Download From Tuesday

Noteworthy
• Book sections not on the exam: 2.8,2.9,3.9,3.10
but these will be covered later.
• Problems in the text worth reading( all of the problems in
the text are worth your time to read, the following are just
more worthy):
– 2.92,2.93,2.104,2.110,2.111
– In Chapter 3, it would be a good idea to read all problems 3.58 to
3.77
– 3.120,3.121,3.130
• Tutoring available:
– Chemistry:Room 221 SB1 most days 9-4
– Campus Tutor Program: www.ess.pdx.edu/iasc/TUT_SCHD.HTM
From Tuesday
• Chemical formulas
– mass % composition
– empirical (simplest) vs molecular formula
• you can always determine the empirical formula from the analysis
• the molecular formula requires knowledge of the gmw
– Remember, since the composition of the compound is intrinsic, any
complete set of mass data will yield the formula.
• The chemical equation
– descriptive
– balanced
• Stoichiometry
– all mass/mole relationships produced by the balanced equation are
direct
– complete treatment
• from the mass/moles of one species determine the quantities in terms
of mass and moles of all the others
– focused
• from one get one
Use of general algorithms
• All problems involving the comparison of “conditions(states)” for a
pair of directly related variables fit the following equation:
(__)
* ____  ____
(__)
• The solution of a problem “simply” involves the correct “insertion” of
the data into the equation
1. Place a complete label for the " missing" quantity t o the right
of the equal sign
(__)
* ____  gs of C
(__)
2. Place the known valu e (w dimensions ) for the " incomplete state"
to the right of the "*"
(__)
* 0.67 moles of O  gs of C
(__)
3. In the numerator position, place the value of the variable from the fully defined state
" identical" to the missing variable
(48g of C)
* 0.67 moles of O  gs of C
( )
4. In the denominato r position, place the remaining variable from the fully defined state,
being certain th at its dimensions match thos e written t o the right of the "*"
( 48g of C)
* 0.67 moles of O  gs of C  16g
(2 moles of O)
For the following chemical reaction, provide a complete treatment for the reaction of 9.34g
of NH3
NH3 + CO2 => C3H6 + NO3
1. Balance the equation, if needed
2NH3 + 3CO2 => C3H6 + 2NO3
2. It is useful to have a method of organizing the needed info. One method is to write the
gfws above all of the species in the equation and the reaction masses (stoichiometric
masses) which are the products of the coefficients and the gfws underneath the species.
17
44
42
62
2NH3 + 3CO2 => C3H6 + 2NO3
34
132
42
124
The stoichiometric masses must obey the mass conservation law. This will verify the
validity of the data. For a given problem, you generally will not need all of this.
` 34+132=166=42 + 124
Be certain to convert the “given” into moles
9.34/17.0=.549moles
3. The solution can then be presented in a tabular fashion
Moles
Grams
CO2: (3/2)*0.549=0.823
*44 = 36.2
C3H6:(1/2)*0.549=0.274
*42 = 11.5
NO3:(2/2)*0.549=0.549
*62 = 34.0
The final masses must follow mass conservation
9.34+36.2=45.5 11.5+34=45.5
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Focused Questions
Most questions involving balanced
chemical equations are binary and have the
form:
How many _____ of _____ would be
_______ (by the reaction of ____ ____ of
____) or (if ____ _____ of ____ are
_____)?
The above should look very familiar.
P4 + 16H2O => 4H3PO4 + 10H2
How many moles of H3PO4 would be
produced by the reaction of 94.0g of H2O?
Reality 101:% yield and the limiting reagent
• We are seldom able to perform a reaction in
the precise fashion shown by the balanced
chemical equation.
• Many reactions stop short of completion
and the quantitative isolation of products is
often difficult.
• It is impossible and often not desirable to
mix reactants in perfect stoichiometric
ratios.
% yield
• % yield is defined as:
– (actual yield/theoretical yield)*100=%yield
– another 3 variable equation, though the
theoretical yield is actually determined from
basic stoichiometry
– If the reaction of 12.3 of C2H4 produces 35.6g
of CO2, what is the % yield?
C2H4 + 3O2 => 2CO2 + 2H2O
Limiting Reagent
• If the reagents are not present in exact
stoichiometric ratios, then one will limit
how far the reaction proceeds.
• There are a number of algorithms for
identifying the limiting reagent. What you
are doing is comparing the actual ratio with
the stoichiometric one.
C2H4 + 3O2 => 2CO2 + 2H2O
If 23.4g of C2H4 are reacted with 74g of O2, what is the
limiting reagent?
•
•
•
•
Method 1-mole based
The question is really, how does a mixture of 23.4g to 74g compare with a
mixture of 1 mole to 3 moles.
– convert to like dimensions:
– 23.4/28=0.84moles
– 74/32=2.31moles
– 2.31/0.84 vs 3/1 ?
– If you’re unsure, divide 2.31 by 0.84(2.75) or multiply 0.84 by 3(2.52)
– What’s the limiting reagent?
Method 2-mass based
How many grams of O2 will react with 23.4g of C2H
– Using standard approaches
– (96/28)*23.4=80.2g of O2
– the reaction mixture has 74g. Which is limiting?
Putting it all together
The reaction of 21.4g of PH3 with 17.2g of water yields
34.2g of P2O5. What is the % yield.
PH3 + H2O => P2O5 + H2
What are the steps ( and in what sequence) needed to solve
this question.
1.Balance the equation
2. Identify the limiting reagent
3. Use the limiting reagent to calculate the theoretical yield of
P2O5
4. Calculate the % yield
It is strongly recommended that you do this problem. The
answer will be posted next week.
Solution Basics
• What’s a solution?
– any homogeneous mixture; solutions are not limited to
the liquid phase. A true solution does not scatter light.
• Solvent and solute are commonly used terms with
no real rigid definitions. It’s common to call the
component present in greatest amount the solvent
and all other species solutes. There’s this implicit
“I dissolved x in y” at work.
• When ionic compounds dissolve, it is common
that the ions dissociate. This is the “stoichiometry
of dissolution” and can result in the ions having
different concentrations.
– NaCl(s) =>Na+(aq) + Cl-(aq)
– Ca(NO3)2 => Ca2+(aq) + 2NO3-(aq)
Concentrations
• There are numerous systems for describing the composition of a
solution and a number of these will be discussed as appropriate. Since
the current focus is on chemical reactions with the mole as the key
measure, a system which readily describes the number of moles of
solute (presumed to be the chemically active component) present in a
given quantity of solution is needed.
• Molarity: M=(moles solute)/(L of solution)
• Quite astonishingly, it’s a three variable equation.
• “Problems” in this arena break down into three obvious categories
– What is the molarity of a solution prepared….
– How many _____ of solute are present in a….
– What volume of solution would contain…
• There is a critical issue here. Remember that your solute is most likely
going to be weighed out, as mass is the most precise measure
available. Thus, use of moles=mass/gfw is commonly needed in
conjunction with molarity computations.
• An improperly prepared solution is likely to doom an experiment.
• You should assume that the name of the solute, rather than its formula
is going to be provided.
Dilution
• It is common practice to have available a set of stock
solutions from which solutions of lesser concentrations are
prepared. Further, one often adds a solution of known
concentration into a larger volume as part of a procedure.
This process is known as dilution.
• Consider two solutions, with the second having been
prepared by a quantitative dilution. That means that the
initial M and V are known and the final V is known.
• In the process of adding a solution into pure solvent (or
adding solvent to the solution, what must be conserved?)
• How are M and V related?
• What general equation fits the above?
• M1*V1=M2*V2
• Recalling general algorithms for inversely related pairs:
(__* ___)
 _____
____
There is the usual standard set of problems in this area.
If ____ (L/mL) of ______M solution are diluted to a final
volume of ______(L/mL) what is the resultant M?
What volume in (L/mL) of _____M solution should be
dissolved to a final volume of ______(L/mL) if you want a
_____M solution?
_____(L/mL) of a _____M solution are diluted to a new
concentration of ______M. What is the new volume?
_____(L/mL) of a solution are diluted to a final volume of
_____(L/mL) and the concentration is found to be
_____M. What was the initial concentration?