8.01 Quantities in Chemical Reactions

Size: px

Start display at page:

Download "8.01 Quantities in Chemical Reactions"

Peregrine King
5 years ago
Views:

1 8.01 Quantities in hemical Reactions Mass alculations Using hemical Reactions Dr. Fred O. Garces hemistry 152 Miramar ollege 1 Mass Relationship to hem Rxn

2 A Vol ( L) The Mole Revisited ( ) particle (atomic) 3 ( ) particle (atomic) (l) 1 B Solid Solid 2 Balance equation Moles ## ## ## ## Moles A Stoic. coefficient. B D 4 W Pressure (atm) Temperature (K) Volume (L) B Gas R (.0821 atm L ) (g) Gas R (.0821 atm L ) 2 Pressure ( atm ) Temperature ( K) Volume ( L) 7 Y D 4 9 X Mass Relationship to hem Rxn

from balance equation: 1 molc N 2 3 molc H 2 2 molc NH 3 (Implied information)- 1 mole 3 mole 2 mole How much NH 3 will be produce if 84.

3 Quantitative Stoichiometry Synthesis: hemical reactions are similar to cooking. The proper amount of ingredients and the conditions must be known to achieve the desired product. onsider: N 2 + 3H 2 D 2NH 3 Info. from balance equation: 1 molc N 2 3 molc H 2 2 molc NH 3 (Implied information)- 1 mole 3 mole 2 mole How much NH 3 will be produce if 84.0 g of N 2 reacts with excess H 2? Note that: Mass-Mass basis would give wrong answer. N 2 + 3H 2 D 2NH g 2 (84g)? No... never mass:mass basis Strategy: alculate the number of moles of N 2 in 84.0 g. Relate moles of N 2 to moles of NH 3. onvert moles of NH 3 to grams of NH 3. 3 Mass Relationship to hem Rxn

Meaning of a Balanced Equation Meaning of balance equation: onsider: 1N 2 + 3 H 2 D 2NH 3 1 molc

100 molc' 300 molc' 200 molc' 1 trillion 3 trillion 2 trillion Or by bulk 30 H 2 10 N 2 20 NH 3 1

4 Meaning of a Balanced Equation Meaning of balance equation: onsider: 1N H 2 D 2NH 3 1 molc N 2 3 molc H 2 2 molc NH 3 By individual molecules 3 H 2 1N 2 2NH 3 or 10 molc' 30 molc' 20 molc' 100 molc' 300 molc' 200 molc' 1 trillion 3 trillion 2 trillion Or by bulk 30 H 2 10 N 2 20 NH 3 1 ( ) 3 ( ) 2 ( ) 1 mole 3 mole 2 mole 4 Mass Relationship to hem Rxn

5 Bike Analogy: Only by Parts onsider the following Analogy: 2 Wheels + 1 Body + 1 Handle bar + 1 Gear hain How many bikes can be produce given, 8 wheels, 5 bodies, 6 handle bar and 5 gear chain. 5 Mass Relationship to hem Rxn

6 Stoichiometry: A ar Analogy onsider the following Analogy: 1 ar = 4 Wheels + 1 Engine + 1 Body + 2 Bumper Given 20 wheels, How many cars can be produced? g 5 ars: Dimensional Analysis g 20 W [1ar/4W] = 5 ars 6 Mass Relationship to hem Rxn

7 Stoichiometry: A ar Analogy onsider the following Analogy: 1 ar = 4 Wheels + 1 Engine + 1 Body + 2 Bumper Given 20 wheels, How many cars can be produced? g 5 ars: Dimensional Analysis - 20 W [1ar/4W] = 5 ars Now suppose you are given the mass of the parts. Parts of ar g How many cars? 6560 lb W? lb E? lb Bd? 4860 lb Bu? an t determine number of cars. Need more information. 7 Mass Relationship to hem Rxn

8 ar Analogy: Predicting Amount Produced To solve problem, need to know mass of each part. Parts of ar g Weight of parts Number of parts Theoretical ars 6560 lb W 20 lb/w 328 W 82 ars lb E 600 lb/e 163 E 163 ars lb Bd 300 lb/b 82 Bd 82 ars 4860 lb Bu 30 lb/bu 162 Bu 81 ars Based on the limiting parts, in this case by 162 Bumpers, which can theoretically produce only 81 cars, these many parts limit the projected number of cars, 81 cars, produced with 4W, 82 E and 1Bd left over. The mass of the parts cannot be used to determine the number of cars that can be produce since the mass give no indication of how many parts are in stock. Only after converting mass to parts can this be done. 8 Mass Relationship to hem Rxn

.. L never mass:mass basis Vol ( ) (6.02 10 23 ) particle (atomic) (6.

9 ...on a chemical basis When it comes to chemical reaction; The amount of product can t be determine directly from the mass of the reactant. The mass:mass strategy should never be used because the number of N 2 molecules in 84.0 g is not known yet. 1N 2 + H 2 D 2NH g 2(84g)? No... L never mass:mass basis Vol ( ) ( ) particle (atomic) ( ) particle (atomic) (l) Pressure (atm) Temperature (K) Volume (L) Solid Gas Balance equation Moles A ## ## ## ## Moles B Stoic. coefficient. R (.0821 atm L ) Solid (g) Gas R (.0821 atm L ) Pressure ( atm ) Temperature ( K) Volume ( L) 9 Mass Relationship to hem Rxn

10 ...on a chemical basis When it comes to chemical reaction; The amount of product can t be determine directly given the mass of the reactant. The mass:mass basis can t be determine because the number of N 2 molecules in 84.0 g is not known yet. 1N 2 + H 2 D 2NH g 2(84g)? Strategy: No... never mass:mass basis i) Mass N 2 g Moles N 2 using Molar mass H 2 ii) Moles N 2 g Moles NH 3 using balanced equation Solution: The mass must first be converted to amount (moles) i. onvert 84.0 g N 2 to mol of N 2 (M Wt. of N 2 ; 28.0 g/mol) 84.0 g 1 mol N 2 = 3 mol 28.0 g ii. onvert mol N 2 to mol NH 3 (Balance Equation) 3 moln 2 2 mol NH 3 = 6 mol 1 mol N 2 iii) Moles NH 3 g Mass NH 3 Using Molar mass NH 3 iii. onvert mol NH 3 to mass NH 3 (M Wt. of NH 3 ; 17.0 g/mol) 6 mol 17.0 g NH 3 = g NH 3 10 Mass Relationship to hem Rxn 1 mol NH 3

11 Mapping out Solving Technique Vol ( L) A B Pressure (atm) Temperature (K) Volume (L) ( ) particle (atomic) Solid Gas Balance equation Moles A ## ## ## ## Moles Stoic. coefficient. B D 4 R (.0821 atm L ) 3 ( ) particle (atomic) Solid (g) Gas R (.0821 atm L ) 1 (l) 2 Pressure ( atm ) Temperature ( K) Volume ( L) 11 Mass Relationship to hem Rxn

12 The Same Problem with Different Faces Type of Stoichiometry Questions onsider: aa + bb g c + dd Questions: Products: Given X grams of A, how much or D is produced Given X grams of B, how much or D is produced A Given X grams D of, how much D is produced 1 Given X grams of D, how much is produced 4 B Reactants: Producing X grams of, how much A or B is required Producing X E grams of D, how much A or B is required 2 o-product: Producing X grams of, how much D is also produced Producing X grams of D, how much is also produced F 3 Also problem can ask what is the limiting and excess reagents? Mass Relationship to hem Rxn

Stoichiometry alculation Activity Example A Vol ( L) B (6.02 10 23 ) particle (atomic) Solid Balance equation Moles A ## ## ## ## Moles Stoic. coefficient. B (6.02 10 23 ) particle (atomic) D 4 Solid (l) Pressure (atm) Gas (g) Gas Pressure ( atm ) Example: onsider the following example.

13 Stoichiometry alculation Activity Example A Vol ( L) B ( ) particle (atomic) Solid Balance equation Moles A ## ## ## ## Moles Stoic. coefficient. B ( ) particle (atomic) D 4 Solid (l) Pressure (atm) Gas (g) Gas Pressure ( atm ) Example: onsider the following example. Temperature (K) R (.0821 atm L R (.0821 atm L ) ) Temperature ( K) 2 KMnO Volume (L) Hl g 5 l Kl + 2Mnl 2 Volume + 8H( L) 2 O If 35.0 g of KMnO 4 is mixed with 45.0 g Hl, how much l 2 is theoretically produced? If only 10.3 g of l 2 was collected in the synthesis, what is the % yield? (l 2 MM=70.9 g.mol) i) 35.0 g KMnO 4 (MM=158.0 g/mol) ii) 45.0 g Hl (MM=36.5 g/mol) 13 Mass Relationship to hem Rxn

14 Stoichiometry alculation Activity Example A Vol ( L) Pressure (atm) ( ) particle (atomic) Solid Gas Balance equation Moles A ## ## ## ## Moles B Stoic. coefficient. Solid (g) Gas Temperature (K) R (.0821 atm L R (.0821 atm L ) ) 158g/mol Volume (L) 36.5 g/mol 70.9 g/mol B ( ) particle (atomic) D (l) Pressure ( atm ) Example: How much l 2 theoretically can be produced? 3 Temperature ( K) Volume ( L) 2 KMnO Hl g 5 l Kl + 2Mnl 2 + 8H 2 O 35.0 g 45.0 g mass l 2 14 Mass Relationship to hem Rxn

15 Stoichiometry alculation Activity Example A Vol ( L) ( ) particle (atomic) Solid ( ) particle (atomic) Solid Balance equation Moles A ## ## ## ## Moles D B 4 Stoic. coefficient. onsider Pressure the (atm) following Gas example. (g) Gas 158g/mol 36.5 g/mol 70.9 g/mol B 2 1 (l) Pressure ( atm ) Temperature (K) R (.0821 atm L R (.0821 atm L ) ) Temperature ( K) 2 KMnO Volume (L) Volume ( L ) Hl g 5 l Kl + 2Mnl 2 + 8H 2 O One line calculation: 3 Mass KMnO 4 Moles KMnO 4 Moles l 2 Mass l 2 1 mol KMnO 35.0g KMnO 4 5 mol l g l 2 = 39.3 g l g KMnO 4 2 mol KMnO 4 1 moles mol l 2 Mass Hl Moles Hl Moles l 2 Mass l 2 1 mol Hl 45.0g Hl 36.5 g Hl 5 mol l g l 2 2 = 27.3 g l 2 16 mol Hl 1 moles mol l 2 15 Mass Relationship to hem Rxn

16 Stoichiometry alculations: % Yield The maximum mass of a product that can be obtained from a given amount of reactant is the theoretical yield. The mass of a product that is actually recovered is called the actual yield. The actual yield (from experiment) expressed as a percentage of the theoretical yield (from calculation) is called the percent yield. % Yield = Actual Yield ar Production Revisited: Theod Yield Remember our ar analogy: Suppose a car manufacturing plant is capable of producing 123 cars per day. What is the efficiency of the company ( percent yield) if only 63 cars are produced each day? Actual Yield - 63 cars Theoretical Yield cars % Yield = Actual Yield = = 51% X 100 Theod Yield 123 Aspirin Production: Suppose a theoretical yield of aspirin is calculated to be g but only gram of aspirin is recovered during the experiment, what is the % yield? Actual Yield g Aspirin X 100 Theoretical Yield g Aspirin % Yield = Actual Yield X 100 = = 76.6 % Theod Yield Mass Relationship to hem Rxn

17 Stoichiometry alculation: % Yield Expanding the problem- 2 KMnO Hl g 5 l Kl + 2Mnl 2 + 8H 2 O If only 10.3 g of l 2 was collected in the synthesis, what is the % yield? Mass Hl Moles Hl Moles l 2 Mass l 2 Theoretical Yield Theoretical yield is the calculated amount form based on 100% conversion 1 mol Hl 45.0g Hl 36.5 g Hl 5 mol l g l 2 2 = 27.3 g l 2 16 mol Hl 1 moles mol l 2 Percent Yield: % Yield = Actual Theoretical 100 % Yield = 10.3 g l g l = 37.7 % 17 Mass Relationship to hem Rxn

18 Stoichiometry alculations (Another example) Use the balance equation shown to answer the questions below. 4 KOH (s) + 3 O 2 (g) g 4 KO 2 (s) + 2 H 2 O (l) A Given 392. g of KOH how many moles does this represent for KOH? B What is the mass (in grams) of 5.00 moles of oxygen If 392 g of KOH reacts with 5.00 moles of O2, what mass of KO2 is produce? D How many molecules of water is produce from this reaction? G What volume of water is produce from the reaction in part D? E If only 284 g of KO 2 is produce, what the % yield of the reaction? F How many oxygen atoms in 284 g of KO 2? 18 Mass Relationship to hem Rxn

19 A B Stoichiometry alculations (Another example) Vol ( L) Pressure (atm) ( ) particle (atomic) Solid Gas Balance equation Moles A ## ## ## ## Moles B Stoic. coefficient. ( ) particle (atomic) Solid D 4 (g) Gas Use the balance equation shown to answer the questions below. (l) Pressure ( atm ) Molar mass Temperature (g/mol) (K) R (.0821 atm L R ( atm L ) ) Temperature ( K) Volume (L) Volume ( L) 3 4 _4_ KOH (s) + 3_3_ O 2 (g) g 4 _4_ KO 2 (s) + 2 _2_ H 2 O (l) 2 1 A Given 392. g of KOH how many moles does this represent for KOH? B g D 392g%%KOH% % 1mol%KOH 56.11g%KOH %=%6.986%mol%=%7.00%mol%%%KOH 19 Mass Relationship to hem Rxn

20 A B Stoichiometry alculations (Another example) Vol ( L) Pressure (atm) ( ) particle (atomic) Solid D Gas Balance equation Moles A ## ## ## ## Moles B Stoic. coefficient. ( ) particle (atomic) Solid (g) Gas Use the balance equation shown to answer the questions below. (l) Pressure ( atm ) Molar mass Temperature (g/mol) (K) R (.0821 atm L R (.0821 atm L ) ) Temperature ( K) Volume (L) Volume ( L) 3 4 _4_KOH (s) + 3_3_O 2 (g) g 4 _4_KO 2 (s) + 2 _2_H 2 O (l) 4 1 B What is the mass (in grams) of 5.00 moles of oxygen D g B 5.00 mol O g O 2 1mol O 2 =160.0 g O 2 = 160. g O 2 20 Mass Relationship to hem Rxn

21 A Stoichiometry alculations (Another example) Vol ( L) B Pressure (atm) ( ) particle (atomic) Solid D Gas Balance equation Moles ## ## ## ## Moles A Stoic. coefficient. B ( ) particle (atomic) Solid (g) Gas Use the balance equation shown to answer the questions below. (l) Pressure ( atm ) Molar mass Temperature (g/mol) (K) R (.0821 atm L ) Temperature ( K) R (.0821 atm L ) Volume (L) Volume ( L) 3 4 _4_KOH (s) + 3_3_O 2 (g) g 4 _4_KO 2 (s) + 2 _2_H 2 O (l) If 392 g of KOH reacts with 5.00 moles of O2, what mass of KO2 is produce? KOH: B g D g 4 O 2 : D g 4 4 g 2 KOH: 7.00 mol KOH 4mol KO 2 4mol KOH = 7.00 mol KO 2 Mass KO 2 : O 2 : 5.00 mol O 2 4 mol KO 2 3 mol O 2 = 6.66 mol KO 2 21 Mass Relationship to hem Rxn 6.66 mol KO g KO 2 1mol = g = 474 g

22 Stoichiometry alculations (Another example) A B Vol ( L) Pressure (atm) ( ) particle (atomic) Solid D Gas Balance equation Moles A ## ## ## ## Moles B Stoic. coefficient. ( ) particle (atomic) Solid (g) Gas Use the balance equation shown to answer the questions below. (l) Pressure ( atm ) Molar mass Temperature (g/mol) (K) R (.0821 atm L R (.0821 atm L ) ) Temperature ( K) Volume (L) Volume ( L) 3 4 _4_KOH (s) + 3_3_O 2 (g) g 4 _4_KO 2 (s) + 2 _2_H 2 O (l) D How many molecules of water is produce from this reaction? O 2 : D g 4 g mol O 2 2 mol H O molecules H 2 O = molecules H 2 O 3 mol O 2 1 mol H 2 O 22 Mass Relationship to hem Rxn

23 Stoichiometry alculations (Another example) A Vol ( L) B Pressure (atm) ( ) particle (atomic) Solid D Gas Balance equation Moles A ## ## ## ## Moles B Stoic. coefficient. ( ) particle (atomic) Solid (g) Gas Use the balance equation shown to answer the questions below. (l) Pressure ( atm ) Molar mass Temperature (g/mol) (K) R (.0821 atm L R (.0821 atm L ) ) Temperature ( K) Volume (L) Volume ( L) 3 4 _4_KOH (s) + 3_3_O 2 (g) g 4 _4_KO 2 (s) + 2 _2_H 2 O (l) E If only 284 g of KO 2 is produce, what the % yield of the reaction? % Yield = Actual Theoretical 100 % Yield = 284 g KO g KO = 59.9 % 23 Mass Relationship to hem Rxn

Stoichiometry alculations (Another example) Use the balance equation shown to answer the questions below. Molar mass (g/mol) 56.11 32.00 71.10 18.

24 Stoichiometry alculations (Another example) Use the balance equation shown to answer the questions below. Molar mass (g/mol) _4_KOH (s) + 3_3_O 2 (g) g 4 _4_KO 2 (s) + 2 _2_H 2 O (l) F How many oxygen atoms in 284 g of KO 2? B g D g Y g X 284 g KO 2 1 mol KO g KO 2 2 mol O 1mol KO atoms O 1mol O = atoms O 24 Mass Relationship to hem Rxn

25 A Mapping out Solving Technique 3 1 B 2 D W B 2 7 Y D 4 9 X Mass Relationship to hem Rxn

26 The History of 2-(acetyloxy)-benzoic acid or salicylic acid 4B Hippocrates- The precursor of aspirin was first documented by Hippocrates. He wrote that the tea from willow bark had medicinal use. Edmund Stone - Discovered that the yellow crystals from willow bark, (Salix abla, Salicin) were the active ingredient to cure headaches. linical test of Salicin showed upset stomach and nausea side effects. Salicin + NaOH yields salt which was more mild but was still acidic. Felix Hoffman - In the next evolution of the drug known as aspirin, Hoffman explored structurally modifying Salicin to help his father s upset stomach side effect. In the discovery of the synthesis, the drug could not be patented, but the synthetic method was. Hoffman worked for the German company Bayer. Toxicity of aspirin show lethal levels of g at one time. It is recommended that patients take about mg /4 hours. Some side effect include nausea, abdominal pain, heartburn, ringing in the ears, increase blood clotting- time and ulcers. Benefits: Nonaddicting, antipyretic (fever), analgesic (pain), anti-inflammation. Aspirin is produced by the reaction of salicylic acid and acetic anhydride. 7 H 6 O 3(s) + 4 H 6 O 3 (l) D 9 H 8 O 4 (s) + H 3 O 2 H (aq) salicylic acid acetic anhydride acetylsalicylic acid acetic acid O OH OH + H 3 O O O H 3 O OH O O H 3 + H3 O OH 26 Mass Relationship to hem Rxn

27 Why you get headaches Prostaglandin - hormone produced in your tissues and body fluids. Responsible for: Sensation of pain Fever Inflammation Swelling Structural representation of these chemicals will be explain in Organic hemistry chapter later in the semester Acetylsalicylic acid (aspirin) Anti-inflammatory Analgesic Antipyretic O OH O H3 Tylenol Acetaminophen...hospitals recommend most. O NH H3 Advil ibuprofen H3 O H OH O H2 H 27 Mass Relationship to hem Rxn OH H3 H3

28 Stoichiometry alculations Aspirin is produced by the reaction of salicylic acid and acetic anhydride. 7 H 6 O 3(s) + 4 H 6 O 3 (l) g 9 H 8 O 4 (s) + H 3 O 2 H (aq) salicylic acid acetic anhydride acetylsalicylic acid acetic acid aspirin A B D 1. Given X grams of A (salicylic acid) and B, how many moles of A and B is this? Strategy: grams-a g moles-a One step [onversion factor - MM ] 2. Given X grams of A (salicylic acid), how many moles of is produced? Strategy: grams-a g moles-a g moles-b Two steps [onversion factor - MM, Bal equation] 3. Given X grams of A (salicylic acid), what mass of is produced (in grams)? Strategy: grams-a g moles-a g moles-b g grams-b Three step [onversion factor - MM-A, Bal equation, MM-B ] 28 Mass Relationship to hem Rxn

29 Stoichiometry alculations Aspirin is produced by the reaction of salicylic acid and acetic anhydride. 7 H 6 O 3(s) + 4 H 6 O 3 (l) g 9 H 8 O 4 (s) + H 3 O 2 H (aq) salicylic acid acetic anhydride acetylsalicylic acid acetic acid aspirin A (138 g/mol) B (102 g/mol) ( g/mol) D 1. Given 100 grams of A (salicylic acid), how many moles of A (salicylic acid) is this? Strategy: grams-a g moles-a One step [onversion factor - MM ] 2. Given 100 grams of A (salicylic acid), how many moles of (aspirin) is produced? Strategy: grams-a g moles-a g moles- Two steps [onversion factor - MM, Bal equation] 3. Given 100 grams of A (salicylic acid), what mass of is produced (in grams)? Strategy: grams-a g moles-a g moles- g grams- Three step [onversion factor - MM-A, Bal equation, MM- ] 29 Mass Relationship to hem Rxn

30 1. Stoichiometry alculations: Mass (Reactants) to Moles (Reactants) Aspirin is produced by the reaction of salicylic acid and acetic anhydride. 7 H 6 O 3(s) + 4 H 6 O 3 (l) D 9 H 8 O 4 (s) + H 3 O 2 H (aq) salicylic acid acetic anhydride aspirin acetic acid If you mix 100. g of each of the salicylic acid and acetic acid (reactants), how many moles of each reactant (salicylic acid and acetic acid) does this represent? Strategy: Mass to Moles M Wt. 7 H 6 O 3 = 7 (12.0) + 6 (1.0) + 3 (16.0) = M Wt. 4 H 6 O 3 = 4 (12.0) + 6 (1.0) + 3 (16.0) = g mol g mol Mass 7 H 6 O 3 Moles 7 H 6 O 3 : 1 mol 100g 7 H 6 O 3 = 0.725mol 7 H 6 O g 7 H 6 O 3 Mass 4 H 6 O 3 Moles 4 H 6 O 3 : 1 mol 100g 4 H 6 O 3 = mol 4 H 6 O g 4 H 6 O 3 Extra Q: How many moles of product is produced from each. 30 Mass Relationship to hem Rxn

31 2. Stoichiometry alculations: Mass (Reactants) to Mole (Products) Aspirin is produced by the reaction of salicylic acid and acetic anhydride. 7 H 6 O 3(s) + 4 H 6 O 3 (l) D 9 H 8 O 4 (s) + H 3 O 2 H (aq) salicylic acid acetic anhydride aspirin acetic acid If you mix 100. g of salicylic acid (reactants), how many moles of aspirin can be theoretically obtain? From Previous problem MWt: 7 H 6 O 3 = g/mol 100g 7 H 6 O 3 is mol i) Mass 7 H 6 O 3 Moles 7 H 6 O 3 : 100g 7 H 6 O 3 1 mol g 7 H 6 O 3 = mol 7 H 6 O 3 ii) Moles 7 H 6 O 3 Moles 9 H 8 O 4 : mol 7 H 6 O 3 1 moles 9 H 8 O 4 1 moles 7 H 6 O 3 = mol 9 H 8 O 4 One - line alculation Mass 7 H 6 O 3 Moles 7 H 6 O 3 Moles 9 H 8 O 4 100g 7 H 6 O 3 1 mol 7 H 6 O g 7 H 6 O 3 1 moles 9 H 8 O 4 1 moles 7 H 6 O 3 = moles aspirin = mol 9 H 8 O 4 31 Mass Relationship to hem Rxn

Needed information MWt: 7 H 6 O 3 = 138.0 g/mol MWt: 9 H 8 O 4 = 180.0 g/mol One - line alculation i) Mass 7 H 6 O 3 Moles 7 H 6 O 3 : 100g 7 H 6 O 3 32 Mass Relationship to hem Rxn 1 mol 138.

32 3. Stoichiometry alculations: Mass (Reactants) to Mass (Products) Aspirin is produced by the reaction of salicylic acid and acetic anhydride. 7 H 6 O 3(s) + 4 H 6 O 3 (l) D 9 H 8 O 4 (s) + H 3 O 2 H (aq) salicylic acid acetic anhydride aspirin acetic acid If you mix 100. g of salicylic acid (reactants), what mass of aspirin can be produced? Needed information MWt: 7 H 6 O 3 = g/mol MWt: 9 H 8 O 4 = g/mol One - line alculation i) Mass 7 H 6 O 3 Moles 7 H 6 O 3 : 100g 7 H 6 O 3 32 Mass Relationship to hem Rxn 1 mol g 7 H 6 O 3 = mol 7 H 6 O 3 ii) Moles 7 H 6 O 3 Moles 9 H 8 O 4 : mol 7 H 6 O 3 1 moles 9 H 8 O 4 1 moles 7 H 6 O 3 = mol 9 H 8 O 4 iii) Moles 9 H 8 O 4 Mass 9 H 8 O 4 : mol 9 H 8 O g 9H 8 O 4 1 moles 9 H 8 O 4 = g 9 H 8 O 4 Mass 7 H 6 O 3 Moles 7 H 6 O 3 Moles 9 H 8 O 4 Mass 9 H 8 O 4 100g 7 H 6 O 3 1 mol 7 H 6 O g 7 H 6 O 3 1 moles 9 H 8 O 4 1 moles 7 H 6 O g 9 H 8 O 4 1 moles 9 H 8 O 4 = mass aspirin = g 9 H 8 O 4

Stoichiometry alculations: Mass (Reactants) to Molecules (Products) Aspirin is produced by the reaction of salicylic acid and acetic anhydride.

33 Stoichiometry alculations: Mass (Reactants) to Molecules (Products) Aspirin is produced by the reaction of salicylic acid and acetic anhydride. 7 H 6 O 3(s) + 4 H 6 O 3 (l) D 9 H 8 O 4 (s) + H 3 O 2 H (aq) salicylic acid acetic anhydride aspirin acetic acid If you mix 100. g of each of the reactants, how many molecules of aspirin are obtained? i) Mass 7 H 6 O 3 Moles 7 H 6 O 3 : 1 mol 100g 7 H 6 O 3 = mol 7 H 6 O g 7 H 6 O 3 ii) Moles 7 H 6 O 3 Moles 9 H 8 O 4 : mol 7 H 6 O 3 1 moles 9H 8 O 4 1 moles 7 H 6 O 3 = mol 9 H 8 O 4 iii) Moles 9 H 8 O 4 Molecules 9 H 8 O 4 : mol 9 H 8 O H 8 O 4 1 moles 9 H 8 O 4 = Molecules 9 H 8 O 4 One - line alculation Mass 7 H 6 O 3 Moles 7 H 6 O 3 Moles 9 H 8 O 4 Mass 9 H 8 O 4 100g 7 H 6 O 3 1 mol 7 H 6 O g 7 H 6 O 3 1 moles 9 H 8 O 4 1 moles 7 H 6 O H 8 O 4 1 moles 9 H 8 O 4 = Molecules aspirin = Molecules 9 H 8 O 4 33 Mass Relationship to hem Rxn

34 Stoichiometry alculations (Litmiting Reagen) Ethanol is produced by the reaction of ethanol from syn-gas. Given 100 g O and 100 g H 2 O, How much ethanol is produced? How many molecules O 2? 6 O (g) + 3 H 2 O (l) g 1 2 H 6 O (l) + 4 O 2 (g) ethanol A (28 g/mol) B (18 g/mol) (46.0 g/mol) D (44.0 g/mol) 1a. Given 100 grams of A (O), how many moles of A (O) is this? 1b. Given 100 grams of B (H 2 O), how many moles of B (H 2 O) is this? Strategy: grams-a g moles-a, grams-b g moles-b One step [onversion factor - MM ] 2a. Given 100 grams of A (O), how many moles of (ethanol) is produced? 2b. Given 100 grams of B (H 2 O), how many moles of (ethanol) is produced? Strategy: grams-a g moles-a g moles-, grams-b g moles-b g moles- Two steps [onversion factor - MM, Bal equation] 3. Given 100 grams of A or B (O, H 2 O) ), what mass of is produced (in grams)? Strategy: grams-a or B g moles-a or Bg moles- g grams- Three step [onversion factor - MM-A, Bal equation, MM- ] 34 Mass Relationship to hem Rxn

A text book example: Stoichiometry alculations: Mass (Reactants) to Mass (Products) & Limiting Reagent Methanol, H 5 OH, an excellent fuel, can be made by the reaction of O and hydrogen.

35 A text book example: Stoichiometry alculations: Mass (Reactants) to Mass (Products) & Limiting Reagent Methanol, H 5 OH, an excellent fuel, can be made by the reaction of O and hydrogen. O (g) + H 2(g) D H 5 OH (l) carbon Hydrogen methanol monoxide Suppose 356 g of O are mixed with 65.0 g of H 2. Which is the limiting reactant? What is the maximum mass of methanol that can be formed? What mass of the excess reactant remains after the limiting reactant has been consumed? 35 Mass Relationship to hem Rxn

36 Percent Yield Revisited The maximum mass of a product that can be obtained from a given amount of reactant is the theoretical yield. The mass of a product that is actually recovered is called the actual yield. The actual yield (from experiment) expressed as a percentage of the theoretical yield (from calculation) is called the percent yield. % Yield = Actual Yield ar Production Revisited: Theod Yield Remember our ar analogy: Suppose a car manufacturing plant is capable of producing 123 cars per day. What is the efficiency of the company ( percent yield) if only 63 cars are produced each day? Actual Yield - 63 cars Theoretical Yield cars % Yield = Actual Yield = = 51% X 100 Theod Yield 123 Aspirin Production Revisited: Suppose in our aspirin analogy, only gram of aspirin is recovered during the experiment, what is the % yield? Actual Yield g Aspirin X 100 Theoretical Yield g Aspirin % Yield = Actual Yield X 100 = = 76.6 % Theod Yield Mass Relationship to hem Rxn

Summary The key to success in stoichiometry is the stoic map and dimensional analysis Vol ( L) A B Pressure (atm) Temperature (K) Volume (L) (6.

37 Summary The key to success in stoichiometry is the stoic map and dimensional analysis Vol ( L) A B Pressure (atm) Temperature (K) Volume (L) ( ) particle (atomic) Solid Gas Balance equation Moles A ## ## ## ## Moles Stoic. coefficient. B D 4 R (.0821 atm L ) 3 ( ) particle (atomic) Solid (g) Gas R (.0821 atm L ) 1 (l) 2 Pressure ( atm ) Temperature ( K) Volume ( L) 37 Mass Relationship to hem Rxn

Limiting Reagent Synthesis of Aspirin Thomas M. Moffett Jr., SUNY Plattsburgh, 2007.

Limiting Reagent Synthesis of Aspirin Thomas M. Moffett Jr., SUNY Plattsburgh, 007. Aspirin (acetylsalicylic acid) is the most common medicinal drug in use today. Aspirin is an analgesic (pain reliever),