# Gen Chem Problem

Discussion in 'MCAT: Medical College Admissions Test' started by dxu, Mar 12, 2007.

1. ### dxu the great one 10+ Year Member

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First off, please do not provide me with the answer to the problem (I need to know how to do these things... )

The problem is:

By titration, 15.0 mL of 0.1008 M NaOH is required to neutralize a 0.2053 g sample of an organic acid. What is the molar mass of this acid? An elemental analysis of the acid indicates that it is composed of 5.89% H, 70.6% C, and 23.5% O by mass. Determine the molecular formula of the acid.

I think I can figure out the second portion of the problem with out determining the first part by assuming I have a 100g sample and thus have 5.89g H, and so on. But the first part is throwing me off a little b/c I need to know the number of moles of each element to find the molar mass, no?

Any help would be greatly appreciated.

and yes I went to the tutor but he just had this dumb look on his face and no real answers.

dxu

2. ### Davjc2009 2+ Year Member

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.015L * Molarity of NaOH = Mols of NaOH used

For every 1 mol of NaOH used= 1 mol of organic acid titrated

.2053g organic acid / mol of organic acid titrated = molar mass

You said you could do the next part... I won't doubt your skills

3. ### punkindrublic calls shenanigans 7+ Year Member

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...and make sure you know the difference between molecular and empirical formulas.

4. OP

### dxu the great one 10+ Year Member

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While I think I have done this correctly I still am going to run this whole problem by the board, hoping I did it right.

.015L x 0.1008 M NaOH = 1.512x10-3 M

Since the ratio of NaOH to Organic Acid is 1/1, there are 1.512x10-3 M of Organic Acid

.2053g of Organic Acid / 1.512x10-3 M = 135.78g/mol

This allowed me to figure out the second portion

H = 5.89% x 135.78 = 7.997 ~ 8

C = 70.6% x 135.78 = 95.86 ~ 96

O = 23.5% x 135.78 = 31.91 ~ 32

Divide each subscript by the smallest subscript, which is 8, and you get formula --> HC12O4

I think I am doing this problem correctly but it still seems I am missing something. Any thoughts?

5. ### punkindrublic calls shenanigans 7+ Year Member

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Again, make sure you know the difference between empirical and molecular formulas: the process you're using now is calculating for empirical formula. You don't care about stoichiometry here, you've already figured out your molecular mass and know the mass ratios of the atoms in the molecule.

If you work out exactly what the numbers from your calculations in the second portion mean, you should be able to see how to get to the correct solution (i.e. label your units!).

6. OP

### dxu the great one 10+ Year Member

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Okay well I know the empirical formula is figured out by the steps I have already completed. The empirical formula for this problem is H(C3O)4

Now in order to get the molecular formula I need to do molar mass/empirical molar mass...

what am I missing though...sometimes cant the empirical formula be the same as the molecular formula?

Does this work:

8g of H x 1 mol compound/135.78g compound = .06 mol of H in compound

96g of C x 1 mol compound/135.78g compound = .71 mol of C in compound

32g of O x 1 mol compound/135.78g compound = .24 mol of O in compound

after this, I would need to multiply each of these numbers by a whole number to get it close to a whole number, but doesn't each one have to be multiplied by the same number?

sorry for taking up so much time...

7. ### dshnay 10+ Year Member

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That's weird, my emperical formula was C4H4O

5.9g H * 1 mol/1g = 5.9 mol
70.6g C * 1mol/12g = 5.9
23.5g O * 1/mol/16g = 1.5 mol

so 6/1.5 = 4
H4C4O

8. ### punkindrublic calls shenanigans 7+ Year Member

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9. ### CATallergy 2+ Year Member

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When you get to 8,96,32, your units are still grams. you have to divide by (1,12,16) to get to moles. then your answer is similar to the C4 H4 O that the others got. You might want to write out the units on this type of problem until you get the hang of it.

10. ### Davjc2009 2+ Year Member

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Now boys and girls let's get out the IR, Mass Spec, and Proton/C-13 NMR and assign structures! (Really wishes there was a noose "smiley")

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