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Chemistry Stoichiometry Help :S

  1. Jan 8, 2010 #1
    URGENT Chemistry Stoichiometry Help :S

    1. The problem statement, all variables and given/known data

    Suppose that you use 1.110g of iron in this experiment. What is the minimum volume of 1.0M copper sulfate solution that you should add?

    2. Relevant equations

    Fe(s) + Cu2+(aq) --> Fe2+(aq) + Cu(s) (1)

    2Fe(s) + Cu2+(aq) --> 2Fe3+(aq) + 3Cu(s) (2)

    *****Since I do not know at this stage whether equation (1) or (2) is the appropriate one, I must base my calculation on whichever of the equations would require the most copper sulfate for a given quantity of iron*****

    3. The attempt at a solution

    Here is what I believe to be a very poor attempt at solving the problem....any help will be greatly appreciated!

    1.110g of Fe x (1 mol / 59.0 g Fe) = 0.018813559 moles Fe

    0.018813559 mol Fe x (3 mol Cu2+ / 2 mol Fe) = 0.028220338 mol Cu2+

    AND this is where I become stuck since the answer I need uses litres but all I have are the moles of Cu2+. Also, I don't know where the whole 1.0M from the question comes into play. I really need help. Thanks in advance!
     
  2. jcsd
  3. Jan 8, 2010 #2

    Borek

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    Re: URGENT Chemistry Stoichiometry Help :S

    Fe2+ it is.

    [tex]C = \frac n V[/tex]

    Do you know how to solve for n?
     
  4. Jan 10, 2010 #3
    Re: URGENT Chemistry Stoichiometry Help :S

    Hello! First of all, thanks so much for replying to me! I am in dire need of help.

    C = n/v makes sense. I would solve for "V", I have "C" (= 1.0 M) and "N" is the number of moles, correct?

    Now, solving for "n" would give me the moles of copper sulfate? Do I obtain this number by using molar ratios? If so, how would I go about doing that? I think I solved for the moles of Cu2+ above but I don't know how I would get the moles of copper sulfate....HELP! :)
     
  5. Jan 10, 2010 #4

    Borek

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    Re: URGENT Chemistry Stoichiometry Help :S

    You have used wrong reaction equation so your result was off, but the approach was correct. Just note that molar mass of iron is not 59.

    If there is one atom of copper per one molecule of copper sulfate, how many moles of copper sulfate per one mole of copper atoms (or copper ions)?

    --
    methods
     
  6. Jan 10, 2010 #5
    Re: URGENT Chemistry Stoichiometry Help :S

    Again, thanks for helping me out. Also, I like how you don't just give me the answer, you "walk me through it" which is actually what I need, better learning experience!

    So...

    Using the correct reaction equation and the correct molar mass for iron now:

    1.110 g Fe x (1 mol / 56.0 g Fe) = 0.018813559 mol Fe

    0.018813559 mol Fe x (1 mol Cu2+ / 1 mol Fe) = 0.018813559 mol Cu2+

    and the ratio between moles of Copper Sulfate and copper atoms is 1:1

    -Since the question is asking the volume of copper sulfate needed, I'm assuming my answer should be in litres? Is that where the c=n/v formula comes in? would this be correct?

    C = n/v ------> V= c x n

    V = 1.0M x 0.018813559
    V = 0.018813559 L

    Somehow this doesn't seem right...I'm so sorry to bother you but I'm just not good at chemistry no matter how hard I try and I have nobody else to help me. Thanks a lot!
     
  7. Jan 10, 2010 #6

    Redbelly98

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    Re: URGENT Chemistry Stoichiometry Help :S

    Borek may be logged off for the night, as he lives in Europe, so I'll try to help out until he returns.

    It looks like you still used 59 when you did this calculation. Redo it using 56.

    Yes.

    Actually, doing the algebra gives ---> V = n/C
    But, since C=1.0M, the numbers work out the same!
    Actually V = 0.01881 mol / 1.0M = 0.01881 L
    or round it off to 0.019 L, since the 1.0M is given to just 2 significant figures.
    Looks pretty close to me :smile:, but you'll need to redo it using the correct 56 g/mol for iron.
     
  8. Jan 10, 2010 #7
    Re: URGENT Chemistry Stoichiometry Help :S

    Hello Redbelly! Oh my! Yes, you are right! I did use the 59 again!!!!!!!!

    THANK YOU SOOOOOOO MUCH!!!!!!! I really appreciate all the help from both Borek and yourself! You guys have made all the difference for me!

    I never took high school chemistry, but now I need to pass first year university chemistry so that I can graduate this summer. Basically, I just taught myself (not very well by the way) in the last 3 months all of high school chemistry and now I'm taking this university course and it's so difficult for me! Anyways, thanks sooooo much guys and i'll probably be on here asking for your help many more times. I hope you all don't get tired of me. Thanks a lot! :)

    -K.
     
  9. Jan 10, 2010 #8
    Re: URGENT Chemistry Stoichiometry Help :S

    So here it is again...I think this time it's correct....

    (1) Fe(s) + Cu2+(aq) --> Fe2+(aq) + Cu(s)

    1.110 g Fe x (1 mol Fe / 56.0 g Fe) = 0.019821428 mol Fe

    0.019821428 mol Fe x (1 mol Cu / 1 mol Fe) = 0.019821428 mol Cu

    C = n/V ----> V = n/C
    V= 0.019821428 mol Cu / 1.0M
    V= 0.019821428
    V = 0.020 L

    I hope this is right! :) Again, thanks so much!
     
  10. Jan 11, 2010 #9

    Borek

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    Re: URGENT Chemistry Stoichiometry Help :S

    20 mL or 0.020 L it is.

    Note: molar mass of iron is 55.845 g/mol. While it usually doesn't matter much and in most stoichiometry questions using 56 g/mol will be OK, stating it is 56.0 g/mol is incorrect. 55.8 g/mol if you want 3 siginificant digits.

    --
     
  11. Jan 13, 2010 #10
    Re: URGENT Chemistry Stoichiometry Help :S

    Thank you so much for all your help!

    -K.
     
  12. Jan 13, 2010 #11
    Hello! I have another question. Please let me know if it's o.k, and if not, could you point out what is wrong? Thanks!

    Q: How many Cu (copper) atoms are present in a piece of sterlingsilver jewelry weighing 33.74g? (Sterling silver is a silver–copper alloy containing 92.5% Ag (silver) by mass.)


    A: (My attempt at an answer) So...if 92.5% of the 33.74g sample of jewelry is made up of silver, 7.5% of the same piece of jewelry must be made of copper. So...

    33.74 x (92.5 / 100) = 31.2

    33.74 - 31.2 = 2.54

    2.54 g Cu x (1 mol Cu / 63.546 g Cu) = 0.03997 mol Cu

    0.03997 mol Cu x (6.022 x 10^23 / 1 mol Cu) = 2.41 x 10^22 atoms of Cu

    Please let me know what you think ! :)

    -K.
     
  13. Jan 13, 2010 #12
    Re: URGENT Chemistry Stoichiometry Help :S

    I did 33.74 g(.075) = 2.53 g and got 2.40×1022. Not much difference, but I think this way is more straightforward. :smile:
     
  14. Jan 13, 2010 #13
    Re: URGENT Chemistry Stoichiometry Help :S

    Hi! o.k sounds good! Thanks so much!!!!! :)
     
  15. Jan 14, 2010 #14

    Borek

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    Re: URGENT Chemistry Stoichiometry Help :S

    Remember to not round down intermediate results. If you show them - use rounded down, but do not use those rodunded down to calculate anything.

    --
     
  16. Jan 14, 2010 #15
    Re: URGENT Chemistry Stoichiometry Help :S

    Hello! Thanks, I will keep that in mind.
     
  17. Jan 14, 2010 #16
    Re: URGENT Chemistry Stoichiometry Help :S

    Hello everyone! I am back, as I had predicted. This time, I would really appreciate it if anyone could just go through these 3 questions and let me know if they think they're alright. Also, for the last part of the very last question, I have no idea how to solve it so if anyone can help me I would REALLY appreciate it, I am learning a lot here! :)

    1.Q: // A solution contains 11.00% sucrose (cane sugar) by mass. What mass of the solution, in grams, is needed for an application that requires 2.80kg sucrose?

    A: // I assumed a 100g sample of the sucrose solution.
    2.80 kg sucrose x ( 1g / 1000kg ) = 0.0028g sucrose
    0.0028g sucrose x (100g solution / 11.00g sucrose) = 0.0255g solution

    2.Q: // Magnesium (Mg2+) occurs in seawater to the extent of 1.4g magnesium per kilogram of seawater. What volume of sea water, in cubic meters, would have to be processed to produce 1.30 x 10^5 ton of magnesium? Assume a density of 1.025g/mL for seawater. Remember, 1 ton = 2000 lb. Give your answer to two significant figures.

    A: // 1.30 x 10^5 tons Mg2+ x (2000 lb / 1 ton) = 260000000 lb Mg2+
    260000000 lb Mg2+ x (453.59237 g / 1 lb) = 1.18 x 10^11 g Mg2+
    1.18 x 10^11 g Mg2+ (1 kg seawater / 1.4 g Mg2+) = 8.43 x 10^10 kg seawater
    8.43 x 10^10 kg seawater (1 g / 1000 kg) = 84300000g
    84300000g x (1 mL / 1.025g) = 82243902.44 mL
    82243902.44 mL x (1m^3 / 1000000 mL) = 82 m^3

    3.Q: // This question will be based on your knowledge of the law of multiple proportions.
    i.)There are 2 different compounds of sulfur and fluorine. In SF_6, the mass of fluorine /g of sulfur is 3.55g, in the other compound , SF_x, the mass of fluorine /g of sulfur is 1.18g. What is the value of “x” for the second compound? (answer must be an integer).

    ii.) Samples of 3 different compounds were analyzed and the masses of each element were determined as follows
    Compound #1: Mass of Nitrogen 5.6g, Mass of Oxygen 3.2g
    Compound #2: Mass of Nitrogen 3.5g, Mass of Oxygen 8.0g
    Compound #3: Mass of Nitrogen 1.4g, Mass of Oxygen 4.0g

    If you were John Dalton and had never heard of a mole, which of the following would you think were possible sets of formulas for the compounds #1,2 and 3, respectively? Chose all that apply

    A.) NO_4, NO_10, NO_5
    B.) NO, NO_2, NO_4
    C.) N_2O, N_2O_4, N_2O_5
    D.) NO_2, NO_8, NO_10

    iii.)In one experiment, the burning of 0.318g sulfur produced 0.635g sulfur dioxide as the sole product of the reaction. In a second experiment, 0.844g sulfur dioxide was obtained. What mass of sulfur must have been burned in the second experiment?

    A:// i.) SF_2 (But I think I guessed, I don’t know how to actually come up with the answer)
    ii.) I chose “C” based on ratios I came up with...but don’t know if that’s the right
    procedure.
    iii.)This one, I have absolutely no idea how to solve it, if someone could help me I would most sincerely appreciate it!
     
  18. Jan 15, 2010 #17

    Borek

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    Re: URGENT Chemistry Stoichiometry Help :S

    It is 1000 g per 1 kg, not the other way around. k stands for kilo, that means thousand.

    3rd can be done using ratios.

    http://www.chembuddy.com/?left=balancing-stoichiometry&right=ratio-proportions

    Note that you don't have to set up ratio using reaction equation. Take a look at "If you can buy 2 eggs for $1 how many eggs can you buy if you have $2?" Written as ratio it is

    [tex]\frac {2_{eggs}} {1USD} = \frac {N_{eggs}} {2USD}[/tex]

    Solve for unknown number of eggs (Neggs).

    --
    methods
     
  19. Jan 15, 2010 #18
    Re: URGENT Chemistry Stoichiometry Help :S

    Hello Borek,

    Thanks a lot for that, I corrected my mistake and I solved the third question and they were right. However, my professor (or rather her computer) said that question #2 is wrong...I can't find the mistake...is it because I didn't cube the centimetres in the last step? or is it because of another reason...I can't figure it out...if you spot anything could you let me know? Thanks again for everything!

    -K.
     
  20. Jan 15, 2010 #19

    Borek

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    Re: URGENT Chemistry Stoichiometry Help :S

    First of all - question is ambiguous, as ton can be either a metric ton (1000 kg) or short ton (2000 lb) and it is not clearly stated which one it is. Could be that's local thing, but I am using metric in everydays life so for me that's 1000 kg, period.

    Your problems started at 84300000g - that's not the correct mass. But I have already told you that.

    --
     
  21. Jan 28, 2010 #20
    Re: URGENT Chemistry Stoichiometry Help :S

    Hi Borek,

    Again, thanks so much for your help in the last question it really helped me out. Nonetheless, I have another question and if you have a chance to help me out I'd appreciate it.

    Q: 1.0560 g of antacid is weighed and mixed with 75.00mL of excess 0.1126 M HCL. The excess acid required 4.68mL of 0.1008 M NaOH for back titration. Calculate the neutralizing power of the antacid in terms of mmol H+ per gram of antacid.

    A: So here is my approach. I figured I would figure out the moles of H in HCl in the following way 0.075 L x (0.1126 mol / 1 L) = 0.008445 mol HCl
    0.008445 mol HCl x (1 mol H / 1 mol HCl) = 0.008445 mol H (since 1:1 ratio)...

    Now I would use the balanced chemical equation and molar ratios to determine the moles of antacid that would react ...problem is though that I wouldn't know how to get a balanced chemical equation since I don't know the chemical formula for "antacid"...do you know of any other approach that I could use to solve this problem? Thanks a lot!!!
     
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