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Chemistry will be the death of me - Dreamer567 - 03-16-2009 09:33 AM

Can someone please help me with these problems? It would be much appreciated. Smile


Write the empirical formula of each of the following substances:
a. C6H6O3
b. C6H12O6
c. B6H10
d. C6Cl6
e. C24H16O12


Re: Chemistry will be the death of me - maxh - 03-16-2009 11:08 AM

That's easy, greatest common factor.

a) C2H2O
b) CH2O
c) B3H5
d) CCl
e) C6H4O3


Re: Chemistry will be the death of me - Dreamer567 - 03-16-2009 11:15 AM

Thank you! ^.^


Re: Chemistry will be the death of me - Dreamer567 - 03-23-2009 09:28 AM

Some more questions. Did I mention I HATE chemistry? :-p

1. How many moles are in 120 grams of Cu?
2. How many atoms of Cu are in 120 grams of Cu?
3. How much does 0.125 mole of Cu(NO3)2 weigh?
4. How many moles of Br2 molecules are in 15 mL of Br2? How many Br atoms?
5. What is the mass of one formula unit of Mg3(PO4)2?
6. What is the mass of 2.25 moles of CO2?
7. How many grams of SiO2 would contain 3.5 x 10 to the 15th power molecules?


Re: Chemistry will be the death of me - Aviator - 03-23-2009 11:34 AM

Quote:1. How many moles are in 120 grams of Cu?
2. How many atoms of Cu are in 120 grams of Cu?
3. How much does 0.125 mole of Cu(NO3)2 weigh?
4. How many moles of Br2 molecules are in 15 mL of Br2? How many Br atoms?
5. What is the mass of one formula unit of Mg3(PO4)2?
6. What is the mass of 2.25 moles of CO2?
7. How many grams of SiO2 would contain 3.5 x 10 to the 15th power molecules?

1) 120/63.546 = 1.888 moles.
2) 120/65.546 = 1.888 moles X 6.02X10^23 = (plug into calculator)
3) 0.125 * 1.8880 + 28.014 + 96 = 15.738g Cu(NO3)2
4) 15/22.4 = 0.669moles = 0.669*6.02X10^23 = (plus into calculator)

AND... my knowledge ends there....


Re: Chemistry will be the death of me - Dreamer567 - 03-23-2009 12:40 PM

Thank you! ^.^


Re: Chemistry will be the death of me - Aviator - 03-23-2009 02:42 PM

Your welcome! Here's how I did it, in case you're interested:

Moles to grams conversion = Moles X Molar Mass
Grams to Moles conversion = Grams / Molar Mass

Moles to Formula Units = Moles X 6.02X10^23
Formula Units to Moles = Forumular Units / 6.02X10^23

Moles to Liters = Moles X 22.4
Liters to Moles = Liters / 22.4

If you ever have to convert something from Grams to Liters, do this ->

Grams -> Moles -> Liters
Grams/Molar Mass = Moles -> Moles X 22.4 = Liters

So basically, convert everything into moles, and then convert that into the unit you need it to be in. It's fairly straightforward. I wish I could give you the notes I took on that when we were covering it in Chemistry, as that explains it perfectly, but, the scanner doesn't work. None

I'll make an attempt at the last 3 problems, too, in a minute.


Re: Chemistry will be the death of me - Aviator - 03-23-2009 02:54 PM

Quote:5. What is the mass of one formula unit of Mg3(PO4)2?
6. What is the mass of 2.25 moles of CO2?
7. How many grams of SiO2 would contain 3.5 x 10 to the 15th power molecules?

5) 1 FU = 6.02X10^23/6.02X10^23 = 1 Mole X 198.851 = 198.851grams
6) I'm going to (educated) guess is 88.022g
7) 3.5X10^15/6.02X10^23 = moles X 60.055 = grams (put into a calculator).

Those are educated guesses, but may not be correct. (I'm pretty sure about #7 though).


Re: Chemistry will be the death of me - Dreamer567 - 03-24-2009 01:14 PM

This totally helps. My chem teacher can't explain this stuff for his life. Thanks for simplifying it.


Re: Chemistry will be the death of me - Aviator - 03-24-2009 01:36 PM

Your welcome! Don't be afraid to ask if you need any furthur explanations or any more help. Biggrin


Re: Chemistry will be the death of me - Dreamer567 - 04-01-2009 01:14 PM

Hehe I'm back! Can someone please explain percent yields to me? I guess my teacher has finally realized how terrible he is at teaching because now he doesn't even bother to explain things. He just gives us homework on concepts we've never worked with before and tells us when it's due.

Here's a problem from my homework:

Carbon disulfide is produced by the reaction of carbon and sulfur dioxide.

5C + 2SO2 ----> CS2 + 4CO

What is the percent yield for the reaction if 40 grams carbon produces 36 grams carbon disulfide?


Re: Chemistry will be the death of me - maxh - 04-01-2009 02:57 PM

This is fairly easy. percent yield == (actual yield) / (theoretical yield) * 100.
Theoretical yield is found by the stoichiometric calculation. Actual yield is found by experiment, or, in problems like this, given to you.

First just run a mass-mass conversion:
40g C (1mol C/12.01g C)(1mol CS2/5mol C)(76.13g CS2/1mol CS2)= 50g CS2

That's the theoretical yield, that is, the greatest mass of product that could be produced from the given amount of reactant.

But that almost never happens. There's usually contaminants in the reaction no matter how hard you try to avoid it, so some of the reactant does not make the expected product. That's what percent yield is for.

This is the same as any other percentage equation, so:
36g CS2/50g CS2 * 100 = 72% yield.

Obviously, you'll have to re-run this with the periodic table your teacher gave you and sig figs/rounding rules your teacher uses, because that changes the value slightly (but still enough to make it wrong in some cases).

Teal Deer: The answer would be 72% yield if you were in my class, but your teacher probably won't like that answer.

Oh, and enjoy having to join this with limiting reactants. It's about a page per problem.

EDIT: Fucker, I'm supposed to be on break!


Re: Chemistry will be the death of me - Dreamer567 - 04-03-2009 12:14 PM

Thank you ^.^


Re: Chemistry will be the death of me - monkey - 05-04-2009 01:54 PM

This is all beyond me. I have no idea what you guys are talking about. Uhoh Is this what high-school is like?


Re: Chemistry will be the death of me - Bob Dole - 05-04-2009 02:10 PM

--monkey666-- Wrote:This is all beyond me. I have no idea what you guys are talking about. :uhoh: Is this what high-school is like?
Yeah. You get used to it.


And, oh man, limiting reactants. Don't remind me. Fucking ten problems and my goddamn arm was hurting.


Re: Chemistry will be the death of me - psychopath - 05-04-2009 02:10 PM

moles confuzzle me


Re: Chemistry will be the death of me - Bob Dole - 05-04-2009 02:17 PM

One mole is as many of atoms/molecules/electrons/ions/etc. as are found in 12 g of carbon-12. So about 6.02*10**23. In short, one mole of a substance is 6.02*10**23 atoms/molecules/etc.