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For question #9 on exam 3 from this summer, how does the balanced equation that you give on the answer key match up with the stoichiometry that is also given? In the equation there are 2NaOH(aq) but in the stoich problem under it, there are 2mol KOH. I guess I am not understanding that relationship.

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Typo. It should be KOH in the equation.

One more...

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I have a question on #5b on the practice test. You posted the answer as

H3AsO4(aq) + 2 KOH(aq)  2 H2O(l) + K3HAsO4(aq)
(0.02500L H3AsO4(aq)) (0.127 M H3AsO4(aq)) (2mol KOH / 1mol H3AsO4) ( 1/0.03868L KOH(aq)) = 0.164M KOH(aq)

I am confused at where the .127 M H3AsO4 came from.

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The 0.127M came from the first part of the question. In 5b, you're using the arsenic acid solution you determined the concentration of in 5a to titrate a new KOH(aq) solution that has an unknown concentration.

Email questions...

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1. For question #14 on exam 1 from this summer, how is the "i" value 3? Also, do you have to add 100 to get the bp for these types of problems, but if it were a fp question would you subtract the value from zero?
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When potassium sulfate dissolves in water, it forms 2 potassium ions and 1 sulfate ion for a total of 3 particles. Remember, when you're calculating these numbers, you are most often calculating a change in freezing point or boiling point. Boiling point is elevated in solution, so the change you calculate is above the boiling point of the pure solvent; freezing point is the opposite direction.

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2. For question #18 on exam 1 from this summer, I understand that the order of [CH3I] is 1st and that [F2] is 0. But, if a concentration is 0 order, does that make it not part of the rate law expression? I just don't see why the [F2] isn't part of the problem after you say that its 0 order on the answer key to the test. Say if it was 1st or 2nd order, how would the problem be different?
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We could explicitly include a "[F2]^0" term in the rest of the problem, but any number raised to the zero power is equal to 1. Since we're only multiplying and dividing, including an extra term that's the equivalent of "1" will not affect the answer. If it was 1st or 2nd order, we would have to include that term. This would impact the units of "k" as well as changing the numerical value.

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3. How many questions will be on the exam? and how much time will we have on the test? Will it mostly be problems to work through or will there be some multiple choice as well?

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30-40 questions, you will have the full 2 hour 10 minute class period. This exam will have a slightly higher proportion of multiple choice questions than most of your previous exams, maybe over half multiple choice. Although there are more questions, the questions will be similar to the type of questions you've seen on previous exams.

Email me any other questions, I'll be checking in throughout the rest of the day and this evening. I will plan to be in my office (HA407H) by 7am tomorrow, if you have other questions you can stop in early. Good luck.

Questions

A couple email questions...

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Hello professor Bodwin,
I have two questions about the material for exam four. When assigning oxidation numbers, would diatomic ions have a charge of 0, like I2? And on problem set #10 number 3, for the second reaction, on the answer key you have that 3 e- needed to be added to both sides of the reduction half-rxn. I am confused why it is added to both sides and not just the reactant side.

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In diatomic molecules like I2, each iodine has an oxidation number of 0. This is iodine in its neutral, uncombined {with other elements} form. All the elements that are diatomic molecules (H2, N2, O2, halogens) are oxidation number zero when they are their uncombined diatomic molecule.

On problem set #10... oops, that's a mistake. I was copying and pasting reactions and I must have forgotten to delete some of those electrons. It should be:

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Reduction half-rxn: 2( 3 e^- + Cr³⁺(aq) ó Cr(s) )

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Let me know if there are more questions.

PS6 and E2 keys...

Blanks and keys are posted on my mnstate.edu page.

I won't be in my office this afternoon, I have an appointment off campus. Email if you have any questions.

Error in PS#05 Key

There was an error in the key I posted yesterday for PS#05, it should be corrected in the version that is online now. In the first problem, there are 3 gas particles on the reactant side forming 2 gas particles on the product side. The error said 2 and 2. Sorry about that, please let me know when you find errors (or things you might think are errors) so I can correct them.

PS#05 and key posted...

On my mnstate.edu page.

If you have questions, I'll be in my office (HA407H) tomorrow morning, probably until the early afternoon. You can also email, I'll answer questions here.

Have a good weekend.

Keys posted

The keys from Exam 1 and Problem Set #4 are posted on my mnstate.edu page.

http://www.mnstate.edu/bodwin/

And for today's problem set...

http://www.mnstate.edu/bodwin/courses/genchem2x/c210nps2k.pdf

Problem Set 1...

Here's the key for today's problem set.

http://www.mnstate.edu/bodwin/courses/genchem2x/c210nps1k.pdf

Make sure you try to work through the problem before checking the key, if you just read through the key you will think you understand the problems, but when faced with a blank page you might not know where to start.

----Question----

On the Spring 2010 exam, for number 17, it says to redo experiment 3 at 16.31 degrees, the rate is 7.53x10-4. How do you calculate the new k for this problem. The answer is 7.17x10-5

----Answer----

The new k is calculated from the concentrations used in experiment 3 and the new rate. When the temperature of a reaction changes, the mechanism doesn't change (over small temperature changes) so the rate law expression is the same, you can plug everything in and use the same orders you determined at the original temperature.

----Question----

I'm having trouble with #14 on the Spring 2008 exam. How do you solve it?

14. You have found the following value in a table of equilibrium constants at 25ºC:

Cu2+(aq) + 4 NH3(aq) 􀀧 [Cu(NH3)4]2+(aq) Kc = 1.7x10-13

What is the equilibrium constant for the reaction:

2 [Cu(NH3)4]2+(aq) 􀀧 2 Cu2+(aq) + 8 NH3(aq)

----Answer----

This is similar to the lead bromide question... To get from the first equilibrium reaction to the second, we have to: 1)reverse the reaction; 2)multiply by 2. To convert the equilibrium constant, that means we have to: 1)take the inverse; 2)raise it to the 2nd power. So:

K{new} = ( 1 / (1.7E-13))^2 = 3.46E25

----Question----

17. You have found the following value in a table of equilibrium constants at 25ºC:

PbBr2(s) 􀀧 Pb+2(aq) + 2 Br-(aq) Kc = 6.29x10-6

What is the equilibrium constant for the reaction:

½ Pb+2(aq) + Br-(aq) 􀀧 ½ PbBr2(s)

----Answer----

To get the second reaction, the first reaction must be: 1)reversed; 2)multiplied by 0.5. Therefore, the equilibrium constant must be: 1)inverted; 2)raised to the 0.5 power. So the "new" equilibrium constant value is:

( 1 / 6.29E-6 )^0.5 = 399

----Question-----

On the exam -summer 2007 (exam 2) #9, I wasn't sure how you got the second grams (16.246 g / 331.208) because in the periodic table it said pb is 207.2, so what do I have to do to get it to 331.208?

----Answer----

The source of the lead(II) is lead(II) nitrate. When you're looking at equilibrium problems (or kinetics problems, or any stoichiometry problems) you can usually just use the net ionic equation, but you need to include spectator ions when you're weighing out reactants.

----Question-----

On exam - fall 2004 (Exam 1), I wasn't sure how to calculate #12. For #14 I converted it to grams then to the concentration but I don't seem to be getting the right answer either.

----Answer----

That's an old exam... #12 has an error, so there is not correct answer listed, but here's how to do it. The rate of oxygen consumption is:

0.433mols/8.5oL/min = 0.0509 M/min oxygen consumed

From the balanced chemical equation, for every 5 mols of oxygen consumed in the reaction 4 moles of ammonia is consumed, so the rate of ammonia consumption is:

(0.0509 M/min oxygen)(4 moles ammonia / 5 moles oxygen) = 0.0408M/min

Last minute questions...

----Question----

I can't figure out the ppm question when it asks what is the concentration in ppm of a solution made by dissolving 14.18mg of dinitrotoluene (182.13g/mol) in 4.00L of water. I thought you took .01481g/182.13 than you took that answer divided by 4kg and than take that times 1 million but its not working out for me.

----Answer----

"ppm" is (by convention) usually thought of as a mass-mass unit, so it's part of the mass fraction family.

ppm (dinitrotoluene, DNT) = [ {mass DNT} / {total mass of sample} ] * 10^6 =

[ {0.01418g DNT} / {4000g + 0.01418g} ] * 10^6 = 3.55ppm DNT

NOTE: because the volume of water only known to 3 sig figs and is so large compared to the mass of DNT, the denominator simplifies to just 4000g.

----Question----

You don't need to know molar mass of a substance to find what: molarity, molality, normality, mole fraction, or mass percent?

----Answer----

The only reason you would need the molar mass would be to calculate moles of the substance, so anything that includes moles will require the molar mass. Because mass percent is just a ratio of masses, you can calculate it without knowing the molar mass of the substances involved. This is related to the previous problem above.

Questions...

A few email questions came in. I may be able to answer a few more of these tonight, but by about 8 or 9pm I'll be offline until morning, so if you have questions, get them in sooner rather than later.

--Question--------------

I just had a question for the freezing point depression problems..for example on the Spring 2009 exam 1b #13, as far as the calculations where do you get the #mols particles/mols ? In that problem it says (2 mols particles/mol LiNO3) how do you get that part?

--Answer--------------

Quite a few people are having trouble with this. The number of particles per solute (the van't Hoff Factor) is a measure of how many pieces each formula unit of solute breaks into when it dissolves. For molecular solutes, the solute formula is a single piece in solution; for example, when a sugar molecule dissolves in water, it's still a single sugar molecule, it doesn't break down into individual carbon, hydrogen and oxygen atoms. When ionic solutes dissolve in water, they break down into the ions that make up the formula. In the LiNO3 example given above, when LiNO3 dissolves in water it does NOT float around as LiNO3 units in solution, it breaks down into lithium ions in solution and nitrate ions in solution, so for every 1 LiNO3 unit, there are 2 particles in solution.

--Question--------------

I am not really understanding the formula for the ppm problems, for ex. (Winter 2006, Exam 1), you have times 10^6, how did you get to the sixth power?

--Answer--------------

10^6 is a million. When you're converting a fraction (mass fraction or volume fraction) to ppm {or ppm(v)}, you have to multiply by a million, 10^6.

--Question--------------

I wanted to make sure I did my work right for the mol fraction questions. If the question asks, what is the mol fraction of sugar in a saturated aqueous sugar solution at 25 C. (solubility of sugar in water is 211.4 g/ 100 mL.

Would it be 211.4 - 180.1548 (total grams of sugar) / 180.1548 = .1744

--Answer--------------

This is essentially a unit conversion problem. The given solubility means that 211.4g of sugar will dissolve in 100mL of water. The mols of sugar is:

211.4g / 180.1548g/mol = 1.173mols sugar

Mols of water in the system:

(100mL)(1g/1mL) / 18.015g/mol = 5.55mols water

So the mol fraction of sugar in this solution is:

(mols of sugar) / (total mols) = (1.173mols sugar) / (1.173mols sugar + 5.55mols water) = 0.1745

--Question--------------

I know we did our lab on molar mass today, but I am not sure how to calculate question #16 on exam 1a (spring 2009). It has atm added to it.

16. A newly discovered protein has been isolated from seeds of a tropical plant and needs to be characterized. A total of 0.137g of this protein was dissolved in enough water to produce 2.00mL of solution. At 31.68°C the osmotic pressure produced by the solution was 0.134atm. What is the molar mass of the protein? (20pts)

--Answer--------------

In lab we were looking at one colligative property (freezing point depression) to determine the molar mass of an unknown. This problem is looking at a different colligative property (osmotic pressure) to determine the mass of an unknown. The expression for osmotic pressure is:

P = MRTi

OK, we have to make an assumption here, we're going to assume that the protein is a single particle in solution, which makes the van't Hoff Factor (i) equal to 1. After that, we can start plugging in the info from the problem.

0.134atm = M(0.08206 L.atm/mol.K)(31.68+273.15K)(1)

M = 0.0053569 mols protein/L solution

We've made 2.00mL of solution so:

(0.0053569 mols protein/L solution)(0.00200L solution) = 0.000010714mols of protein

So the molar mass of the protein is:

(0.137g protein)/(0.000010714mols of protein) = 12800g/mol

Exam #1 next Friday

We've gone through Chapters 10 (Gases) and 11(Solids and Liquids), we'll look at Chapter 15 (Solutions) next. The first exam is next Friday, let me know when you have questions, I'll post answers here.

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