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The Mole and Stoichiometry

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1 The Mole and Stoichiometry

2 understand the Avogadro number and mole (of particles)
be able to carry out calculations involving quantities of substances expressed in moles be able express solution concentrations in mol dm-3 recall and understand the use of the ‘molar volume’ recall and be able to use the ideal gas equation know how to balance equations know how to balance ionic equations understand the purpose of, be able to carry out, and be able to carry out calculations involving, titration understand the use of empirical and molecular formulas be able to calculate empirical and molecular formulas be able to calculate theoretical yield and percentage yield of reactions understand and be able to calculate atom economy

3 Mole and Stoichiometry

4 FIND MASS Ex. If we need mole Ca3(PO4)2 for an experiment, how many grams do we need to weigh out? Calculate MM of Ca3(PO4)2 3 × mass Ca = 3 × g = g 2 × mass P = 2 × g = g 8 × mass O = 8 × g = g 1 mole Ca3(PO4)2 = g Ca3(PO4)2 What do we want to determine? 0.168 g Ca3(PO4)2 = ? Mol Fe Start End

5 Learning Check: Using Molar Mass
So, we need 0.168 = g Ca3(PO4)2

6 EX: 2 FIND MASS The expression relating mass and number of moles is: Mass = mol) × Molar mass (or Mr ) Mass (g) = n(mol) × M(gmol-1) Example Calculate the mass in grams in 0.75mol of sodium hydroxide, NaOH Step 1: Find the molar mass of the compound Na: gmol-1 Mr: gmol-1 O: gmol-1 H: gmol-1 Step 2: Substitute into the above expression Mass of sample = 0.75mol × gmol-1 = 30g

7 Your Turn! How many grams of platinum (Pt) are in 0.475 mole Pt? 195 g
Molar mass of Pt = g/mol = 92.7 g Pt

8 Find MOLES Example Number of moles = mass of sample (g)
molar mass (gmol-1) Example Convert 25.0g of KMnO4 to moles Step 1: Calculate the molar mass K Mn O 1 × gmol-1 1 × gmol-1 4 × gmol-1 39.10 gmol-1 54.93 gmol-1 64.00 gmol-1 Mr = gmol-1 Step 2: Substitute into above expression 25.0g . 158.03gmol-1 No. of moles = = mol

9 FIND MOLES moles = mass / M
Calculate the mass in grams present in: (a) 0.57mol of potassium permanganate (KMnO4) Answer: Molar mass KMnO4 = gmol-1 Mass in grams = 0.57mol × gmol-1 = g (b) 1.16mol of oxalic acid (H2C2O4) Answer: Molar mass H2C2O4 = gmol-1 Mass in grams = 1.16mol × gmol-1 = g (c) 2.36mol of calcium hydroxide (Ca(OH)2) Answer: Molar mass Ca(OH)2 = 74.1 gmol-1 Mass in grams = 2.36mol × 74.1 gmol-1 = g

10 Your Turn! = 0.227 mol CO2 How many moles of CO2 are there in 10.0 g?
Molar mass of CO2 1 × g = g C 2 × g = g O 1 mol CO2 = g CO2 = mol CO2

11 Balancing Equations AgNO3(aq) + Na3PO4(aq)  Ag3PO4(s) + NaNO3(aq)
Count atoms Reactants Products 1 Ag Ag 3 Na Na Add in coefficients by multiplying Ag & Na by 3 to get 3 of each on both sides 3AgNO3(aq) + Na3PO4(aq)  Ag3PO4(s) + 3NaNO3(aq) Now check polyatomic ions 3 NO3 NO3 1 PO43 PO43 Balanced

12 Balance by Inspection __C3H8(g) + __O2(g)  __CO2(g) + __H2O(ℓ)
Assume 1 in front of C3H8 3C 1C  3 8H 2H  4 1C3H8(g) + __O2(g)  3CO2(g) + 4H2O(ℓ) 2O  5 =10 O = (3  2) + 4 = 10 8H H = 2  4 = 8 1C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(ℓ)

13 Your Turn! Balance each of the following equations.
What are the coefficients in front of each compound? __ Ba(OH)2(aq) +__ Na2SO4(aq) → __ BaSO4(s) + __ NaOH(aq) 1 1 1 2 ___KClO3(s) → ___KCl(s) +___ O2(g) 2 2 3 __H3PO4(aq) + __ Ba(OH)2(aq) → __Ba3(PO4)2(s) + __H2O(ℓ) 2 3 1 6

14 Using Balanced Equations: Reaction Stoichiometry
Critical link between substances Gives relationship between reactants used & products formed Numeric coefficient tells us MOLES

15 Stoichiometric Ratios
Consider the reaction N2 + 3H2 → 2NH3 Could be read as: “When 1 molecule of nitrogen reacts with 3 molecules of hydrogen, 2 molecules of ammonia are formed.” Molecular relationships 1 molecule N2  2 molecule NH3 3 molecule H2  2 molecule NH3 1 molecule N2  3 molecule H2

16 Stoichiometric Ratios
Consider the reaction N2 + 3H2 → 2NH3 Could also be read as: “When 1 mole of nitrogen reacts with 3 moles of hydrogen, 2 moles of ammonia are formed.” Molar relationships 1 mole N2  2 mole NH3 3 mole H2  2 mole NH3 1 mole N2  3 mole H2

17 Using Stoichiometric Ratios
Ex. For the reaction N2 + 3 H2 → 2NH3, how many moles of N2 are used when 2.3 moles of NH3 are produced? Assembling the tools 2 moles NH3 = 1 mole N2 2.3 mole NH3 = ? moles N2 = 1.2 mol N2

18 Your Turn! If mole of CO2 is produced by the combustion of propane, C3H8, how many moles of oxygen are consumed? The balanced equation is C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) 0.575 mole 2.88 mole 0.192 mole 0.958 mole 0.345 mole Assembling the tools 0.575 mole CO2 = ? moles O2 3 moles CO2 = 5 mole O2 = mol O2

19 Using Balanced Equation to Determine Stoichiometry
Ex. What mass of O2 will react with 96.1 g of propane (C3H8) gas, to form gaseous carbon dioxide & water? Strategy 1. Write the balanced equation C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) 2. Assemble the tools 96.1 g C3H8  moles C3H8  moles O2  g O2 1 mol C3H8 = 44.1 g C3H8 1 mol O2 = g O2 1 mol C3H8 = 5 mol O2

20 Using Balanced Equation to Determine Stoichiometry
Ex. What mass of O2 will react with 96.1 g of propane in a complete combustion? C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) 3. Assemble conversions so units cancel correctly = 349 g of O2 are needed

21 Your Turn! 2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(ℓ)
How many grams of Al2O3 are produced when 41.5 g Al react? 2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(ℓ) 78.4 g 157 g 314 g 22.0 g 11.0 g = 78.4 g Al2O3

22 Calculating the number of particles
Avogadro’s Number One mole of any object means × 1023 units of those objects. For example, 1 mol of H2O contains × 1023 molecules 1 mol of NaCl contains × 1023 formula units Calculating the number of particles Avogadro’s number is used to convert between the number of moles and the number of atoms, ions or molecules. Example 0.450mol of iron contains how many atoms? Number of atoms = number of moles × Avogadro’s number (NA) Therefore No. of atoms = (0.450mol) × (6.022 × 1023) = 2.7 × 1023 atoms

23 Using Avogadro’s Number
What is the mass, in grams, of one molecule of octane, C8H18? Molecules octane  mol octane  g octane 1. Calculate molar mass of octane Mass C = 8 × g = g Mass H = 18 × g = g 1 mol octane = g octane 2. Convert 1 molecule of octane to grams = × 10–22 g octane

24 The Mole How many atoms in 1 mole of 12C ?
1 mole of 12C = × 1023 atoms = g Avogadro’s number = × 1023 atoms = NA 1 mole Xe = 6.022×1023 Xe atoms, I’ll round to 6 × 1023 1 mole NO2 = 6.022×1023 NO2 molecules. QUESTION: How many ATOMS does it, 1 mole of NO2, have? Ans: 18 × 1023 atoms Question: How many atoms are in 36.04g of water? Answer: 12 × 1023 or 1.2× 1024 Question: Given 2.4× 1025 atoms of diphosphorous pentoxide, how may moles is this? Answer : 5.70 mole

25 Macroscopic to Microscopic
How many silver atoms are in a 85.0 g silver bracelet? What do we know? g Ag = 1 mol Ag 1 mol Ag = 6.022×1023 Ag atoms What do we want to determine? 85.0 g silver = ? atoms silver g Ag  mol Ag  atoms Ag = 4.7 × 1023 Ag atoms

26 Example: Determine the # of molecules
How many molecules are there in 4 moles of hydrogen peroxide (H2O2)? No. of molecules = no. of moles × Avogadro’s number (NA) = 4mol × (6.022 × 1023 mol1) = 24 ×1023 molecules = 2.4 × 1024 molecules Questions How many atoms are there in 7.2 moles of gold (Au)? Answer: 4.3 × 1024 atoms The visible universe is estimated to contain 1022 stars. How many moles of stars are there? Answer: 1022 stars = = 0.17 mol. 6.022×1023

27 Example Calculate the mass of one molecule of ammonium carbonate [(NH4)2CO3] Step 1: Calculate the molar mass 2 Nitrogen atoms 8 Hydrogen atoms 1 Carbon atom 3 Oxygen atoms 2 × 14.01gmol-1 8 × gmol-1 1 ×12.01gmol-1 3 × gmol-1 = gmol-1 = gmol-1 = gmol-1 = gmol-1 Total = gmol-1 Step 2: Employ Avogadro’s Number, NA Mass of one molecule = 96.09 gmol-1 . 6.022×1023mol-1 = 1.59 × 10-22g Questions Calculate the mass of one molecule of: Ethanoic acid (CH3COOH) Methane (CH4) Potassium dichromate (K2Cr2O7) 9.96 × g 2.66 × g 4.89 × g

28 Learning Check: Mole Conversions
Calculate the number of formula units of Na2CO3 in 1.29 moles of Na2CO3. How many moles of Na2CO3 are there in 1.15 x 105 formula units of Na2CO3 ? = 7.77×1023 particles Na2CO3 = 1.91×10–19 mol Na2CO3

29 Your Turn! How many atoms are in 1.00 x 10–9 g of U (1 ng)? Molar mass U = g/mole. 6.02 x 1014 atoms 4.20 x 1011 atoms 2.53 x 1012 atoms 3.95 x 10–31 atoms 2.54 x 1021 atoms = 2.53 x 1012 atoms U

30 Mole-to-Mole Conversion Factors
Can use chemical formula to relate amount of each atom to amount of compound In H2O there are 3 relationships: 2 mol H ⇔ 1 mol H2O 1 mol O ⇔ 1 mol H2O 2 mol H ⇔ 1 mol O Can also use these on atomic scale 2 atom H ⇔ 1 molecule H2O 1 atom O ⇔ 1 molecule H2O 2 atom H ⇔ 1 molecule O

31 Stoichiometric Equivalencies
Within chemical compounds, moles of atoms always combine in the same ratio as the individual atoms themselves Ratios of atoms in chemical formulas must be whole numbers!! These ratios allow us to convert between moles of each quantity Ex. N2O5 2 mol N ⇔ 1 mol N2O5 5 mol O ⇔ 1 mol N2O5 2 mol N ⇔ 5 mol O

32 Your Turn! Calculate the number of moles of calcium in moles of Ca3(PO4)2 2.53 mol Ca 0.432 mol Ca 3.00 mol Ca 7.59 mol Ca 0.843 mol Ca 2.53 moles of Ca3(PO4)2 = ? mol Ca 3 mol Ca  1 mol Ca3(PO4)2 = 7.59 mol Ca

33 Internal Mass-to-Mass Calculations
Chlorophyll, the green pigment in leaves, has the formula C55H72MgN4O5. If g of Mg is available to a plant for chlorophyll synthesis, how many grams of carbon will be required to completely use up the magnesium? Analysis g Mg ⇔ ? g C g Mg → mol Mg → mol C → g C Assembling the tools g Mg = 1 mol Mg 1 mol Mg ⇔ 55 mol C 1 mol C = g C

34 Ex. Mass-to-Mass Conversion
1 mol C ⇔ 12.0 g C 1 mol Mg ⇔ 24.3 g Mg g Mg → mol Mg → mol C → g C 1 mol Mg ⇔ 55 mol C = g C

35 Your Turn! How many g of iron are required to use up all of g of oxygen atoms (O) to form Fe2O3? 59.6 g 29.8 g 89.4 g 134 g 52.4 g = 59.6 g Fe mass O  mol O  mol Fe  mass Fe 25.6 g O  ? g Fe 3 mol O  2 mol Fe

36 Percentage Composition
Way to specify relative masses of each element in a compound List of percentage by mass of each element Percentage by Mass Ex. Na2CO3 is 43.38% Na 11.33% C 45.29% O What is sum of % by mass? 100.00%

37 Ex. Percent Composition
Determine percentage composition based on chemical analysis of substance Ex. A sample of a liquid with a mass of g was decomposed into its elements and gave g of carbon, g of hydrogen, and g of oxygen. What is the percentage composition of this compound? Analysis: Calculate % by mass of each element in sample Tools: Eqn for % by mass Total mass = g Mass of each element

38 Ex. % Composition of Compound
For C: For H: For O: % composition tells us mass of each element in g of substance In g of our liquid 60.26 g C, g H & g O = 60.26% C = 11.11% H = 28.62% O Sum of percentages: 99.99%

39 Your Turn! A sample was analyzed and found to contain g nitrogen and g oxygen. What is the percentage composition of this compound? 1. Calculate total mass of sample Total sample mass = g g = g 2. Calculate % Composition of N 3. Calculate % Composition of O = 25.94% N Percent composition 25.94% N & 74.06% O = 74.06% O

40 Percent Compositions & Chemical Identity
Theoretical or Calculated % Composition Calculated from molecular or ionic formula. Lets you distinguish between multiple compounds formed from the same 2 elements If experimental percent composition is known Calculate Theoretical % Composition from proposed Chemical Formula Compare with experimental composition Ex. N & O form multiple compounds N2O, NO, NO2, N2O3, N2O4, & N2O5

41 Calculating mass percentage from a chemical formula
The mass percentage composition allows you to determine the fraction of the total mass each element contributes to the compound. Example Ammonium nitrate (NH4NO3) is an important compound in the fertiliser industry. What is the mass % composition of ammonium nitrate? Step 1: Calculate the molar mass of ammonium nitrate Two N atoms: gmol-1 Four H atoms: gmol-1 Three O atoms: gmol-1 Molar mass NH4NO3 = gmol-1

42 Step 2: Determine the mass % composition for each element
Nitrogen: g N in one mol of ammonium nitrate Mass fraction of N = g 80.05g Mass % composition of N = g × 100% 80.05g = 34.99% ≈ 35% Hydrogen: 4.032g H in one mol of ammonium nitrate Mass fraction of N = g 80.05g Mass % composition of H = g × 100% 80.05g = 5.04% ≈ 5% Oxygen: 48.00g O in one mol of ammonium nitrate As above, the mass % composition of O is found to be 60%

43 Therefore, the mass % composition of ammonium nitrate (NH4NO3) is:
% Nitrogen: 35% % Hydrogen: 5% % Oxygen: 60% To check your answer, make sure it adds up to 100% Question What is the mass % composition of C12H22O11? Answer: % Carbon: 42.1% % Hydrogen: 6.5% % Oxygen: 51.4%

44 Your Turn If a sample containing only phosphorous & oxygen has percent composition 56.34% P & 43.66% O, is this P4O10? Yes No 4 mol P  1 mol P4O10 10 mol O  1 mol P4O10 4 mol P = 4  g/mol P = g P 10 mol O = 10 16.00 g/mol O = g O 1 mol P4O10 = g P4O10 = % P = % O

45 Determining Empirical & Molecular Formulas
When making or isolating new compounds one must characterize them to determine structure & Molecular Formula Exact composition of one molecule Exact whole # ratio of atoms of each element in molecule Empirical Formula Simplest ratio of atoms of each element in compound Obtained from experimental analysis of compound 4.3 | Determining Empirical and Molecular Formulas Empirical formula CH2O glucose Molecular formula C6H12O6

46 Three Ways to Calculate Empirical Formulas
From Masses of Elements Ex g sample of which g is Fe and g is O. From Percentage Composition Ex % P and % O. From Combustion Data Given masses of combustion products Ex. The combustion of a g sample of a compound of C, H, and O in pure oxygen gave g CO2 and g of H2O.

47 Strategy for Determining Empirical Formulas
Determine mass in g of each element Convert mass in g to moles Divide all quantities by smallest number of moles to get smallest ratio of moles Convert any non-integers into integer numbers. If number ends in decimal equivalent of fraction, multiply all quantities by least common denominator Otherwise, round numbers to nearest integers

48 1. Empirical Formula from Mass Data
When a g sample of a compound was analyzed, it was found to contain g of C, g of H, and g of N. Calculate the empirical formula of this compound. Step 1: Calculate moles of each substance 3.722  103 mol C 1.860  102 mol H 3.723  103 mol N

49 1. Empirical Formula from Mass Data
Step 2: Select the smallest # of moles. Lowest is x 10–3 mole C = H = Step 3: Divide all # of moles by the smallest one Mole ratio Integer ratio 1.000 = 1 4.997 = 5 N = 1.000 = 1 Empirical formula = CH5N

50 Empirical Formula from Mass Composition
One of the compounds of iron and oxygen, “black iron oxide,” occurs naturally in the mineral magnetite. When a g sample was analyzed it was found to have g of Fe and g of O. Calculate the empirical formula of this compound. Assembling the tools: 1 mol Fe = g Fe 1 mol O = g O 1. Calculate moles of each substance mol Fe mol O

51 1. Empirical Formula from Mass Data
2. Divide both by smallest #mol to get smallest whole # ratio. =1.000 Fe × 3 = Fe =1.33 O × 3 = 3.99 O Or Empirical Formula = Fe3O4

52 A sample of a brown gas, a major air pollutant, is found to contain 2
A sample of a brown gas, a major air pollutant, is found to contain 2.34 g N and 5.34g O. Determine a formula for this substance. require mole ratios so convert grams to moles moles of N = 2.34g of N = moles of N 14.01 g/mole moles of O = g = moles of O 16.00 g/mole Formula:

53 Determining Molecular Formulas
Empirical formula Accepted formula unit for ionic compounds Molecular formula Preferred for molecular compounds In some cases molecular & empirical formulas are the same When they are different, & the subscripts of molecular formula are integer multiples of those in empirical formula If empirical formula is AxBy Molecular formula will be An×xBn×y

54 Determining Molecular Formula
Need molecular mass & empirical formula Calculate ratio of molecular mass to mass predicted by empirical formula & round to nearest integer Ex. Glucose Molecular mass is g/mol Empirical formula = CH2O Empirical formula mass = g/mol If molecular mass & empirical formula mass are same, then molecular for Molecular formula = C6H12O6

55 Learning Check The empirical formula of a compound containing phosphorous and oxygen was found to be P2O5. If the molar mass is determined to be g/mol, what is the molecular formula? Step 1: Calculate empirical mass Step 2: Calculate ratio of molecular to empirical mass = 2 Molecular formula = P4O10

56 Your Turn! The empirical formula of hydrazine is NH2, and its molecular mass is What is its molecular formula? NH2 N2H4 N3H6 N4H8 N1.5H3 Molar mass of NH2 = (1×14.01)g + (2×1.008)g = g n = (32.0/16.02) = 2 and (B)

57 Balancing Equations Use the inspection method
Step 1. Write unbalanced equation Zn(s) + HCl(aq)  ZnCl2(aq) + H2(g) unbalanced Step 2. Adjust coefficients to balance numbers of each kind of atom on both sides of arrow. Since ZnCl2 has 2Cl on the product side, 2HCl on reactant side is needed to balance the equation. Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g) 1 Zn each side 2 H each side So balanced

58 Limiting Reactant Excess reactant
Reactant that is completely used up in the reaction Present in lower # of moles It determines the amount of product produced For this reaction = ethylene Excess reactant Reactant that has some amount left over at end Present in higher # of moles For this reaction = water

59 Limiting Reactant Calculations
Write the balanced equation. Identify the limiting reagent. Calculate amount of reactant B needed to react with reactant B Compare amount of B you need with amount of B you actually have. If need more B than you have, then B is limiting If need less B than you have, then A is limiting mass reactant A have mol reactant A mol reactant B Mass reactant B need

60 Limiting Reactant Calculations
Calculate mass of desired product, using amount of limiting reactant & mole ratios. mass limiting reactant mol limiting reactant mol product mass product

61 Ex. Limiting Reactant Calculation
How many grams of NO can form when 30.0 g NH3 and 40.0 g O2 react according to: 4 NH3 + 5 O2  4 NO + 6 H2O Solution: Step 1 mass NH3  mole NH3  mole O2  mass O2 Assembling the tools 1 mol NH3 = g 1 mol O2 = g 4 mol NH3  5 mol O2 Only have 40.0 g O2, O2 limiting reactant = 70.5 g O2 needed

62 Ex. Limiting Reactant Calculation
How many grams of NO can form when 30.0 g NH3 and 40.0 g O2 react according to: 4 NH3 + 5 O2  4 NO + 6 H2O Solution: Step 2 mass O2  mole O2  mole NO  mass NO Assembling the tools 1 mol O2 = g 1 mol NO = g 5 mol O2  4 mol NO Can only form 30.0 g NO. = 30.0 g NO formed

63 Your Turn! If 18.1 g NH3 is reacted with 90.4 g CuO, what is the maximum amount of Cu metal that can be formed? 2NH3(g) + 3CuO(s)  N2(g) + 3Cu(s) + 3H2O(g) (MM) (17.03) (79.55) (28.01) (64.55) (18.02) (g/mol) 127 g 103 g 72.2 g 108 g 56.5 g 127 g CuO needed. Only have 90.4g so CuO limiting 72.2 g Cu can be formed

64 Reaction Yield In many experiments, the amount of product is less than expected Losses occur for several reasons Mechanical issues – sticks to glassware Evaporation of volatile (low boiling) products. Some solid remains in solution Competing reactions & formation of by-products. Main reaction: 2 P(s) + 3 Cl2(g)  2 PCl3(ℓ) Competing reaction: PCl3(ℓ) + Cl2(g)  PCl5(s) By-product 4.6 | Theoretical Yield and Percentage Yield

65 Theoretical vs. Actual Yield
Theoretical Yield Amount of product that must be obtained if no losses occur. Amount of product formed if all of limiting reagent is consumed. Actual Yield Amount of product that is actually isolated at end of reaction. Amount obtained experimentally How much is obtained in mass units or in moles.

66 Percentage Yield Useful to calculate % yield. Percent yield
Relates the actual yield to the theoretical yield It is calculated as: Ex. If a cookie recipe predicts a yield of 36 cookies and yet only 24 are obtained, what is the % yield?

67 Ex. Percentage Yield Calculation
When 18.1 g NH3 and 90.4 g CuO are reacted, the theoretical yield is 72.2 g Cu. The actual yield is g Cu. What is the percent yield? 2NH3(g) + 3CuO(s)  N2(g) + 3Cu(s) + 3H2O(g) = 80.7%

68 Learning Check: Percentage Yield
A chemist set up a synthesis of solid phosphorus trichloride by mixing 12.0 g of solid phosphorus with g chlorine gas and obtained 42.4 g of solid phosphorus trichloride. Calculate the percentage yield of this compound. Analysis: Write balanced equation P(s) + Cl2(g)  PCl3(s) Determine Limiting Reagent Determine Theoretical Yield Calculate Percentage Yield

69 Learning Check: Percentage Yield
Assembling the Tools: 1 mol P = g P 1 mol Cl2 = g Cl2 3 mol Cl2 ⇔ 2 mol P Solution Determine Limiting Reactant But you only have 35.0 g Cl2, so Cl2 is limiting reactant = 41.2 g Cl2

70 Learning Check: Percentage Yield
Solution Determine Theoretical Yield Determine Percentage Yield Actual yield = 42.4 g = 45.2 g PCl3 = 93.8 %

71 Your Turn! When 6.40 g of CH3OH was mixed with 10.2 g of O2 and ignited, 6.12 g of CO2 was obtained. What was the percentage yield of CO2? 2CH3OH + 3O2  2CO H2O MM(g/mol) (32.04) (32.00) (44.01) (18.02) 6.12% 8.79% 100% 142% 69.6% =9.59 g O2 needed; CH3OH limiting = 8.79 g CO2 in theory

72 Conclusions Continued
We can use the following flow chart for mole calculations:

73 Dilution of Solutions

74 Concentration = # of moles volume (L) V = 1000 mL V = 1000 mL
n = 8 moles [ ] = 32 molar V = 1000 mL V = 1000 mL V = 5000 mL n = 2 moles n = 4 moles n = 20 moles Concentration = 2 molar [ ] = 4 molar [ ] = 4 molar

75 Making Molar Solutions
…from liquids (More accurately, from stock solutions)

76 Concentration…a measure of solute-to-solvent ratio
concentrated vs dilute “lots of solute” “not much solute” “watery” Concentration of a solution describes the quantity of a solute that is contained in a particular quantity of solvent or solution Knowing the concentration of solutes is important in controlling the stoichiometry of reactant for reactions that occur in solution A concentrated solution contains a large amount of solute in a given amount of solution. A 10 mol/L solution would be called concentrated. A dilute solution contains a small amount of solute in a given amount of solution. A 0.01 mol/L solution would be called dilute. Add water to dilute a solution; boil water off to concentrate it.

77 Making a Dilute Solution
remove sample moles of solute initial solution same number of moles of solute in a larger volume mix Making a Dilute Solution diluted solution

78 Concentration “The amount of solute in a solution”
A. mass % = mass of solute mass of sol’n B. parts per million (ppm)  also, ppb and ppt – commonly used for minerals or contaminants in water supplies C. molarity (M) = moles of solute L of sol’n – used most often in this class % by mass – medicated creams % by volume – rubbing alcohol MOLARITY - Most common unit of concentration Most useful for calculations involving the stoichiometry of reactions in solution Molarity of a solution is the number of moles of solute present in exactly 1 L of solution: moles of solute molarity = liters of solution Units of molarity — moles per liter of solution (mol/L), abbreviated as M Relationship among volume, molarity, and moles is expressed as VL M Mol/L = L (mol) = moles (L) There are several different ways to quantitatively describe the concentration of a solution, which is the amount of solute in a given quantity of solution. 1. Molarity – Useful way to describe solution concentrations for reactions that are carried out in solution or for titrations – Molarity is the number of moles of solute divided by the olume of the solution Molarity = moles of solute = mol/L liter of solution – Volume of a solution depends on its density, which is a function of temperature 2. Molality – Concentration of a solution can also be described by its molality (m), the number of moles of solute per kilogram of solvent – Molality = moles of solute kilogram solvent – Depends on the masses of the solute and solvent, which are independent of temperature – Used in determining how colligative properties vary with solute concentrations 3. Mole fraction – Used to describe gas concentrations and to determine the vapor pressures of mixtures of similar liquids – Mole fraction () = moles of component total moles in the solution – Depends on only the masses of the solute and solvent and is temperature independent 4. Mass percentage (%) – The ratio of the mass of the solute to the total mass of the solution – Result can be expressed as mass percentage, parts per million (ppm), or parts per billion (ppb) mass percentage = mass of solute  100% mass of solution parts per million (ppm) = mass of solute  106 parts per billion (ppb) = mass of solute  109 – Parts per million (ppm) and parts per billion (ppb) are used to describe concentrations of highly dilute solutions, and these measurements correspond to milligrams (mg) and micrograms (g) of solute per kilogram of solution, respectively – Mass percentage and parts per million or billion can express the concentrations of substances even if their molecular mass is unknown because these are simply different ways of expressing the ratios of the mass of a solute to the mass of the solution M = mol L D. molality (m) = moles of solute kg of solvent

79 Glassware

80 Glassware – Precision and Cost
beaker vs. volumetric flask When filled to 1000 mL line, how much liquid is present? beaker 5% of 1000 mL = volumetric flask 50 mL 1000 mL mL Range: 950 mL – 1050 mL Range: mL– mL imprecise; cheap precise; expensive

81 Markings on Glassware Beaker 500 mL + 5% Range = 500 mL + 25 mL
Graduated Cylinder 1000 mL mL Range = mL + 5 mL 475 – 525 mL Volumetric Flask 500 mL mL Range = – mL TC 20oC “to contain at a temperature of 20 oC” 22 TD “to deliver” T s “time in seconds”

82 Measure to part of meniscus w/zero slope.
~ ~ water in grad. cyl. mercury in grad. cyl. ~ ~ ~ ~ Measure to part of meniscus w/zero slope. Copyright © 2007 Pearson Benjamin Cummings. All rights reserved.

83 How to mix solid chemicals
Lets mix chemicals for the upcoming soap lab. We will need 1000 mL of 3 M NaOH per class. How much sodium hydroxide will I need, for five classes, for this lab? M = mol L 3 M = ? mol 1 L ? = 3 mol NaOH/class x 5 classes 15 mol NaOH How much will this weigh? 1 23g/mol + 16g/mol g/mol MMNaOH = 40g/mol To prepare a solution that contains a specified concentration of a substance, it is necessary to dissolve the desired number of moles of solute in enough solvent to give the desired final volume of solution. Solute occupies space in the solution so the volume of the solvent that is needed is less than the desired volume of solution. To prepare a particular volume of a solution that contains a specified concentration of a solute, calculate the number of moles of solute in the desired volume of solution and then covert the number of moles of solute to the corresponding mass of solute needed. 40.0 g NaOH X g NaOH = mol NaOH = 600 g NaOH 1 mol NaOH FOR EACH CLASS: To mix this, add 120 g NaOH into 1L volumetric flask with ~750 mL cold H2O. Mix, allow to return to room temperature – bring volume to 1 L.

84 How to mix a Standard Solution
Wash bottle Volume marker (calibration mark) Weighed amount of solute Use a VOLUMETRIC FLASK to make a standard solution of known concentration Step 1> add the weighed amount of solute in the volumetric flask Step 2> add distilled water (about half of final volume) Step 3> cap volumetric flask, and shake to dissolve solute completely Step 4> add distilled water to volume marker (calibration mark) The solution process may be exothermic (release heat). This may cause the liquid to show a larger volume than is real. Allow the solution to return to ambient (room) temperature and check volume again. Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 480

85 How to mix a Standard Solution
An aqueous solution consists of at least two components, the solvent (water) and the solute (the stuff dissolved in the water). Usually one wants to keep track of the amount of the solute dissolved in the solution. We call this the concentrations. One could do by keeping track of the concentration by determining the mass of each component, but it is usually easier to measure liquids by volume instead of mass. To do this measure called molarity is commonly used. Molarity (M) is defined as the number of moles of solute (n) divided by the volume (V) of the solution in liters. It is important to note that the molarity is defined as moles of solute per liter of solution, not moles of solute per liter of solvent. This is because when you add a substance, perhaps a salt, to some volume of water, the volume of the resulting solution will be different than the original volume in some unpredictable way. To get around this problem chemists commonly make up their solutions in volumetric flasks. These are flasks that have a long neck with an etched line indicating the volume. The solute (perhaps a salt) is added to the flask first and then water is added until the solution reaches the mark. The flasks have very good calibration so volumes are commonly known to at least four significant figures.

86 Process of Making a Standard Solution from Liquids
Solutions can be made using liquids or solids (or gases). To make a 5% solution v/v (volume to volume) This means to add 5 mL of solute in 95 mL of solvent. The total is 5 mL / 100 mL or 5%. For the diagram add 25 mL of liquid solute and add water to bring volume to 500 mL (about 475 mL water). SAFETY NOTE: Always add acid concentrate to water…never add water to concentrated acid. If you’ve seen what happens when water or ice crystals hit hot oil…a similar phenomenon occurs when water is added to concentrated acid. The addition of water to concentrated dissipates a large amount of heat. This heat rapidly boils the acid and causes it to spatter. If however, you start with a large volume of water and slowly add acid, the same amount of heat is generated. This time, the large volume of water is capable of absorbing the heat. The solution will not splatter. Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 483

87 How to mix a dilute solution from a concentrated stock solution
A solution of a desired concentration can be prepared by diluting a small volume of a more-concentrated solution, a stock solution, with additional solvent. – Calculate the number of moles of solute desired in the final volume of the more-dilute solution and then calculate the volume of the stock solution that contains the amount of solute. – Diluting a given quantity of stock solution with solvent does not change the number of moles of solute present. – The relationship between the volume and concentration of the stock solution and the volume and concentration of the desired diluted solution is (Vs) (M s) = moles of solute = (Vd) (M d). Copyright © 2007 Pearson Benjamin Cummings. All rights reserved.

88 Reading a pipette 4.48 - 4.50 mL 4.86 - 4.87 mL 5.00 mL
Identify each volume to two decimal places (values tell you how much you have expelled) mL mL 5.00 mL

89 MConc.VConc. = MDiluteVDilute
Dilution of Solutions Solution Guide Formula Weight Specific Gravity Molarity Reagent Percent To Prepare 1 Liter of one molar Solution Acetic Acid Glacial (CH3COOH) 60.05 1.05 17.45 99.8% 57.3 mL Ammonium Hydroxide (NH4OH) 35.05 0.90 14.53 56.6% 69.0 mL Formic Acid (HCOOH) 46.03 1.20 23.6 90.5% 42.5 mL Hydrochloric Acid (HCl) 36.46 1.19 12.1 37.2% 82.5 mL Hydrofluoric Acid (HF) 20.0 1.18 28.9 49.0% 34.5 mL Nitric Acid (HNO3) 63.01 1.42 15.9 70.0% 63.0 mL Perchloric Acid 60% (HClO4) 100.47 1.54 9.1 60.0% 110 mL Perchloric Acid 70% (HClO4) 1.67 11.7 70.5% 85.5 mL Phosphoric Acid (H3PO4) 97.1 1.70 14.8 85.5% 67.5 mL Potassium Hydroxide (KOH) Sodium Hydroxide (NaOH) 40.0 19.4 45.0% Sulfuric Acid (H2SO4) 98.08 1.84 18.0 50.5% 51.5 mL This chart quickly shows you the amount of concentrated acid needed to make 1 liter of a 1 M solution. If you need a 5 M solution, add 5x the amount of acid in the same volume. MConc.VConc. = MDiluteVDilute

90 **Safety Tip: When diluting, add acid or base to water.**
Acids (and sometimes bases) are purchased in concentrated form (“concentrate”) and are easily diluted to any desired concentration. Dilutions of Solutions  **Safety Tip: When diluting, add acid or base to water.** C = concentrate D = dilute Dilution Equation: Concentrated H3PO4 is 14.8 M. What volume of concentrate is required to make L of M H3PO4? VC = L = 845 mL

91 How would you mix the above solution?
1. Measure out L of concentrated H3PO4 . 2. In separate container, obtain ~20 L of cold H2O. 3. In fume hood, slowly pour [H3PO4] into cold H2O. 4. Add enough H2O until L of solution is obtained. Be sure to wear your safety glasses!

92 You have 75 mL of conc. HF (28.9 M); you need 15.0 L of
0.100 M HF. Do you have enough to do the experiment? MCVC = MDVD 28.9 M (0.075 L) = M (15.0 L) Yes; we’re OK. mol HAVE > 1.50 mol NEED

93 What do you call a tooth in a glass of water? - A one molar solution.
One mole, in solution. What do you call a tooth in a glass of water?        - A one molar solution.

94 Dilution Preparation of a desired solution by adding water to a concentrate. Moles of solute remain the same.

95 Dilution GIVEN: M1 = 15.8M V1 = ? M2 = 6.0M V2 = 250 mL WORK:
What volume of 15.8M HNO3 is required to make 250 mL of a 6.0M solution? GIVEN: M1 = 15.8M V1 = ? M2 = 6.0M V2 = 250 mL WORK: M1 V1 = M2 V2 (15.8M) V1 = (6.0M)(250mL) V1 = 95 mL of 15.8M HNO3

96 Preparing Solutions How to prepare 500 mL of 1.54 M NaCl solution
mass 45.0 g of NaCl add water until total volume is 500 mL 500 mL volumetric flask 500 mL mark 45.0 g NaCl solute

97 Stoichiometry Summary


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