gen chemistry question

so i decided today that i would just get myself familiar with chemistry again - so i am starting out at the basics…

i was doing a problem, and just wanted to see if someone may review it for me:

infer the simplest chemical formula for potassium copper fluoride, which has the analysis by weight:

K â€“ 35.91%

Cu â€“ 29.19%

F â€“ 34.90%

KCu2F

Key for below:

Element - Atomic Ratio, Atomic Relative, Relative Weight, and Weight Percent

Potassium - 1, 39.098g, 1.089 (does this have a unit of measurement, like mole or something?), 35.91%

Copper - 2, 63.546g, 2.177, 29.19%

Fluoride - 1, 18.998g, 0.544, 34.90%

I came up with relative weight by dividing the atomic relative by the weight percent.

Then, to come up with the atomic ratio for each element I divided 1.089 (K’s relative weight) by 2.177 (Cu’s relative weight) which is about .5 - therefore (assuming this is a stoichiometric compound)

1 K

-

2 Cu

And did the same thing for finding F’s atomic ratio - but I divided 2.177 (Cu) and 0.544 (F)

which is about 2 - so,

1 Cu

-

1 F

_________________________ _________________________

Since I have not had ANY chemistry or exposure to it since 2001 - I am not tooooo sure at all about all this LOL but figure to go through some material and get the basics back prior to the semester cannot hurt.

I know that when I did this same problem for Aluminum Oxide (analysis was given as 53% aluminum and 47% oxygen by weight) I did it right… at least I think.

Key for below:

Element - Atomic Ratio, Atomic Relative, Relative Weight, and Weight Percent

Aluminum - 2, 26.98g, 1.9161, 52.92%

Oxygen - 3, 16.00g, 2.943, 47.08%

Relative weights obtained the same way, the Weight Percent divided by the Atomic Relative.

Since the ratio between Al and O is .66 (1.9161/2.943), there are 2 Al atoms and 3 O atoms in Aluminum Oxide.

I also noticed that if i flip the order in which I try and find the atomic ratio, it ends up the same. 2.943/1.9161 = 1.54 (since this is about whole numbers, we cannot have .4 of an atom, i round up to 1.5, if it were 1.56 then would i round to 1.6??) which would be 3 over 2 …

_________________________ _________________________

So I guess I have 5 questions here.

1. Does the relative weight, that is calculated from the ratio of weight percent and atomic relative, have a unit of measurement?

2. How do I figure out the simplest Chemical Formula, from just the percentage of each element by weight, when there are more than TWO atoms in the compound?

3. When figuring out the atomic ratio, is it kosher to flip which element’s relative weight you put first? (it seems okay to me, but not really sure if there can ever be an exception to the rule)

4. Since most chemical compounds are stoichiometric compounds, what are the cases of non-stoichiometric compounds I need to worry about?

5. If I understand the process right, metals that are used as catalysts in the oxidation process temporarily create non-stoichiometric compounds (as the vacancy of oxygen is only temporary and replaced by O2)?

Thanks everyone!

new question - chemical reactions with gases…

law of conservation of mass does not apply, so the reactants don’t always have to equal the products for gases.

i am having a bit of difficulty getting how to figure out this problem:

How many liters of oxygen (O2) are required for the complete oxidation of 1 gram of methane?

CH4(g) + 2O2(g) --> CO2(g) 2H2O(g)

1 mole of gas occupies 22.4 liters - so the products occupy 67.2 liters - the reactant CH4 occupies 22.4 liters as well… i would figure that this should require 44.8 liters of O2, but the law of conservation of mass does not have to apply here… soooo, what am i missing here?

the atomic weight of each molecule doesn’t matter right? or do i have to really get into the nitty gritty there and determine the atomic weight for each element in CH4, and consider each part a mole?? cause 1 mole of carbon + 1 mole of molecular hydrogen = 1 mole of methane…

I’m guessing you need to look at the entire problem and note changes in pressure if any and changes in temperature. In theory your products could equal 22.4 liters and nothing but pressure change, but I never used a lot of dimensional analysis with gas problems.

sorry forgot to mention that it was done at 0 degrees Celsius and at 1 atmosphere of pressure

Well, welcome back to the fantastic world of chemistry! 1st of all, you are totally over analyzing this problem. I am referring to the smallest chemical formula of KCu2F. OK… you have

K = 35.91%

Cu = 29.19%

F = 34.90%

If you’ll notice, the sum of these percentages equals 100%, therefore we are able to simply say that this is the same as 100 grams, because we are dealing in grams. For example, 35.91% of 100g = 35.91 g. Thus,

K = 35.91 g

Cu = 29.19 g

F = 34.90 g

I hope that made sense. Anyways, from there you convert each element into mols using their given atomic weight. Therefore,

35.91g K x 1 mol/39.09g = 0.9186 mol of K

29.19g Cu x 1 mol/63.54g = 0.4593 mol of Cu

34.90g F x 1 mol/18.998g = 1.837 mol of F

From here you divide each of the values by the smallest amount of mols.

0.9186/0.4593 = 2

0.4593/0.4593 = 1

1.837/0.4593 = 3.999 = 4

Now using the ratio of K:Cu:F = 1:2:1, multiply the number just calculated to the ratios, accordingly. Thus,

K has a ratio # of 1 , times 2 mols = K2

Cu has a ratio of 2, times 1 mol = Cu2

F has a ratio of 1, times 4 mols = F4

Now you have your simplest chemical formula --> K2Cu2F4

Now, if only two percentages are given, and you still need to find another, simply find the difference needed to reach 100%. This makes sense because no matter how many grams you use, the sum will always be 100 percent. For instance, if you have a total of 7 grams being used, and you are given 2 masses ( 1g and 2g), you know that 4g of the unknown mass is needed to reach that 100%. ( 1g + 2g + 4g = 7g => 100%)

I hope this helped you more than confused you! I will get back to you on the other problem, I gotta brush up on that one. Good luck!

ohhhh that makes SO much more sense!! i suppose that i just got lucky with the aluminum oxide equation working out the way i did it - thank you!!

i figured it would be better for me to start all this now than try and get a grasp on it during the semester - most ppl in gen chem will be coming right out of HS chem class so this will all be fresh in their heads…

• The Truth Said:
new question - chemical reactions with gases...

law of conservation of mass does not apply, so the reactants don't always have to equal the products for gases.

i am having a bit of difficulty getting how to figure out this problem:

How many liters of oxygen (O2) are required for the complete oxidation of 1 gram of methane?

CH4(g) + 2O2(g) --> CO2(g) 2H2O(g)

1 mole of gas occupies 22.4 liters - so the products occupy 67.2 liters - the reactant CH4 occupies 22.4 liters as well... i would figure that this should require 44.8 liters of O2, but the law of conservation of mass does not have to apply here.... soooo, what am i missing here?

the atomic weight of each molecule doesn't matter right? or do i have to really get into the nitty gritty there and determine the atomic weight for each element in CH4, and consider each part a mole?? cause 1 mole of carbon + 1 mole of molecular hydrogen = 1 mole of methane...

I'll take this one -

First off I am not sure why you are saying that conservation of mass does not apply since this is not an antimatter/quantum chemistry rxn where matter is being created or annihilated.

Cardinal rule of chemistry - always make sure your stoichometry is consistent and this roughly translates to "convert everything into moles".

1g of CH4 = 1g/(16g/mol) = .0625 mol

1 mol of CH4 requires 2 mol O2 for complete oxidation/combustion.

Thus 0.0625 mol CH4 requires 0.1250 mol O2.

1 mol O2 = 22.4L at STP,

0.1250 mol O2 = 2.8L O2 at STP <--- I used simple ratio/proportions here but if you are so inclined, you can use Avogadro's gas law i.e. n1/V1 = n2/V2.

Thus 1 g CH4 requires 2.8L O2 for complete oxidation/combustion.

Your reactants are 1 mol of CO2 and 2 mol of H2O so the 3 mol of gas would indeed occupy 3*22.4 = 67.2 L.

Hope I didn't screw anything up. Note - mass is indeed being conserved. The only thing not being conserved is volume which, of course, is an entirely different quantity than mass.

i was just stating that because volumes do not need to be conserved in reactions involving gases. ie combustion of carbon monoxide - 2CO(g) + O2(g) --> 2CO2(g)

so in this case, the law of conservation of mass does not apply… and in any case when we are dealing with gases and volumes, the law does not have to apply.

and THANK YOU for reminding me about converting everything into moles (which, unfortunately, i did not review using moles as a unit til today lol)

• The Truth Said:
i was just stating that because volumes do not need to be conserved in reactions involving gases. so in this case, the law of conservation of mass does not apply... and in any case when we are dealing with gases and volumes, the law does not have to apply.

Be VERY careful with that statement - it's a test writer's dream come true!

Note the last sentence in my previous reply. Don't confuse mass and volume. Mass is ALWAYS conserved unless you are converting matter to energy and vice versa via the famous E=mc2!

Assuming the simplest chemical formula is the empirical formula, I would solve as follows:

1. assume 100g of product

2. calculate GAW of reactant

K â€“ 35.91% = 35.91g

Cu â€“ 29.19% = 29.19g

F â€“ 34.90% = 34.90g

3. calculate # moles of reactant

K - 35.91g/39.10gpm = 0.9184m

Cu - 29.19g/63.55gpm = 0.4593m

F - 34.90g/ 18.10gpm = 1.9282m

4. divide all quantities by smallest (0.4593m)

K - 0.9184m/0.4593m = 2.0000

Cu - 0.4593m/0.4593m = 1.0000

F - 1.9282m/0.4593m = 4.1981

5. find a multiplier (usu. between 1 and 5) to obtain an integer for the ratio of F (5) and multiply all ratios to determine the empirical formula

K = 10

Cu = 5

F = 20.9905 or 21

Hence K10Cu5F21 would be the empirical

@dazed - yeah the wording i used there was pretty bad; i totally get that mass isn’t created or destroyed in the reaction of that problem…

@perdido - is there a difference between empirical formula and simplest chemical formula??

cause once you get the atomic ratios down, everything falls into place… but i think your solve for this problem is different from mine and dazed’s answer because we are solving for the simplest form of this compound… in that, this is assumed to be a stiochiometric compound (having integral atomic ratios). anyone have any thoughts on this?

The empirical formula is the simplest whole number ratio of a compound; however I made an rounding error with the GAW of F. I meant to use 19.00gpm which would give a formula K2Cu1F4. Sorry about that.

lol i was going to say i felt a little more lost now cause when i saw you multiplying the atomic ratio by 5… man did that confuse me (as my text doesn’t, well at least yet, say to do that)

especially when the answer is right there when you first got the ratios (part 4 of your solve)…

i suppose that brings up this question just to clarify…

when we are dealing with stiochiometric compounds, we assume that they are integral numbers (rounding to the nearest whole number)… so, in the case of Fluorine in this problem, because the atomic ratio is 4.1981 - this is nowhere near 5, therefore we round-down to 4 because this is dealing with the simplest whole number (stiochiometric compound)?

what happens in the event of an intermediate number… that has an atomic ratio of, say 1.453?? is there a general threshold at which you round to make a whole number?

If I’m following you correctly, at near 0.5 you multiply by 2 in order to get a whole number (I seem to recall multiplying by three for close to .3 etc.)

It is the Law of Multiple Proportions. Elements combine ratios of small, positive integers.

So, if you find that the ratio of two elements in a substance is 0.33 to 0.50 and the integer ratio is not obvious, typically the simplest step is to first divide each by the smallest number, resulting, in this example, in 1.00 and 1.51. At that point, it is easy to see that doubling the numbers gives you integer values of 2 and 3.

Oooh! I don’t have an answer but just wanted to say your enthusiasm in all your posts TheTruth is really cool…and now you have made me want to take chemistry RIGHT NOW! Although I technically have to wait til Fall to take it. I do like me some chem.

Mehgan

thanks! i can’t take chem until this summer or at the latest fall… i thought i’d be able to squeez myself into a program but haven’t heard back from any schools… so im just assuming that i will be admitted for the summer/fall sessions… right now i am taking two courses in my grad program (haven’t closed it out yet cause i haven’t done my administrator-in-training program)… but i still feel like i want to get a jump on all the prereqs

in the whole process of trying to find great places to review at… i came across this:

http://oedb.org/library/beginning-onl ine-learning/…

and from there, a course offered by MIT, principles of chemical science

http://ocw.mit.edu/OcwWeb/Chemistry/5-11 1Fall-2005…

so i plan on using this resource to get a solid grasp on chemistry, biology, physics, and calc… hopefully it pays off!!