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Can anyone help me with my chem homework?

Discussion in 'Off Topic [BG]' started by Kibuddy, Nov 12, 2006.


  1. Kibuddy

    Kibuddy

    Apr 30, 2005
    Alright all of you chem buffs. Here's a question for you.

    I have a 100mL solution of SrCl[SUB]2[/SUB]*6H[SUB]2[/SUB]O. To make the solution, we took 4.76g of the hydrate and added enough water to dissolve the powder and create the 100mL solution. Now I need to calculate the molarity.

    I wound up with a molarity of .179M, but I'm not sure I did it correctly. Can anyone double check my work for me?

    Here's my work to the best of my typing ability (it's supposed to be dimensional analysis):

    4.76g | 1mol | 1000mL|

    100mL | 266.64g | 1L |


    = .179M
     
  2. Mike Money

    Mike Money In Memoriam

    Mar 18, 2003
    Bakersfield California
    Avatar Speakers Endorsing Hooligan
    Sure.

    Why not.


    Dude, honestly, Chemistry class is one of those classes that on the last day of school, when you leave that class for the last time... you don't remember ANY OF IT.

    something about avacados number and uh... moles. ya.
     
  3. Kibuddy

    Kibuddy

    Apr 30, 2005


    Well, let's just say that I enjoy the class and am considering going to college to major in chemistry. So, it may not have been worthwhile for you, but it sure as hell is to me.
     
  4. Avogadro's Number... haha, I remember enough of chemistry to atleast know that. Unfortunatey Kibuddy, I can't help you :( I remember a bit about moles, but I can't help you on that problem.
     
  5. Christopher

    Christopher

    Apr 28, 2000
    New York, NY
    It's been a really long time since I did this, so take it with a grain of salt, but the atomic mass of the hydrate should be around 267 by my count, so a mole of the stuff should be 267 g. You have 4.76 g of the stuff, which works out to 0.0179moles in 100 mL, or 0.179 moles in 1L.

    Looks right to me.
     
  6. I can't believe, I took AP chem last year and I don't even remember how to do this kind of problems! Ugh1!!
     
  7. Kibuddy

    Kibuddy

    Apr 30, 2005

    Awesome, thanks a bunch!
     
  8. Ain't that the truth? I took multiple chemistry classes in high school, but I remember next to nothing about them.
     
  9. bah chemistry
     
  10. Kibuddy

    Kibuddy

    Apr 30, 2005
    Here's another one:


    We reacted sodium carbonate and strontium chloride hexahydrate and got strontium carbonate as a product. Why is it safe to pour strontium carbonate down the sink?

    I really have no idea. I would think it has something to do with it being neutralized, but I'm not sure. At all.
     
  11. Demon

    Demon

    Mar 17, 2006
    Sweden, Stockholm
    Put black powder in, always works.
     
  12. dlloyd

    dlloyd zzzzzzzzzzzzzzz

    Apr 21, 2004
    Scotland
    Yep, that's right, but you'd be better off calling that 179 mM
     
  13. dlloyd

    dlloyd zzzzzzzzzzzzzzz

    Apr 21, 2004
    Scotland
    Check your solubility rules.
     
  14. Kibuddy

    Kibuddy

    Apr 30, 2005

    Oh man. I feel like an idiot. :scowl:


    Thanks for the help!
     
  15. Kibuddy

    Kibuddy

    Apr 30, 2005
    Okay, this should be my last question:

    I have 5.3g sodium carbonate and 4.76g strontium chloride hexahydrate and got 1.09g strontium carbonate (really bad yield, I'm aware).

    How do I determine how much of the sodium carbonate was not used (the hydrate is the limiting reactant)? I vaguely remember something to do with ratios, but I can't remember how to do it. Could someone explain the process to me (and hopefully not give away the answer)?
     
  16. Poop-Loops

    Poop-Loops Inactive

    Mar 3, 2006
    Auburn, Washington
    Don't you have a chem book or something? This stuff is pretty rudimentary. Limiting reactant I think is correct, but I also don't remember the exact process to find this stuff, so just check your book. It should have a detailed explanation.
     
  17. Kibuddy

    Kibuddy

    Apr 30, 2005

    My book is in my locker :)scowl:). And keep in mind I'm a first year chem student, so I'm still trying to get a grasp on this stuff.
     
  18. Poop-Loops

    Poop-Loops Inactive

    Mar 3, 2006
    Auburn, Washington
    Read the book. It's that easy. Books will have explanations in them with examples. If then you cannot understand how to do a problem, ask away. But it just seems like you have no idea what you are doing, because you don't know the concept yet. This is the hardest part in learning something, but that's why it's the most important, too.

    If you don't have the book with you (shame on you! :mad: ) then Google is your friend.

    http://www.chem.tamu.edu/class/majors/tutorialnotefiles/limiting.htm
    http://wine1.sb.fsu.edu/chm1045/notes/Stoich/Limiting/Stoich07.htm
     
  19. Kibuddy

    Kibuddy

    Apr 30, 2005

    Here's the thing: it's not that I don't understand stoichiometry. I can do basic stoichiometric equations with ease. This question just flipped things around in a direction that I've never seen before. The only stoich I've been doing so far is theoretical yield, actual yield, molarity, etc. I've not yet been shown how to determine how much of a certain reactant was actually used. Maybe it was just a case of not knowing how to apply it yet. I don't know. All I know is that this was a new scenario for me, and I didn't know how to go about solving it.

    So, maybe it is a very rudimentary concept for you, and if it annoys you that I asked, then I am sorry. But I AM a first year student. These are all brand new concepts to me. I've never even seen this particular problem at hand until now.

    And I take some offense to your accusation that I don't know what I'm doing. It was a simple question about an application I had not seen. That doesn't mean that I don't know anything about stoichiometry. I'll stop here, though, as I'm starting to get all defensive, and nobody likes a defensive, angsty teen. :p


    And thank you for the links. They've been very helpful to me.
     
  20. steve21

    steve21 Inactive

    I'm in AP Chem and I don't know this. :(
     

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