#chem111

[blanketdiaries]

Trying to bring my chem grade up by actually going to lecture and taking notes. #college #chemistry #chem111 #lecture #notes (at Colorado State University)

[blanketdiaries]

Chem notes #inclassnotes #chemistry #chem111 #colostate #college #notes #studygram (at Colorado State University)

[science-of-anthropology]

Themochemistry - Part 2

In this bit, I'll cover the other two types of thermo problems you might see (part one is here.).  But first, I'm going to add something that I forgot in the previous entry and meant to have.  

Specific heat - the amount of heat necessary to move one amount of a substance one degree.  So, water's specific heat is 4.18 J/(g*C).  That means that it takes 4.18 Joules to move one gram of water one degree.  Sometimes you will see this with kJ or kG as the unit.  It takes .00418 kilojoules to move one gram of water one degree.  The units here trip me up on a regular basis, so just make sure you read the question very carefully and give the answer in the right form.  

Heat Capacity - The amount of heat required to raise a sample of something one degree.  The official formula for it is q = CΔT.  HehehehEHEHehe....it looks like cat.  C = a given value of specific heat (IE, the 4.18J/(g*C).  This isn't a thing that's used a whole lot, but I figured I'd toss it in.  Specific heat is much more important.  

First, we'll do coffee cup calorimetry.  I hate these problems.  They're boring and needlessly complicated and serve virtually NO purpose except for to help you do your coffee cup calorimetry lab.  You probably won't see them again after this.  

Ok, so the things you'll need to be able to do a CCC problem: mass, the ability to calculate molar weight, and a change in temperature (which is designated as Tfinal - Tinitial, or Δt.  In physics this symbolism gets confusing when you deal with tension, but this isn't physics class, YAY!).  There's no real way for me to explain it except to do it, so here's an example problem: 

When a 5.87-g sample of solid ammonium nitrate dissolves in 90.77 g of water in a coffee-cup calorimeter, the temperature drops from 23.0oC to 18.4oC. Calculate ΔH (in kJ/mol NH4NO3) for the solution process assuming specific heat of the solution is the same as that of pure water*. 

*The reason is because there's not enough of the solute to effect the specific heat of the water.  Highly concentrated solutions don't follow this rule, but this is chem 111 so almost everything you see will.  If they don't, you'll be given a different specific heat.  

Another note about symbols: Because ΔH is already taken by enthalpy, changes in heat are noted by writing q.  Why the thing that starts with an H didn't get to claim ΔH, well, that's beyond me.  Chemists are super weird about names.  

Ok, so on to figuring this out.  Here's what's important: 

q/amount of solute is going to give us the answer.  In this case, we're being asked for the answer in kJ/moles, so we'll need to switch the grams to moles later.  

Find q: Mass of the solution x Specific heat of the solution x Δt

Mass = Ammonium Nitrate + Water, so in our case: 90.77g + 5.87g = 96.64g

Δt = Tf - Ti, so 18.4C-23.0C = -4.6.  Note that negatives are ok, and that this reaction is endothermic, as the temperature of the water was LOWERED 4.6 degrees.  That means the reaction absorbed 4.6 degrees of heat.  

96.64 x 4.184 x -4.6 = -1859.972 J.  We're asked for the answer in the form of kJ/moles, so we change this to -1.9kJ.  

Also, because we're being asked the amount the reaction absorbed and this number signifies how much the water lost, we flip the sign.  Because of the laws of thermodynamics, the amount of heat (energy) in a system is constant.  What the water lost the reaction MUST have gained.  That's why we can simply flip the sign.  

1 mole of NH4NO3 weighs 80g, so the conversion (if you missed how to do that, I wrote it up here.) results in 7.33x10^-2 moles.  

The next step is division.  Divide q by whatever amount you're being asked for.  If you're being asked for one mole/g/kG/whatever obviously you don't need to divide.  For us here:  1.9/7.33x10^-2 = 26kJ/mole

And that's it, basically.  Interesting sidebar: These are called "coffeecup calorimetry" problems because coffee cups are really good insulators.  It's a shorthand for "something that keeps heat from getting out into the general environment".  The calorimetry part is derived from the fact that calories used to be the standardized measure of heat.  This is how they figure out how many calories are in food.  They burn it and measure the change in temp in a closed environment.  The temperature change is then converted into calories.  A calorie is equal to the amount of heat needed to raise one gram of water one degree.  So...calories are a measure of heat, not accurately of energy, and they say absolutely jack all about what your body does with it the food you eat.  

Ok, the third type of problem are Hess's law problems.  

Hess's Law: The enthalpy accompanying a chemical reaction is independent of the pathway between the initial and final states (Stolen from wikipedia: http://en.wikipedia.org/wiki/Hess_law)

That means, essentially, that if you know ΔH for a couple of similar reactions you can rearrange them (and their ΔH) to figure out the enthalpy for a similar reaction.  Now, this isn't a terribly practical thing to know.  Not like actual enthalpy problems, but I kinda like them because they're like a puzzle.  Here's the problem: 

Calculate the standard enthalpy change, ΔHo, for the formation of 2 moles of sulfur trioxide given the following information:

2 S(s)  +  3 O2 (g) --> 2 SO3(g)      ΔH = ?

Given:

S(s)  +  O2 (g) --> SO2 (g)   ΔH = -297 kJ

2 SO2(g)   +  O2 (g) --> 2 SO3 (g)    ΔH = -198 kJ  

So, the goal in manipulating these problems is to have the product be the same as the equation for whom you are trying to find the ΔH. The two given formulas will be added after the manipulation, and anything that is the same on opposite sides of the arrow will cancel.  So our goal is to end up with 2S and 3O2 on the left, and 2SO3 on the right.  The goal is always to manipulate the given formulas in a way so that when they add, and the stuff cancels, they look like the goal formula.  There are two ways of manipulating the given eqs.  They are: 

1. Multiply.  The top one of our two given eq. has Sulfur, and the bottom one doesn't.  Since our source eq. has two sulfer, we multiply the entire eq - INCLUDING THE ΔH - by two, which gives us: 

2S + 2O2 --> 2SO2  ΔH = -594

If you'll notice, our source eq. doesn't actually have any SO2, so that's one of the things we're going to want to cancel.  Luckily, there's now 2SO2 in given 1, and in given 2.  

2. If something you want to be on the left side is on the right side (or visa versa), flip the eq.  If you do this, just make sure you change the sign of the ΔH.  In this particular problem, we don't need to do that.  This isn't math.  You don't need to keep things exactly as they were.  This is chemistry...WE CAN DO WHAT WE WANT!  ;) 

After you've done these things (not necessarily in this order, the order is irrelevant.), then add the results: 

2S + 2O2 --> 2SO2      | ΔH = -594 kJ 2 SO2 +  O2 --> 2 SO3 | ΔH = -198 kJ   -----------------------------|-------------------- 2S + 3O2 --> 2SO3      | ΔH = -792 kJ

Note: These problems DO follow one rule of math: only like terms can be added.  That means you cannot add 3O to 3O2.  You can't add things that are in different states (gas and liquid, for example.  I left them out of the problem, but the state notations are in the question.).  You can't add joules and kilojoules, you need to convert if they're given to you in different units.  

And, that's basically everything.  Hopefully I didn't forget anything, or leave anything super necessary out.  There are probably things I've chosen to leave out that your teacher might want, but I think I've covered the important stuff.  :)  

[science-of-anthropology]

Thermochemistry - Part 1

Alright, here we go.  Thermochemistry.  Thermo is, well, it's a thing.  You will come across this again and again (which is weird, because practically the purpose seems mostly to be to make sure things aren't going to blow up.).  It's in physics, all kinds of chem, and a tiiiiiiny bit in bio.  It's weird, if you're in the sciences long enough you kind of pick up on the competition between them.  There's a saying that Physics is applied math, chem is applied physics, and bio is applied chem.  Which to a certain extent is true, but it's often used to apply hierarchy and that's stupid (math isn't a science anyway, it's a tool to describe how things work.).  Thermo is one of those places where you start to see crossover in the sciences.  I did it in chem and in physics, and I liked the chem version better.  I actually thought chem described what was really happening better than physics did.  

Thermochemistry (or thermodynamics, which is very similar, it's just taught in physics and applies to different types of objects.) is basically the study of how heat moves.  Heat is created when energy enters or leaves a system.  Energy excites (or slows) atoms, which creates heat.  The measurement of that energy is what is covered in thermo.  Both are subject to the laws of thermodynamics (http://en.wikipedia.org/wiki/Laws_of_thermodynamics), but you won't learn about those in chemistry.  In chemistry you'll learn only that the total energy of a system doesn't change.  That's important, and we'll get into it in more detail when we get to Hess' law.  

For now, there are three really important terms that we need to cover: 

Exothermic - A reaction that gives off heat and energy into the environment.  Exo is a greek prefix that means out of.  The energy is going out of the reaction.  Sometimes you'll hear me (or maybe others) say, sarcastically, that something that explodes is slightly exothermic.  For example, putting alkali metals in water is *slightly* exothermic: 

Endothermic - A reaction that removes heat from the environment, taking the energy into itself.  Endo is a greek prefix meaning into.  So, we're humans, we have endoskeletons! Bugs have exoskeletons.  

Enthalpy - Written as H, it indicates the change of energy in a system.  This property can change dependant on the ambient pressure, temperature, or amount of the substance.  Obviously it takes more heat to boil a gallon of water than it does a cup.  

Delta - The greek letter delta, written as Δ, is used to mean "a change in".  ΔH is a change in enthalpy, or heat energy.  

Heat of formation - The heat given off when a specific compound is created.  For elements, it's always 0.  They aren't compounds, they exist in a natural state.  Everything else is a standard based on lots of experimenting and is either given by your teacher in the problem, or you can look it up in the CRC or your textbook sometimes.  It's represented by the symbol Hf, where the f is a subscript.  

For the purposes of chem 111, almost all problems for thermo are done at STP (standard temperature/pressure.).  Meaning you're not going to have to worry about pressure changes until later, and almost always the temperature of your environment will start at 20-odd C, or room temp.  

There are several units of measurement in thermo.  The one you're probably most familiar with is the calorie.  When you see the amount of calories per serving on a food item, what they've done is burned it and measured the amount of heat given off.  You'll probably do something really similar to that in lab, actually, only not with food.  It's imprecise and we don't like it in actual science classes, so we use the Joule instead.  Joules are used in both physics and chem, so learn to love 'em.  Interestingly, volts could also kind of be a unit for thermo too because they measure electron flow and electrons = heat a lot of the time, but it's not a typical thing.  Joules are going to be what you're asked for most of the time.  

In chemistry, there's generally three types of thermo problems that you'll do: Enthalpy, Hess' law, and coffee cup calorimitry problems.  Also, themo is really fiddly about signs.  You need to read the problems carefully and take note of what you're being asked for.  

A negative ΔH is exothermic and a positive ΔH is endothermic because enthalpy is a measurement of the heat of the reaction! Exothermic reactions give off heat, they lose it, they *subtract* the heat from the reaction.  Endothermic reactions gain heat, they take it from the environment to make themselves warmer.  

Enthalpy Problems:

These are the easiest, most practically used, and my personal favorite of the three methods.  They all follow this formula: 

ΔH = H(products) - H(reactants) 

This formula format is actually kind of a standard thing in chem, and you'll see it again when you get to entropy and Gibbs free energy (Gibbs'll blow your mind.).  To be able to do these problems you need to be given the heats of formation for the various compounds.  Because enthalpy is dependent on the amounts present, you will need to multiple by molar values.  Typically, the above formula tends to look more like this: 

ΔH = [moles(Hf prod 1) + moles(Hf prod 2)] - [moles(Hf react 1) + moles(react2)]

Provided that none of those products are elements, you MUST be given the Hf for them to do the problem.  You also must balance the eq. first, or you'll get the wrong answer.  Here's an example problem: 

Calculate the heat given off when one mole of B5H9 reacts with excess oxygen according to the reaction below: 

2B5H9 + 12O2 --> 5B2H3 + 9H2O

Hf:  

B5H9 - 73.2kj

O2 - 0 (It's an element)

B2H3 - -1272.77kj

H2O - -241.82kj

Find ΔH.  

We set the problem up according to what I wrote above, so: 

ΔH = [5(-1272.77) + 9(-241.82)] - [2(73.2) + 0]

ΔH = -8686kj

HOWEVER, because when we did the math there were 2 moles of the stuff, we divide the answer by 2: 

-4343kj of heat is given off.  

You may be asking, "could I have just used one mole in the problem?" NO.  It would give you -8613.4, which is the wrong answer.  Why is it the wrong answer? Well, because we're looking at the heat of formation for the entire problem, right, so it has to be balanced before you can do that.  If you just stuck in a 1 instead of a two, it wouldn't be balanced anymore.  If you wanted to you could divide the entire problem by two (so instead of multiplying by 5 and 9, you'd multiply by 5/2, and 9/2), but that's needlessly complicated.  Just divide the answer by two, it's the same thing.  

Alright, since this is really really long at this point, I'm going to stop now, post, and put the other two methods in their own post.  

[science-of-anthropology]

Limiting Reagents

Ok, so this is closely related to the previous entry.  You have to be able to do stoichiometry to be able to do the math associated with this concept.  Here's the general concept though: 

When you are doing a reaction, the reaction will only last as long as you have reagents.  The reagent you run out of first is called the limiting reagent, because it limits the output of a reaction.  Now, in function (IE, lab) this is going to be pretty easy to spot because you'll have 10g of one thing and 3g of the other and obviously you'll run out of the 3g thing first.  However, mathematically it's different.  

Let's say you have exact same weight amounts of the two reagents.  How do you know which one is the limiting reactant? Or what if you need to know how much of them to react to get the amount of product you want? Well, let's consider the following (balanced) reaction:  

4KO2 + 2H2O --> 4KOH + 3O2

KOH is the main product, and the O2 is a byproduct, but we can use either to test.  Let's say we have 71.10g KO2 and 22.50g H2O, which one is the limiting reagent? You'd THINK the 22.50g H2O, but because of the ratios involved not so much.  Basically you do conversions to see how much of either O2 or KOH you get with each product, and then whichever is less is the limiting reagent.  Let's do the math and see how much O2 is produced: 

71.10 g KO2  x  1 mole KO2 x 3 Mole O2 

         1                  71g KO2     4 mole KO2

Because limiting reagents talk about molar values, it's ok not to convert back to grams.  The answer here is .75 moles O2 are made with 71.10g of KO2.  

22.50 g H2O x 1 mole H2O x 3 mole O2

        1              18 g H2O       2 mole H2O

You come out with 1.875 moles of O2 produced.  That makes the KO2 the limiting reagent, despite having more of it.  

Mostly this is a concept you'll probably deal with in lab.  Chem 111 labs love to ask "which of these was the limiting reagent".  Now you know how to figure it out, instead of half-assing it like I did the whole semester and guessing based on which I used less of.  ;)  

[science-of-anthropology]

The Basics: Balancing equations and Stoichometry.

Ok, this is going to be a long entry, I'm pretty sure.  BUT! This is the most fundamental tool for chemistry that exists.  If you can get this down, you can do chemistry.  It's THAT important.  

Chemists have their own system of measuring that's based on ratios.  It allows them to recreate experiments exactly. It's called the mole.  It is a unit based on the weight of carbon.  Honestly, in function that part is somewhat irrelevant and I never quite "got" it, so I'm not going to bother with it.  It's not going to help you one way or another.  

Remember how I told you in the converting entry that it would be important going forward? Well here we are.  It's important.  You can easily convert grams to moles (and often will be asked to do so.).  The conversion factor is the atomic weight of the element.  So, for example. 16g O = 1 mole O.  39.03g of Potassium = 1 mole of potassium.  You'll often see these types of questions: 

How many moles are in 652.89g of Na? 

Set it up like this: 

652.89g Na                        1 mole

--------------       x      -------------------------

         1                             22.99g Na

The units "g Na" cancel (which is how you know you're doing it right), and you're left with (652.89 x 1mole)/22.99 = 28.40 moles of Na

You'll often see a lot of questions where you are asked something like, "How many moles of ____ are needed to make ____ g of _____ in the following reaction?"  However, because the mole is a measurement of things in proportion to each other you first need to figure out what the proportions ARE before you an answer that question.  You need to know how many moles of one substance are needed to bind to how many moles of another substance to create the compound.  To get THAT answer, you need to know how to balance an equation.  Usually, this is the first step you take in any stoichometry problem.  

The end goal of balancing an equation is to make the numbers of stuff on one side match the numbers of stuff on the other side.  The numbers I'm choosing to use are *arbitrary*.  I'm choosing to use them because I like them and they're small and they make sense.  The only rule here is that you can't change the amounts of molecules in a compound.  So if you have H2O, you can't make it H3O just to make the balancing easier.  

I'm going to show you the way _I_ do it.  This is not necessarily the fastest or easiest, but it works for me.  There are a lot of ways to do this, some more "math-y" than others, and you need to find whichever one makes sense to you and stick with it.  I found myself often forgetting how much of each thing I had as I worked through it, so this is what I came up with.  

C6H6O + O2 --> CO + H2O

I like to make a little table, like this: 

     Left                                     Right

C    6                                        1

H    6                                         2

O    3                                         2

Ok, so I basically start with the most obviously easy thing.  Hint: if something is by itself, like O2, leave it till last or you'll be constantly adjusting it.  

In this case I'm going to put a 3 in front of the water to make the Hs match.  Then I draw a line through the count, and change it.  

C6H6O + O2 --> CO + 3H2O

     Left                                     Right

C    6                                        1

H    6                                         2 6

O    3                                         2 4

Yay! H's match....for now.  Let's do carbon next.  Since there's 6 on the left, I'll adjust the CO.  

C6H6O + O2 --> 6(CO) + 3(H2O)

The count becomes: 

     Left                                     Right

C    6                                        1 6

H    6                                         2 6

O    3                                         2 4 9

Oh no! Os are unbalanced again.  But that's really easy to fix: 

C6H6O + 4(O2) --> 6(CO) + 3(H2O)

     Left                                     Right

C    6                                        1 6

H    6                                        2 6

O    3 9                                     2 4 9

Yay! They're the same! Nerd party! 

Now, remember, those numbers I chose were arbitrary.  You could use any multiple of those numbers.  It would always work because it's based on a ratio, but I find that the smaller the number, the better.  So, if you want to practice this, here's a decent site to let you do that (I may or may not have stolen that example off of it): http://education.jlab.org/elementbalancing/

Ok, so we ended up with this: 

C6H6O + 4(O2) --> 6(CO) + 3(H2O)

Those numbers in front of them? Those are molar quantities.  So in this equation, there's 1 mole of C6H6O, 4 moles of O2, 6 moles of CO, and 3 moles of water.  The gram values of each are as follows (based on the atomic weights, as I discussed before): 

Phenol (C6H6O): 

C x 4 = 12*4 = 48

H x 6 = 1*6 =   6

O x1 = 1*16 = 16

48+6+16 = 70g

Oxygen:  

O x 2 = 16* 2 = 32g

Carbon Monoxide: 

C = 14

O = 16

14 + 16 = 30g

Water: 

O = 16

H x 2 = 2

16+2 = 18g.  

So, since we have 1 mole of phenol, we have 70g.  Since we have 4 moles of O2, we have 4*32 = 128g of O2, 6*30 = 180g CO, and 3*18 = 54g of water.  All of that comes from simply balancing the equation.  So you don't necessarily NEED to be told how much of anything you have, it is part of the reaction equation.  

What it also tells you is that for every 1 mole of Phenol you need 4 moles of oxygen to make 6 moles of CO and 3 moles of water.  Also, as you might of noticed, this means phenol is nasty stuff.  ;P  It also tells us that if phenol, like every other organic chemical under the sun, wasn't completely and utterly flammable...it would make a pretty nice way of putting out fires because it consumes oxygen when it sublimates (<--fancy word meaning to go directly from solid to gas.).  

Anyway, I digress.  Knowing those ratios really helps you figure out how many you need of what things.  Let's say you got the following question: 

"If you have 238.53g of Phenol, how many grams of carbon monoxide will be created when combined with Oxygen? "

By itself, that is scary looking.  It looks like you've been given no information at ALL.  Except...we have.  I just told you everything you need to know.  So you balance the reaction, which we already did above: 

C6H6O + 4(O2) --> 6(CO) + 3(H2O)

After that, it's just a ratio and you set it up like this: 

238.53g Phenol        1 mole Phenol       6 mole CO      30 g CO

-------------- x ------------- x ----------- x --------------

             1                70g Phenol           1 mole phenol    1 mole CO

             A                             B                         C                     D

Alright, let's explain the columns: 

A. This is the amount you're given.  Over a 1 to make things pretty.  

B. This is where you're converting that amount into moles.  The 70g comes from what we did above, and it goes on the bottom to make sure it cancels the "g phenol".  The thing you want to convert *into* goes on the top.  

C. This is our proportion, taken directly from the balanced reaction.  The phenol goes on the bottom because we want to cancel those terms.  We we *asked* for grams of CO, meaning that's the one we want.  It goes on top, and we cancel the "mole phenol" term since we don't care about that.  The 1 here isn't a constant "to make it pretty" number.  If I'd had 3 moles when balancing the equation, I would have written 3 mole phenol.  

D. This is us converting back into g.  Moles go on the bottom because we don't want that term, we want grams.  The 30 is, to repeat what I said above, the grams of CO in 1 mole of CO, taken from the molecular weights on the periodic table.  

Now you just math, and you come up with 631.36g of CO.  The strike throughs are the things that cancel, which is how you end up with your desired unit: g of CO.  I've said this before, but I'm going to reiterate, DO THE PROBLEMS WITH THE UNITS STILL ATTACHED.  At least when you're setting them up.  If everything cancels correctly, then you know that you've set it up right.  

That's about it.  It's long, but if you can do it you've got a huge leg up in chemistry.  Do practice problems until your brain melts out your ears.  :)  

Also practice the balancing of equations, because if you go on to chem 112 you do a more complicated version of this involving electrons, and if you don't understand this you will be utterly lost and confused.  

[science-of-anthropology]

The Basics: Unit Conversion

Alright, this'll probably be a short-ish one, but I think it's useful.  Today I'm going to teach you how to convert units like a boss.  This is the way you should do it, even when it's easy, until you get the hang of it.  That's because this is how, later on, you're taught to calculate moles.  That much-dreaded unit in chemistry that tells you how much of some crap you need to use.  

HOKAY.  Here's how you set it up: 

Amount of thing                   1(unit you want the answer to be) -------------------     x      -----------------------              1                     Conversion factor

So, for example, you want to turn feet into meters, right? Here you go: 

15.658 ft               1 m ------------    x   --------------        1                 3.28084 ft

Because the ft are on the top and bottom, the units cancel.  This is why, especially when you're starting out doing this, you leave the units on.  It's a check.  If somehow the only unit left was ft, then I'd know I did it wrong.  This is especially true in physics, so don't be a cheat like me (sometimes) and just ignore the units, do the math, and tack the unit onto the end in whatever I know it's supposed to be.  x.x  Why do we put the ones there? Because I said so.  As place holders that don't really effect the outcome of the numerical answer.  We just pulled that part out of air to make the problem easier to look at.  

Anyway, in this case you're left with: 15.658m/3.28084 = 4.7726

NOTE I USED THE RIGHT SIG FIGS.  I will do a sig fig entry next cause it turns out they're not as hard as I thought.  Also no one uses them in premed orgo.  So, take heart.  Hell, even in 112 they were mostly lax.  

So, now that you have this basic idea down, you can make it more complicated.  Let's say you need to go from inches --> feet --> meters.  You can do that.  Yeah, it'll be easier to go right from inches to meters, but this is an example and I don't want to pull some weird ass units like picometers or something out on you plus I don't have my thingee that tells me wtf a pico is on hand and I don't want to google more conversion factors, so you get inches.  Shhhhhhhh.  Sometimes this'll be hard.  

69.72in               1ft                  1m

--------      x     ----------   x   -----------------

     1                   12in              3.28084ft

So you cancel the inches and the ft and you end up with meters.  What you end up with is this:  (69.72/12)/3.2808 = 1.771.  

You can put together long strings of these things and do the conversions all at once.  You start with the first number and the go down the line dividing if it's on the bottom and multiplying if it's on the top.  Easy-peasy.  

Now.  Calculator trick!  If you feel really, really comfortable with this method and you feel like you've learned it and that doing this for conversions is a waste of time and paper, here's what you can do.  Most of us should, at this point, have some sort of scientific calculator.  If you don't have one and you're in college, grow the eff up and go buy one.  Mine is a Ti-83 plus, but I'd imagine that all of the TI calculators have a version of this.  

Push the blue button that says apps.  It's 2nd column, 3rd row.  

Press 4, for SciTools.  

Press any button.

Press 2 for unit converter.  

.......

PROFIT.  

Note, it always gives the answer in scientific notation, so you best know scientific notation.  

Seriously, don't use this until you understand the basic concept involved in converting, because you'll have to do it often for things that don't have direct conversion factors (like moles, and certain units in physics.) and can't just be stuck into your calculator.  You NEED to understand how to do it the regular way.  

Also! Speaking of moles and things you'll have to do later (like buffer problems and other types of enormously terrible acid base chemistry.), my new chem teacher taught us a neat trick.  You probably already knew this, but it never occurred to me because I'm not naturally awesome at math.  I mean, I'm good enough to pretty easily do all of the physics and chem calculations, but only through repetition.   Anyway, if you're silly like me and didn't know this, it will really help you when you get to the "how much of X will you need to make X amount of a compound if you have X amount of this particular starting material, and X amount of this same starting material gives you X amount of product" limiting reagent word problems.  

amount of a given thing               How much X needed?

------------------------------    x   -----------------------------

Amount of product given            Amount of product wanted       

The left half is the numbers that are from the hypothetical original experiment.  It's the thing you already know.  It's the part of the problem that's worded "If X amount of a thing makes X amount of a product..." The right is the product, the part that you want.  It's the part of the word problem that says, "then how much of X do you need to make X amount of product?" Figuring out what goes in those spots is probably the hardest part.  Stupid word problems.  

Anyway, then you cross multiply and divide.  SO, to make this clearer:  If 2.53g of Hg makes 2.73g of HgO, how many grams of Hg do you need to make to form 10g of HgO? You set this up as:

2.53g Hg                       Xg Hg

-------------         x        ---------------------

2.73 HgO                     10.0g Hg

Which becomes: 25.3g = 2.73X, and then you divide by 2.73.  This gives you 9.27g of Hg needed.  Don't worry too much about writing the element in there, it doesn't work like a unit.  I only added it so you'd know what was what.  

Alright, I think that's it for this entry.  I hope you all are having a good semester.  I sorta am, my physics 112 class got cancelled so I'm pretty bummed about that cause now it pushes back my MCAT taking date, possibly until after they change the test.  :|  It also means I can't graduate, and I'm moving on to a 4 year school without my associates, so I'm bummed about that.  Oh well..stuff you didn't need to know.  :)