Calculate the molar mass of the metal

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The discussion revolves around calculating the molar mass of a metal (M) that reacts with phosphorus (P4) to form a compound with the formula M3P. Given that 13.30 g of metal produces 16.81 g of the compound, participants emphasize the importance of unit analysis and the correct application of stoichiometry in solving the problem. A balanced chemical equation is established, and participants engage in deriving relationships between the masses and moles of the reactants and products. One participant suggests using the quadratic formula to isolate the molar mass variable, leading to different calculated values. Discrepancies in results arise from errors in the setup of equations and the application of mole ratios. Ultimately, the correct molar mass is suggested to be around 39 g/mol, while earlier calculations yielded incorrect values due to misunderstandings in the stoichiometric relationships and unit conversions. Participants encourage careful review of the logic and calculations to ensure accurate results.
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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.
 
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Have you tried the question? Please show us what you have done. I always approach these problems using unit analysis.
 
I don't understand it, iam new to this, but don't 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.
 
Start with the balanced equation:
12M+P_{4}\longrightarrow4M_{3}P
Now unit analysis (remember that the equation expresses mole ratios between the reactants and products, not mass ratios):
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
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:
 
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I got 45 g/mol, is it right?? This is just too confusing for me, I don't get it.
 
please share the answer with me and show me step by step
 
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
x=3(16.81)(13.3)\left(\frac{1}{3x+30.97}\right)
which becomes the quadratic equation
3x^{2}+30.97x-670.719[/itex]<br /> if my calculations are correct. Now apply the quadratic formula to find x.
 
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well what do u get please tell fast, i don't have a decent calculator.,
 
ok i got 10.6g/mol and
 
  • #10
Correct. Do you understand the method?
 
  • #11
but its wrong, i plugged it in, and it said it was wrong
 
  • #12
parwana said:
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.

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

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

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

\displaystyle\frac{50.43}{3M+30.97}=\frac{13.30}{M}

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

Take care.
 
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  • #13
thanks so much chem tr and sirus, i understand it much better now.
 
  • #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?
 
  • #15
Sirus said:
Start with the balanced equation:
12M+P_{4}\longrightarrow4M_{3}P
Now unit analysis (remember that the equation expresses mole ratios between the reactants and products, not mass ratios):
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

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.
 
  • #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:

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

To solve for x...

3x^{2}+30.97x-50.43

x=1.430207329molM

Now to find molar mass...

x\frac{g}{mol}M=\frac{13.30gM}{1.430207329molM}=9.29935103\frac{g}{mol}M

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.
 
  • #17
It is very likely that there is something wrong with your setup. Please review the logic behind your formula.
 
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