2 6.1 The Mole and Avogadro’s Number 6.2 Gram–Mole Conversions6.3 Mole Relationships and Chemical Equations6.4 Mass Relationships and Chemical Equations6.5 Limiting Reagent and Percent Yield
3 Goals 1. What is the mole, and why is it useful in chemistry? Be able to explain the meaning and uses of the mole and Avogadro’s number.2. How are molar quantities and mass quantities related?Be able to convert between molar and mass quantities of an element or compound.3. What are the limiting reagent, theoretical yield, and percent yield of a reaction?Be able to take the amount of product actually formed in a reaction, calculate the amount that could form theoretically, and express the results as a percent yield.
6 Counting molecules: Conceptually The reactants in a chemical reaction must be balancedOne molecule of hydrogen (H2) reacts with one molecule of iodine (I2), creating two hydrogen iodide molecules (HI)
7 Of course, we can’t visually count molecules to achieve this one-to-one ratio In fact, the only tool available is an analytical mass balance (using grams!)
8 Molecular WeightBut how do we relate a sample’s mass in grams to the number of molecules it contains?We begin with the concept of ‘molecular weight’ (molecular mass)Let’s start by utilizing the Ch 2 concept of mass by ‘amu’Note: We will be focusing on molecules for a while
9 Recall, atomic weight is the average mass of an element’s isotope atoms (review next slide) By analogy, molecular weight (MW) is the mass of a compound’s moleculesA molecule’s molecular weight is the sum of the atomic weights for all the atoms in the moleculeA salt’s formula weight is the sum of the atomic weights for all the atoms in the formula unit
10 fyi: Atomic Weight Calculation C-12 is 98.89% of all natural carbonC-13 is 1.11% of all natural carbonThe mass of C-12 is: 12 amu (by definition)The mass of C-13 is: amuAtomic weight = [(% isotope abundance) × (isotope mass)][98.89% x 12 amu] + [1.11% x amu] =(0.9889)(12 amu) + (0.0111)( amu) =0.144312.01 amu
11 fyi: Disambiguation For molecules, these terms are identical molar weightmolar massmolecular weightmolecular massFor atoms, these terms are identicalatomic weightatomic mass
13 Molecular weightSo if the mass (atomic weight) of one atom of hydrogen (H) is amuthen the mass (molecular weight) of one hydrogen molecule (H2) is amuIf the atomic weight of one iodine atom (I) is amuthen the molecular weight of one iodine molecule (I2) is amuFinally, the molecular weight of each HI molecule must be:1.008 amu amu = amu
15 Here is where we left our reaction making two HI molecules from H2 and I2
16 We learned in chapter 2 that the iodine atom (I) is approximately 126 times heavier than the hydrogen atom (H)H = amuI = amuSo mathematically, it is necessary that the iodine molecule (I2) is approximately 126 times heavier than the hydrogen molecule (H2)H2 = amuI2 = amu
20 Mass / Number relationship for I2 & H2 So in order control the number ratio of I2 to H2 molecules at 1:1 we can measure out any of the followingso long we keep the mass ratio at 1:126H2(g) + I2(g) → 2HI(g)2.016 amu amu2.016 g g1.00 g gg g
21 But how many molecules is that? We know that:2.016 amu of H2 contains one molecule, andamu of I2 contains one moleculeOur final question is, how many molecules are contained in:2.016 g of H2g of I2
23 The mole (NA) has now been measured A mole is the amount of substance whose mass in grams is numerically equal to its molecular weight2.016 g of H2 contains one mole of H2 moleculesg of I2 contains one mole of I2 moleculesThe mole (NA) has now been measuredOne mole of any molecule contains × 1023 moleculesOne mole of any salt contains × 1023 formula units
24 Getting from amu to grams The number ‘ ,’ typically placed as shown, functions BOTH:as the mass in amu per atomand the mass in grams per mole of atomsSo I has a mass of grams per mole of I atomsI = g/mol
28 fyi: SynonymsThe terms Molecular Weight (section 6.1) and Molar Mass (this section – 6.2) are identicalAlso, the term Molecular Mass is allowedAlso, the term Molar Weight is allowedWhen we defined Molecular Weight in Section 6.1, we had not yet define the Moleso you couldn’t have understood the more common term “Molar” Mass
29 The molar mass of water is 18.02 g/mol So how many moles of water are there in 27 grams?Most importantly, molar mass serves as a conversion factor between numbers of moles and mass in grams for use in dimensional analysis
30 Molar Mass: Mole to Gram Conversion Worked Example 6.3Molar Mass: Mole to Gram ConversionIbuprofen is a pain reliever used in Advil. Its molecular weight is g/mol.If a bottle of Advil contain mole of ibuprofen, how many grams of ibuprofen does it contain?
31 WORKED EXAMPLE 6.3 Molar Mass: Mole to Gram Conversion (Continued)
32 We now have the tools to convert: grams → moles → number of atomsgrams → moles → number of moleculesgrams → moles → number of formula unitsAnd we can go backwards
34 What is the mass in grams of 3.2 × 1022 molecules of water?
35 6.3 Mole Relationships and Chemical Equations Molar coefficients:using Mole ratios asconversion factors
36 Coefficients in a balanced chemical equation tell us the necessary ratio of moles of reactants and how many moles of each product are formed
37 Especially useful, coefficients can be put in the form of mole ratios which act as conversion factors when setting up dimensional analysis calculations
38 Mole to moleThis reaction represents the reaction of A2 (red) with B2 (blue)Chapter 6, Unnumbered Figure 2, Page 174a) Write a balanced equation for the reactionb) How many moles of product can be made from 1.0 mole of A2? From 1.0 mole of B2?
39 How many moles of NH3 can be produced from 5 How many moles of NH3 can be produced from moles of H2 according to the following equation? N2 + 3 H2 2 NH30.67 mol2.00 mol3.33 mol7.50 mol
40 6.4 Mass Relationships and Chemical Equations Putting it alltogether
41 The actual amounts of substances used in the laboratory must be weighed out in grams Furthermore, customers generally want the quantity of product reported in terms of grams
42 We already have the necessary tools Mass to mole conversionsMole to mole conversionsMole to mass conversions
43 Mole to mole conversions (section 6 Mole to mole conversions (section 6.3) are carried out using mole ratios as conversion factorsIf you have 9.0 moles of H2, how many moles of NH3 can you make?
44 2) Mole-to-mass and mass-to-mole conversions (from section 6 2) Mole-to-mass and mass-to-mole conversions (from section 6.2) are carried out using molar mass as a conversion factor
45 3) Mass to mass conversions cannot be carried out directly If you know the mass of A and need to find the mass of Bfirst convert the mass of A into moles of Athen carry out a mole to mole conversion to find moles of Bthen convert moles of B into the mass of B
46 So after collecting the fundamental data, always follow this three step process for mass to mass 1) convert grams of A to moles via molar mass2) convert moles of A to B via mole ratio3) convert mole of B to grams via molar mass
47 Let’s go back to our hydrogen / iodine reaction giving hydrogen iodide
48 10.0 grams of HI requires how many grams of H2 I2 = g/molH2 = g/molHI = g/mol
49 How many grams of oxygen are needed to react with 25 How many grams of oxygen are needed to react with 25.0 g of K according to the following equation? K(s) + O2(g) 2 K2O(s)10.2 g O22.56 g O28.66 g O25.12 g O2
50 6.5 Limiting Reagent and Percent Yield Vocabulary:Reactant = Reagent
51 limiting reactant (reagent) Reactants are not always perfectly balancedWhen running a chemical reaction, we generally ‘overcharge’ one of the reactantsAs our mechanic has ‘overcharged’ tires
52 limiting reactant (reagent) The limiting reactant is the reactant that runs out firstThe reactant that never runs out is called the excess reactantSo which is ‘limiting here?’ Tires or car bodies.
53 limiting reactant (reagent) In order to make this ‘reaction’ balanced stoichiometrically, how many tires should our mechanic actually have on hand?32 tires would perfectly ‘balance’ this ‘reaction’
58 A limiting reactant mixture on the other hand, contains an excess of one of the reactantsThe other reactant is thus limitingN2(g) + 3H2(g) 2NH3(g)Which reactant is inexcess in thisreaction mixture?Which reactant islimiting?
62 10.0 g of Chemical A are reacted with 10.0 g of B Example10.0 g of Chemical A are reacted with 10.0 g of BWhat information do you need to calculate the mass of the product (C) that will be produced?A + 3B 2Cmolar masses:A is 10.0 g/molB is 20.0 g/molC is 25.0 g/mol?
63 10.0 g of Chemical A are reacted with 10.0 g of B Example10.0 g of Chemical A are reacted with 10.0 g of BA + 3B 2C1) Convert known masses of reactants to moles
64 10.0 g of Chemical A are reacted with 10.0 g of B Example10.0 g of Chemical A are reacted with 10.0 g of BA + 3B 2C2) Convert to the number of moles of productSo B is limiting reactant
65 10.0 g of Chemical A are reacted with 10.0 g of B Example10.0 g of Chemical A are reacted with 10.0 g of BA + 3B 2CChoose the least number of moles of product formed as limiting reactant: B is limiting
66 10.0 g of Chemical A are reacted with 10.0 g of B Example10.0 g of Chemical A are reacted with 10.0 g of BA + 3B 2C3) Convert moles of C to grams of C using the molar mass
67 shortcut to limiting reactant How do we determine quickly which reactant is limitingand by deduction, which is in excess?Three step programCalculate the number of moles of each reactantdivide the number of moles of each reactant by its coefficient from the balanced equationthe reactant with the smaller result is limiting
68 If 56 g of K is reacted with 56 g of oxygen gas according to the equation below, indicate the mass of product that can be made and identify the limiting reactant K(s) + O2(g) 2 K2O(s)K2O = 94.2g/mol67g K2O; K is the limiting reactant.270 g K2O; K is the limiting reactant.270 g K2O; O2 is the limiting reactant.67g K2O; O2 is the limiting reactant.
69 Theoretical vs actual yield Theoretical YieldThe maximum amount of a given product that can be formed once a limiting reactant has been completely consumed – it is a calculationActual yieldThe amount of product that is actually produced in a reaction – it is a measurementIt is usually less than the maximum expected (theoretical yield)
70 FYITheoretical yield is found by using the amount of limiting reactant calculated in a mass-to-mass calculationFor the chemist in the lab, the actual yield is found by weighing the amount of product obtained
71 Theoretical Yield Actual yield For our auto mechanic, the theoretical yield is 8 finished carsActual yieldIf one car body was damaged beyond repair, the actual yield would have been 7 finished cars
72 Percent yield is the percent of the theoretical yield actually obtained from a chemical reaction For our car mechanic:
73 WORKED EXAMPLE 6.8 Percent Yield The combustion of acetylene gas (C2H2) produces carbon dioxide and water as indicated in the following reaction 2 C2H2 (g) + 5 O2 (g) → 4 CO2 (g) + 2 H2O (g) When 26.0 g of acetylene is burned in sufficient oxygen for complete reaction, the theoretical yield of CO2 is 88.0 g Calculate the percent yield for this reaction if the actual yield is only 72.4 g CO2 Analysis—The percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100
75 WORKED EXAMPLE 6.9 Mass to Mole Conversions: Limiting Reagent and Theoretical Yield The element boron is produced commercially by the reaction of boric oxide with magnesium at high temperature. B2O3 (l) + 3 Mg (s) → 2 B (s) + 3 MgO (s) What is the theoretical yield of boron when 2350 g of boric oxide is reacted with 3580 g of magnesium? The molar masses of boric oxide and magnesium are 69.6 g/mol and 24.3 g/mol, respectively
76 WORKED EXAMPLE 6.9 Mass to Mole Conversions: Limiting Reagent and Theoretical Yield (Continued)
77 If 28. 56 g of K2O is produced when 25 If g of K2O is produced when g K is reacted according to the following equation, what is the percent yield of the reaction? K(s) + O2(g) 2 K2O(s)87.54%95%94.82%88%
79 What is the mole, and why is it useful in chemistry? A mole refers to Avogadro’s number × 1023 formula units of a substance.One mole of any substance has a mass (a molar mass) equal to the molecular or formula weight of the substance in grams.Because equal numbers of moles contain equal numbers of formula units, molar masses act as conversion factors between numbers of molecules and masses in grams.
80 How are molar quantities and mass quantities related? The coefficients in a balanced chemical equation represent the numbers of moles of reactants and products in a reaction.The ratios of coefficients act as mole ratios that relate amounts of reactants and/or products.By using molar masses and mole ratios in factor-label calculations, unknown masses or molar amounts can be found from known masses or molar amounts.
81 What are the limiting reagent, theoretical yield, and percent yield of a reaction? The limiting reagent is the reactant that runs out first.The theoretical yield is the amount of product that would be formed based on the amount of the limiting reagent.The actual yield of a reaction is the amount of product obtained.The percent yield is the amount of product obtained divided by the amount theoretically possible and multiplied by 100%.
82 The official name for what we are doing is Stoichiometry
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