Questions...

From email...
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On Exam 4, Fall 2007, you have 2 answers highlighted for number 7. I don't understand how Li and P can both be the smallest; can we circle more than one answer on the exam?
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Depending upon your explanation, I would have accepted either of those answers. (That's one of the reasons more recent exams have these comparisons as short answer questions.) If you look at the electron configuration, P has more than a full shell of additional electrons, so it might seem like Li is the smaller atom, but because atomic size decreases left-to-right across the Periodic Table and P is much farther right than Li, P might be smaller. Looking at the actual data (Figure 7.22 in your textbook, page 307), P has a radius of 110pm and Li has a radius of 157pm, so it looks like in this case the left-right trend makes more of a difference than the up-down trend.

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On Exam 4, Fall 2006, you circled A for question number 7 for being the largest ions. I thought the largest ion was the lowest negative charge, and the smallest ion was a positive charge. I am not sure how to figure out what ion is the largest?
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It's not only a matter of charge, we also have to look at the size of the parent atom. For a given element, the higher the charge the smaller the ionic radius and the lower the charge the larger the atomic radius, so, for example, Ge⁴⁺ is smaller than Ge²⁺ which is smaller than Ge which is smaller than Ge^2-. Within a row, this trend is pretty reliable, but as we move far up or down the P.T. things can change. Fr⁺¹ is smaller than Fr, and F^-1 is larger than F, but Fr⁺¹ is much larger than F^-1 because the parent Fr atom is SO huge compared to the parent F atom that the change in size when they form ions doesn't make up for the original difference in size. Looking at the ions in this question, F^-1, Li⁺¹ and Al⁺³ are all definitely small, so it comes down to comparing Pb²⁺ and Br^-1. Pb²⁺ has over a full shell of additional electrons, but it's a cation. Looking at the atomic radii, Pb = 180pm and Br = 115pm, so a Pb atom is bigger than a Br atom, but a Pb²⁺ ion should be smaller than 180pm and a Br^-1 ion should be larger than 115pm. How much smaller and larger will determine the answer to this question... I accepted either answer for this question, but looking at real data, Pb²⁺ has a radius of around 140pm and Br^-1 has a radius of around 180pm, so the correct answer should be Br^-1.

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Have we talked about number 11 and 16 on the Fall 2006 exam or is it just something we should study and know?
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We have addressed these, but maybe not in exactly these terms. #11 is based upon forming stable electron configurations, so being able to write a correct electron configuration and then adding or removing electrons to give full shells, full subshells, or half-full subshells will demonstrate which ions are (relatively) stable. #16 is an application of VSEPR, lone pairs are more "sterically demanding" than bonding pairs so the repulsion in each of these molecules will affect the bond angle. We talked about this comparing methane, ammonia and water bond angles in class.

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Then I also had a question on the most polar bonds. Is the most polar the furthest apart on the periodic table?
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In general, yes, but... The polarity of a bond is a function of the difference in electronegativity of the elements involved, so it's better to look at it from that perspective, with fluorine being the most electronegative. For example, if we're comparing a P-Cl bond to a Si-Cl bond, the Si-Cl bond is a little more polar. If instead we compare a C-S bond to an O-S bond, the O-S bond is quite polar while the C-S bond is barely polar at all, even though C and S are farther apart on the P.T. than O and S.

BTW, atomic and ionic radii numbers came from your textbook and from WebElements.com, it's an interesting website if you haven't checked it out. It's much more information-based than explanation-based, but it's a handy one to keep in mind.

Lewis structures...

We are getting in to Lewis structures and looking at how the electrons are distributed in ionic and molecular/covalent substances. Lewis structures are all about practice. The rules I typically use for Lewis Structures are a little bit different from those listed in the book, so:

Lewis Structures – electron counting method

1. Add up total valence electrons in the molecule or ion

2. Draw a skeleton structure using all single bonds (usually the least electronegative atom is central, hydrogen is NEVER the central atom, some structures have multiple “central” atoms)

3. Fill the octet of all peripheral atoms (hydrogen exception…)

4. Place any extra electrons on the central atom, pair up if possible

5. Check formal charge (find missing or extra electrons…)

6. Minimize formal charge distribution (if possible) by forming multiple bonds (resonance?)

7. Check formal charge

My favorite thing about Lewis Structures is that there are a couple places in the "rules" where you can check yourself and find mistakes early without going through the entire process. Formal charge is an extremely useful tool (perhaps even more useful for those of you who will have to take organic chemistry at some point...) so make sure you're comfortable calculating formal charge.

There's a new OWL assignment posted, due next Wednesday.

Old exam keys

I'm not going to be able to post regular keys to Fall 2008 and Fall 2009 exam 3, so to let you check yourself, here are the answers to the "a" forms...

Fall 2008, Exam 3a

1 = d; 2 = b; 3 = c; 4 = c; 5 = endo,exo,exo,endo; 6 = f; 7 = d; 8 = c; 9 = c; 10 = -143kJ/mol; 11 = 127g; 12 = -39.79kJ/mol

Fall 2009, Exam 3a

1 = d; 2 = b; 3 = c; 4 = endo,exo,exo,endo; 5 = +1184.0kJ/mol; 6 = 35.42kJ; 7 = 1692kJ; 8 = +3810kJ/mol; 9 = -2375.2kJ/mol; 10 = 81.4g; 11 = +6.71kJ/mol

A few people are having problems solving problems like Winter 2006, Exam 3, #13. It sets up as:

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(0.84 J/g•ºC)(2.95x103 g)(Tfinal – (-77.91ºC)) = - (1.015 J/g•ºC)(10.00x103 g)(Tfinal – (20.00ºC))

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Solving this is algebra. I know the units all work out so let me work through this solution just using the numbers...

(0.84)(2950)(x - (-77.91)) = -(1.015)(10000)(x-20.00)

Let's combine all the numerical terms...

(2478)(x + 77.91) = (-10150)(x-20.00)

Distribute the numerical terms...

2478x + 193061 = (-10150)x + 203000

Moving all the "x" terms to one side and all the numerical terms to the other side...

(2478+10150)x = (203000-193061)

12628x = 9939

x = 0.787 = Tfinal

Another...
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How do you figure out problems like #7 on the Fall of '07 exam 3b?
7. Rust (Fe₂O₃) can be converted to iron by the following reaction:
2 Fe₂O₃(s) --> 4 Fe(s) + 3 O₂(g)
What is ΔHºreaction for this process? (ΔHfº = -824.2^kJ/_mol for Fe₂O₃.)
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This is an enthalpy of reaction problem, although it might seem like there's not enough info given. The products here are both uncombined elements in their standard states, so their standard enthalpy is zero. That makes the enthalpy of this reaction:
2(824.2^kJ/_mol) + 4(0^kJ/_mol) + 3(0^kJ/_mol) = 1648.4kJ (or ^kJ/_mol, or ^kJ/_{mol rxn}, or ^kJ/_rxn, see the discussion of this below...)

"Moles of Reaction"

Email question:
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I know you answered this question in class on Friday but I'm still uncertain on how you get mols of rxn? Can you give me a hypothetical on the equation we did on Friday from exam 3a from Fall '08? In this it's 1 mol rxn per 2 mols of haxane.
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OK, this one always causes some trouble, largely because I try to tie together the units that we typically see on enthalpy and the balanced equations. It might be easier to think of it simply as "reaction" rather than "moles of reaction", so we could look at something like:
2 H₃PO₄(aq) + 3 Ca(OH)₂(aq) --> Ca₃(PO₄)₂(s) + 6 H₂O(l)
Calculating {delta}H_rxn for this process, we get...
2(1288.3^kJ/_mol) + 3(542.8^kJ/_mol) + 6(230.02^kJ/_mol) + 1(-4120.8^kJ/_mol) + 6(-285.8^kJ/_mol) = -250.48^kJ/_mol
{Note: These are numbers I pulled from a table similar to the one in your textbook. Change the sign on reactants because these are being consumed in the reaction, not formed.}
Since {delta}H is negative, this reaction is exothermic, but what exactly are those units? Remember when I ran through one of the first enthalpy problems in class I used very complete and expanded units, let's just look at the first term here. The units on the enthalpy of formation for phosphoric acid are "kilojoules per mole of phosphoric acid formed". If we want to properly add these terms together, they have to have the same unit, so we have to convert/relate "moles of phosphoric acid" into something that is consistent throughout this problem. That's where the units on the "2" become important, even through they are often left off. That "2" comes from the balanced chemical equation and is really "2 moles of phosphoric acid per balanced chemical equation" or "2 moles of phosphoric acid per reaction". OK, so we do that for every term in the problem and then add them together to get the final answer with units of "kilojoules per balanced chemical equation" or "kilojoules per reaction" and everything is great... except that these enthalpies of reaction are often reported with units of "kJ/mol". Mole of what? In the above reaction, we could say it's per mole of calcium phosphate because for each "reaction" there is 1 mole of calcium phosphate formed, but that seems to limit us to problems that only deal with calcium phosphate. Here comes the magic unit "mol of reaction". Using the terminology above, we can say that one "balanced chemical equation" is one "mol of reaction".
Many (most? maybe all?) textbooks get around this problem by being quite explicit in the way they present thermochemical reactions, in fact your textbook has a section in Chapter 6 called "Thermochemical Expressions" {Sec. 6.5, p 230) that gives a nice example. If the {delta}H for a chemical equation is shown right next to the balanced chemical equation to which it refers, it can be implied that the {delta}H is valid only for the exact balanced equation shown, so the "per mol" or "per mol rxn" is often omitted and {delta}H is just reported with units of "kJ".

OK, after that LONG explanation, let's try a shorter answer. You can think of "mol of rxn" simply as "rxn". Each time the reaction happens once (as balanced), the calculated heat is liberated or consumed. The reason I tend to use the "mol of rxn" label is because it naturally leads to the question "What reaction?" which means that every enthalpy you calculate MUST be related to a specific balanced chemical equation.

Other questions, let me know...

A couple notes...

#1. If you look at old exams from a couple years ago, you might see some questions about photon energy and deBroglie wavelengths. These are topics from the next chapter and will not be included on this exam.

#2. I may post some answer keys for last year's exams before Monday, but I can't guarantee that I'll have time. I will try.

#3. As I mentioned in class, I misplaced my regular Casio calculator. If you happened to slip it in your bag when you visited my office I'd love to get it back. Thanks.

Email any questions, I'll answer them here on the blog. Have a good weekend, if you're looking for a study break the soccer team is home tomorrow (noon, v. Wayne State) and Sunday (1pm, v. Augustana). This may be the last weekend with nice weather, try to enjoy it. It might snow Tuesday or Wednesday night...

In-class problem

I think a number of people were confused at how to approach the problem we did in class on Wednesday. The problems was:

"fuel" reacts with oxygen to produce carbon dioxide gas and water gas. If 10.00g of "fuel" is burned in excess oxygen and all of the energy is transferred to 5.00L of water initially at 11.24degC, what is the final temperature of the water?

I'll pick a fuel none of you had, benzene, C6H6(l). This is a coupled-systems problem, the benzene will burn to produce/liberate heat in an enthalpy process, then the 5.00L sample of water will absorb the heat in a heat capacity process. Start with a balanced equation:
2 C₆H₆(l) + 15 O₂(g) --> 12 CO₂(g) + 6 H₂O(g)
Now calculate the {delta}H for the reaction from the standard enthalpies of formation found in the table in the back of your book.
2(-49.03^kJ/_mol) + 15 (-0^kJ/_mol) + 12(-393.509^kJ/_mol) + 6(-241.818^kJ/_mol) = -6271.08^kJ/_{mol rxn}
NOTE: change sign on reactants, don't change sign on products.
10.00g of benzene does not represent a "mol of reaction", so we need to scale that number to the amount of fuel being used:
(10.00g C₆H₆) / (78.113 ^g/_mol) = 0.1280mols benzene
(0.1280 mols C₆H₆) (1 mol rxn / 2 mols C₆H₆) = 0.06401 mols rxn
(0.06401 mols rxn) (6271.08^kJ/_{mol rxn}) = 401.4kJ of energy released by the reaction

OK, now the heat capacity part of the problem. I'm putting 401400J of heat into this 5.00L sample of water and changing its temperature. Use the units on heat capacity to set up the problem correctly:
(401400J) (1 g.degC / 4.184J) (1 / 5000g) = 19.19degC
This is the change in temperature, so the final temperature of the water must be (11.24+19.19)degC = 30.43degC

Other questions, let me know...

This week and new OWL

This (short) week we have started Chapter 6, Thermochemistry. Thermochemistry is all about heat: where it is, where it's moving, how much is moving, what it does when it moves, etc. As with everything, these problems become MUCH easier with some practice, so make sure you work some examples. To help you with that, there's a new OWL assignment posted. I've included a few more simulations and other rather visual modules in this set of assignments, if you're struggling with some of these energy issues it will help to work through some of these, be sure to take advantage of these opportunities rather than just clicking through to get the right answer.

Our next exam will cover ONLY chapter 6 and is just over a week away (October 25th). Do not wait until the last minute to study. I expect a significantly better average on Exam 3, but that won't happen if you don't start working on this material this weekend.

If you need a study break, there's a football game AND a volleyball game at home today (Saturday). What happens to the kinetic energy of a wide receiver and a linebacker when they collide? When the libero gets an awesome dig and the volleyball goes straight up in the air, how are the kinetic and potential energy of the ball changing? What kind of energy is represented by the glucose molecules in the energy drinks that the players drink during the game? See? There are even fun energy problems to consider when you're watching sports...

More questions...

A couple question...

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I was just wondering about the oxidation number. example: PH3 P has 3+ and H has 1- so all you need to do is subract the two or do i have the method wrong. thanks
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I'd start from the other direction on this, hydrogen is almost always oxidation number +1, so if there are 3 hydrogens at +1, and the molecule is neutral overall, then the phosphorus must be oxidation number -3.

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I was doing one of the "Conceptual Exercise" problems and can't figure out how they came up with the answer that they did. It is 5.4 on page 175 part B. I came up with H^+ + OH^- yields H_2_0. It looks like that is the answer but then it has another equation for an answer as well. I know I am forgetting something simple but I can't figure out what it is and it's driving me nuts. Please help me out and thanks in advance.
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These are a bunch of acid-base net ionic equations, so I'll address them all.

HCl(aq) + KOH(aq) --> H₂O(l) + KCl(aq)
H⁺(aq) + Cl^-(aq) + K⁺(aq) + OH^-(aq) --> H₂O(l) + K⁺(aq) + Cl^-(aq)
Chloride ions and potassium ions don't change, so they are spectators and the net ionic equation is:
H⁺(aq) + OH^-(aq) --> H₂O(l)

H₂SO₄(aq) + Ba(OH)₂(aq) --> 2 H₂O(l) + BaSO₄(s)
This one should look familiar...
2 H⁺(aq) + SO₄^2-(aq) + Ba²⁺(aq) + 2 OH^-(aq) --> 2 H₂O(l) + BaSO₄(s)
Since water is a molecule and barium sulfate is a precipitate, these both stay together in the full ionic equation, so the net ionic equation doesn't have any spectator ions here...
2 H⁺(aq) + SO₄^2-(aq) + Ba²⁺(aq) + 2 OH^-(aq) --> 2 H₂O(l) + BaSO₄(s)

CH₃COOH(aq) + NaOH(aq) --> H₂O(l) + NaCH₃COO(aq)
You could also represent acetic acid and acetate ions as CH₃CO₂H/CH₃CO₂^- or HC₂H₃O₂/C₂H₃O₂^-. Acetic acid is a weak acid, so it should not be split up in the ionic equation:
CH₃COOH(aq) + Na⁺(aq) + OH^-(aq) --> H₂O(l) + Na⁺(aq) + CH₃COO^-(aq)
The sodium ion is a spectator, so the net ionic is:
CH₃COOH(aq) + OH^-(aq) --> H₂O(l) + CH₃COO^-(aq)

Other questions, let me know, I'll be checking email all evening. I'll also post answers first thing in the morning. Good luck...

From email...

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On the exam, will you write the chemical formula as words or shorthand?
for example, hydroiodic acid as opposed to HI

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Yes. You will most likely see both, depending upon the goal of the problem. In some cases I want to see if you can write balanced formulas and balanced reactions, so I will give you the name of the chemical and expect you to write the correct formula. In other cases, I want to see if you can do the stoichiometry from a given chemical reaction so I'll use a chemical formula. In "special" cases, there are only a few names I expect you to know. You should know the names and formulas of all strong acids, ammonia, etc.

Almost exam time...

We've finished up chapters 4 and 5, exam next Wednesday. Let me know if you have any questions or anything specific you'd like to review on Monday in class. I've posted problem set #4 and the answer key on my mnstate.edu page, let me know if there are problems.

Have a good weekend and good luck preparing for the exam.

Titrations are stoichiometry problems

Today we talked about titrations. Titrations are (usually) used to determine the concentration of an acid or base in solution, although they can be used in a LOT of places other than acid-base reactions. Treat them the same way you approach all stoichiometry problems.

The extra SI session next week will be next Monday at 6:30 in BR269. We have an exam next Wednesday.

Catch up...

Yikes, it's been a while...

Since the last exam, we've been looking at stoichiometry and various classes and types of reactions. This is chapter 4 & 5 material from the textbook. The heart of every stoichiometry problem is the mol-to-mol conversion made using the coefficients from the balanced chemical equation. The rest of the problem is all about getting into and out of moles.

We're currently in the middle of reaction types, we've talked about precipitations and molecule-forming reactions so far. We'll finish up acids and bases on Monday and move on to everyone's favorite reaction type, redox!

Have a good weekend, the weather's supposed to be nice. If you need a study break, the soccer team is at home this weekend on both Saturday and Sunday at 1pm. What could be better than a beautiful fall day and a little soccer?

Another question...

From email...
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Just wondering if you could tell me what I am doing wrong for #8 on last years chem 150 test. The question is "What is the formula weight of nickel(II) nitrate?"
Heres what I did:
Nickel is 58.69 and there are 2 so I took (2)58.69. Nitrate is NO3 so Nitrogen is 14.01 and Oxygen is 16.00 and there are 3 so it would be 14.01+3(16). This leads me to:
2(58.69)+14.01+3(16) which gives me 179.39. This is the wrong answer...on the answer key the answer is 182.70. Just wondering what I did wrong. Thanks!
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A lot of people trip up on this one. Remember, when there's a roman numeral after a metal, that tells you the charge of the metal cation, it does not tell you how many of that cation are in the balanced formula. For this one, the nickel has a +2 charge. Nitrate has a charge of -1, so to balance the charge of the formula, we need two NO₃^-1 for each Ni⁺², Ni(NO₃)₂, so the formula weight of nickel(II) nitrate should be:
(58.69g/mol) + 2(14.007g/mol) + 6(15.999g/mol) = 182.70g/mol

Other questions, let me know...

Exam questions...

Already a couple questions...

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Is the empirical formula just the smallest whole number ratio? What if it came out C=1.5 H=2.5 O=3 or something like that? Is that still the empirical formula? or would the empirical formula be C=3 H=5 O=6, and then work from there to get your molecular formulas?!

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Yes, the empirical formula is the smallest whole number ratio, so in your example the most correct way to report the empirical formula would be C₃H₅O₆. The molecular formula would be some multiple of that and you'd have to be given more information in the problem to determine the correct molecular formula.

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For percent composition questions with multiple elements, will be expected to have the elements in the correct order in the final answer? For example: KMnO₄ instead of say MnKO₄....Or will the main concern be that we achieved the correct amount of each element?

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Quite a few people have asked me about this and in general the order doesn't matter. The only place your should pay attention to the order and groupings is in the formulas of ionic compounds and polyatomic ions. When writing the formula for an ionic compound it is usually best accepted practice to list the cation first, followed by the anion, and you should always write polyatomic ions as their common formula is written. In your example, since permanganate is a polyatomic ion, it should always be written together as "MnO₄^-". Since this is an ionic compound, the cation {potassium ion} should also be written first, so this should be written KMnO₄. That's not a result of it being a percent composition problem, that's the naming convention for ionic compounds.

If you have other questions, let me know, I'll post answers to the blog ASAP.

Suggested problems from the text...

I've posted some suggested problems from the textbook on my web page. These are not required problems, they will not be collected, they will not be graded, they are merely problems that I think are good ones from the end of each chapter if you'd like some additional practice.

Almost exam time...

Sara was unable to get a room, so there will not be an extra SI session tonight.

Next Monday is your first exam, be sure to look at the old exams on my webpage to get an idea of what to expect. Also, make sure that if you intend to use a calculator on the exam it cannot be a graphing/programmable calculator, or a cell phone calculator, or an iPod/iPad calculator, or any other networked or interactive "calculator experience". Sit with at least 1 open chair between everyone.

Today we worked through another percent composition question, this time using percent composition to determine the formula of a salt that someone carelessly neglected to label completely.

If you have questions you'd like to go over in class on Friday, let me know in advance or just bring them to class and we'll go through as many as we have time for.

The cusp of the Labor Day weekend...

There is (was?) a problem with the time in OWL, it has been set to east coast time. I've contacted a few people at OWL about this and it should be fixed soon if it's not already corrected. If it's not corrected and you happen to be doing your OWL assignment between 11:00pm and 11:59pm, please continue to do the assignment even if you get a "past due" message. When the clock is corrected, the system will automatically re-grade your assignment.

There's new OWL posted, due next Friday.

Lab hand-ins turned in by 3:00pm on Mondays get a bonus point, BUT next Monday is Labor Day, so the university will not be open (at least not completely). Therefore, for this assignment only you will get a bonus point as long as your assignment is in by noon Tuesday. The regular deadline, Wednesday at 3:00pm, is still in place.

Today in class we looked at ionic formulas again and checked out how to calculate percent composition from a chemical formula, and a chemical formula from percent composition. One question a few people had about the example I did was "Where did that 78g/mol number come from?" That's a number that would have to be given in the problem for you to use, it's not something you would calculate in that specific example.

If you have any input about the SI schedule, please let Sara know. If none of the current times fit in your schedule, or if some other time would work better in your schedule, she is open to adjusting the SI times to better serve as many student as possible. Keep in mind that Sara is a student, so she also has a lot of time commitments for classes and study time and other activities (and she probably likes to sleep an hour or two every night...).

Enjoy your 3-day weekend, be safe and I'll see you on Wednesday.

Announcements and ionic compounds

Today we had LOTS of announcements, so:

Dr. Wallert announced the bi-weekly (that's every other week, not twice a week...) Biochemistry and Biotechnology Seminar Series. There are some very interesting talks and topics planned, so keep them in mind. There are poster hanging around the building, look for the ear of corn with some funny looking kernels for more details.

Tonight is the first meeting of the Chemistry and Biochemistry Club. SL104, 7pm. CBC is a great opportunity to get involved on campus and in the community, and gives you an opportunity to get to know other chemistry and biochemistry majors. There will be pizza and soda...

Keep an eye on the schedule, our first exam is coming up on September 13th...that's less than 2 weeks away. It will cover Chapters 1-3. Don't forget about the calculator rules: no graphing calculators, no sharing calculators, no cell phones, no iPods/iPads. If it's more complex than a basic $15 scientific calculator, it's probably not allowed. If you're not sure, check with me BEFORE the exam.

Today in class we looked at ionic compounds. You will be required to know any polyatomic ion listed in Table 3.7, page 88 of the textbook.

SI Information

There was an error in the room posted for SI, sorry for the confusion. The correct rooms are now posted.

We are pretty much through chapters 1 & 2, getting a good start on chapter 3. I will post some suggested problems from the textbook as soon as I have a chance to look through them. These will not be required problems, but are good examples if you'd like a little more practice.

Welcome to Fall 2010!

Welcome to the Fall semester! I'm looking forward to a great semester. There is already an OWL assignment (actually 2) posted, take a look at them as soon as possible. The first assignment is sort of an introduction to the OWL system. The second assignment is a math review. The math review assignments use applied examples, but if you look at the problems carefully you'll see that you really don't have to understand anything about pH or radioactive decay to do the problems, they're just algebra problems that you have to plug values into and solve. They're also good practice using your calculator.

The direct link to initially register in OWL is:

http://www.cengage.com/owl/

Once you're registered, you can use the link in the upper left corner of this page to get to the login page. Good luck and let me know if you're having any problems.

See you Wednesday.

PS keys are all posted

I have all the problem set keys posted and I have a couple exam keys posted as well on my mnstate.edu page. Let me know if you have questions, I will be out of town this weekend but I should be able to check email and update questions to the blog.

PS keys posted

The problem sets and keys from this week are posted on my mnstate.edu web page. Let me know of you have questions.

Also a note, you should know strong acids and strong bases, there aren't many of them.

Strong acids: perchloric, nitric, sulfuric (first Ka), hydrochloric, hydrobromic, hydroiodic

Strong bases: any soluble hydroxide (alkaline metal hydroxides, somewhat alkaline earth hydroxides) that can be at least ~1M. {Look at Ksp...}

If it's not one of these "strong" species, you can probably assume that it's weak, unless something in the problem tells you otherwise. For example, if you are told that an acid has a Ka = 200, it's strong even if it doesn't appear on "the list".

PS04 and PS05 posted

The problem set answer keys are posted. Question? Let me know...

Email questions...

A few questions have arrived in my email box. Don't forget to check the answer keys that are posted in my mnstate.edu web page. Here are the questions...

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Practice set one:
- on #4, I know that it is a three-step question. However, I can't get from 100C to 136.19. The H/fusion H/vaporization has me confused.
- #5 and 6, I don't know how to set them up. I'm sure I can figure it out, but I don't know from where to start.

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On #4, you need to do a heat capacity problem to heat the liquid water up to 100degC, then a {delta}H(vaporization) problem to convert all the liquid water to gaseous water, then another heat capacity problem to get from 100degC to 136.19degC. I think when I wrote the answers up on the board I had a math error in the steam part, check the answer key for the correct numbers.

For #5 and #6, start with the reactant or product that you have enough information about to calculate the rate, then figure out the rate with respect to the other reactants and products using the stoichiometry of the balanced chemical equation. For example, in #5 you react 3 mols of hydrogen gas for each mol of nitrogen gas, so the rate of hydrogen consumption should be 3x the rate of nitrogen consumption.

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Practice set two:
- #1, I don't understand what to do with the grams and how to make it into a reaction order.

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Rate laws relate concentrations to rates, so you will (at some point) need to calculate initial concentrations based upon the grams of starting materials given. You can actually figure out the orders of the rate law expression using grams, but to get a proper value of "k" with concentration units, you'll need to convert.

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I'm not very good at conversions and I am stumped on how you would convert mL or L to kg?..for problems like finding molality? In particular question 2 on the first problem set?

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Volume and mass are related to one another by the density of a substance. Densities are often reported in units of "g/mL", so work through whatever unit conversions you need to get through the density. For example, the density of chloroform is 1.5 g/mL, so if I want to know the mass of 2.0L of chloroform in kg...

(2.0L)(1000mL/1L)(1.5g/1mL)(1kg/1000g) = 3.0kg

As with most of the problems we'll work with, if you're having trouble getting things set up correctly, make sure you keep an eye on the units. It won't always be magic, but you'll have a better chance of setting things up correctly if your units work out.

Good luck and let me know of there are other questions.

Problem Set answer keys

I've posted the answer keys for this week's problem sets on my mnstate.edu web page. If you have questions about the exam, let me know and check here for answers...

Summer 2010

Welcome to Chem 210 - General Chemistry II, Summer 2010

Check back for class announcement, answers to questions, and other info.

Exam 3 keys

The keys for Exam #3 are fixed and should work now, let me know if there are still problems. Also, I think the blank exam #3's that were posted were old copies that still had a typo in the big titration problem (asking for the concentration of selenic acid instead of potassium sulfite...). I posted the corrected blank exam #3's.

Let me know if there are other questions, I'll be in my office for at least part of the day today...

A couple process questions...

Questions:

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First off, for some reason, and this might be something just happening at this time, exam 3a and 3b answer keys are not available online.

Also, I was wondering if we are going to be given any formulas on the final in addition to what was listed on exam 4a with the periodic table.

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Yep, there's something not right with the keys for 3a and 3b, I'll get them fixed first thing tomorrow morning when I get to my office.

The cover page for your final exam will contain the same information as the cover page for Exam 4. If there are any other constants or equations you would like to see on the cover page, let me know and I'll consider it.

OK, everybody concentrate...

Email question...

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I am studying for the exams, and I just had a random question. Why are there Molarity and Molality? Aren't they almost always going to be the same number since 1 g H2O = 1 mL H2O? I guess I'm just curious of when you would choose one measurement over the other.

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In practice, yes, they are almost always the same (to our ability to measure them...), but the reasons we need them both are:

1. Molality is mols of solute per kilogram of solvent. Since the mass of a substance does not depend upon its temperature, molality is a concentration unit that is independent of temperature.

2. The density of pure water at (somewhere in the neighborhood of) 17degC is exactly 1g/mL, with many zeros of sig figs after the decimal point. For most dilute aqueous solutions at moderate temperatures, it's pretty safe to assume that the density of the solution is pretty close to 1g/mL, but there aren't nearly as many sig fig zeros after the decimal point. If you're working with a solution that's not dilute or not relatively close to that 15-20degC temperature, the density will start to vary significantly. That means the molarity (M) will change, but the molality (m) should not, therefore the conversion between these two units will not be 1:1. For an example you can try at home (or Kise...), take a little bit of water and add sugar or table salt to it until you get a very concentrated solution. Carefully (and slowly) pour that solution down the inside wall of a glass of "pure" water... the solution is more dense than the pure water and will sink to the bottom without mixing (if you're very careful).

3. What if your solvent is not water? In your freezing point depression experiment, your solvent was cyclohexane, so there was absolutely NOT a 1:1 relationship between the M and m of the solutions you were making.

Again, in many "real" situations the difference between M and m in an aqueous solution you're using will be so small that it's ignored. Similarly for other concentration unit conversions, we tend to simplify them in practice because the solutions we are most often working with in a lab (and especially a biology or biochemistry lab) are probably aqueous and probably relatively dilute. As long as you understand the assumptions you're using to simplify your in-lab calculations, you should be able to adjust to different situations or solvents...

Redox...

A question from email:

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Hey Dr. Bodwin I had a question on balancing equations. I can't tell if this equation is a redox or not

2 C2H2(g) + 5 O2(g) 􀀧 4 CO2(g) + 2 H2O(g)

I was looking for some guidelines on how to tell if an equation is redox or not, and if it is I am also looking for help on how to write the half reactions.

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OK, I'll answer the easier part of this question first. If a reaction is redox, then the oxidation numbers of some (or all) of the elements have to change. Let's look at ox#'s here...

In C2H2(g), let's use the "rules"... if each of the H's is +1 and the molecule is neutral, then the two C's must total -2. (The sum of all the oxidation numbers has to equal the charge of the molecule or polyatomic ion.) That means each C must have ox# = -1. {I'm assuming that the two C's are identical.}

Since O2(g) is an uncombined element in it's natural/standard state, the ox# = 0.

For CO2(g) it's back to the rules... if each oxygen is -2, then the carbon must be +4.

For H2O(g), the rules pretty much do all the work for us, H is +1, O is -2.

So in this reaction, carbon is going from -1 to +4 and oxygen is going from 0 to -2, there are changes in ox#'s, therefore this is a redox reaction. {Which one is reduction and which one is oxidation? Hmm...}

Now for the harder part of this specific reaction... writing half reactions for a combustion reaction like this is a little more involved than writing half reactions for metal/metal redox reactions because the oxygen is incorporated into both products. That means that our stepwise procedure for balancing redox reactions doesn't work all that great here... fortunately, hydrocarbon combustion reaction can usually be balanced by inspection/trial-and-error. If you have to balance a redox reaction (and write out half reactions) on the final exam, it will not be a hydrocarbon combustion. You should still be able to assign oxidation numbers and identify which element is being oxidized or reduced, but these are probably easier to balance by inspection.

Other questions? Let me know...

Strong acids and bases

The strong acids and bases you will be expected to know for the exam are:

Strong acids: perchloric, hydrochloric, hydrobromic, hydroiodic, sulfuric (first Ka), and nitric

Strong bases: any very soluble hydroxide can be considered a strong base, so any first-column hydroxide (LiOH, NaOH, KOH, etc). The second-column hydroxides are usually soluble enough to be considered strong (magnesium hydroxide, calcium hydroxide, barium hydroxide), but I don't expect to make any fine distinctions on those, if I want to use a strong base on the exam it will be either NaOH or KOH.

Keys posted...

The answer keys are posted for last year's exam 3.

Someone asked about the pH range for a good buffer (part of #5 on last year's exam). To be able to make a good buffer, the pH of the buffer should be within 1 unit of the pKa of the weak acid component of the buffer. If the pKa of the weak acid is 5.62, then you can make an effective buffer at any pH between 4.62 and 6.62.

Titrations and flood preparation...

We're into titrations and just touched on buffers. More on that after break. The Tuesday SI session after break has been cancelled. To address the potential flood issues, the University has asked all instructors to provide an addendum to the syllabus for each class. The syllabus addenda for Chem 210 and 210L are below. If you have any questions, let me know. Have a good and safe Spring Break.

Chem 210

In the event of class cancellation or campus closure due to flood, material will be presented online using assorted methods (videos, online notes/lectures, etc.). For information, refer to the class blog (http://msumgenchem.blogspot.com/) and/or Dr. Bodwin’s website (www.mnstate.edu/bodwin). There will continue to be Mastering Chemistry assignments, the exams may be rescheduled or reformatted. Dr. Bodwin will remain in contact via email (bodwin@mnstate.edu).

Chem 210L

In the event of class cancellation or campus closure due to flood, material will be presented online using assorted methods (videos, online notes/lectures, etc.). For information, refer to the class D2L page (https://mnstate.ims.mnscu.edu/shared/login.html) and/or Dr. Bodwin’s website (www.mnstate.edu/bodwin). If more than 1-2 weeks are lost to flood, “at home” experiments will be posted and quizzes/assignment will be posted in D2L. Dr. Bodwin will remain in contact via email (bodwin@mnstate.edu).

Acids and Bases...

We're into acids and bases...

There's a new Mastering Chemistry assignment posted due next Friday.

Reaching equilibrium...

When you study the kinetics of a reaction for a long enough time, the reaction reaches equilibrium. We've hit the high points of kinetics (rates, rate laws, integrated rate laws, activation energies, mechanism) and now we have reached equilibrium. Equilibrium is the state where the forward and reverse rates of reaction are equal, so although those reactions both continue to occur, the concentration of reactants and products remains constant.

Looking at the schedule, our next exam is coming up Wednesday Feb. 24th. Because of my schedule, it is extremely unlikely that I will ever be able to grade a Wednesday exam by class time on Friday, and given that kinetics and equilibrium are both pretty large topics, I think it would be best to move that exam back to Friday Feb. 26th. If anyone has a significant conflict with that day (other exams, plans to be out of town, etc.), let me know. At this point, the exam will still be scheduled for Wednesday Feb 24th, but if I don't hear of too many conflicts we will probably move it to Friday Feb 26th.

Arrhenius says...

We looked at a few examples of rate law problems and have determined rate law orders by comparing experiments with differing initial concentrations of reactants.

Kinetics is all about probability, and that probability is dependent upon the activation energy required to get a reaction started. Activation energy is calculated using the Arrhenius equation. We looked at the compact form of the Arrhenius equation (very elegant in its simplicity, but not always very practically useful) and derived a couple variations that are more useful, the comparative and linear forms.

As with so many things, kinetics problems become easier with practice, so keep up on your Mastering Chemistry and take a look at some textbook problems. The new MC assignment is posted.

Have a good weekend.

There oughta be a (rate) law!

We looked at average rates and instantaneous rates, almost did some calculus but not quite. The only unique instantaneous rate is the initial rate and we looked at Rate Laws as a relationship between the initial rate of a reaction and the concentrations of the reactants. Rate laws give us insight into the molecule-scale pathway of the reaction.

I handed back exams, if you didn't pick yours up I will have them with me in class on Wednesday and Friday.

Kinetics began, slowly...

Today we started talking about chemical kinetics, the study of the rates and pathways of chemical reactions. Kinetics can be explained using Collision Theory, and we're in the middle of looking at different types of rates.

Have a good weekend, I'll see you on Monday and I'll have the exams graded.

Exam Wednesday...

Today we talked about colloids a little and reviewed for the exam. I problem came up that I made a mistake on, but it wasn't the mistake I thought it was. We went through a quick osmotic pressure problem with approximately 1M MgCl2 in water and the answer I calculated on the board was around 74atm. That answer is correct. I was thinking about a vapor pressure problem that someone had asked about, and for a solution to have a vapor pressure of 74atm would be just goofy, at least under any reasonable conditions. An osmotic pressure of 74atm is also pretty high, but that's an indication of just how strong a force osmosis can be.

Because of the university closing, tonight's SI exam review session has been cancelled. If you have questions, feel free to email me, I will answer to the blog.

Stay warm, stay safe, and enjoy(?) the snow.

Colligativity...

Today we wrapped up the colligative properties and concentration units we're going to talk about before next Wednesday's exam.

There is a new Mastering Chemistry assignment posted, due Tuesday.

More IMFs...

Today we looked at how intermolecular forces can be used to predict some properties of matter like viscosity, surface tension, and capillary action. We also looked at vaporization and vapor pressure and discussed heating/cooling curves and phase diagrams.

We will not be meeting for lab this week BUT there is an assignment up on D2L covering the syllabus and course policies. It's due by Friday at 11:59pm, but you should be sure to look at it before then, perhaps during the 3 hours of open time you have tomorrow because we are not meeting for lab...

Day One

Today we started looking at intermolecular forces, we'll continue on Wednesday.

Don't forget to sign on to MasteringChemistry and take a look at the current assignments as soon as you have a chance.