Calculate the molar mass of the metal

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Discussion Overview

The discussion revolves around calculating the molar mass of a metal that reacts with phosphorus to form a compound, specifically focusing on the reaction and the necessary calculations involved. The participants explore various methods and approaches to solve the problem, including unit analysis and the quadratic formula.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents the reaction formula and asks for the molar mass of the metal based on given masses of the metal and the compound.
  • Another participant suggests using unit analysis to approach the problem, emphasizing the importance of mole ratios in the balanced equation.
  • A participant expresses confusion about needing the molar mass of phosphorus and the grams of P4 to solve the problem.
  • Multiple participants attempt to derive the molar mass using different methods, leading to varying results, such as 45 g/mol and 10.6 g/mol.
  • One participant proposes a quadratic equation derived from the reaction and suggests using the quadratic formula to find the molar mass.
  • Another participant provides an alternative method involving mole ratios and calculations, leading to a different result of approximately 39 g/mol.
  • Several participants express uncertainty about their calculations and seek clarification on errors in their methods.
  • Participants discuss potential mistakes in their setups and calculations, particularly regarding the ratios used in their equations.

Areas of Agreement / Disagreement

There is no consensus on the correct method or the resulting molar mass, as participants present multiple competing views and results, leading to ongoing confusion and discussion.

Contextual Notes

Participants highlight various assumptions and steps in their calculations, but there are unresolved issues regarding the setup of equations and the application of unit analysis. Some participants acknowledge errors in their approaches but do not reach a definitive resolution.

parwana
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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:
[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:
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
[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]<br /> if my calculations are correct. Now apply the quadratic formula to find x.[/tex]
 
Last edited:
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 [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:
  • #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:
[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]

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:

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

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