Presentation on theme: "Unit 6 Chapter 12 Chemical Quantities or"— Presentation transcript:
1Unit 6 Chapter 12 Chemical Quantities or "Our Friend the Mole"
2Stoichiometrystoichiometry—using balanced chemical equations to obtain info
3Particles of matter are too small and numerous to count 12.1 Counting Particles of MatterParticles of matter are too small and numerous to countSI unit of chemical quantity =the mole (abbreviated mol)The mole is a counting unit
4How do you measure how much? You can measure mass = gramsor volume = litersor you can count pieces = MOLES
5Moles 1 mole = 6.02 x 1023 representative particles Treat it like a very large dozen6.02 x is called Avogadro's number
6Representative particles = The smallest pieces of a substancerepresentative particles =ATOMSIONSMOLECULESFORMULA UNITS
7Representative particles For a molecular (covalent)compounds it is a moleculecompound with all nonmetalsCO, BF3 , Cl21 mole = 6.02 x 1023 molecules
8Representative particles For an element it is an atomUnless it is diatomicone symbol, no charge: Br,Cs1 mole = 6.02 x 1023 atoms
9Representative particles For an ionic compound it isa formula unitcompound with metal andnonmetal - KI, Na2SO41 mole = 6.02 x 1023 formula units
10Representative particles For ions it is 1 mol ionsone symbol with charge (monatomic) ormore than one symbol with charge (polyatomic: Na+ , N3- , (C2H3O2) –1 mole ions = 6.02 x 1023 ions
11Molar Mass = The mass of 1 mole of an element in grams. We can make conversion factors from these.Example - We can write this as g C = 1 molTo change grams of a compound to moles of a compound.Or moles to grams
12Molar Masses the atomic masses on the periodic table have a unit of amu (atomic mass unit)GAM = gram atomic mass= the atomic mass (listed on the periodic table) written in grams1 atom Xe = uGAM of Xe = g
13molar mass—the mass, in g, of 1 mole of a substance Add up the gram atomic mass of all elements in compound to calculate molar mass of a substance
14Find the molar mass of methane, CH4. CH4 = 1(12.0) + 4(1.0) = = 16.0 gFind the molar mass of calcium hydroxide, Ca(OH)2.Ca(OH)2 = 1(40.1) + 2(16.0) + 2(1.0) = 74.1 g
152. What is the molar mass of a molecule of H2? 1. What is the molar mass of 1 mole of phosphorus?2. What is the molar mass of a molecule of H2?3. What is the molar mass of a formula unit of MgCl2?30.97 g2.016 g94 g
16MOLAR VOLUME of any gas at STP: 22.4 L = 1 mol Molar Volume: volume-to-mole andmole-to-volume conversionsAt STP, all gases occupy the same amount of space:MOLAR VOLUME of any gas at STP:22.4 L = 1 mol
17dimensional analysis steps for success: = using the units (dimensions) to solve problemssteps for success:1) identify unknown (read carefully)2) identify known (read carefully)“Play checkers” with the units, moving them diagonally, canceling when appropriate. All units should cancel except those of the desired answer.3) plan solution4) calculate5) check (sig.figs., units, and math)
18CONVERSION FACTOR SUMMARY: 6.02 x 10^23 representative particles1 MOLE&
19CONVERSION FACTOR SUMMARY: MOLAR MASS (g) & MOLE1 MOLE MOLAR MASS (g)for a gas at STP:22.4 L & 1 MOLE1 MOLE L
20Calculation questionHow many molecules of CO2 are the in 4.56 moles of CO2 ?4.56 moles CO2 X x 1023 molecules CO2mole CO2= x 1023 molecules CO2
21For example How many moles is 5.69 g of NaOH? 5.69 g NaOH x 1 mole= mol NaOHgNaOHneed to change grams to moles for NaOH1mole Na = 22.99g 1 mol O = g 1 mole of H = 1.01 g1 mole NaOH = g
22Examples = 28.10 g C How much would 2.34 moles of carbon weigh? 2.34 mol C X g Cmol C= g C
23Calculation questionHow many moles of salt is 5.87 x formula units?5.87 X 1022 formula units NaCl X mole NaCl_x 1023 f.u. NaCl= moles NaCl
24Examples 4.61 g Mg X _1 mol Mg 1 24.3 g Mg = 0.1897 mol Mg How many moles of magnesium in 4.61 g of Mg?4.61 g Mg X _1 mol Mgg Mg= mol Mg
25exampleWhat is the volume, in L, of mol of NO2 gas at STP?0.495 mol NO2 x L NO2 = L NO2mol NO2
26How many moles are found in 84 L of neon gas at STP? 84 L Ne x mol Ne = mol NeL Ne
27There are many types of mole problems: 1 step: r.p mol & mol r.p.mass mol & mol mass2 step: mass r.p. & r.p massmass volume & volume massr.p volume & volume r.p.
28Moles of Compounds1 mole of a compound contains as many moles of each element as are indicated by the subscripts in the formula for the compound.Example:1 mole of ammonia (NH3) has1 mole of nitrogen atoms and3 moles of hydrogen atoms.
29Percent compositioncomposition of a compound is the percent by mass of each element in the compound.Find the mass of each elementdivide by the total mass of compound.
30Determine the percent composition of calcium chloride (CaCl2). Step 1: determine the molar masses of each element in compound and the compoundCa = gCl2 = 2 x g = gCaCl = g
31****make sure they add up to 100 Step 2: divide the molar mass of each element by the molar mass of the compound and times by 100****make sure they add up to 100
32Empirical FormulaThe lowest whole number ratio of elements in a compound.CH2H2O
33You can determine the empirical formula from percent composition and mole ratios *** percent means “parts per hundred”assume you have 100 g of the compoundStep 1: change the percent sign to g (grams)Calculate the empirical formula of a compound composed of % C, % H, and %N
34Step 2:calculate the number of moles for each element by converting grams to moles using 1 mole = molar mass38.67 g C x 1mol C = mole C g C16.22 g H x 1mol H = 16.1 mole H g H45.11 g N x 1mol N = mole N g N
35So empirical formula is CH5N Step 3: divide each number of moles by the smallest number of moles3.220 mole C = 1 mol C3.22016.1 mole H = 5 mol H3.220 mole N = 1 mol NSo empirical formula is CH5N
36Determine the empirical formula of a compound with percent composition of 38.4 % Mn, 16.8 % C, 44.7 % O38.4 g Mn X 1mol Mn = mole Mn g Mn16.8 g C X 1mol C = mole C g C45.7 g O X 1mol O = mole O g O
37So formula is MnC2O4 0.699 mole Mn = 1 mol Mn 0.699 1.339 mole C = 2 mol C2.798 mole O = 4 mol OSo formula is MnC2O4
38For many compounds, the empirical formula is not the true formula A molecular formula tells the exact number of atoms of each element in a molecule
39Example: acetic acidmolecular formula = C2H4O2empirical formula = CH2OThe molecular formula for a compound is always a whole-number multiple of the empirical formula.
40To determine molecular formula you need to know the molar mass of the compound Divide the actual molar mass by the molar mass of the empirical formula.Multiply the empirical formula by this number.molar mass of compoundmolar mass of empirical formula
41Example: A compound has an empirical formula of ClCH2 and a molar mass of g/mol. What is its molecular formula?molar mass of ClCH2 = g/mol= 2.01X (ClCH2)Empirical formula = ClCH2Molecular formula = Cl2C2H4
42ExampleA compound has an empirical formula of CH2O and a molar mass of g/mol. What is its molecular formula?
43ExampleIbuprofen is % C, 8.80 % H, % O, and has a molar mass of about 207 g/mol. What is its molecular formula?
44Unit 6 Chemical Quantities Continued Chapter 12.2“Using the Mole"
45remember our friend the mole? Recall that the mole should be treated like a dozen1 dozen = 12 pieces &1 mole = pieces (or 6.02 X 1023 r.p)
46remember our friend the mole? You still need to know how and when to use: Avogadro’s number (6.02 x 1023), representative particles Molar mass, grams in one mole, Molar volume of a gas at STP (22.4 L)
47Conversion factor review 1 mole = 6.02 x 1023 r.p 1 mole = molar mass (g) 1 mole = 22.4 L
48So what about this silly mole? Our useful friend the mole allows us to do calculations called stoichiometry—using balanced chemical equations to obtain info ex: 2C + O2 2CO
49Mole - Mole (MOL – MOL) Conversions new conversion factor –# of mol A = # of mol B# = coefficients in balanced equation
50I.Mole - Mole (MOL – MOL) Conversions the most important, most basic stoich calculationB. uses the coefficients of a balanced equation to compare the amounts of reactants and productsC. coefficients are mole ratios
51D. the way to go from substance A to substance B E. mol – mol is the only time the mole number in the conversion is not automatically 1. (Avogadro’s #, molar mass, and 22.4 L are all = to 1 mol)MOL – MOL : # mol A # mol B# mol B # mol A# = coefficients
52***always start with a balanced chemical equation MOL – MOL Conversions# mol A # mol B# mol B # mol Aex: How many moles of carbon monoxide are produced when mol of oxygen reacts with carbon?2C + O CO***always start with a balanced chemical equation
53How many moles of carbon monoxide are produced when 0 How many moles of carbon monoxide are produced when mol of oxygen reacts with carbon? C + O COStep 1; determine given and unknown given = mol O2 unknown = ? mol CO
54How many moles of carbon monoxide are produced when 0 How many moles of carbon monoxide are produced when mol of oxygen reacts with carbon? C + O COStep 2; setup conversion using mole ratios (coefficients) in balanced equation0.750 mol O2 X 2 mol CO = 1.5 mol COmol O2
55Step 1; determine given and unknown given = 0.661 mol Al2O3 Find the number of moles of the reactants, given mol of product is formed Al + 3O Al2O3Step 1; determine given and unknowngiven = mol Al2O3unknown = ? mol Al= ? mol O2
564Al + 3O Al2O3Step 2; setup conversion using mole ratios (coefficients) in balanced equation0.661 mol Al2O X 4 mol Al = mol Almol Al2O30.661 mol Al2O3 X 3 mol O2 = mol O2mol Al2O3
57II. MASS – MASS Conversions – Using molar mass in stoich problems to predict masses of reactants and/or productsa balanced chemical equation can be used to compare masses of reactants and productsB. mass – mass cannot change which substance you are dealing with; only mol – mol can do that
58PT = periodic table, molar mass # = coefficients, chemical equation MASS – MASS ConversionsGIVEN g A X 1 mol A X # mol B X PT g BPT g A # mol A mol BPT = periodic table, molar mass# = coefficients, chemical equation
59How many grams of hydrochloric acid are made from the reaction of 0 How many grams of hydrochloric acid are made from the reaction of g of hydrogen gas with excess chlorine gas? H2 + Cl2 2HCl Known = 0.500g H2 Unknown = ? g HCl
60H2 + Cl2 2HCl GIVEN g A X 1 mol A X # mol B X PT g B 0.500 g H2 X 1 mol H2 X 2 mol HCl X 36.5 gHCl g H2 1 mol H2 1 mol HCl = 18 g HClH2 + Cl HClGIVEN g A X 1 mol A X # mol B X PT g BPT g A # mol A mol B
61Calculate the numbers of grams of oxygen formed when 25 Calculate the numbers of grams of oxygen formed when 25.0 g of sodium nitrate decomposes into sodium nitrite and oxygen.2NaNO NaNO2 + O225.0 g NaNO3 X 1 mol NaNO3 X 1 mol O2 X 32.0 g O2g NaNO3 2 mol NaNO3 1 mol O2= 20.3 g O2
62“mole – mass” (mass – mole) calculations “MASS – MOLE”: GIVEN g A X 1 mol A X # mol BPT g A # mol A“MOLE – MASS”:GIVEN mol A X # mol B X PT g B# mol A mol BPT = periodic table, molar mass# = coefficients
63How many g of water are produced from the complete combustion of 0 How many g of water are produced from the complete combustion of mol of C2H2?2C2H2 + 5O CO2 + 2H2Omol C2H2 X 2 mol H2O X g H2Omol C2H2 1 mol H2O= 12.3 g H2O
64Using the equation 2C2H2 + 5O2 4CO2 + 2H2O how many moles of O2 would be needed to produce g of CO2? g CO2 X 1 mol CO2 X 5 mol O g CO2 4 mol CO2 = 1.59 mol O2
65“mass – volume” (volume – mass) calculations – “MASS – VOLUME”: STP) GIVEN g A X 1 mol A X # mol B X 22.4 L B 1 PT g A # mol A 1 mol B “VOLUME – MASS”: STP) GIVEN L A X 1 mol A X # mol B X PT g B L A # mol A 1 mol B PT = periodic table, molar mass # = coefficients
66How many L of hydrogen are produced from the decomposition of 3 How many L of hydrogen are produced from the decomposition of 3.50 g of water at STP?2H2O H2 + O23.50 g H2O X 1 mol H2O X 2 mol H2 X 22.4 L H2g H2O 2 mol H2O 1 mol H2= 4.36 L H2
67Using the equation 2C2H2 + 5O2 4CO2 + 2H2O how many liters of water vapor are produced when 5.02 g of C2H2 undergoes complete combustion?5.02 g C2H2 x 1 mol C2H2 x 2 mol H2O x 22.4 L H2Og C2H mol C2H mol H2O= 4.32 L H2O
68“volume – volume” calculations GIVEN L A x 1 mol A x # mol B x 22.4 L BL A # mol A mol B# = coefficients (SHORT CUT: compare coefficients!)
69How many L of sulfur trioxide are produced from the reaction of 36 How many L of sulfur trioxide are produced from the reaction of 36.1 L of oxygen with sulfur dioxide at STP?2SO2 + O SO336.1 L O2 x 1 mol O2 x 2 mol SO3 x 22.4 L SO3L O2 1 mol O mol SO3= 72.2 L SO3SHORTCUT: coefficient of O2 = 1 coefficient of SO3 = 2 so 36.1 L x 2 = 72.2 L SO3
70SHORTCUT: coefficients = 1 mol HCl to 1 mol CO2. How many liters of CO2 are produced from L of HCl reacting with NaHCO3?NaHCO3 + HCl NaCl + CO2 + H2O0.252 L HCl x 1 mol HCl x 1 mol CO2 x 22.4 L CO2L HCl mol HCl mol CO2= L CO2SHORTCUT: coefficients = 1 mol HCl to 1 mol CO2.so L HCl = L CO2
71mass –particle (particle – mass) calculations “MASS – PARTICLE”: GIVEN g A x 1 mol A x # mol B x 6.02 x 1023 r.p. BPT g A # mol A mol B“PARTICLE – MASS”:GIVEN r.p. A x 1 mol A x # mol B x PT g Bx 1023 r.p. A # mol A mol BPT = periodic table, molar mass# = coefficients
72How many molecules of NH3 are produced from reacting 2 How many molecules of NH3 are produced from reacting 2.07 g of H2 with excess N2?N2 + 3H NH32.07 g H2 x 1 mol H2 x 2 mol NH3 x 6.02 x 1023 NH3g H mol H mol NH3= 4.2 x 1023 molecules NH3
73Limiting ReactantsRarely are the reactants in a chemical reaction present in the exact mole ratios specified in the balanced equation.Usually, one or more of the reactants are present in excess, and the reaction proceeds until all of one reactant is used up.
74The reactant that is used up is called the limiting reactant. The left-over reactants are called excess reactants.
75Limiting Reactant To figure out the limiting reactant: You must do the stoichiometric calculation with all reactants.The one that produces the least amount of product is the limiting reactant.
76Ex: 4Al + 3O2 2Al2O3What is the limiting reactant when 35 g of aluminum reacts with 35 g of oxygen? How much aluminum oxide is formed in this reaction?35 g Al X 1 mol Al X 2 mol Al2O3 =g Al mol Al35 g O2 X 1 mol O2 X 2 mol Al2O3 = 1.1 mol Al2O3g O mol O20.65 mol Al2O3Aluminum is limiting reactant0.65 mol Al2O3 X g Al2O = g Al2O3mol Al2O3
77Percent Yield theoretical yield —amount of product predicted by the math (theory) actual yield —amount of product obtained in lab
78percent yield —percentage of product recovered; comparison of actual and theoretical yields% YIELD = ACTUAL YIELD X 100THEORETICAL YIELD
7935. 0 g of product should be recovered from an experiment 35.0 g of product should be recovered from an experiment. A student collects 22.9 g at the end of the lab. What is the percent yield?22.9 g X 100 = 65.4%35.0 g
80= THEORETICAL YIELD = 3.19 g NaCl % YIELD = 2.89 g NaCl X 100 = 90.6% What is the percent yield if 2.89 g of NaCl is produced when 1.99 g of HCl reacts with excess NaOH? Water is the other product.HCl + NaOH NaCl + H2OActual yield = 2.89 g NaClTheoretical yield = ?1.99 g HCl x 1 mol HCl x 1 mol NaCl x 58.5 g NaClg HCl 1 mol HCl mol NaCl= THEORETICAL YIELD = g NaCl% YIELD = 2.89 g NaCl X = 90.6%3.19 g NaCl