May 31, 2020, 06:05:42 PM
Forum Rules: Read This Before Posting


Topic: percent composition by mass  (Read 6641 times)

0 Members and 1 Guest are viewing this topic.

Offline ainoko_hikaru

  • Regular Member
  • ***
  • Posts: 18
  • Mole Snacks: +0/-0
  • Gender: Female
percent composition by mass
« on: July 19, 2008, 09:15:38 AM »
I don't know if I'm on the right track. please help.
       A mixture of Na2CO3 and MgCO3 of mass 7.63g is reacted with excess hydrochloric acid. The CO2 gas generated occupies a volume of 1.67L at 1.24 atm and 26 degrees Celcius. From these data, calculate the percent composition by mass of Na2CO3 in the mixture.

I have already made a balanced equation:

Na2CO3 + MgCO3 + 2HCl -> 2NaCl + MgCl2 + 2CO2 + 2H2O

after that, I computed for the number of moles of CO2 using the ideal gas equation and arrived with 0.0860 moles CO2. I then calculated for the mass of carbon dioxide, then using this I calculated for the number of moles of Na2CO3 which turned out to be 4.29 x 10-2. After computing for the number of moles I then computed for the mass of Na2CO3 which turned out to be 4.55 grams. My solution for percent composition involved using the computed mass of Na2CO3 divided by the given mass of Na2CO3 and MgCO2 multiplied by a hundred resulting to 59.6%. Am I correct?

Thanks!

Offline sjb

  • Global Moderator
  • Sr. Member
  • ***
  • Posts: 3451
  • Mole Snacks: +217/-42
  • Gender: Male
Re: percent composition by mass
« Reply #1 on: July 19, 2008, 11:04:56 AM »
I'm not sure if you're right, but the approach seems wrong.

Essentially you have 2 things going on here

Na2CO3 + 2HCl -> 2CO2 + 2NaCl + 2H2O, and
Mg2CO3 + 2HCl -> 2CO2 + MgCl2 + 2H2O.

So you have mass of the solid, which is equal to the mass of Na2CO3 + the mass of Mg2CO3, and the amount of CO2, which comes from the decomposition.

Two equations, two unknowns.

What I think you have done is calculated the mass of Na2CO3 if there were an equimolar amount of Mg2CO3 and Na2CO3 present.

Offline enahs

  • 16-92-15-68 32-7-53-92-16
  • Retired Staff
  • Sr. Member
  • *
  • Posts: 2179
  • Mole Snacks: +206/-44
  • Gender: Male
Re: percent composition by mass
« Reply #2 on: July 19, 2008, 03:00:38 PM »
sjb is correct, your answer is wrong.

However, I do not like the two equation approach for this because it puts more emphasis on math and not thinking about what is going on behind it, in my opinion. But yes, math is required.

You have a mixture of two components, producing the same thing. Think about it at a molar level first.


1 mol Na2CO3 + 1 mol MgCO3 -> 2 mol CO2

You know the value for the total mols of CO2 produced (and I would use more significant figures in your calculations and do not round until the end, it is off a decent amount).

So you know

x mols Na2CO3 + y mols MgCO3 -> 0.084 mols CO2

Correct?
But you only know the mass, but how can you relate mass to mols?
Molecular Weight!

So, call one arbitrarily x and the other 7.63-x (because you know you have 7.63 grams total)

So now you have

   x g Na2CO3             
         +         7.63 -x MgCO3              =  0.084
 105.99 g Na2CO3 /mol                     84.32 g /  MgCO3

Which the units cancel and leave mols + mols = 2 mols, which is what your balanced equation said.
You use the molecular weight to relate the mass to the amount of mols, as dictated by stoichmetry, directly into the balanced equation with the appropriate coefficient. It is no longer a system of equations and just a math problem, but the math is now displaying the chemical relationships.

Remember, balanced equation relates mols, not mass!

Now it is just the basic algebra and solving for x, and once you know x you know the other as well.

Hope that makes sense and helps.




Offline ainoko_hikaru

  • Regular Member
  • ***
  • Posts: 18
  • Mole Snacks: +0/-0
  • Gender: Female
Re: percent composition by mass
« Reply #3 on: July 20, 2008, 04:29:11 AM »
oh, ok, thanks for pointing out my mistake. I'll try out the solution. Thanks for the help, enahs and sjb! :)

Sponsored Links