Chapter 3 Formulas, Equations and Moles
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1 hapter 3 Formulas, Equations and Moles
2 Balancing hemical Equations What does mother nature lets you change and what doesn t she? O 2 2 O + O O oefficients in black (bold) are changeable (numbers appearing before a substance) Subscripts in bold red are not changeable: hanging the subscript changes the chemical that is formed or used and you can t do that! Example: The combustion of butane: O 2 O O
3 Remember: conservation of mass
4 Equivalent symbols: (means goes to ) can be replaced or is equivalent to a mathematical = sign O 2 = O O Application of the conservation of mass for carbon and hydrogen results in: O 2 = 4O O Balancing the oxygens: /2O 2 = 4O O or O 2 = 8O O Notice that we made an assumption here, that we had as much oxygen as we needed. In reality, this may not be the case. Therefore you also need to know the limiting reagent.
5 Limiting reagent: the reagent present in the least amount based on the stoichiometry of the reaction. The stoichiometry of the reaction is the ratio of reagents that are needed for the balanced reaction: O 2 = 8O O
6 Sometimes by changing the coefficients in front of the reactants, you can change the chemistry, but not control it except superficially. oefficients are changed by changing the relative amounts of reagents. In the case of the reaction we just looked at, if O 2 is the limiting reaction, the product composition may change: In a more limited amount of air (O 2 ) O 2 = O + 2 O + O 2 = 4O O /2O 2 = 4O O O 2 = 8O O Less than the required amount of oxygen will lead to incomplete combustion and a mixture of O and O 2, carbon soot, etc.
7 + 2 O = O + 2 water gas O O 2 = O O + heat water gas.mov
8 K + 2 O KO + 2 K + 2 O KO + 1/2 2 2K O 2KO + 2 Na + 2 O NaO + 2 2Na O 2NaO + 2 Li + 2 O LiO + 2 2Li O 2LiO + 2 movie MF h 6 07
9 Fermentation of glucose by yeast to yield ethyl alcohol O 6 O O 6 12 O 6 2O O Fermentation of sucrose O 11 O O O 11 4O O? What else is present when we ferment grape juice or grains? 2 O O O 4O O
10 Atomic weight = sum of the number of protons and neutrons Molecular weight = sum of the number of protons and neutrons that make up the molecule Gram atomic weight = weight in grams of 6.02x atoms Gram molecular weight = weight in grams of 6.02x molecules or formula units (used for for ionic solids, for example: Nal)
11 White willow tree A wonder drug? The classic white willow tree, Salix alba, provides more than just shade and shelter for natures animals. Salicin is the key ingredient that is isolated from the tree and can be converted into one of our most reliable and heavily used drugs, aspirin! Aspirin is used as an antipyretic-fever reduction. It is also used as an analgesic-pain reliever, and as an anti-inflammatory agent. 2 O O O O O salicyn O O 2 O 10.1
12 O 2 OO 3 Aspirin What is the gram molecular weight of aspirin = 9 8 O? 4 1 molecule has a mass of 9*(mass of )+8*(mass of ) + 4*(mass of O) Mass of atom*6.02x10 23 = 12 g/mol Mass of atom*6.02x10 23 = 1 g/mol Mass of O atom*6.02x10 23 = 16 g/mol Total = 9*12+8+4*16 = 180 g /mol
13 Aspirin is made by reacting salicylic acid ( 7 6 O 3 ) with acetic anhydride ( 4 6 O 3 ) according to the following reaction: 7 6 O O O O 2 ow many grams of acetic anhydride ( 4 6 O 3 ) are needed to react with 4.5 g of salicylic acid ( 7 6 O 3 ) and how many grams of aspirin would be formed? gmw aspirin ( 9 8 O 4 )= 180 9*12+8*1+4*16 = 180 g/mol gmw acetic anhydride ( 4 6 O 3 ) = gmw salicylic acid ( 7 6 O 3 ) = gmw acetic acid ( 3 O 2 ) = 4*12+6+3*16 = 102 g/mol 7*12+6+3*16 = 138 g/mol 2*12+4+2*16 = 60 g/mol Note that = 240; = 240 ; mass is balanced One last question! Is the reaction balanced?
14 Aspirin is made by reacting salicyclic acid ( 7 6 O 3 ) with acetic anhydride ( 4 6 O 3 ) according to the following reaction: 7 6 O O O O g/mol 102g/mol = 180g/mol 60g/mol mass of reactants = mass of products ow many grams of acetic anhydride are needed to react with 4.5 g of salicylic acid and how many grams of aspirin would be formed? 4.5g/138g/mol = mol salicylic acid Since the coefficients in front of both salicyclic acid, acetic anhydride and aspirin are 1, mol of aspirin should be theoretically formed; this corresponds to x/180g/mol = mol or x = 5.87 g 5.87 is called the theoretical yield of salicylic acid. Suppose you actually isolated 3.0 g when you ran this reaction, the actual yield would be 3.0/5.87x100=51%
15 ow many grams of acetic anhydride are needed to react with 4.5 g of salicylic acid? 7 6 O O O O g/mol 102g/mol 180g/mol 60g/mol salicylic acid + acetic anhydride aspirin + acetic acid 4.5g/138g/mol = mol 7 6 O mol 4 6 O 3 are also needed or mol x 102g/mol =3.33 g Usually an excess of one reagent is used; the other reagent is called the limiting reagent
16 Glucose: 6 12 O 6 What is the % composition of glucose (the %, and O by mass)? Suppose we had a mole of glucose, what is the molecular weight? = 6*12 = 72g/mol = 12*1 = 12g/mol O = 16*6 = 96 g/mol Total = 180 g/mol = 72/180 = % = 12/180 = % O = 96/180 = %
17 Lets examine a series of reactions to make alum: KAl(SO 4 ) O What is Alum? The most common form, potassium aluminum sulfate, or potash alum, is one form that has been used in food processing. Another, sodium aluminum sulfate, is an ingredient in commercially produced baking powder. (ave you never noticed the faint metallic taste in baking powder? It comes from the alum.) The potassium-based alum has been used to produce crisp cucumber and watermelon-rind pickles as well as maraschino cherries, where the aluminum ions strengthen the fruits' cell-wall pectins. Alum is approved by the U.S. Food and Drug Administration as a food additive, but in large quantities well, an ounce or more it is toxic to humans. Alum's antibacterial properties contribute to its traditional use as an underarm deodorant. It has been used for this purpose. Today, potassium alum is sold commercially for this purpose as a "deodorant crystal," often in a protective plastic case. Alum in powder or crystal form, or in styptic pencils, is sometimes applied to cuts to prevent or treat infection. Powdered alum is commonly cited as a home remedy for canker sores.
18 Lets examine the following series of reactions to make alum KAl(O) SO O KAl(SO 4 ) O Al + KO + 2 O KAl(O) Let s start by balancing the reaction: Al and K are balanced; ow about O and? 2O vs 4O and 3 vs 6 If we change the coefficient of KO, the K is not balanced; changing the coefficient on 2 O doesn t effect anything else Al + KO O KAl(O) Al + 2KO O 2 KAl(O) Now suppose we add dilute sulfuric acid to this compound KAl(O) SO 4 KAl(SO 4 ) O KAl(O) SO 4 KAl(SO 4 ) O Remember Alum is: KAl(SO 4 ) O
19 2Al + 2KO O 2 KAl(O) KAl(O) SO O KAl(SO 4 ) O 2Al + 2KO O 2 KAl(O) step 1 2 KAl(O) SO O 2KAl(SO 4 ) O step 2 Overall the reaction is: 2Al +2KO O SO 4 = 2KAl(SO 4 ) O +3 2
20 This equations represents the overall stoichiometry of the reaction. It was generated by a process of mass balance, so that we can use this reaction to calculate the overall yield of the reaction. Suppose we started with 0.5 g of Al and used an excess of all the remaining reagents. What would be the theoretical yield of alum (KAl(SO 4 ) O)? 2Al +2KO O SO 4 = 2KAl(SO 4 ) O +3 2 Al = 27 g /mol KAl(SO 4 ) O = 474 g/mol 2 SO 4 = 98g/mol 0.5g/27g/mol = mol Al ow many mol of Al is required to produce 2 mol of alum? 2 mol Al produces 2 mol of alum ow many mol of alum is produced from mol Al? mol alum; What is the theoretical yield? 474g/mol x mol = 8.77 g alum
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