16. A newly discovered protein has been isolated from seeds of a tropical plant and needs to be characterized. A total of 0.137g of this protein was dissolved in enough water to produce 2.00mL of solution. At 31.68°C the osmotic pressure produced by the solution was 0.134atm. What is the molar mass of the protein? (20pts)
Osmotic pressure is calculated using the equation:
Pos = cRTi
Where:
Pos = osmotic pressure
c = concentration, usually in units of molarity, M
R = Universal gas constant, 0.08206 L.atm/mol.K in this case
T = absolute temperature in K
i = van't Hoff factor, the number of solute particles per formula unit
For this problem, the osmotic pressure and temperature are given. We'll assume that this "protein" is a single molecule, so i = 1. Plugging in:
0.134atm = M(0.08206 L.atm/mol.K)(304.73K)(1)
Solving for molarity, M = 0.005359 mol/L
We have 2.00mL of solution, that's 0.00200L, so our sample must contain (0.005359 mol/L)(0.00200L) = 1.0717x10-5 mols of protein. We measured 0.137g of protein, so the molar mass must be:
0.137g / 1.0717x10-5 mols = 12780 g/mol
2009-05-13
Question...
From Exam 3:
2. Explain why each of the following does not result in an effective buffer? (15pts)
1.24mol NH4NO3(aq) + 0.03mol NH3(aq)
0.38mol HI(aq) + 0.38mol NaI(aq)
1.28mol Na2CO3(aq) + 0.64mol NaOH(aq)
2. Explain why each of the following does not result in an effective buffer? (15pts)
1.24mol NH4NO3(aq) + 0.03mol NH3(aq)
0.38mol HI(aq) + 0.38mol NaI(aq)
1.28mol Na2CO3(aq) + 0.64mol NaOH(aq)
What makes an effective buffer solution? 1) an approximately equimolar combination of a weak conjugate acid and its weak conjugate base; 2) at concentrations at least 100x the Ka of the weak conjugate acid.
The first combination listed above is indeed a combination of a weak conjugate acid (NH4+) and its weak conjugate base (NH3), but their concentrations are not approximately equimolar. If the concentration of conjugate acid and conjugate base are not within a factor of ~10, the buffer will not be able to effectively control pH. This could be an effective buffer if an extra mol of NH3 was added.
The second combination is equimolar, but HI(aq) is a strong acid, therefore it would not make an effective buffer. This could be an effective buffer if HF and NaF were used instead of HI and NaI.
The third combination is a mixture of a weak base with a strong base, and the resulting solution does not contain an equilmolar combination of a weak conjugate acid and its weak conjugate base, so it is not an effective buffer. This could be an effective buffer if 0.64mols of HCl(aq) was added instead of 0.64mol NaOH(aq), because the resulting solution would contain 0.64mols of HCO3-(aq) {a weak conjugate acid} and 0.64mols of CO32-(aq) {its weak conjugate base}.
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I am in my office right now, I plan to be here all morning so if you have questions feel free to stop in...
2009-05-11
Email question...
Can you explain problem 11 on Spring 2009 Exam 2b
Methane (CH4) reacts with oxygen to form carbon dioxide and water. under some set of conditions at some point in time, you find that 6.495g of methane react every minute in a 600.0 mL vessel.
a. what is the rate of methane consumption?
b. what is the rate of oxygen consumption?
c. what is the rate of carbon dioxide production?
d. what is the rate of water production?
e. what is the rate of the reaction?
Reaction rates are expressed as (change in concentration)/(change in time), so for this reaction under these conditions, the rate of methane consumption is:
{(6.495g CH4)/(16.043g/mol)/(0.6000L)}/(1min) = 0.6748 M/min
For the rest of the rates, we'll need a balanced reaction:
CH4(g) + 2 O2(g) --> CO2(g) + 2 H2O(g)
Average rates are determined by the stoichiometry of the reaction. For every mol of CH4 consumed, 2 mols of O2 are consumed, so the rate of O2 consumption must be twice as fast as the rate of CH4 comsumption. 2x(0.6748 M/min)=1.350 M/min
Every mol of CH4 that reacts produces 1 mol of CO2, so the rate of CO2 production is the same as the rate of CH4 consumption, 0.6748 M/min.
Every mole of CH4 that reacts produces 2 mols of H2O, so the rate of H2O production is twice the rate of CH4 consumption, 1.350 M/min. {Which is the same as the rate of O2 consumption. Note that things that have the same coefficient in the balanced reaction have the same rate.}
The rate of the reaction is also related to the stoichiometry of the reactants. Take the rate of consumption or production of anything in the reaction and divide it by its coefficient, to get the rate of the reaction, 0.6748 M/min.
Other questions, let me know...
Methane (CH4) reacts with oxygen to form carbon dioxide and water. under some set of conditions at some point in time, you find that 6.495g of methane react every minute in a 600.0 mL vessel.
a. what is the rate of methane consumption?
b. what is the rate of oxygen consumption?
c. what is the rate of carbon dioxide production?
d. what is the rate of water production?
e. what is the rate of the reaction?
Reaction rates are expressed as (change in concentration)/(change in time), so for this reaction under these conditions, the rate of methane consumption is:
{(6.495g CH4)/(16.043g/mol)/(0.6000L)}/(1min) = 0.6748 M/min
For the rest of the rates, we'll need a balanced reaction:
CH4(g) + 2 O2(g) --> CO2(g) + 2 H2O(g)
Average rates are determined by the stoichiometry of the reaction. For every mol of CH4 consumed, 2 mols of O2 are consumed, so the rate of O2 consumption must be twice as fast as the rate of CH4 comsumption. 2x(0.6748 M/min)=1.350 M/min
Every mol of CH4 that reacts produces 1 mol of CO2, so the rate of CO2 production is the same as the rate of CH4 consumption, 0.6748 M/min.
Every mole of CH4 that reacts produces 2 mols of H2O, so the rate of H2O production is twice the rate of CH4 consumption, 1.350 M/min. {Which is the same as the rate of O2 consumption. Note that things that have the same coefficient in the balanced reaction have the same rate.}
The rate of the reaction is also related to the stoichiometry of the reactants. Take the rate of consumption or production of anything in the reaction and divide it by its coefficient, to get the rate of the reaction, 0.6748 M/min.
Other questions, let me know...
2009-05-10
Questions....
First a question that was posted before Exam #4. I'm sorry I didn't see this one before the test.
I was wondering about adding for delta H, and delta S. Sometimes I mix the numbers up and put positives where there should be negatives. for example number 11 on summer 2007 test. for the delta H I got all the numbers right except for PH3(g). How come it is positive not negative?
Calculate the following values for the unbalanced reaction listed at 25ºC.
Reaction 1: P4O10(s) + H2(g) -> PH3(g) + H2O(g)
The values listed in the table are (in order of the reaction):
{delta}Hf : -2984 ; 0 ; +5.4 ; -241.818
So : +228.86 ; +130.684 ; +310.23 ; +188.825
{delta}Gf : -2697.7 ; 0 ; +13.4 ; -228.572
If something is a reactant, change the sign listed in the table; if it's a product, don't change it. PH3(g) is a product in this reaction, so use the values directly from the table, in this case they all happen to be positive numbers. You can think of these numbers sort of like prices. Let's say you look up the price of a calculator and it's $10. If you're buying the calculator, the balance in your checking account will decrease by $10. If you're selling the calculator, the balance in your checking account will increase by $10. If you're forming something in a reaction, the values listed in the table represent what you have; if you're consuming something in a reaction, the values in the table represent what you have lost.
Someone also asked about a Mastering Chemistry question from Assignment 5. It has 2 parts...
The rate constant for a certain reaction is k = 3.30×10−3 s-1. If the initial reactant concentration was 0.250M, what will the concentration be after 13.0 minutes?
Integrated rate laws are all about choosing the correct order and then doing some algebra. In this case, the order of the rate law expression must be determined by the units of the rate law constant. Remember that:
Rate0 = k[reactant]x
Since the rate is expressed as "M/time", then this must be a first-order rate law because (1/s)(M) = M/s. That means we have to use the first-order integrated rate law expression
ln[reactant]t = -kt + ln[reactant]0
ln[reactant]13.0min = -(3.30x10-3 s-1){(13.0min)(60s/min)} + ln(0.250) = -3.96
[reactant]13.0min = 0.0191M
A zero-order reaction has a constant rate of 4.30×10−4 M/sec. If after 40.0seconds the concentration has dropped to 5.00×10−2 M, what was the initial concentration?
This one tells you it's a zero-order process, but it might trip you up by giving you the "constant rate" rather than a "rate constant". We can plug the rate and the concentration at 40.0sec in to the regular rate law expression to get the rate law constant:
Rate0 = k[reactant]0
(4.30x10-4 M/sec) = k (5.00x10-2 M)0
Since this is a zero-order process, it turns out that k is equal to the constant rate. Again, plug values into the correct integrated rate law expression and have some algebra fun. For a zero-order process, the IRL is:
[reactant]t = -kt + [reactant]0
(5.00x10-2 M) = -(4.30x10-4 M/sec)(40.0sec) + [reactant]0
Solving, the initial concentration of reactant was 0.0672M.
Other questions, let me know...
2009-05-04
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