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Calculate the molar mass of the metal

  1. Oct 23, 2004 #1
    A metal M reacts with phosphorus ( P4 ) to form a compound having the formula M3P.
    13.30 g of the metal produce 16.81 g of the compound. Calculate the molar mass of the metal.


    Enter a numeric answer only, no units.
     
  2. jcsd
  3. Oct 23, 2004 #2
    Have you tried the question? Please show us what you have done. I always approach these problems using unit analysis.
     
  4. Oct 23, 2004 #3
    I dont understand it, iam new to this, but dont i need the molas mass of one to find the molar mass of the other???? Also I need some grams of P4as well, please tell me how to do this, or i'll fail the exam.
     
  5. Oct 23, 2004 #4
    Start with the balanced equation:
    [tex]12M+P_{4}\longrightarrow4M_{3}P[/tex]
    Now unit analysis (remember that the equation expresses mole ratios between the reactants and products, not mass ratios):
    [tex]x\frac{g}{mol}M=16.81gM_{3}P\times\frac{1molM_{3}P}{(3x+30.97)gM_{3}P}\times\frac{12molM}{4molM_{3}P}\times13.3gM[/tex]
    Do you see how the units cancel to give me the ones I want for the answer? That's unit analysis. Usually, you don't have a variable on both sides of the equation, but in this case you do. That's not a problem; just isolate x, and that is your answer.

    PS: normally I don't give away this much, but you have an exam. :smile:
     
    Last edited: Oct 23, 2004
  6. Oct 23, 2004 #5
    I got 45 g/mol, is it right?? This is just too confusing for me, I dont get it.
     
  7. Oct 23, 2004 #6
    please share the answer with me and show me step by step
     
  8. Oct 23, 2004 #7
    Because of the way the equation is set up, you need to use the quadratic formula here. You can simplify the equation, dropping the units, to
    [tex]x=3(16.81)(13.3)\left(\frac{1}{3x+30.97}\right)[/tex]
    which becomes the quadratic equation
    [tex]3x^{2}+30.97x-670.719[/itex]
    if my calculations are correct. Now apply the quadratic formula to find x.
     
    Last edited: Oct 23, 2004
  9. Oct 23, 2004 #8
    well what do u get please tell fast, i dont have a decent calculator.,
     
  10. Oct 23, 2004 #9
    ok i got 10.6g/mol and
     
  11. Oct 23, 2004 #10
    Correct. Do you understand the method?
     
  12. Oct 24, 2004 #11
    but its wrong, i plugged it in, and it said it was wrong
     
  13. Oct 24, 2004 #12

    chem_tr

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    Hello Parwana, Sirus has done much to help you, but maybe I can be of a little more help.

    Let me consider that [tex]\displaystyle\frac{13.30}{M}[/tex] moles of metal is reacted with tetraphosphorus to give [tex]\displaystyle\frac{16.81}{3M+P}[/tex] moles of compound. Here, M and P denotes the molar masses of metal and phosphorus, respectively.

    We also know that [tex]\displaystyle\frac{3*16.81}{3M+P}=\frac{13.30}{M}[/tex], as you understand from the reaction Sirus wrote. By taking 30.97 grams/mol for phosphorus, we get this:

    [tex]\displaystyle\frac{50.43}{3M+30.97}=\frac{13.30}{M}[/tex]

    where [tex]50.43M=39.90M+411.901[/tex]. You can find M here. I hope this is settled now.

    Take care.
     
    Last edited: Oct 24, 2004
  14. Oct 24, 2004 #13
    thanks so much chem tr and sirus, i understand it much better now.
     
  15. Oct 24, 2004 #14
    I am curious as to why my method gave a different answer. From what chem_tr did, I get about 39 g/mol, but my method gave about 10.6 g/mol. What was I doing wrong?
     
  16. Oct 24, 2004 #15

    chem_tr

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    I think the error is hidden in 12/4=3 ratio, as in the first fraction of your equation; I did the same error and corrected.
     
  17. Oct 24, 2004 #16
    Hmm. I now realize one error I made...I should not have multiplied by 13.3 g at the end to get proper units (not g*mol). Therefore the calculations should be as follows:

    [tex]xmolM=16.81gM_{3}P\times\frac{1molM_{3}P }{(3x+30.97)gM_{3}P}\times\frac{12molM}{4molM_{3}P}}}[/tex]

    To solve for x...

    [tex]3x^{2}+30.97x-50.43[/tex]

    [tex]x=1.430207329molM[/tex]

    Now to find molar mass...

    [tex]x\frac{g}{mol}M=\frac{13.30gM}{1.430207329molM}=9.29935103\frac{g}{mol}M[/tex]

    Why am I getting a different answer than you? I'm not sure what error you are getting at regarding the 12/4=3 ratio.
     
  18. Oct 24, 2004 #17

    chem_tr

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    It is very likely that there is something wrong with your setup. Please review the logic behind your formula.
     
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