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Chapter 3 Mass Relationships in Chemical Reactions.

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Presentation on theme: "Chapter 3 Mass Relationships in Chemical Reactions."— Presentation transcript:

1 Chapter 3 Mass Relationships in Chemical Reactions

2 Atomic Mass  What is atomic mass?  Mass of an atom in atomic mass units (AMU)  1 AMU is the mass of 1/12 of a carbon-12 atom  Also called atomic weight

3 Relative Atomic Mass  What does the relative atomic mass of an element represent?  Average mass of all known existing isotopes of that element  It is “unitless”  Proportioned based on abundance  Ex. Carbon has a mass of which is an average of all of the carbon isotopes

4 Isotope Proportions A certain sample of element Z contains 60% of 69 Z and 40% of 71 Z.  What is the relative atomic mass of element Z in this sample?

5 Isotope Proportions Chlorine exists as two isotopes, 35 Cl and 37 Cl. The relative atomic mass of chlorine is  Calculate the percentage abundance of each isotope.

6 Molar Mass and N A  What unit is used to measure the number of atoms/molecules?  Moles (mol) – the number of atoms in 12 grams of a carbon-12 sample  1 mole = 6.02 x  Avogadro's number (N A )  We use it like we would a pair (2), dozen (12), or a gross (144) to simplify our math because it is freaking huge

7 How Big Is A Mole? 1. If you would distribute 1 mole of pennies among the current population of the Earth, how much would everyone get? 2. How long would it take to spend this amount if you would spend $1,000/day? 3. How many people could you make from 1 mole of cells? 4. How long is 1 mole of seconds? 5. If we took a mole of regular 8.5 x 11 inch sheets of paper and started laying them out side by side on the planet’s surface. What percentage of the Earth would be covered?

8 How Big Is A Mole? Answers 1. $900 million dollars 2. 2,500,000 years billion ,000 x age of the universe (14 billion yrs) % …and it would be 17 million sheets thick

9 Molar Mass  How is molar mass different from relative atomic mass?

10 Molar Mass  What is the molar mass (MM) of the following elements?  Lithium  Plutonium  Bromine

11 Molar Mass  How do we determine the molar mass of a molecule?  What is the molar mass of water?

12 Molecular Vs. Molar  What is the difference between atomic, molecular and molar mass?

13 Molar Mass  What is the molar mass of each of the following species:  KBr  FeCl 3  Sodium Hydroxide  Lead (II) Oxide

14 Navigating  How can we navigate between mass (m), moles (n) and the number of atoms, molecules or particles?  We use molar mass (MM) and Avogadro's number (N A )

15

16 Double Circles

17 Double Triangles N A Avogadro's number m mass n moles MM molar mass n moles N number

18 Mass-Mole Conversions 1. How many moles in 28 grams of CO 2 ? 2. What is the mass of 5 moles of Fe 2 O 3 ? 3. Find the number of moles of argon in 452 g of argon. 4. Find the grams in 1.26 x mol of HC 2 H 3 O Find the mass in 2.6 mol of lithium bromide.

19 Mole-Number Conversions 1. How many moles of magnesium is 3.01 x atoms of magnesium? 2. How many molecules are there in 4.00 moles of glucose, C 6 H 12 O 6 ? 3. How many moles are 1.20 x atoms of phosphorous? 4. How many atoms are in moles of zinc? 5. How many oxygen atoms are in moles of N 2 O 5 ?

20 To the right, To the right  How many atoms are in 20.0 grams of gold?  Mass  20.0 g  Divide by molar mass to get to moles  moles  Multiply by Avogadro's number to get to atoms  6.14 E 22 Au atoms

21 To the left, To the left  What is the mass of 1.00 E 24 carbon atoms?  Number  1.00 E 24 atoms  Divide by Avogadro's number to get to moles  1.66 moles  Multiply by molar mass to get mass  19.9 grams

22 Combined 1. Find the mass in grams of 2.00 x molecules of diatomic fluorine. 2. Find the number of molecules in 60.0 g of dinitrogen monoxide. 3. How many molecules are in 5 mg of aspartame, C 14 H 18 N 2 O 5 ?

23 Combined

24 PRELAB 1 Gum

25 PRELAB 1 Sugar in Chewing Gum  How can we test for the mass of sugar in chewing gum?

26 PRELAB 1 Sugar in Chewing Gum

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28  Compare and contrast your measured percentage of sugar, and the actual percentage from the nutrition label for the brand of gum that you used.  What are some possible reasons for any differences in these values, and how you might increase the accuracy of your data collection?

29 PRELAB 1 Sugar in Chewing Gum What mathematical relationship/formula did you use to determine percent by mass? Why? If you were given one molecule of sucrose, C 6 H 12 O 6, what information do you think you would need to figure out the percent mass of carbon? Why?

30 Percent Composition  What is percent composition?  What two pieces of info do we need?

31 Percent Composition  How do we calculate it?

32 Percent Composition  What is the percent mass of hydrogen in water?  2 hydrogens  2(1.01) = 2.02 g/mol  Water is H 2 O  g/mol  2.02/18.02 x100 =  11.2% (tenths place)

33 Percent Composition  Calculate the percent composition of oxygen in sodium sulfate.  Na 2 SO 4  g/mol  4 oxygens  4(16.00) = g/mol  (64.00/142.04) x 100 =  45.1%

34 Percent Composition  Tartrazine, or yellow dye # 5, is commonly found in many foods. Its molecular formula is C 16 H 9 N 4 Na 3 O 9 S 2. What is the percent composition of Carbon and Sulfur? The molar mass of Tartrazine is g/mol.

35 Working in Reverse  If we start with %’s, what can that possibly lead us to?  Working the formula in reverse and using mole ratios, we can get the empirical and possibly the molecular formula

36 Determination of the Empirical Formula  How do we determine it? 1. Assume that you have 100% of the element and set equal to mass 2. Divide each element by its molar mass 3. Divide each by the smallest amount of moles 4. If numbers aren’t close to whole, multiply all of them by a coefficient

37 Butane’s Empirical Formula  Butane is made up of only carbon and hydrogen. If hydrogen makes up 17.34% of butane’s mass and carbon the remaining 82.66%, what is the empirical formula of butane? C2H5C2H5  What is the molecular formula for butane if its molecular mass is g/mol?  Multiply each subscript by 2  C 4 H 10

38 Determine the Empirical Formula  Determine the empirical formula of a compound having the following percent composition by mass…  K: 24.75%  Mn: 34.77%  O: 40.51% Empirical Formula: KMnO 4  What is the molecular formula if its molecular mass is g/mol? ANSWER: KMnO 4 (potassium permanganate)

39 Chemical Reactions and Equations  What is the difference between a chemical reaction and equation?  A chemical reaction is the process of making a new substance(s)  Ex. Iron oxidizing  A chemical equation is a symbolic representation of the reaction  4Fe + 3O 2 → 2Fe 2 O 3

40 Chemical Reactions and Equations  How do we read a chemical equation?  4Fe(s) + 3O 2 (g) → 2Fe 2 O 3 (s)  Reactants (left)  Products (right)  → Goes in one direction  ↔ Reversible reaction  Coefficients represent the mole ratios  Notations (s, l, g, aq) represent phases/states  Conservation of mass  Balanced

41 Chemical Reactions and Equations  What is a balanced equation?  The number of atoms on the left has to equal the number on the right  Use coefficients to balance them  Unbalanced:  CH 4 + O 2 → CO 2 + H 2 O  Balanced:

42 Chemical Reactions and Equations  How do we balance the equations?  KClO 3 → KCl + O 2  2KClO 3 → 2KCl + 3O 2  Al + O 2 → Al 2 O 3  4Al + 3O 2 → 2Al 2 O 3  C 2 H 6 + O 2 → CO 2 + H 2 O  2C 2 H 6 + 7O 2 → 4CO 2 + 6H 2 O

43 Chemical Reactions and Equations Balance the following chemical equations:  K 2 CO 3 + HCl  KCl + H 2 O + CO 2  K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2  CaCO 3 + HNO 3  Ca(NO 3 ) 2 + H 2 O + CO 2  CaCO 3 + 2HNO 3  Ca(NO 3 ) 2 + H 2 O + CO 2  Pb(NO 3 ) 2 + NaI  PbI 2 + NaNO 3  Pb(NO 3 ) 2 + 2NaI  PbI 2 + 2NaNO 3  Al 2 (SO 4 ) 3 + NaOH  Al(OH) 3 + Na 2 SO 4  Al 2 (SO 4 ) 3 + 6NaOH  2Al(OH) 3 + 3Na 2 SO 4

44 Amounts of Reactants and Products  What is stoichiometry?  The study of any numerical observation within a chemical reaction  Using the mole method to determine how much reactant is needed (or product created) based on the number of moles

45 Amounts of Reactants and Products  What can be said about the reaction:  H 2 + O 2 → H 2 O  Reactants: Hydrogen and Oxygen  Products: Water  Unbalanced  Balanced: 2H 2 + O 2 →2H 2 O  Proportion: 2 moles of hydrogen and 1 mole of oxygen create 2 moles of water

46 Amounts of Reactants and Products  How many moles of water should be formed with 8 moles of oxygen?  What happens when our measurements are not in moles?  1:2 Ratio  8 moles of oxygen is equivalent to 16 moles of water  Convert them into moles before starting using molar mass

47 Amounts of Reactants and Products

48  I have the formula: Mg + HCl → MgCl 2 + H 2  What is the mass of MgCl 2 that is formed when 20.0 g of HCl reacts with Mg? 1. Balance it  Mg + 2HCl → MgCl 2 + H 2 2. Convert into moles  (20.0 g)/(36.45 g/mol) = mol HCl 3. Use the mole ratio method  2:1 ratio → mol MgCl 2 4. Convert back into mass  mol MgCl 2 → 26.2 g Amounts of Reactants and Products

49  What mass of silver nitrate as a solution in water would need to be added to 5.0 g of sodium chloride to ensure a complete precipitation of the chloride?  AgNO 3 (aq)+ NaCl (aq)  AgCl(s) + NaNO 3 (aq) Amounts of Reactants and Products

50  What mass of barium sulfate would be produced from 10.0 g of barium chloride in the following reaction?  BaCl 2 + H 2 SO 4  BaSO 4 + HCl  Balanced: BaCl 2 + H 2 SO 4  BaSO 4 + 2HCl Amounts of Reactants and Products

51  What volume of ammonia gas would be needed to produce 40. g of ammonium nitrate in the following reaction.  NH 3 (g) + HNO 3 (aq)  NH 4 NO 3 (aq) Amounts of Reactants and Products

52  In a fermentation reaction, glucose is converted to alcohol and carbon dioxide according to the following equation. What mass of alcohol and carbon dioxide would be produced from 10. g of glucose?  C 6 H 12 O 6  C 2 H 5 OH + CO 2  Balanced: C 6 H 12 O 6  2C 2 H 5 OH + 2CO 2 Amounts of Reactants and Products

53  What is the difference between theoretical and actual yield?  What is its purpose?  Theoretical is the predicted amount of product via stoichiometry  Actual is how much product you obtained experimentally  To determine the efficiency of the procedure and/or experiment Reaction Yield

54  What is the % yield of CO 2 if you started with g of O 2 obtained g of CO 2 ?  CO + O 2  CO 2  Balanced:  2CO + O 2 → 2CO 2

55 Limiting Reagent  What is a limiting reagent?  The reactant that is depleted first during a reaction is limiting  Prevents a reaction from achieving 100% yield  Any reactants remaining after the reaction completes are called excess reagents  i.e. anything in excess cannot be a limiting reagent

56 Limiting Reagent  124 g of aluminum react with 601 g of iron (III) oxide. Calculate the mass of Al 2 O 3 formed. Al + Fe 2 O 3 → Al 2 O 3 + Fe 2Al + Fe 2 O 3 → Al 2 O 3 + 2Fe 1. Balance! 2. Determine the number of moles of each reactant. 3. Divide by the molar coefficient to determine the limiting reagent 4. Use that number of moles to determine the amount produced using proper stoichiometry

57 Limiting Reagent Table Reaction2AlFe 2 O 3 Al 2 O 3 2Fe M olar Mass (g/mol) I nitial Mass (g) I nitial Moles (mol) L imiting Reagent (mol) F inal Moles (mol) F inal Mass (g)

58 Limiting Reagent Table Reaction2AlFe 2 O 3 Al 2 O 3 2Fe Molar Mass (g/mol) Initial Mass (g) Initial Moles (mol) Limiting Reagent (mol) Final Moles (mol) Final Mass (g)

59 Limiting Reagent Table Reaction2AlFe 2 O 3 Al 2 O 3 2Fe Molar Mass (g/mol) Initial Mass (g) Initial Moles (mol) Limiting Reagent (mol) Final Moles (mol) Final Mass (g)

60 Limiting Reagent Table Reaction2AlFe 2 O 3 Al 2 O 3 2Fe Molar Mass (g/mol) Initial Mass (g) Initial Moles (mol) Limiting Reagent (mol) Final Moles (mol) Final Mass (g)

61 Limiting Reagent Table Reaction2AlFe 2 O 3 Al 2 O 3 2Fe Molar Mass (g/mol) Initial Mass (g) Initial Moles (mol) Limiting Reagent (mol) Final Moles (mol) Final Mass (g) Note: The mass of reactants equals the approximate mass of the products

62 Limiting Reagent  How much chlorine is produced in the following reaction when g of manganese (IV) oxide reacts with 48.2 g of hydrochloric acid? MnO 2 + 4HCl → MnCl 2 + Cl 2 + 2H 2 O ReactionMnO 2 4HClMnCl 2 Cl 2 2H 2 O Molar Mass (g/mol) Initial Mass (g) Initial Moles (mol) Limiting Reagent (mol) Final Moles (mol) Final Mass (g)

63 Limiting Reagent  How much chlorine is produced in the following reaction when g of manganese (IV) oxide reacts with 48.2 g of hydrochloric acid? MnO 2 + 4HCl → MnCl 2 + Cl 2 + 2H 2 O ReactionMnO 2 4HClMnCl 2 Cl 2 2H 2 O Molar Mass (g/mol) Initial Mass (g) Initial Moles (mol) Limiting Reagent (mol) Final Moles (mol) Final Mass (g)

64 Limiting Reagent  How much chlorine is produced in the following reaction when g of manganese (IV) oxide reacts with 48.2 g of hydrochloric acid? MnO 2 + 4HCl → MnCl 2 + Cl 2 + 2H 2 O ReactionMnO 2 4HClMnCl 2 Cl 2 2H 2 O Molar Mass (g/mol) Initial Mass (g) Initial Moles (mol) Limiting Reagent (mol) Final Moles (mol) Final Mass (g)

65 Limiting Reagent  How much chlorine is produced in the following reaction when g of manganese (IV) oxide reacts with 48.2 g of hydrochloric acid? MnO 2 + 4HCl → MnCl 2 + Cl 2 + 2H 2 O ReactionMnO 2 4HClMnCl 2 Cl 2 2H 2 O Molar Mass (g/mol) Initial Mass (g) Initial Moles (mol) Limiting Reagent (mol) Final Moles (mol) Final Mass (g)

66 Limiting Reagent  How much chlorine is produced in the following reaction when g of manganese (IV) oxide reacts with 48.2 g of hydrochloric acid? MnO 2 + 4HCl → MnCl 2 + Cl 2 + 2H 2 O ReactionMnO 2 4HClMnCl 2 Cl 2 2H 2 O Molar Mass (g/mol) Initial Mass (g) Initial Moles (mol) Limiting Reagent (mol) Final Moles (mol) Final Mass (g)

67 Limiting Reagent  How much chlorine is produced in the following reaction when g of manganese (IV) oxide reacts with 48.2 g of hydrochloric acid? MnO 2 + 4HCl → MnCl 2 + Cl 2 + 2H 2 O ReactionMnO 2 4HClMnCl 2 Cl 2 2H 2 O Molar Mass (g/mol) Initial Mass (g) Initial Moles (mol) Limiting Reagent (mol) Final Moles (mol) Final Mass (g)

68 Excess Reagent  If you started with g, how much manganese (IV) oxide is left in excess after the reaction is completed?  – 28.7 = 18.6 g of MgO 2 excess ReactionMnO 2 4HClMnCl 2 Cl 2 2H 2 O Molar Mass (g/mol) Initial Mass (g) Initial Moles (mol) Limiting Reagent (mol) Final Moles (mol) Final Mass (g)

69 Limiting Reagent

70 Practice Problem – pg. 93

71 Limiting Reagent Practice Problem – pg. 95

72 Limiting Reagent Practice Problem – pg. 96

73 Reaction Yield


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