2016-01-11

States of Matter - Balancing energy

We've all thought about solids, liquids, and gases since we were pretty young. Even before we probably knew the terms "solid", "liquid", and "gas", we probably had a reasonable grasp on the concept that hard things were hard, watery things were wet, and… OK, the concept of a gas was probably a bit more abstract than our toddler brains were capable of understanding. But why do those states of matter behave the way they do? And more importantly, how can we understand states of matter in a way that will allow other states (plasmas, liquid crystals, supercritical fluids, etc) to exist?

As with many things scientific, there are multiple levels of understanding that we can use to explain our observations. One of the more basic definitions of states of matter relies on macroscopic observations of those states of matter:

  • Solid - definite shape, definite volume
  • Liquid - variable shape (assumes the shape of its container), definite volume
  • Gas - variable shape, variable volume (assumes the shape and volume of its container)

Those are great functional definitions and they are sufficient in many situations, but WHY do they work? That requires a little more detail. Most of the time when we want to answer a "why does stuff do what it does?" question, we have to look at the interaction of matter and energy, so in this case, let's look at the matter and energy considerations in states of matter.

Why is a solid a solid? On a molecular level, we could say that the atoms (or ions, or molecules) interact very strongly. In gases, the atoms (or ions, or molecules) do not interact. In liquids, the atoms (or ions, or molecules) interact moderately. Why do atoms (or ions, or molecules) "interact"? Well, because of intermolecular forces (IMFs)! The IMFs between particles in a solid are strong, liquids are weaker, gases are weaker still. IMFs to the rescue!

That's great when we are differentiating between rocks and water and helium, but it still falls a little short. What about water? Water can be a solid, liquid, or gas under pretty achievable conditions. Do the IMFs between water molecules change? Not really. So what does change when water changes state? In a solid, the molecules don't really move, but in a gas, they move very quickly. Motion is described by kinetic energy… so states of matter are really determined by the balance between the IMFs present in a substance and the Ekin of the atoms (or ions, or molecules) in the substance. If the IMFs are significantly stronger than the Ekin, then the atoms (or ions, or molecules) will tend to stick together and be solid; if the IMFs are significantly weaker than the Ekin, the material will be a gas. So we can make water (or pretty much anything) change from a solid to a liquid to a gas by increasing the Ekin of the molecules to the point where they overcome the energy of the IMFs holding those molecules together.

There are still some simplifications and assumptions in that explanation, but it's a good "next step" for us to use as we try to understand and explain the differences in states of matter.



2015-07-19

It's elementary!

Question from email...
------
Hello Dr. Bodwin,

I am going over the practice sheet you handed us in class on Thursday and I am stuck on question #10. Where it asks why certain reactions cannot be elementary? And explain why? I'm wondering how to figure that out! If you could clarify that would be awesome!! 

------

Elementary reactions or elementary steps describe the individual collisional events that occur in each step of a reaction. When a chemical reaction occurs, the reacting species have to interact with each other (Collision Theory). If we think about those interactions or "collisions", there is a pretty low probability that a collision will take place that is properly oriented and at the correct energy if that collision involved more than two components. From a Gen Chem perspective, this means that we can pretty safely say that an elementary reaction cannot have more than 2 reacting particles.

Because elementary reactions describe molecular level collisions, we can also say that the elementary reaction must be reversible, meaning that if we reverse the reactions (exchange products and reactants), it must also be elementary. So an elementary reaction should have no more than two reactants, but it should also have no more than two products.

Reaction mechanisms and elementary reactions can have a lot of subtle little twists and turns that make them fascinating to study and we are treating elementary reactions very simply at this point. As you explore more an more chemistry, you will find that reaction mechanisms can be a fascinating puzzle to work through. Enjoy!

2014-05-02

Preparing for a Final Exam

What's the best way to prepare for a cumulative Final Exam? That can be a little ominous, and the "best" way to study can vary considerably from person to person. At this point, you've (hopefully) found something that works for you, whether that means a quiet corner in the newly remodeled library or a lounge with a little more "white noise". Go with the method that works for you!

With that as a foundation, what should you study? It's probably not a good idea to just sit down with the textbook and start reading on page 1. Use the tools you have available to focus and prioritize your studying time and energy!

1. Start with the exams you've already taken!
Look back at your exams. There are things that you knew very well when you took Exam 1 that have gotten a little foggy over the past few months. The good thing is that you knew that material fairly recently, so it'll probably just take a little review to freshen up those concepts and problems.
What about things you didn't do well on earlier exams? Chemistry (and many other fields...) is a cumulative subject. We tend to look at things from a bunch of different directions, and we often approach a concept or problem 2 or 3 different ways over the course of the semester and year. In the past few months, we might have looked at something differently in a way that suddenly makes perfect sense to you.
Use your exams to jot down and prioritize things to study. It won't be a perfect list, but it will give you a good starting point.

2. Review notes from class
Once you've identified the things you need to review, look back at your notes from class.(Sometimes this serves as a reminder to take better notes in the future!) For some topics, a brief reminder from your notes will be all that you need to bring these concepts and problems up from the cold, dark storage room of your memory. For others, it's a good way to once again identify and prioritize your study topics.

3. Use the book
For topics that are still fuzzy, look up key words in the index to help you find a good place to start. Not sure you remember integrated rate laws? Look it up in the index and hit those pages!
It's also not a bad idea to skim over sections that you're pretty sure you understand. Once you're fairly comfortable with a topic, it becomes easier to pick up some of the more subtle points.
End-of-chapter problems can also be helpful because they present information a little bit differently than I do. I'm not talking "better" or "worse" here, the book just uses different wording to ask the same questions I ask. At the end of my class, I don't just want you to be good at answering questions that I write, I want you to understand chemistry, even if someone else is asking the questions.

If you get through all of those steps and still have time, energy, and a thirst for more knowledge, use your favorite tool to search the World Wide Interwebs Net or stop in and chat with me. This isn't a perfect list of study tips, but it's a good place to start. Identify topics, prioritize your studying time and energy, look for connections and common themes in the material, and good luck. You can do it!

2014-04-01

More Email Questions...

A couple more email questions:

-----Question-----
On exam b spring 2011 #5 why should this be changed to Ka instead of Kb? Also on exam b spring 2010 #2 the 3rd line its says that HCl  works as an effective buffer. I thought an effective buffer had to be a weak conjugate acid/base?
-----Answer-----
Here's question #5
This is Ka because ammonium ion is a conjugate acid. You could also do this as a Kb equilibrium, but you'd be starting with products and shifting to the left to form reactants. Either way should give the same answer.

Here's #2:
Correct, an effective buffer is an approximately equimolar combination of a weak conjugate acid and its weak conjugate base. In the 3rd line here, HCl(aq) is protonating the carbonate in solution 1.5 times so the resulting mixture is 0.64mols HCO3-1(aq) and 0.64mols H2CO3(aq). This is an equimolar mixture of a weak acid and its conjugate base, so it should be a good buffer. {NOTE: because carbonic acid decomposes and the resulting CO2 can escape from solution, this might not be the best buffer in the real world, but it works fine as a sample problem.} This is similar to how you are making the carbonate buffer that you will measure in lab this week.

2014-03-30

Pre-Exam 3 email questions...

A few questions have come in by email, here they are:

----Question-----
I have a question regarding problem 9 on exam 3a from spring 2013. I understand that the calculated x value doesn't fit under the assumptions. I see where the first two values come from. I was wondering where the -1.238 x 10^-3 came from?

{(x)(x)} / (0.516 – x) = 2.40 x 10^-3
0.516 is the initial concentration and 2.40 x 10^-3 is Ka.
 x^2 + (2.40 x 10^-3)x + (-1.238 x 10^-3 ) = 0
----Answer-----
In order to use the quadratic formula, we have to solve the equation to the form:
ax2 + bx + c = 0
The (-1.238x10-3) term comes from (2.40x10-3)(0.516).


----Question-----
I was going through some of the old exams and the problems that use the Henderson - Hasselbalch. In those examples when it asked for the concentration of the conjugate base over the concentration of conjugate acid, but in the key only the moles of both are put in those places. Why? 
----Answer-----
This is a little mathematical shortcut. Since both the conjugate acid and conjugate base are in the same total volume of solution, the volumes mathematically cancel so I left them out. For example, if we had a buffer made from 0.65mols of HA and 0.55mols of A-1 in 800.0mL of buffer solution, that last part of the Henderson-Hasselbalch would look like:
You can always keep the volume in there and calculate the actual concentrations of each component, you should get exactly the same answer either way.


----Question-----
I dont understand how you can derive pH values from pka's given in the question. I also dont understand how pH can be calculated at each eq point in the titration curve as in number 11 on spring 2013 where it asks what indicator to use.. it says use the 2 pkas...how does this help?
----Answer-----
There are a couple ways that pKa (or pKb) can lead to a pH. One possibility is in a question like "What is the expected pH of a 0.618M solution of ammonium nitrate solution?" In this question, you can set up a Ka-type equilibrium for ammonium ions and use the Ka of ammonium to calculate [H3O+] and pH. This is similar to the problem I posted yesterday (http://chemistryingeneral.blogspot.com/2014/03/neutral-salts.html). This method can be used to calculate the initial pH for a titration. It would also work to approximate the pH of an equivalence point. Let's think about that...
For the titration of phosphite ions with hydrochloric acid, we can calculate the initial pH by setting up a Kb-type equililbrium and using the Kb of phosphite ion to calculate [OH-1] and pOH and pH. At the first equivalence point in this titration, we have a solution that we can think of as HPO3-2(aq) because we have added just enough acid to complete the following equation exactly once:
PO3-3(aq) + H+(aq)  <=> HPO3-2(aq)
Between equivalence points, we have buffering regions of the titration curve... at the mid-point of this buffering region, the pH is equal to the pKa of the weak acid of the mixture. If we know the pKa (and therefore the pH) on either side of the equivalence point we're interested in, we can get a pretty reliable estimate of the pH of that equivalence point. On the titration curve below, if the pKas that bound the equivalence point are 6 and 9, the pH of the equivalence point should be right between them at pH = 7.5.

Keep an eye on the weather... Unless MSUM officially closes campus tomorrow, we will have class and the exam as planned. I'll be there at 7:30.

2014-03-29

Neutral salts

The conjugate of a strong acid or strong base is neutral. That's just something we tend to accept, memorize, and move on. But why? Those two words are a big part of the reason I'm a chemist.

Let's take a look at a strong acid and see if we can make sense of this. Of the typical strong acids, nitric is usually the weakest, and nitric is also the only one that might have a Ka listed in standard tables. The stronger strong acids have really useful Ka values listed in the tables like "large" or "strong"... I don't have a "large" button on my calculator, so let's just use nitric acid and we'll hopefully see why the other strong acids would follow the same trend if we had a value for their Ka.

The Ka for nitric acid is usually listed at around 25. That means the Kb for nitrate ions is:
That's a REALLY weak Kb, but we can go ahead and calculate the pH of a solution just like in any other situation. How about the problem: What is the expected pH of a 0.500M solution of sodium nitrate? {By the way, we could make similar arguments to show that the sodium ions don't affect the pH, but we'll save those for another day...} As with all good equilibrium problems, it's probably not a bad idea to start with a table:
Now we can set up the Kb expression and plug in the numbers we have:
We should be able to simplify that with some assumptions... Let's assume that "x" is much smaller than 0.500 and much larger than 10-7. That gets us the simplified expression:
Solving this expression, we get x = 1.41x10-8, which gives us a pOH = -log(1.41x10-8) = 7.85, and pH = 6.15. Hmm, that's not neutral, that's acidic. The whole point of this was to prove that nitrate was a neutral ion. This is a disaster.

BUT WAIT!

We made some assumptions. We didn't check our assumption after we solved for "x". This is why I always tell you to check assumptions... We assumed that "x" would be much smaller than 0.500, which it is, but we also assumed that "x" was much larger than 10-7, which it absolutely is not! So the assumptions we made were an oversimplification of the problem and that's where we entered the danger zone. Looking back at our Kb expression, we can only simplify it to:
That's still going to require the quadratic formula to solve. I'll let you work out the details, but the result should be that x = 7.96x10-9. That means:
[OH-1]eq = 10-7 + (7.96x10-9) = 1.08x10-7 M
pOH = -log(1.08x10-7) = 6.9667
pH = 14 - 6.9667 = 7.0333
That's not exactly 7.0000000000 neutral, but it's pretty darn close, especially if we're thinking about this in terms of selecting a visual acid-base indicator for a titration. 

2014-01-19

Quiz 2 Hint

Quiz 2 has a heat capacity problem, and the heat capacity value given in the problem is in units of "joules per (mole Kelvin)". Remember, for heat capacity problems, we're looking at changes in temperature. Whether it's Celsius or Kelvin, the ΔT is the same because 1 degree C and 1 K are exactly the same size. You can use either, but make sure you're doing everything in the right order. A temperature that changes from 3.45°C to 12.71°C is changing by:
ΔT = 12.71°C - 3.45°C = 9.26°C
If you prefer Kelvin, convert BOTH individual temperatures to K, THEN subtract:
ΔT = 285.86K - 276.60K = 9.26K
Other questions, let me know...

Quiz 1 questions...

I've gotten a few questions about Quiz 1. Here are a few hints...

Question 1: How many moles of nitrogen atoms are in a given mass of iron(III) nitrate?
Start with a balanced chemical formula. How many moles of N are in each mole of iron(III) nitrate? Use the mole ratio to get moles of N from the moles of iron(III) nitrate.

Question 2: How many moles of chlorine atoms are in a given volume of a given concentration vanadium(IV) chlorate solution?
This is almost the same as question 1, but you need to use concentration and volume to determine the moles of vanadium(IV) chlorate instead of mass and molar mass.

Two other helpful hints:
1. Don't wait until the day before a quiz is due to work on it. This is especially true because...
2. When you have a question, show me your work. If you have the problem all set up, it's much easier for me to look at what you've done and help you work through the problem correctly. It's always easier to help you when you can give me an idea of where you're getting confused.

2014-01-12

Spring Semester 2014 is here!

We're 18 hours away from the first Gen Chem II class of 2014! Watch this blog for general info about the topics we'll be exploring and answers to specific questions I get by email. I'll also be tweeting links and daily summaries using #GenChem2014. And I might try a few other things this semester…

A couple tips that are always useful:
1. If there's a way to calculate moles of a substance, that might be a good start.
2. Balance your chemical formulas. Then balance your chemical reactions. Balance is the key to chemistry. And skiing. If you don't balance your formulas and reactions, you're gonna have a bad time.
3. Be curious. That what science is all about.
4. Ask questions. Unless you're the only person in the room, it's pretty likely that someone else in the room has the same question you do.
5. Be bold. It's OK to answer a question incorrectly in class, we're learning chemistry here not juggling chainsaws. The other side of this is to be supportive of your classmates, we're all here to learn (including me).

Enjoy the beautiful weather today and I'll see you tomorrow.

2013-12-12

Email question 2013-12-12

------------------------------
Hello Dr. Bodwin. I've been working on some old exams for studying and on the exam 1a from fall 2011, the last question is asking about the empirical formula, and I was just wondering when solving for each part, where do the last two numbers come from?
For example,
C -> (71.98 g)/(12.011 g/mol) = 5.993 mols -> 4.5 -> 9
------------------------------

Let me pull up the whole problem from the exam key:

{from:  http://www.drbodwin.com/teaching/exams/c150fe1ak.pdf}

When determining empirical formulas from percent composition data, the first step is to assume that you have 100g of sample. You can assume any amount of sample you like, but 100g simplifies things a little because if I have 100g of a sample and I know that 71.98% of that sample is carbon, then there must be 71.98g of carbon in the sample. Using the percentages given in the problem, we know that the "100g" sample contains 71.98g of carbon, 6.711g of hydrogen, and 21.31g of oxygen. Grams are great, but we want to count the number of different atoms, so we need to convert grams to moles... That's the first step in the calculation that is shown.
Once we have moles, we know the relative amounts of each element present in the sample, and we can write a balanced chemical formula:
C5.993H6.658O1.332
Hmm, that doesn't look quite right... But at this point, we have the correct relationship between the moles of each element, so we can force those relationships to be whole numbers by dividing all of them by the smallest one. Essentially, we're saying "what if 1.332 actually represents 1 oxygen atom?" Dividing them all gives the forumla:
C4.5H5O1
Still not perfect, but it's a lot more "normal" looking than the first formula. Again, since we know that the relationship between moles is correct here, we can multiple all the subscripts by something that gives us a nice, round, whole number ratio. If we double everything, we get:
C9H10O2
And now we have a good empirical formula for this elemental analysis.

Other questions? Let me know.

2013-10-26

Suggested problems posted

Chapter 5 (and 4, sort of) suggested problems are posted:
http://www.drbodwin.com/teaching/genchem.php

Enjoy. I also tweaked the calendar on that page to reflect the changes we've had.

2013-10-10

Problem Set 2

There's a key posted for the problem set we did in class that was titled "Problem Set #2" on my Gen Chem webpage:
http://www.drbodwin.com/teaching/genchem.php
Direct link to the key:  http://www.drbodwin.com/teaching/problemsets/c150gps02k.pdf

Don't forget, there are also old exams posted at:
http://www.drbodwin.com/teaching/examarchive.php

Other questions, let me know…

2013-09-14

In-Class Problem Set #01

If you want to check your answers to the first problem set we looked at in class, here's the link:

http://www.drbodwin.com/teaching/problemsets/c150gps01k.pdf

I hope studying is going well, let me know if you have any additional questions. And don't forget, if you need a little study break tomorrow afternoon, the Dragon Soccer team has a home game at 3pm. After today's rain, tomorrow looks like it will be a lovely day to watch some soccer.

2013-09-13

Carbons in a propane sample

Today in class we looked at a problem that some of you didn't quite get to the end of by the time class ended. Here it is. If you haven't already worked it through, give it a good try before you jump ahead to the answer...

How many carbon atoms are in a 37.43L sample of propane gas at 17.52°C and 1.472atm?

This starts out as an Ideal Gas Law problem with a single set of conditions.
PV = nRT
Plugging in the values from the problem:
(1.472atm)(37.43L) = n(0.08206L.atm/mol.K)((17.52+273.15)K)
n = 2.3099mols of C3H8(g)
{NOTE: I'm in the middle of the problem, so I'm not rounding for significant figures yet, 
but it looks like 4 sig figs would be good at this point…}
Each propane molecule contains 3 carbon atoms, so each mole of propane molecules contains 3 moles of carbon atoms.
(2.3099mols C3H8) (3mols C/1mol C3H8) = 6.92974mols C
(6.92974mols C) (6.022x1023 C atoms/mol C) = 4.173x1024 C atoms in the sample.

DISCLAIMER: This problem assumes that propane is behaving as an ideal gas under these conditions. There's a pretty good chance that it would not be ideal under these conditions, but for the purposes of this problem let's assume it is.

2013-08-26

Welcome to Fall 2013!

Welcome to the first day of Fall 2013 at MSUM!

 If you're looking for the textbook for class, here's the link on Amazon. http://www.amazon.com/Chemistry-Science-Context-Third-Edition/dp/0393934314/ref=sr_1_1?ie=UTF8&qid=1377519676&sr=8-1&keywords=chemistry+3rd+edition+gilbert
This is not an endorsement or recommendation of amazon.com, it's just the quickest way I could find a fairly universal link to the correct textbook. The book is available at the MSUM Bookstore and many other places.

2013-07-18

In-Class problems 2013-07-18

Today was buffers day in class so we looked at 294 different ways to use the Henderson-Hasselbalch equation. Remember, the Henderson-Hasselbalch equation is just a rearrangement of the Ka expression we use for understanding any acid... You can always plug directly into the Ka expression and get the same result you will get using Henderson-Hasselbalch. So the problems...

1. Combine 12.642g of chlorous acid with 15.372g of lithium chlorite, dilute to 500.0mL. What is the expected pH of this buffer? (Chlorous acid Ka = 1.12x10-2)
Plug into the Henderson-Hasselbalch equation...
And solve...

2. Prepare 500.0mL of a 0.650M HN3/N3-1 buffer at pH = 5.10 from HN3(s) and NaN3(s). (Hydrazoic acid Ka = 1.93x10-5)
Start off by solving for the ratio of conjugate acid to conjugate base using either the Henderson-Hasselbalch equation or just an unmodified Ka expression. I'll use H-H...
[N3-1] / [HN3] = 2.4297
[N3-1] = 2.4297[HN3]
Now that we know the ratio of these concentrations, we can solve for the actual concentrations by using the relationship...
0.650M = [HN3] + [N3-1]
0.650M = [HN3] + 2.4297[HN3] = 3.4297[HN3]
[HN3] = 0.1895M
[N3-1] = 0.650 – [HN3] = 0.650 – 0.1895 = 0.4605M
To make a solution that's 0.1895M HN3 at 500.0mL, we need...
(0.5000L)(0.1895M) = 0.09475mols HN3
(0.09475mols HN3)(43.029g/mol) = 4.077g HN3(s)
To make a solution that's 0.4605M N3-1 at 500.0mL, we need...
(0.5000L)(0.4605M) = 0.23025mols N3-1
(0.23025mols N3-1)(65.011g/mol) = 14.969g NaN3(s)
So we should be able to make the target buffer by combining 4.077grams of HN3 and 14.969g of NaN3 in enough water to make 500.0mL of solution.
NOTE: If you're actually making a buffer, be very careful about the order of addition of the components. Whenever you're combining a solid or concentrated solution with a solvent, it's usually a good practice to add the solid or concentrated stock to the larger volume of solvent slowly with very good mixing. Dissolving and/or mixing can liberate a LOT of heat in some cases that could be dangerous if the order of addition is reversed.
NOTE2: Hydrazoic acid is not a solid at room temperature, and the pure liquid is a non-trivial safety risk... We can talk about this type of a buffer on paper, but there's very little chance you (or I) will ever prepare or use a hydrazoic acid-based buffer.

3. What is Ka of a weak acid, “HA”, if a solution made by dissolving 0.316mol HA and 0.327mol A-1 in water and diluting to 750.0mL has a pH of 9.374?
Plug in to Henderson-Hasselbalch or the generic Ka expression...
9.374 = pKa + log (0.327 / 0.316)
pKa = 9.359
Ka = 4.374x10-10

4. What is Kb of a weak base, “B”, if a solution made by dissolving 0.143mol B and 0.158mol HB+1 in water and diluting to 400.0mL has a pH of 5.975?
Similar to the previous problem, plug in to Henderson-Hasselbalch or the generic Ka expression...
5.975 = pKa + log (0.143 / 0.158)
pKa = 6.018
pKb = 14 – 6.018 = 7.892
Ka = 1.043x10-8

Good luck.

2013-07-03

Rate Law and Integrated Rate Law Problem {2013-07-03}

Under some set of conditions, ammonia gas and fluorine gas react to form nitrogen trifluoride gas and hydrogen gas at 8.73ºC. What is the correct rate law expression (including the rate law constant) given the following data:
Rxn #
[NH3]0
[F2]0
Rate (M/min)
1
0.274
0.218
1.841x10-2
2
0.274
0.436
7.362x10-2
3
0.822
0.436
2.209x10-1
The rate law expression for this reaction is:
Rate0 = k[NH3]0x [F2]0y
Comparing Rxn #1 and #2:
y = 2, the reaction is 2nd order with respect to fluorine concentration
Comparing Rxn #3 and #2:
x = 1, the reaction is 1st order w.r.t. ammonia concentration
Plugging in values from Rxn #1 and solving for k...
1.841x10-2 M/min = k (0.274M)1 (0.218M)2
k = 1.42 M-2min-1

In Rxn #2, how much time must pass before [F2] = 0.324M?
Plugging in to the 2nd order integrated rate law expression...
t = 0.558 minutes

In Rxn #1, how much time must pass before [NH3] = 0.217M?
Plugging in to the 1st order integrated rate law expression...
t = 0.164 minutes


2013-07-02

Know Your Polyatomic Ions

I expect you to know polyatomic ions. They're part of the vocabulary of chemistry and if you have to look them up every time one pops up it will slow everything down. I was about to type up a post when I thought "Hmm, this seems like something I would have typed up before..." Sure enough just about a year ago...
http://chemistryingeneral.blogspot.com/2012/06/polyatomic-ions.html

Enjoy!

Colligative properties and gas law problems 2013-07-02

1. 12.64g of sodium sulfate is dissolved in 400.0mL of water. What are the boiling point and freezing point of the solution?
Na2SO4 = 142.041g/mol
12.64g / 142.041g/mol = 0.0889884mols Na2SO4
{NOTE: don't round sig figs in the middle of a problem...}
(0.0889884mols Na2SO4) / 0.4000kg water = 0.22247m Na2SO4
{NOTE: density of water is 1.0000g/mL...}
ΔTbp = (0.512 ºC/m)(0.22247m)(3mol particles / 1mol Na2SO4) = 0.342ºC
{NOTE: I think I used a different value for the boiling point elevation constant in class. This one is correct.}
{NOTE: When Na2SO4 dissolves in water, the result is 2 Na+(aq) ions and 1 SO4-2(aq) ion. 3 particles.}
Tbp = 100.000ºC + 0.342ºC = 100.342ºC

ΔTfp = (1.858 ºC/m)(0.22247m)(3mol particles / 1mol Na2SO4) = 1.240ºC
Tfp = 0.000ºC – 1.240ºC = -1.240ºC

2. A weather balloon is filled with 23.65L of an ideal gas at 28.73ºC and 1.042atm pressure. It is released and rises to an altitude where the temperature is -6.35ºC the pressure is 0.842atm. How many moles of gas are in the balloon and what is its volume when it reaches the described altitude?
Plugging in to the Ideal Gas Law:
PV = nRT
(1.042atm)(23.65L) = n(0.08206 L.atm/mol.K)(301.88K)
n = 0.9948 moles
{NOTE: Convert to kelvins! If you pay attention to the units on R you'll be less likely to forget...}

Now that we know how many moles of gas are present, the next part could be solved by plugging in to the regular Ideal Gas Law again:
PV = nRT
(0.847atm)V = (0.9948mols)(0.08206 L.atm/mol.K)(266.80K)
V = 25.7L
Or by using the comparative form of the Ideal Gas Law:
P1V1/n1T1 = P2V2/n2T2
{NOTE: Since “n” is not changing, we can drop it from the equation...
(1.042atm)(23.65L) / 301.88K = (0.847atm)V2 / 266.80K
V2 = 25.7L


Practice, practice, practice...

2013-06-30

Begin again...

Summer 2013 Gen Chem II starts bright and early tomorrow morning! Kinetics, equilibrium, acids & bases, thermodynamics, redox, and nuclear chemistry all wrapped up in a neat little 5-week package. Let the games begin!
http://www.drbodwin.com/teaching/genchem.php

2013-05-31

Moderation...

For the past couple weeks, I've been getting hammered by spam in the comments so I have turned on comment moderation (for now). If you have any comments that are on-topic, please be sure that they will be allowed, I'm just trying to avoid a flood of spam-bot garbage in the comments section of my posts. Hopefully this will be a short-term problem...

Have a great day.

2013-05-14

Email question - van't Hoff factor

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Can you please explain how to determine the factor.  Now matter how much I read the book and look at examples I can't figure it out.  Thanks,
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There are a couple things that could be "the i factor", but I'll assume this question is about the van't Hoff factor that's used when determining changes due to colligative properties. Remember, colligative properties are properties of a solution that depend only upon the number of solute particle, not the identity of those solute particles. It doesn't matter if those solute particles are molecules, or cations, or anions, or a big random mixture of them all. When an ionic compound dissolves in water, it dissociates (at least somewhat) into its component ions. The number of ions (particles) that each formula unit breaks into when it dissolves is the van't Hoff factor. For example, if NaCl(s) is dissolved in water, it forms 1 Na+1(aq) ion and 1 Cl-1(aq) ion, so each formula unit (NaCl) forms 2 particles (ions) in solution; the van't Hoff factor is "2". For polyatomic ions, the ions don't break down into their individual atom, they stay polyatomic ions, so if K2SO4(s) is dissolved in water, it forms 2 K+1(aq) ions and 1 SO4-2(aq) ion in solution; the van't Hoff factor is "3".



Exam strategies...

How do you take an exam? The most important thing is, of course, knowing the material, but there are also some "tricks" that can work in your favor, especially on a standardized multiple-choice exam. Some of this comes down to game theory, so it can be useful in settings outside of exams.

1. Know the rules - In a standardized exam, there can be different rules and formulas used for calculating your score. Some of the formulas have the effect of penalizing you for guessing if you truly don't know the answer. Always check to determine whether or not there is a penalty for guessing. Personally, I prefer to just take the total number of correct answers and not mess around with an elaborate formula that penalizes you for guessing.

2. Improve your odds - When you look at a series of multiple choice answers, there are often answers included that have to be obviously wrong if you think about them a little bit. If you're looking at a 4-option multiple choice question, your odds of randomly guessing the correct answer without even reading the answers is 25%. Eliminate one wrong answer, and your odds increase to 33%. Eliminate two, you're at 50%. Keep this in mind when you get to a question that has you stumped; rather than hunt for the right answer, try to eliminate wrong answers. Be like Sherlock Holmes; when you have eliminated all other possibilities, the remaining possibility, no matter how unlikely, must be correct.

3. Build momentum - Don't get hung up on a question you don't know when there are others you know the answer to. Answering a few questions correctly builds confidence and helps you tackle the more challenging questions. This is a challenge for "linear" test-takers who naturally want to start with #1 and work through in order to the end of the exam. {I was a very linear test-taker, and to some extent I still am...} It's OK to skip a question and come back to it later. This also helps with time management on the exam.

4. Answer all the questions - This is related to #1 and #2 above. If there is no penalty for guessing, then make sure you answer every question, even if you have no idea what the correct answer is. For a 5-option multiple choice question, there's a 1 in 5 (20%) chance that you'll random guess it correctly. If you don't answer the question at all, there's a 100% chance that you'll get it wrong.

Good luck.

2013-05-04

Redox Lab question...

A few people have asked about the chemical reactions for the Redox lab hand-in {Redox Hand-in}, and this is often a source of confusion for students, so let me answer it here. First of all, I think a number of people get confused because you're trying to over-think the question and make it more complex than it really is.
So you need to draw a voltaic cell... you can draw this by hand, or you can draw it electronically. DO NOT just find one online and copy-paste it into your hand-in, if that's all I wanted I would have pasted this {voltaic cell} into the hand-in myself before I posted it.
The part that causes some confusion is the "...write a correctly balanced net ionic equation for the spontaneous process..." Don't over-read that! The metal cation solutions you were using in lab were probably nitrate salts, but nitrate (or whatever anion might have been present) was a spectator in all of your reactions. Net ionic equations are actually easier than full-formula equations because they're not cluttered up with with a bunch of extra stuff, and net ionic equations actually just describe the CHEMISTRY that's happening rather than distracting you with a bunch of spectator ions and species.
Now, I'm not going to write out a net ionic equation that's the exact answer to one that you have to write, but here's an example. Let's say I made up the voltaic cell Fe|Fe+3||Cd+2|Cd with the black/negative lead of my meter hooked up to the Fe(s) electrode and the red/positive lead connected to the Cd(s). The potential I measure is -0.32V. Because the measured potential is negative, the cell is running backwards, so the spontaneous cell reaction is Cd|Cd+2||Fe+3|Fe. Translating that into a reaction, we can write the two half-reactions as:
Cd(s) <=> Cd+2(aq) + 2 e- 
3e- + Fe+3(aq) <=> Fe(s)
Adding those up gives the overall (or "net") reaction:
3Cd(s) + 2 Fe+3(aq) <=> 2 Fe(s) + 3 Cd+2(aq)
{Remember to multiply each half-reaction by an appropriate integer to make all the electrons cancel...}
That's a "correctly balanced net ionic equation for the spontaneous process" in this case. Now go do that for all the cells you measured in the experiment.



2013-04-28

Working with ammonia

From my perspective, aqueous ammonia is a fascinating reagent to use in the lab. As a type of matter, it is a gas dissolved in a liquid, which seems pretty wild. In many cases, aqueous ammonia is just a fairly typical weak base that's nothing all that special as long as you're using the Bronsted-Lowry definition of a base. Those are great features of aqueous ammonia, but they really pale in comparison to what we see when we start combining aqueous ammonia with metal ions, especially transition metal ions. The key to thinking about aqueous ammonia in these situations is to remember that aqueous ammonia is always involved in a Kb-type equilibrium:
NH3(aq) + H2O(l) <=> NH4+1(aq) + OH-1(aq)
This means that in any solution of aqueous ammonia, there are both ammonia molecules and hydroxide ions. If we think about the Lewis definitions of acids and bases, this means that floating around in every solution of aqueous ammonia, there are nitrogen-based lone pairs of electrons on ammonia molecules and oxygen-based lone pairs of electrons on hydroxide ions. Different metal ions have different affinities for different types of lone pairs, so sometimes when a metal ion is added to aqueous ammonia it forms complexes with ammonia while other times it forms complexes with hydroxide.
How do we tell which is which? Whenever possible, by comparison with know reactions. If the observed reaction between a metal ion and aqueous ammonia looks identical to the reaction of that same metal ion with a known hydroxide source {like NaOH(aq)}, then the metal ion is probably more attracted to oxygen lone pairs and is reacting with the hydroxide ions in the aqueous ammonia. If, however, the observed reaction between a metal ion and aqueous ammonia is different from the reaction of that same metal ion with NaOH(aq), then the metal ion is probably reacting with the ammonia molecules in the aqueous ammonia solution.
Differential affinities between metal ions (Lewis acids) and different Lewis base donors is a very diverse field and was a driving force in chemistry before newer instrumental methods were developed. It's still an important consideration in chemistry and physics and biology... Biology? That's right! Every biological system that contains metal ions (especially transition metal ions) relies heavily upon differential binding affinities to function correctly. And that's just one of the many reasons why biologist need to understand chemistry...

2013-04-24

Qualitative analysis of metal cations

This week in lab, you'll be using the chemical tests you observed last week to separate (and identify) the metal cations in an unknown mixture of cations. It's probably better to think about this as a separation rather than just an identification because you will be given 1 sample and through a series of chemical (or physical) steps you will end up with up to 5 different metals in 5 different container. Some tips:
1. Flow chart - You have to organize your procedure to do well. The logical way to do this is with a flow chart for this type of a problem... a sequence of steps with decision points and branches along the way. You might not be a "flow chart person", but it makes it MUCH easier to follow through on a logical set of tests if you become a bit of a flow chart person for this experiment. Here's an example of a flow chart for separating and identifying anions if you're looking for an example: http://chemlab.truman.edu/chemlab_backup/CHEM131Labs/QualFiles/Figure2.gif
2. Assume you have all 5 metals - OK, there's not much chance that you'll get the sample that has all 5 metals, but you might. Even if you don't, design your flow chart and approach so that it will work for any possible unknown in this experiment.
3. The first step is the key! - Any step in which you are potentially making 2 or more precipitates represents a potential problem because you need to be able to separate those solids from one another. If you can't separate the solids from one another, it's not a useful test. For example, chromate makes a BUNCH of precipitates with the metals we are using, but there's no way (in our list of chemical tests) to separate those solids from one another. Adding chromate would be a horrible first step in your flow chart, but it might be handy later on when you only have to test for the presence of a single metal cation.
4. Positive tests - You need to have a positive confirmation test for all the metals, do not try to infer the presence or absence of a specific metal cation based upon negative results.
5. Chemical equations - If you come to lab prepared with a good plan, the "wet" part of this lab will not take three hours. This is a lab report experiment and as part of your lab report, you will have to write out balanced chemical equations for ALL of the tests you performed in the first week of the experiment. {quit rolling your eyes, it's not that hard and it's good practice!} Take advantage of your time in lab to make sure you understand the chemical equations and ask your instructor and lab assistant for feedback.

Good luck on your unknowns. One final note on your flow chart: this is one of those experiments where there is not a single correct answer. There are probably half a dozen or more variations on the "correct" flow chart that all work well.

2013-04-08

Class this week

As was mentioned in class, I am at a conference for part of this week so we willnot be having class today (Monday, 2013-April-08) or Wednesday (2013-April-10). We will also not have lab this week. I will be back on Friday, see you then.

2013-04-04

What about that ratio in the Henderson-Hasselbalch equation?

A few people have asked about some buffer questions that use the Henderson-Hasselbalch equation. In the proper Henderson-Hasselbalch equation derived from Ka, the pH and pKa are related by "log of the concentration of conjugate base over concentration of conjugate acid". Great, that works. When we're actually using the H-H equation, we can get away with a little shortcut in the calculation BECAUSE we're talking about a single buffer solution that has some conjugate acid and some conjugate base. Here's an example from an old exam:

The set-up calls for concentration {implied by the square brackets}, but when I worked through the calculation, I used moles. THAT'S CHEATING!!!! Well, not really... If we wanted to convert both of those moles to molarity, we'd divide each of them by 0.5000L, and you can go ahead and do that if you like, BUT since we're dividing the numerator and the denominator by the same volume, those volumes cancel out (mathematically) and give the same ratio as when we use moles directly. Let's look at a simpler example of this using simple whole numbers so I don't have to find a calculator... Let's say we have 2 liters of buffer solution that contains 4 moles of conjugate base and 2 moles of conjugate acid. The concentration of conjugate base is (4moles/2L = 2M) and the concentration of conjugate acid is (2moles/2L = 1M), so the ratio of conjugate base concentration to conjugate acid concentration is (2M/1M = 2). If I just use moles, the ratio is (4moles/2moles = 2). Same ratio.
So when we're using the Henderson-Hasselbalch equation correctly for buffer calculations, we can often just use the ratio of moles rather than the ratio of concentrations because it gives the same answer.



Exam questions in my email...

A couple questions have trickled in to my email box...
--Question 1------------
Quick question, what does it mean to be diprotic or monoprotic? And how do you know if something is diprotic monoprotic?
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This refers to how many acidic protons (H+) there are on an acid, how many protons can be donated. Something like HCl(aq) only has 1 H+ to donate so it's monoprotic. Phosphoric acid has 3 acidic H+ so it's triprotic. Usually it's just a matter of "count the H's" in the formula, but for organic acids (and some others...) there are H's that are not acidic. Think of the acetic acid you used in lab... CH3COOH(aq)... the 3 H's that are connected to carbon are not acidic, only the H attached to oxygen is acidic, so this is a monoprotic acid.
I will often use shorthand terminology and refer to bases as being "monoprotic" to mean that the base can accept only 1 proton or "diprotic" if the base can accept 2 protons.


--Question 2------------
On one of the practice exams there is a question that asks for a detailed titration curve. What exactly would you be looking for in something like that as far as details to include?
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You caught me... I usually don't include those drawings in the keys because I type the keys and those titration curves are a little hard to draw electronically... I've drawn a couple and they look horrible. When drawing a titration curve like this, you should label any equivalence points, indicate the species present in solution, label any pH's you can reasonably estimate (or might know...), maybe something like this:

You don't have to label all of those pH values, but again, if you know the pH values for some of those points, you might as well use them to draw a titration curve that's a little more quantitatively accurate.

Other questions, let me know.





2013-04-03

Titration question...

Email question:
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I am having trouble with this question from a previous exam: You find that 25.00mL of phosphorous acid requires 41.39 mL of Sodium hydroxide to reach the second equivalence point. What is the concentration of the phosphorous acid solution?

I understand the process of what to do in this question, but am really struggling understanding why we are supposed to use the equation that you used in the explanation: H3PO3(aq) +2 OH-1(aq)†(一)2O(l) +HPO3-2(aq)
Earlier in this problem, we broke down this into three different equations, and I don't understand why we don't use 
H2PO3-1 + OH-1 --> H2O + HPO3-2?
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The equation you list gets us from the first equivalence point to the second equivalence point. In the titration described in the equation, we're going from phosphoric acid all the way to the second equivalence point, so we're doing 2 steps of the potential 3 step process. To get to the second equivalence point, have to go through the following steps:
H3PO3(aq) + OH-1(aq)(aq) <=> H2O(l) + H2PO3-1(aq)

H2PO3-1(aq) + OH-1(aq) <=> H2O(l) + HPO3-2(aq)
So the whole 2-step process to get from the beginning of the titration to the second equivalence point has the net equation:
H3PO3(aq) + 2 OH-1(aq) <=> 2 H2O(l) + HPO3-2(aq)
That's where the 2:1 stoichiometry comes from.



2013-03-03

Questions 2013-03-03

Email questions...

"I am having trouble with determining orders of reactions.  I am pretty confused on the entire concept of them.  I am confused on questions like 3-5 on the practice tests.  I also don't understand what it means when the reaction increases by factors."
Some of this confusion may be a result of unfamiliar terminology. First, remember that for our class at this point, the only orders we are going to use are 0, 1, & 2. We determine those orders by changing the concentration of one reagent and seeing how the observed initial rate of the reaction changes. If we double the concentration of "A" {that's changing it by a factor of 2}, the observed initial rate of the reaction will either be unchanged {change by a factor of 20} if the reaction is zero-order with respect to "A", or it will double {change by a factor of 21} if the reaction is first-order w.r.t. "A", or it will quadruple {change by a factor of 22} if the reaction is second-order w.r.t. "A". Textbooks tend to really like just doubling concentrations, but there's nothing magic about multiplying by 2. You could determine the orders of a reaction by dividing the concentrations by 2 {this is also a change by a factor of 2, it's just dividing instead of multiplying}, or multiplying/dividing the concentrations by a factor of 3 or 4 or 72, it should all work the same way.

"I was looking at last spring's old chem exam 2a and for problem number 27 I got a different answer. I believe I did the math right but maybe I didn't. To get t, I did:  ln(1.03)-ln(1.67)/(-3.63x10^-2) and my answer was 14.2 and not 13.3 like you got."
This is a good calculator warning. Most importantly, when you are answering exam questions, show your work clearly and as completely as possible. If you have a calculation clearly set up correctly and just make a math/calculator error, you won't lose as many points as if you don't have your equation clearly set up. For this specific question, be sure to use parentheses on your calculator to make sure the math is being done in the order you intend. In the absence of parentheses, your calculator will evaluate multiplication/division before addition/subtraction, so if I put in the implied parentheses:
ln(1.03) - {ln(1.67)/(-3.63x10^-2)} = 14.2
But we really want that to be:

{ln(1.03) - ln(1.67)} / (-3.63x10^-2) = 13.3
By the way, if you see something on a posted answer key that seems incorrect, please let me know. I think I have everything done correctly on the keys, but there definitely could be some mistakes.


2013-03-02

Lab questions...

A couple email questions about lab reports that are due this week...

"Could you give more specifics on how to write the introduction for the lab report. "
The Lab Report Format sheet has been updated on my website to include a bit about Introductions (http://www.drbodwin.com/teaching/genchemlab/labrep2013a.pdf).
The Introduction eases your reader into the content of the lab report. Some instructors and some disciplines don't use Introductions in their lab reports, I like them because they give me an indication of your understanding of the theory behind the experiment rather than just the (sometimes robotic) procedure you performed.

"I am a little confused how to report my final K value.  I got an answer of 176.  Using range over 2, I calculated my error as 57.  Does this mean I should report my K value as 170 +/- 50 because we should only have one sig fig in our error?  I'm also running in to the same problem for my "El".  For example, one of my average El's was 4528.5 +/- 130. My understanding then, is I should report that as 4500 +/- 100?  Also, should I show the work of how I got my original and then round because of significant figures?"
Good attempt at error, you're very close. For the K value, you're good except for the rounding... 57 should round to 60. Your "error digit" here is in the "tens" digit, so you should round your result to the "tens" digit, 176 should round to 180. The result you should report is 180 +/- 60.
Your result for the Beer's Law constant is a good example of an exception to the "rules" we used for the error on K. Let's look at the error part first... your range-over-2 error was 130... for the sake of this discussion, let's say it was really 131.8274 when it displayed on your calculator. We want to round the error to a single digit EXCEPT when that digit is "1". Why is that? Well, every time you round any number, you are accepting or introducing some error in that reported value. When we round a calculated error, we are introducing error into our error. Yikes, this could get out of control! As an example, let's look at 2 cases:
Case 1: You have calculated an error of 944. Rounding that to a single digit (sig fig) gives you 900. You've essentially thrown away 44 out of 944, that's a little under 5%, and we can probably live with that.
Case 2: You have calculated an error of 144. Rounding that to a single digit (sig fig) gives you 100. Now you've thrown away 44 out of 144, that's over 30%. Not good.
So the "rule" for rounding error is "Always round error to a single digit (sig fig), unless that single digit is "1", in which case you should keep two digits (sig figs) of error." Looking back at Case 2, if we round our calculated error of 144 to two digits, we get a rounded error of 140. We've still thrown away 4 out of 144, but that's under 3%, so I think we can live with it. Depending upon the exact values of the numbers and the type of numbers they are, many scientists will advocate for keeping two digits of error if the first digit is "1" or "2"... If you work through a series similar to Case 1 & 2 for values 244, 344, 444, 544, 644, 744, 844 you'll see that the whole idea of "error on error" can be very interesting and can certainly lead to debate... For our purposes in Gen Chem, I think we can stick with "Always round error to a single digit (sig fig), unless that single digit is "1", in which case you should keep two digits (sig figs) of error."
Getting back to your original question, the error in {el} should be rounded to 130, and the reported value should be rounded to the same place the error is rounded, in this case, the "tens" digit again, so this should be reported as 4530+/-130. You ARE planning to include units on that number, right? As for the "show your work" part of your question, the answer is pretty much always yes, although for common (and hopefully trivial) things like taking an average or adding up a series of numbers, you can just say "the average is XXXX" or "the total is XXXXX". Show a sample of how you calculated one of the {el} values (with an appropriate number of sig figs, but don't worry too much about rounding at that point) and then you can report "Average {el} = 4530+/-130". Remember, it's always better to show a little too much detail of your calculation than not enough... When in doubt, show it.
Error (or uncertainty, or variability) is an important tool in our understanding of the experiments we do and it takes practice. Gen Chem is a good place to start practicing. With practice, error becomes very natural and automatic and you'll find that you start estimating error and including error without even thinking about it.

Other questions, let me know...

2013-02-28

Lab Reports

The exact requirements for a lab report will vary from field to field, class to class, even instructor to instructor. If you're looking for an example of a "good" lab report for my Gen Chem class, try this one:
http://www.drbodwin.com/teaching/genchemlab/iodinationlabreport12a.pdf
The most common problem I see in lab reports is that students don't always explain the experiment and its results in a way that makes it (somewhat) clear that the concepts behind the experiment are understood. The purpose of a Gen Chem experiment is almost never "We collected a bunch of numerical data, made some observations, and calculated/determined this result". What does that result mean? How is that result related to the concepts we talked about in class? How can that result help inform the exercises and exam questions you'll see in the classroom?
One of the harder things for students to do is get a feel for "reasonable" answers because Gen Chem level students don't have a lot of experience looking at these answers. Activation energy is a great example of this. If you have no feeling for how activation energy relates to the observed rate of a reaction, you might calculate an activation energy of 25 J/mol for some problem. Is that a fast reaction or a slow reaction? If you've only every done on-paper activation energy problems, that might be a hard question to answer. The advantage of doing experiments is that you have personally observed what happened, you've gained experience that will help you make some of these judgement calls. For the iodination of acetone experiment, the reaction is fast enough to easily observe, but it's not so fast that it blows up in your hand. The activation energy for the iodination of acetone is somewhere around 80-100 kJ/mol. If 80-100kJ/mol is the activation energy for a reaction that's "kinda fast, but not super fast", what do you think about that 25 J/mol activation energy reaction? {Pay attention to units.}
Lab experiments are a great way to build your knowledge base. When you're writing a lab report, think about the bigger picture and show the reader that you've recognized the link between classroom exercises and the first-hand experience you've had in the lab.


2013-02-23

Keeping track of concentrations


When doing kinetics and equilibrium problems, especially when you're doing an actual hands-on experiment, there are a lot of different concentrations to keep track of. The key to keeping them straight is mostly careful reading and organization, but there are a couple common definitions or descriptions that can help.
Stock Concentration
This is the one that comes up most common in a lab experiment. "Stock" refers to the large samples of reagent from which smaller amounts are taken for individual experiments. When you come to lab, the big bottles or carboys of solution that are on the side benches or in the dispensing hood are "stock" solutions and should have a "stock concentration" listed on the bottle. Most data analysis in lab begins with stock concentrations.
Initial Concentration
In either kinetics or equilibrium problems or experiments, we will often come upon something called an initial concentration. "Initial concentration" is (to me at least) quite fascinating because it's one of those things that we can do on paper that's just not physically possible to do in the real world. The "initial concentration" in a problem or experiment is the concentration of reactants after mixing everything together but before any reaction is allowed to take place. It's as if there was a little "start reaction" button on the side of the beaker, and nothing reacted until we pushed that button. In the real world, as soon as reactant solutions come in contact with each other, they begin to react, so the "initial concentration" is never the actual concentration we might observe in a reaction mixture. The initial concentration is most commonly calculated as a dilution of the stock concentration.
{In some specific reactions, we can probably observe an initial concentration because either the reaction is SO slow that we can mix the reactants before any measurable reaction has occurred, or because there's some external stimulus (like light or heat) required to make the reaction start. These reactions aren't that rare, but they're not reactions that we're likely to use very often in Gen Chem.}
Partial Pressures
When we're working with gases, we can often use Molarity to express the concentration of reactants, but we can also use partial pressures of the gaseous components. Partial pressures are a way to measure the number of moles of a specific gas in a mixture of gases. {Dalton's Law of Partial Pressures} That sounds an awful lot like a concentration... As with the vast majority of data analysis in chemistry, counting moles is the key to figuring out relationships between reacting species. If we know the concentration and volume of a liquid solution, we'll probably be calculating moles at some point. If we know the partial pressure of a gas, we'll probably also be calculating moles at some point. Chemistry is all about the mole!

The most important thing to do when approaching these problems is organization. This is especially true of equilibrium problems; if we can organize the information given in the problem, we'll be much more successful.

Did you hear about the chemist who was thinking very hard about removing excess solvent from a solution? She was concentrating.

Suggested Problems posted

A few people have asked about some suggested end-of-chapter problems from your book. I've picked some from our current chapters (http://www.drbodwin.com/teaching/genchem.php), this should be a good list for you to start with. Use these to identify the areas that you need to study/learn/review a little more, work through additional problems as needed. And always ask questions if you're unsure.


2013-02-16

Gen Chem Pre-Lab poll

For the past few years, I have used video pre-labs (posted on YouTube) as a way to introduce students to the week's experiment in General Chemistry Labs. There were  number of reasons for this, and now that I've done it for a while, I'd like to get some feedback. There's a poll on this blog (look left, right above the fish...) to collect your opinions/preferences. If you'd like to leave additional comments, leave them here, that's why I put up a post as well as the poll. I'm not trolling for praise, I really just want to know how useful the video pre-labs are compared to some other options. Setting up, shooting, editing and posting the videos takes a little bit of time and effort, and I kind of enjoy doing it, but if students don't consider it a positive contribution to the lab experience, I could probably re-direct my effort toward other projects.



2013-02-09

Exam manners

I'd like to start by apologizing for my behavior at the end of the exam. I admit that I got a little grumpier than I should have, and I will do my best to never let it happen again. Part of the problem is that I did not make my expectations clear about what I expect when students take exams in my class, so let me try to clarify.

As your instructor, I feel it is my responsibility to provide you the best possible environment in which to take your exams. We take our exams in a large room so that everyone has space. I try to be as quiet as possible while I'm walking around the room. I write time reminders up on the board for those of you who may not be able to see the clock from your seat. I try to provide an environment that is as free from distractions as possible without putting everyone in their own individual, noise-proof rooms.

In return, I expect a few things from you.
1. Arrive on time. It is disruptive to your classmates to have people trickling in, opening and closing doors, plopping down in their seats, and otherwise making noise that is distracting. I understand that there are legitimate reasons for being late, but I doubt that everyone who wandered in late had a good reason. If you KNOW you might be late for a legitimate reason, let me know in advance.
2. Turn OFF your phone. You're not going to answer it during the exam, are you? Many phones vibrate rather loudly when you're in a (hopefully) quiet room, so just turning off the ringer can still be very disruptive to the students sitting near you, trying to concentrate and perform well on the exam.
3. When you finish the exam, either leave or be silent. If there are still people taking the exam and you decide to have a chat with your friend in the exam room, you are being disruptive. In fact, you are being an inconsiderate jerk. In the future, if I hear or see anyone talking in the exam room, I will assume you are trying to cheat in some way and you will fail the course. Even if you've already handed in your exam, I will assume you are trying to cheat and you will fail the course. I understand that many of you have class in SL104 right after our class, so it seems natural to just stay in the room, but if you want to talk to anyone other than me, do it in the hallway.

All of these things can be pretty easily summed up as "Don't do anything that is intentionally disruptive to the other students in the room". That doesn't seem like an unreasonable request to me. I would hope that all of your classmates would behave similarly. If everyone is just a LITTLE considerate of the rest of the people in the room, then EVERYONE can have a more positive and productive exam experience. A couple other tips fall under the "Be prepared and responsible for your own success" category:
1. If you're taking an exam, there's a pretty good chance that you'll need something to write with. I do not use Scantrons for the regular exams in my Gen Chem classes, so it doesn't matter whether you use pencil or pen for regular Gen Chem exams. Pencils and pens are not all that expensive, so it's not a bad idea to have one of each with you. In fact, it's not a bad idea to have 3 or 4 of each with you.
2. If you're feeling a little sniffly or prone to coughing, bring a small pack of tissues and/or some cough drops with to class. {NOTE: If you're using tissues in class, please dispose of them properly. Every semester there's at least one person who throws used tissues on the floor or under their seat in SL104 and the janitors have to clean them up. That's gross. Don't do it.} If your cough or throat tickle is persistent, you might even bring a bottle of water to sip during the exam or class. Speaking of which...
3. Think before you drink. This applies to evening parties, but it's also good advice for early mornings. For most people, if you drink a quart of coffee or soda or juice or water or Red Bull, there's a pretty good chance that all of that liquid will require a stop in a restroom. It's best if this does not occur in the middle of an exam, especially when the exam only lasts 50 minutes. A quick stop in the restroom before the exam starts will let you think about the exam without wondering if your teeth really are literally floating out of your mouth.
4. Bring an appropriate calculator. My calculator policy is in the syllabus and is there thanks to one of your predecessors who downloaded most of the textbook into his TI-84 graphing calculator. In the middle of an exam, I stood looking over his shoulder as he scrolled through the text. I might have a "loaner" calculator with me at the exam, but don't expect it. Similar to the pen/pencil comment above, it wouldn't be the wackiest idea in the world to think about bringing 2 calculators to class, especially since the calculators you are allowed to use on my exams can cost less that $10. I noticed a couple people using extremely basic check-book style calculators during the exam that can't do anything more than add, subtract, multiply and divide. These probably cost $1-2 and would be a GREAT "backup" calculator to have in your bag.
5. Don't ignore the clock. But don't obsess about the clock for the whole 50 minutes. I typically don't audibly announce how much time is left until the 10 minute mark because it's disruptive to students. When I announce "10 minutes left", it means you have some time left to work through a problem or two. When I announce "2 minutes left", that means you might have time to finish the problem you're working on and give a quick look through the exam to make sure your name is on every page and you've answered any multiple choice questions that might be on the exam (even if you have to guess). If I have to announce "30 seconds left", that means you need to get out of your seat and bring me your exam NOW. There is another class waiting to come in the room, and I need to clear out to allow the next class to start on time. I will not come to you and take the exam out of your hands, I will just leave the room and you will earn a score of zero on the exam.

For those of you who have read this far, congratulations, you must actually care about your success in my class and in college and in life. In many cases, you are probably also not the students who really need to read this because you are already considerate of your classmates and doing your best to be prepared and responsible. The vast majority of you are good students, and I thank you for everything you do. You make class better by being there. Have a good weekend and I'll see you on Monday.


2013-02-07

Improving your lab scores

A few trends are emerging that are causing people to lose points in lab.Hopefully this will help.

Pre-Lab Quizzes:
1. Do them. One of the fastest ways to fail Gen Chem lab is to skip quizzes.
2. Look at them early enough to ask questions. Pre-lab quizzes are due by 8:00am on Thursdays this semester. If you email me a question at 11:45pm on Wednesday night, I will not see it or answer it in time to be helpful for you. Please do not interpret this as "Don't ask questions." I WANT you to ask questions if you don't understand a problem, but it's more effective to ask those questions on Monday or Tuesday.
3. Read the questions carefully. This actually applies to any situation in any class. There are ALWAYS little clues in the way a question is asked. They may be subtle at times, but they're there. For example, "assume volumes are additive" probably means that you might have to add some volumes together at some point. There may be times that I include information that is NOT necessary to solve a problem, but I definitely don't go out of my way to put a bunch of non-essential information in a problem, especially when it's an online quiz question.

Hand-In Assignments:
1. Circle, highlight, or otherwise clearly indicate who your Lab Assistant is on everything. When I'm trying to sort 100+ assignments, it's much easier if you've obviously done this. On future assignments, if you do not clearly and obviously indicate your Lab Assistant, you will lose points.
2. All assignments must be typed. This includes sample calculations. There are only 2 (maybe) exceptions to this: experimental set-up diagrams in lab reports, and questions that ask you to label things on a printed graph. In both of these cases, it would be BETTER to do everything electronically so you have a back-up copy to print. ("I lost my only copy of the graph as I was walking to Hagen to hand in my assignment 5 minutes before it was due.") If you don't know how to use an equation editor in MSWord or other word processing programs, look into it. They're great tools and can make really nice looking sample calculations.
3. Read the questions carefully and answer them completely. If your answer to a question takes fewer words than the actual question, you're probably not answering the question completely. Especially in lab, there are very few questions that have "Yes" or "No" as an answer. Always explain at least a little bit why the answer is "Yes" or "No".

Look back through old posts on this blog; this isn't the first time I've posted tips on how to improve scores. Use the resources you have available to you.


2013-01-23

Stoichiometry (Calcium iodate pre-lab quiz)

Lots of people with questions about the Calcium Iodate Pre-Lab Quiz. These are stoichiometry problems and require practice, practice, practice. If you didn't get enough practice in Gen Chem I, you should practice, practice, practice some more, then practice again.

How to do a stoichiometry/limiting reagent problem:
1. Write a balanced chemical equation.
2. Convert a known amount to moles.
3. Using the relationships in the balanced equation, convert the moles of the known substance to moles of the substance you're looking for.
4. Convert the moles of the substance you're looking for into whatever quantity you'd like to know.

Those are the same 4 steps you'll use for every stoichiometry problem you will ever do. The details will change, but the basic steps are the same.

12.5grams of nitrogen gas reacts with 11.6g of hydrogen gas to form ammonia gas. What is the theoretical yield of ammonia gas in grams?
Step 1:  N2(g) + 3 H2(g) -->  2 NH3(g)
If nitrogen is the limiting reagent...
Step 2:  (12.5g N2) / (28.014g/mol) = 0.4462mols N2(g)
Step 3:  (0.4462mols N2(g)) (2 mols NH3(g) / 1 mol N2(g))  =  0.8924mols NH3(g)
Step 4:  (0.8924mols NH3(g)) (17.031g/mol)  =  15.2g NH3(g)
But we don't really know if nitrogen is the limiting reagent, so...

If hydrogen is the limiting reagent...
Step 2:  (11.6g H2) / (2.0158g/mol) = 5.7545mols H2(g)
Step 3:  (5.7545mols H2(g)) (2 mols NH3(g) / 3 mol H2(g))  =  3.8364mols NH3(g)
Step 4:  (3.8364mols NH3(g)) (17.031g/mol)  =  65.3g NH3(g)
Since the nitrogen gas is all used up by the time we've made 15.2g of ammonia, there's no way we could make 65.3g of ammonia, so nitrogen must be the limiting reagent and the theoretical yield is 15.2g of ammonia gas.


What can you do besides practice, practice, practice stoichiometry problems to get better at them? Always, always, always write out your units and make sure they cancel out correctly. If your units to not cancel out correctly, it is exceedingly unlikely that you are setting up the problem correctly. Units are a wonderful little gift that lets you check your own work, I use them for every problem I set up.

Good luck and practice, practice, practice, practice, practice, practice, practice, practice, practice.

2012-12-04

Analysis of Iron Ore - question

Question from email:
I'm working on the hand in right now and I'm just a little stuck on the % Fe question. I have all of the molarities calculated out. Do I simply take the molarity of the samples and divide them by .100 L for the moles and then go to grams to percent? My percentages just seem so small. I'm getting numbers such as .22% Fe. Or do I use the .050 L instead of .100.

If you break this calculation down into its parts, it's not quite as intimidating as it might seem.
Step 1 - Using your Beer's Law plots, calculate the iron concentration in the orange solutions you made in the second week of lab.
Step 2 - How did you make those orange solutions? You should have made them from 1.00mL of the yellow iron solution you made by dissolving your iron ore sample (red powder) in acid. You can use C1V1=C2V2 to determine the concentration of iron in the yellow solution.
Step 3 - Once you know the concentration (moles of iron per liter of solution) of the yellow iron solution, you can calculate how many moles of iron are present in that solution. You made 100.0mL of the yellow solution from your red iron ore sample.
Step 4 - If you know how many moles of iron were present in the yellow iron solution, you know how many moles of iron were present in the red iron ore powder. {Hint: they just might be the same...}
Step 5 - You know how many moles of iron were present in the red iron ore sample, using the Periodic Table you can calculate how many grams of iron were in the red iron ore sample.
Step 6 - If you know how many grams of iron were present in the red iron ore sample, and you know the total mass of the red iron or sample, then you should be able to calculate the percent (by mass) of iron that was present in the red iron ore sample.

For the iron ore samples you all analyzed, the percent iron could have been as low as 9-11% or as high as 33-35%. If you get something lower than 9% or higher than 35%, double check your calculations.