only coefficient you can still freely change is coefficient on O2. taking x as unknown coefficient, we have 11 + 2x = 35 C12H22O11 + 12 O2 12 CO2 + 11 H2O Types of Chemical Reactions 17 Decomposition reactions Single replacement reactions REACTIONS Synthesis reactions Double replacement reactions 18 Synthesis Reactions In Synthesis (aka Combination) reactions, two (or more) reactants combine together to make one product – simpler substances combining together 2 CO + O2 2 CO2 2 Mg + O2 2 MgO P4 + O2 ---> P4O10 19 Decomposition Reactions • In Decomposition, Decomposition there’s a single reactant producing 2 or more products » a large molecule is broken apart into smaller molecules or its elements 2 FeCl3 2 FeCl2 Cl2 2 HgO Hg+ O2 Single Replacement Reactions • In single replacement, one of the reactants is an element, and it replaces another element from its compound A + XY AY + X Sub-types of single replacement: • Metal displaces hydrogen from acid; • Metal displaces hydrogen from water; • Metal displaces another metal; • Nonmetal displaces another nonmetal 20 21 21 Single Single Replacement Replacement Reactions Reactions Metal displaces hydrogen from acid Mg(s) + 2 HCl(aq) H2(g) + MgCl2(aq) 22 22 Single Single Replacement Replacement Reactions Reactions Metal displaces hydrogen from water 2 K + 2 H2O --> 2 KOH + H2 Single Single Replacement Replacement Reactions Reactions 23 23 Zn(s) CuCl2 (aq) Cu(s) ZnCl 2 (aq) More reactive metal (Zn) displaces less reactive metal (Cu) Sample Problem 3 PROBLEM: 24 Identifying the Type of Reaction Classify each of the following reactions as a synthesis, decomposition, single replacement, or double replacement reaction, and write a balanced molecular equation for each: (a) magnesium(s) + nitrogen(g) (b) hydrogen peroxide(l) magnesium nitride (aq) water(l) + oxygen gas (c) aluminum(s) + lead(II) nitrate(aq) aluminum nitrate(aq) + lead(s) SOLUTION: (a) 3 Mg(s) + N2(g) Mg3N2 (aq) Synthesis (b) H2O2(l) 2 H2O2(l) H2O(l) + 1/2 O2(g) 2 H2O(l) + O2(g) Decomposition (c) Al(s) + Pb(NO3)2(aq) 2Al(s) + 3Pb(NO3)2(aq) Al(NO3)3(aq) + Pb(s) 2Al(NO3)3(aq) + 3Pb(s) Single Replacement or 25 Types of Chemical Reactions • In double displacement, two ionic compounds exchange ions Cation always links with an anion (never with another cation)! AX + BY AY + BX Pb(NO3)2(aq) + 2 KI(aq) ----> PbI2(s) + 2 KNO3 (aq) Types of reactions • A precipitation reaction involves formation of an insoluble compound — which is called precipitate. Pb(NO3)2(aq) + 2 KI(aq) 2 KNO3(aq) + PbI2(s) Solid (not soluble) 26 Driving Forces of Chemical Reactions 27 Precipitation Reactions (formation of solid) Gas-Forming Reactions REACTIONS Acid-Base Neutralization Reactions (formation of water) Formation of a Weak Acid or Base 28 Predicting Products Predict products and balance the equation for: NaI + Pb(NO3)2 Double displacement, but not quite Na(NO33)22 and PbI charges carry over from left side, but subscripts don’t NaI + Pb(NO3)2 PbI2 + NaNO3 2NaI + Pb(NO3)2 PbI2 + 2NaNO3 29 Predicting Products Predict products and balance the equation for: NaCl + F2 NaF + Cl 2 single displacement, but what displaces what? Nonmetal (F) displaces another nonmetal (Cl) 2 NaCl + F2 2 NaF + Cl2 30 Gas-Forming Reactions • Some unstable substances in solution: H2CO3 H2O(l) + CO2(g) H2SO3 H2O(l) + SO2(g) NH4OH H2O(l) + NH3(g) • Example: NH4Cl + NaOH ?NH4OH + NaCl NH3 + H2O + NaCl Ammonia in exchange reactions • In water solutions, ammonia is in equilibrium with ammonium hydroxide: NH3(aq) + H2O(l) <---> NH4OH (aq) • When ammonia acts as a reactant in double replacement reactions, think of it as NH4OH : 2H2O(l) + 2 NH3(aq) + MgCl2(aq) 2 NH4Cl(aq) + Mg(OH)2(s) NH4OH 31 3.3. The Mole MOLE is a word that stands for a particular number. 1 dozen = 12 2 dozen = 24 6.022 x 1023 1 mole = ? Mole is formally defined as: 11 mole mole is is the the number number of of carbon carbon atoms atoms in in 12C. exactly exactly 12.0 12.0 g g of of pure pure 12 C. 32 Particles in a Mole 33 Avogadro’s Number 6.022 x 1023 11 mole mole of of any any substance substance contains containsAvogadro’s Avogadro’s number number of of particles. particles. Amedeo Avogadro 1776-1856 AA“particle” “particle”here heremeans: means: ••an anatom atomfor forelements elements(Mg) (Mg) •a moleculefor formolecular molecularcompounds compounds(H (H22O) O) •amolecule ••aaformula formulaunit unitfor forionic ioniccompounds compounds(NaCl) (NaCl) 34 Each of these samples contains 1 mole 35 MOLAR MOLAR MASS MASS • MOLAR MASS of an element is the mass, in grams, of 1 mol of that element. we know it from periodic table. ATOMIC ATOMIC MASS MASS AND AND MOLAR MOLAR MASS MASS For example, atomic mass of Na is 22.99 amu. molar mass of Na is 22.99 g g/mol Preferred units of molar mass: g/mol (works as a conversion factor) 22.99 g 1 mol 1 mol 22.99 g 36 37 PROBLEM: PROBLEM: How How many many moles moles of of Mg Mg are are in in 0.200 0.200 g? g? How How many many atoms atoms of of Mg Mg are are there? there? 1) Mg has a molar mass of 24.31 g/mol. #moles = 0.200 g • 1 mol = 8.23 x 10 ­3 mol 24.31 g 2) How many atoms are in this piece of Mg? 23 6.022 x 10 atoms ­3 8.23 x 10 mol • 1 mol = 4.95 x 1021 atoms Mg Sample Problem 4 PROBLEM: Calculating the Mass and the Number of Atoms in a Given Number of Moles of an Element 38 (a) How many grams of silver (Ag) are in 0.0342 mol of Ag? (b) How many atoms of iron (Fe) are in 95.8g of Fe? PLAN: (a) To convert moles of Ag to grams of Ag we use molar mass M of Ag, in g/mol. SOLUTION: 0.0342mol Ag x 107.9 g Ag = 3.69g Ag mol Ag PLAN: (b) To convert grams of Fe to atoms we first have to find # of moles of Fe, and then convert moles to atoms. SOLUTION: 95.8g Fe x mol Fe = 1.72mol Fe 55.85g Fe 6.022x1023atoms Fe = 1.04x1024 atoms 1.72mol Fe x Fe mol Fe moles of Ag multiply by M of Ag (107.9g/mol) mass(g) of Ag mass(g) of Fe divide by M of Fe (55.85g/mol) moles of Fe multiply by 6.022x1023 atoms/mol atoms of Fe 3.4. 3.4. MOLAR MOLAR MASS MASS OF OF COMPOUNDS COMPOUNDS Molar mass of a compound is the mass (in grams) of 1 mole of that compound. • We find it from periodic table – add atomic masses of all the atoms in a chemical formula, change units to g/mol 39 Sample Problem 5 PROBLEM: Calculating the Molar Mass of a Compound Using the data in the periodic table, calculate the molar mass of the following compounds: (a) tetraphosphorus trisulfide PLAN: SOLUTION: 40 (b) ammonium nitrate Write the formula and then multiply the number of atoms (shown by subscript) by the respective atomic masses. Add the masses for the compound. (a) P4S3 formula = (4xatomic mass of P) mass + (3xatomic mass of S) (b) NH4NO3 formula = (2xatomic mass of N) mass + (4xatomic mass of H) = (4x30.97amu) + (3x32.07amu) = 220.09 amu molar mass = 220.09 g/mol + (3xatomic mass of O) = (2x14.01amu)+ (4x1.008amu) + (3x16.00amu) = 80.05 amu molar mass = 80.05 g/mol Sample Problem 6 PROBLEM: Calculating the Molar Mass of a Compound Calculate the molar masses of: a) Zn3(PO4)2; b) CuSO4•5H2O SOLUTION: a) Molar mass of Zn3(PO4)2 = 3M(Zn) + 2M(P) + 8M(O) = 3x65.3 + 2x31 + 8x16 = 386 g/mol; b) Molar mass of CuSO4•5H2O = M(Cu) + M(S) + 4M(O) + 5M(H2O)= 63.6 + 32 + 4x16 + 5x18 = 249.6 g/mol 41 42 What is the molar mass of ethanol, C2H6O? 1 mol of ethanol contains 2 mol C (12.01 g C/1 mol) = 24.02 g C 6 mol H (1.01 g H/1 mol) = 6.06 g H 1 mol O (16.00 g O/1 mol) = 16.00 g O TOTAL = molar mass = 46.08 g/mol 43 How many molecules of ethanol are there in a can of beer if it contains 21.3 g of C2H6O? (a) Molar mass of C2H6O = 46 g/mol (b) Calculate # of moles of ethanol 1 mol 21.3 g • = 0.463 mol 46 g (c) Calculate # of molecules of ethanol 23 molecules 6.022 x 10 0.463 mol • 1 mol = 2.79 x 1023 molecules 44 How many atoms of C are there in a can of beer if it contains 21.3 g of C2H6O? There are 2.79 x 1023 molecules. Each molecule contains 2 C atoms. Therefore, the number of C atoms is 2.79 x 1023 molecules • 2 C atoms 1 molecule = 5.58 x 1023 C atoms Summary of the mass-molenumber relationships 45 MASS(g) MASS(g) of ofsubstance substance Conversion factor is Molar mass (g/mol) ##of ofmoles moles of ofsubstance substance Conversion factor is Avogadro’s number (particles/mol) ##of ofparticles particles of ofsubstance substance Sample Problem 7 46 Calculating the Molar Mass If 0.30 moles of unknown substance weigh 19.2 grams, what is its molar mass? PROBLEM: SOLUTION: What is the mass of 1 mole? 0.30 moles --- in 19.2 grams 1 mole --- in x grams 0.30 19.2 1 x 19.2 x 64 g mol 0.30 3.5. 3.5. Percent Percent Composition Composition MASS PERCENT COMPOSITION – proportion by mass of an element within a compound Ethanol, C2H6O 52.13% C 13.15% H 34.72% O 47 Percent Percent Composition Composition Mass percent of an element: mass of element in compound mass % 100% mass of compound Consider NO2 What is the mass percent of N and of O in NO2? 14.0 g N mass % N 100% 30.4% 46.0 g NO2 2 16.0 g O mass % O 100% 69.6% 46.0 g NO2 48 Mass Mass Percent Percent What is the mass percent of water in CuSO4•5H2O? mass of water mass % water 100% mass of compound Imagine a sample of 1 mol CuSO4•5H2O What’s contribution of water to molar mass of CuSO 4•5H2O? M(Cu) + M(S) + 4M(O) + 5M(H2O)= 63.6 + 32.0 + 4x16.0 + 5x18.0 = 249.6 g/mol 5 18.0 g H 2O mass % water 100% 36.1% 249.6 g total 49 3.6. Determining a Formula The EMPIRICAL ( “SIMPLEST”) formula can be derived from mass percent data. What do empirical formulas look like? CH3 C2H4 C3H7 - shows 1:3 ratio between atoms of C and H - not a valid empirical formula; 2:4 ratio can be reduced to simpler form (1:2); correct empirical formula is CH2 - valid empirical formula; 3:7 ratio can not be simplified 50 Types of Chemical Formulas An empirical formula shows the simplest ratio of the atoms of each element in the compound. The empirical formula for hydrogen peroxide is HO. A molecular formula shows the actual number of atoms of each element in a molecule of the compound; it may be the same as empirical formula or a multiple of it. The molecular formula for hydrogen peroxide is H2O2. A structural formula shows the relative placement and connections (bonds) between atoms in the ( molecule. The structural formula for hydrogen peroxide is H-O-O-H. 51 Sample Problem 8 52 Determining the Empirical Formula PROBLEM: Quantitative analysis of sodium perchlorate showed 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. What is the empirical formula? PLAN: Once we find the relative number of moles of each element, we can divide by the lowest mol amount to find the relative mol ratios (empirical formula). mass(g) of each element SOLUTION: 2.82 g Na x mol Na = 0.123 mol Na 22.99 g Na divide by M(g/mol) amount(mol) of each element convert to integer subscripts empirical formula mol Cl 4.35 g Cl x = 0.123 mol Cl 35.45 g Cl 7.83 g O x mol O = 0.489 mol O 16.00 g O 0.123/0.123 = 1 0.123/0.123 = 1 0.489/0.123 = 3.98 4 sodium perchlorate is NaClO4 NaClO4 What is the molecular formula of a compound which53 53 has an empirical formula of CH2 and a molar mass of 126.2 g? Let n = the number of formula units of CH 2. Calculate the mass of each CH2 unit 1 C = 1(12.01 g) = 12.01g 2 H = 2(1.01 g) = 2.02g 14.03g 126.2 g n 9 (empirical formula units) 14.03 g The molecular formula is = C9H18 54 A A compound compound of of B B and and H H is is 81.10% 81.10% B. B. What What is is its its empirical empirical formula? formula? • Because it contains only B and H, it must contain 18.90% H. • In 100.0 g of the compound there are 81.10 g of B and 18.90 g of H. Calculate the number of moles of each element. 1 mol 1 mol = 7.502 mol B 81.10 g B • 81.10 g B • = 7.502 mol B 10.81 g 10.81 g 1 mol 1 mol = 18.75 mol H 18.90 g H • 18.90 g H • = 18.75 mol H 1.008 g 1.008 g 55 A A compound compound of of B B and and H H is is 81.10% 81.10% B. B. What What is is its its empirical empirical formula? formula? Take the ratio of moles of B and H. Always divide by the smaller number. 18.75 mol H 18.75 mol H = 2.499 mol H 2.499 mol H = 2.5 mol H 2.5 mol H = = 7.502 mol B 1.000 mol B 1.0 mol B 7.502 mol B 1.000 mol B 1.0 mol B But we can’t have a fractional # of atoms in a molecule, and 2.5 isn’t very close to either 2 or 3. We need a whole number ratio. 2.5 mol H / 1.0 mol B = 5 mol H to 2 mol B EMPIRICAL FORMULA = B2H5 56 56 A A compound compound of of B B and and H H is is 81.10% 81.10% B. B. Its Its empirical empirical formula formula is is B B22H H55.. The The molar molar mass mass is is 53.3 53.3 g/mol. g/mol. What What is is its its molecular molecular formula? formula? Use the molar mass ( 53.3 g/mol ) Compare with the mass of B2H5 = 26.66 g/unit Find the ratio of these masses. 53.3 g/mol 2 units of B H 53.3 g/mol == 2 units of B22H55 26.66 g/unit of B 1 mol 26.66 g/unit of B2HH5 1 mol 2 5 Molecular formula = Empirical Formula Determination 1. If data is in mass%, base calculation on 100 grams of compound. 2. Determine moles of each element in 100 grams of compound. 3. Divide each value of moles by the smallest of the values. 4. Multiply each ratio by an integer to obtain all whole numbers. 57 Sample Problem 9 58 Determining a Molecular Formula PROBLEM: Quantitative analysis shows that lactic acid contains 40.0 mass% C, 6.71 mass% H, and 53.3 mass% O. The molar mass of lactic acid is 90.08 g/mol. (a) Determine the empirical formula of lactic acid. (b) Determine the molecular formula. PLAN: assume 100g lactic acid and find the mass of each element divide each mass by mol mass(M) amount(mol) of each element convert to integer subscripts empirical formula molecular formula divide molar mass by mass of empirical formula to get a multiplier Sample Problem 9 59 Determining a Molecular Formula continued SOLUTION: 40.0 g C x Assuming there are 100.0 g of lactic acid, the constituents are mol C 12.01g C mol H 6.71 g H x 1.008 g H = 3.33 mol C = 6.66 mol H 3.33 6.66 3.33 3.33 3.33 3.33 molar mass of lactic acid mass of CH2O 53.3 g O x 30.03 g 16.00 g O = 3.33 mol O 1:2:1 90.08 g mol O empirical formula CH2O 3 C3H6O3 is the molecular formula 60 Problem. 1.023 g of FeSO4 • nH2O (where n is unknown) were heated in a crucible, after which 0.558 g of anhydrous FeSO4 remained. Find the number n of crystallized waters. Solution. 1) Find the mass of water in the hydrated FeSO4 • nH2O: mass of hydrate 1.023 g mass of anhydrous FeSO4 0.558 g --------------------------------------------------mass of water 0.465 g Problem. 1.023 g of FeSO4 • nH2O (where n is unknown) were heated in a crucible, after which 0.558 g of anhydrous FeSO4 remained. Find the number n of crystallized waters. 2) Convert masses of FeSO4 and H2O into moles: 0.558 g FeSO4 / 151.8 g/mol = 0.00368 mol FeSO 4 0.465 g water / 18.0 g/mol = 0.0258 mol water 3) Find the ratio of moles: moles water / moles FeSO4 = 7:1 4) There are 7 crystallized water molecules per one FeSO4 so the formula of hydrate is FeSO4 • 7H2O 61 62 3.9. 3.9. STOICHIOMETRY STOICHIOMETRY Stoichiome...
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