[reddagon5]

I need to know if I am doing this correctly:
You distilled a mixture whose composition was known on a vol:vol basis. However, molar composition is usually more useful. Apply the data below to a 15.0 mL sample consisting of: 30% 2-propanol, 30% chloroform and 40% octane, by volume to convert it to molar composition:
Chloroform: formula mass: 119.38 g/mol; density: 1.48 g/mL; bp: 61.7 degrees C
2-propanol: formula mass: 60.11 g/mol; density: 0.786 g/mL; bp: 82.4 degrees C
Octane: formula mass: 114.23 g/mol; density: 0.702 g/mL; bp: 125.7 degrees C

Determine the percent, by mass, of each component in this mixture.


I did this:
Chloroform: 0.3 * 15.0 mL=4.5 mL
4.5 mL*1.48 g/mL= 6.7%

2-propanol: 0.3*15 mL=4.5 mL
4.5 mL* .786 g/mL=3.5%

Octane: 0.4 *15 mL= 6 mL
6mL * 0.702 g/mL =4%

I doubt this is correct.

[TheUnassuming]

You are very close.  So let's look at the chloroform example.  The first equation gave you find the number of ml's of CHCl3 in your distillate.  Now look at your second equation, what are the units for the number you calculated?

[zsinger]

see that all units cancel correctly, and that shall guide you……as stated, you are VERY close to being perfect.
          -Zack

[reddagon5]

So for chloroform, you would have 6.7 g
2-propanol: 3.5 gram
octane: 4.0 grams

so, 14.2 total grams, so 47.2% chloroform, 24.6% 2-prop, 28.2% octane?

[TheUnassuming]

Perfect!  Good job.
When doing calculations there is a fancy name for it (that I don't remember) but you just line up the whole series of calculations and it makes it easier to watch the units.  So for this example you would do:
0.3 * 15mL * 1.48 g/mL = 6.7g
You can do this with much more complex calculations as well.

When doing calculations always make sure your units makes sense in the end, and when calculating percentages always make sure you get a total of 100%.

[zsinger]

That fancy name is "Unit Conversion Standard" sets.
      -Z

[reddagon5]

Thanks!

[reddagon5]

So if you wanted a mole percent composition, for chloroform it would be:
6.7 g *119.38 g/mol=799.846 mol
2-propanol would be 210.385 mol
Octane would be 456.92 mol

And then you could calculate %'s? Right?

[Borek]

6.7 g *119.38 g/mol=799.846 mol

Nope.

6.7 g *119.38 g/mol=799.846 g²/mol

Looking at units I doubt that's what you want.

[reddagon5]

Whoops. So convert 6.7 g to mol, then multiply by 119.38 g/mol? so
6.7 g CHCl3 *6.022*10^23 mol/1 g = 4.03 *10^24 mol?
Wait, I don't think that's right.

[TheUnassuming]

Your units for avagadros number are incorrect.  If you multiply grams of CHCl3 * avagadros number you get a number with the units g/mol.   Avagadro's number is just the number of atoms in a mol, so doesn't help you in this case because you don't have the number of CHCl3 atoms, you have the mass. 
Look at Borek's post again and your earlier post.

[reddagon5]

So chloroform (CHCl3) would be 119.3779 amu. So 6.7 g*1mol/119.3779 g= .056 mols?

[TheUnassuming]

Exactly right.

[reddagon5]

Thanks!