chemistry

[Horseaholic]

:rant: So I'm taking chem - which is supposed to be easy but I'm really struggling. I have a lab at 4 until 9 (school is 45 minutes away and I'm freaking out because I do NOT understand this stuff)

I have to turn this in before I start my lab and I'm just so lost and confused. My teacher is REALLY rude and always says things like "that's an ignorant question" and she'll laugh and shake her head if you ask questions and she uses all her own terms and stuff so I'm worried ab paying a tutor to help because she teaches so weird I'm afraid it would do no good.

We haven't even learned this stuff in class yet but she expects you to know it. I didn't do well in chem in HS. I REALLYYY struggled and I knew this class would kill me but man am I freaking out.

anyone feel like explaining this stuff to me????

i have to convert the following word equations to balanced chemical equations.

a. copper metal (s) + oxygen (g) yields (with heat symbol on top) copper 2 oxide. :headscratch: how do I even begin??

[missyfritz]

I feel your pain. :( I'm in the same boat you are...and I know I SHOULD know how to do that problem...

[Equine_Woman]

Sorry I can't help! AS soon as I finished college I shoved all the Chemistry and Organic Chemistry and Biochemistry into the deep dark recesses of my brain never to be sorted through again. . . .

Go with the tutor. IT will help!! Sounds like you have a bad teacher.

[Horseaholic]

:( I have a terrible teacher and not a single helpful classmate. Not even my lab partner will help me...I do all the labs and she is so reluctant to even let me verify her answers w mine.

[TheJFactor&Fergie<3]

:rant: So I'm taking chem - which is supposed to be easy but I'm really struggling. I have a lab at 4 until 9 (school is 45 minutes away and I'm freaking out because I do NOT understand this stuff)

I have to turn this in before I start my lab and I'm just so lost and confused. My teacher is REALLY rude and always says things like "that's an ignorant question" and she'll laugh and shake her head if you ask questions and she uses all her own terms and stuff so I'm worried ab paying a tutor to help because she teaches so weird I'm afraid it would do no good.

We haven't even learned this stuff in class yet but she expects you to know it. I didn't do well in chem in HS. I REALLYYY struggled and I knew this class would kill me but man am I freaking out.

anyone feel like explaining this stuff to me????

i have to convert the following word equations to balanced chemical equations.

a. copper metal (s) + oxygen (g) yields (with heat symbol on top) copper 2 oxide. :headscratch: how do I even begin??


Hey!

I might be able to help, but I need to know what level you're at (not in American or I won't understand lol) are you expected to know valencies of these (do you have valency tables or are you meant to be able to work it out from the periodic table.

I can explain either, but one is more complicated and possibly unnecessary.

[Horseaholic]

I'm assuming I'm supposed to know this from the periodic table

[Horseaholic]

unfortunately I have to leave for class now :( I can check back on my cell phone though.

[outriding01]

Hey, which professor do you have and what chem is it? I may still have my labs and such....

[Horseaholic]

Its booher and chem 1030 all my labs are out of the catalyst book

[TheJFactor&Fergie<3]

*digs around on desk for periodic table... I know it's here somewhere*

start by substituting the names of the elements for their chemical symbols...

which would give you

Ca + O ----> Ca(II)O

so calcium is in group II of the periodic table (and you've been told it's calcium TWO oxide). Both of these factors mean that calcium ions are 2+.

If you want explanation of this, basically, electrons exist in shells. When shells are full, they're very stable. When they're not full, electrons will either be gained or lost (or shared) from the outer most shell to leave a complete shell. There's a set number of electrons to fill each shell. The first shell holds 2 electrons, the second holds 8. The third also holds 8, but the transition metals do peculiar things, so that's not a hard and fast rule... So if you take Calcium which has a proton number of 20 and therefore has 20 electrons, this has 2 electrons in the first shell, 8 in the second, 8 in the third and 2 in the fourth. This last shell is not full, it would need 6 more electrons to complete it. it takes less energy to lose two electrons than it does to gain 6 (which is logical, right?). So Calcium will lose these 2 electrons. Obviously, it now has more protons than electrons, so it's got a positive charge. The positive charge equal to minus 2 electrons (/two additional protons) thus is Ca2+.

Oxygen is in group VI. It has a proton number of 8, therefore 8 electrons. So that means there's 2 in the first shell and six in the second. To complete this shell, oxygen needs 8 electrons. So it needs to gain 2. In doing so it will become O2-

How convenient that Ca wants to lose 2 and O wants to gain 2! This means that one Ca and one O can balance each other out. This means that your final symbol for Calcium oxide is indeed CaO.

However, oxygen exists as the diatomic gas O2. This is where balancing comes in.

If you have 2 oxygens, you will end up with 2 molecules of CaO at the end. Unless you can make it disappear or something ;). But in order to have 2 molecules of CaO, you need to start with 2Ca.

This would make your equation:

2Ca + O2 ---> 2CaO.

I really hope this makes sense, and I'm not either talking nonsense, or telling you stuff you already know.

I'm gonna keep talking anyway...

If you were dealing with sodium, which is in group one and has eleven electrons, (2, 8, 1) it would lose a single electron and become Na+. If it were to form a sodium oxide salt, it would need 2 sodiums to bond with a single O2-. So this would have a formula of Na2O. Because you'd have O2 on the left hand side of your equation, you would need two of these Na2O molecules, and need 4Na to start with...

4Na + O2 ----- > 2Na2O

The way to look at balancing is that for each individual element, you need to start with the same number of them as you end up with. If you have O2 at the start you need 2 O's on the right hand side. If you have 2Na2, you need 4 on the left. etc.

i really hope this helps!

[outriding01]

Damn, I found my lab book but not my folder with my lab reports :( You're taking the class as well as the lab right? Are you using a book for that or a manual? I have my manual from the professor I had who no longer teaches there and it's written well and really simple and easy to follow with problems and a solution manual to solve with. It might be somewhat helpful if you want it.... I'll keep looking for my reports. Do you guys do Q Sets? Or just the post-lab report? I randomly found the stuff for lab 7 on aqueous solutions if you want that one...

[Oreos Girl]

Horseaholic,
sorry I didn't get on the board yesterday. JFactor&Fergie explained it very well. What is the atomic symbol of Copper? Oxygen is one of 7 diatomic elements which means that it is usually O2 when by itself (otherwise it is ozone O3). You should know those atomic symbols to atomic names backwards and forwards for about the first 90 elements. Some you will find you will use more than others like you will hardly ever need Tantalum (Ta). I suggest notecards for this kind of thing. Put the atomic symbol on one side and the name on the other and carry it around with you so if you have a minute flip through some of them. Yields means the reaction arrow. Copper II oxide. Copper is a transition metal which means it can have more than 1 charge so the charge is written as the number between the atomic names. So copper will have a 2+ charge. Oxygen is in group 6A so to find the charge of it take the group number minus 8. So 6-8 gives you a 2- charge. A 2+ and a 2- cancel each other so you have no subscripts in the compound. So CuO.

Hope this helps even if it was after it was due.

[KittySawrus]

JFactor&Fergie nailed it as far as British syllabuses are concerned. Just to add, Copper is Cu, calcium is Ca.
If you ever need any help with chemistry I'm currently doing British A2-level, so I can probs lend a hand :)

[TheJFactor&Fergie<3]

JFactor&Fergie nailed it as far as British syllabuses are concerned. Just to add, Copper is Cu, calcium is Ca.
If you ever need any help with chemistry I'm currently doing British A2-level, so I can probs lend a hand :)


LMAO i appear to have switched from copper to calcium half way through... that helps huh? sorry;)

i did A Level chem a few years ago and have since done a science degree at uni so between us we can probably help lol :)

[Horseaholic]

Thanks for all of your help. I really appreciate it. I may be coming back for more ;)

[Horseaholic]

I'm back for help here :( This is what I'm having trouble with now! This class makes me feel so...stupid...I feel like I should know this stuff but I just can't get it to stick.

We did a gas lab where we collected gas from Mg + HCl we did 5 trials (had a tube in a graduated cylinder upside down in water) which consisted of collecting 122.0ml , 122.5 ml , 130.2ml , 125.0ml , 127.5 ml.... the barometer in the room read 755.0mmHg and the temp of the water was 22*c We were asked to find the average of all 5 trials which was 125.4ml = V1

my first question of the post lab is the change your average vol of H collected from mL to Liters so i do 125.4ml x (1 L / 1000ml) which equals .1254 L that's correct right?

my second question is - Using the ideal gas law, the temp of your water, and the pressure in the room, calculate the number of moles of hydrogen you actually collected..
So the ideal gas law formula is PV= nRT ....I have T - 22*c - change to Kelvin +273 = 295k, P - 755.0mmHg , V = 125.4L (assuming #1 is correct) R- 62.4(constant) , N - ? that's what we are solving for...

so my formula would be N= PV/TR so... N= (755.0 mmHg) (.124L)/ (273k)(62.4) = 94.677/18408 .. n = .00514

#3 is write the balanced reaction for the Mg plus HCl.
so.. Mg+ HCl --> H2 + MgCl2 ...I have to balance that so the balanced equation is Mg + 2HCl --> H2 + MgCl2

SOOOO number 4 is ---Using stoichiometry, calculate the number of moles of hydrogen you "should" have collected. (Start with the 0.12 of Mg you weighed.)

This is where I get lost...

5. Calculate the % yield comparing the moles of hydrogen from #4 and # 2

I know that the formula for %yield is actual/theoretical x 100 but that's all I know

I'm so confused and so flustered.

[Horseaholic]

anyone :(

[cloedoll]

I wish I could help Michelle...I'm taking chemistry right now and omfg, I hate it. It can go dieee.

[Oreos Girl]

Ok, You everything you have so far is pretty much correct. In the calculation of moles of H2 collected you dropped a 5 it should have been 0.1254 instead of 0.124. That changes the answer slightly.

For the part you need help on. You have to be in moles to compare amounts of a reaction. How do you go from 0.12 g of Mg to moles of Mg? The teacher in me just doesn't want to give answers but make you think through the answers. Once you have moles of Mg, looking at the balanced equation, what is the relationship between Mg and H2. It is 1 to 1 since your balanced equation is Mg + 2 HCl ----> H2 + MgCl2. So the moles of Mg you found is the moles of H2 you should have produced.

Last part. The moles you calculated already and had mostly correct were the actual yield and the amount I helped you will is the theoretical yield. Just to give a hint the % yield is the 112.6%. The error is because you really need to subtract the pressure of the water that was in your graduated cylinder. (All air above water has water vapor in it.) I don't know if your teacher wants you to do this or not. At 22 degrees the vapor pressure of the water is 19.8 mm Hg (a table of information I have). So you would subtract the 755.0-19.8 to give 735.2 mm Hg which would change your moles of H2 to 0.00501 which still gives you over 100% yeild.

Hope this helps and didn't add to the confusion. I will pm an email address but I don't normally check it at night.

Oreo's Girl

[Horseaholic]

Hrm I'm slightly confused. Don't I want to keep my Mg in grams because with the conversion stuff I want it to cancel out. Right??
To change .12g Mg to moles I'd multiply it by by it's molar mas right? .12 x 24.3 which would give me 2.916 but then idk what to do with that? :(

Also with my ratio of moles wouldn't it be 2 to 1 because I have 2 hydrogen and only one Mg on both sides. Ahhh I'm so confused.

[Horseaholic]

Because wouldn't my stoich problem look like this...
.12g Mg (1 mol/ 24.3g Mg)(2 mol H /1 mol Mg)(2.016 g H / 1 mol H) = .0199

So then I do my yield as
(.00514 / .0199)X 100 = 25.83%