Stoichiometry

[parwana]

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.

[Sirus]

Have you tried the question? Please show us what you have done. I always approach these problems using unit analysis.

[parwana]

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.

[Sirus]

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:

[parwana]

I got 45 g/mol, is it right?? This is just too confusing for me, I dont get it.

[parwana]

please share the answer with me and show me step by step

[Sirus]

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
[tex]3x^{2}+30.97x-670.719[/itex]
if my calculations are correct. Now apply the quadratic formula to find x.

[parwana]

well what do u get please tell fast, i dont have a decent calculator.,

[parwana]

ok i got 10.6g/mol and

[Sirus]

Correct. Do you understand the method?

[parwana]

but its wrong, i plugged it in, and it said it was wrong

[chem_tr]

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.

[parwana]

thanks so much chem tr and sirus, i understand it much better now.

[Sirus]

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?

[chem_tr]

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.

[Sirus]

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.

[chem_tr]

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