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:

------

(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.