# Empirical vs Molecular Formulas [closed]

For which compound are the empirical and molecular formulas the same?

A) $\ce{C6H5COOH}$

B) $\ce{C6H4(COOH)2}$

C) $\ce{HOOCCOOH}$

D) $\ce{CH3COOH}$

How can I proceed without knowing the molecular weights of the compounds?

-

## closed as off-topic by Loong, jerepierre, bon, TIPS, ronMar 27 '15 at 22:34

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

Do you know the difference between molecular and empirical formulae? Also, do you know how to use the periodic table? This is a good question, but an answer to it could be quite long, depending on your level of knowledge. – CHM Jul 2 '12 at 2:27
Empirical formulas have subscripts all relatively prime to each other, while molecular formulas don't have to be relatively prime. Molecular formulas can have pairwise co-prime terms, but if they are are all relatively prime to each other, it must be a empirical formula – Frank Jul 2 '12 at 2:32
Alright, got that cleared. The title of your question made me wonder if you knew the difference. Now, do you know how to use the periodic table? Do you know what the numbers present mean? – CHM Jul 2 '12 at 2:34
I do,but can we know the right choice without using the periodic table? – Frank Jul 2 '12 at 2:36
Could you please check out our homework policy and edit your question to fit it? Otherwise, I'm afraid it may get closed. You haven't shown your work, nor have you highlighted the specific issue that is impeding you. – ManishEarth Jul 2 '12 at 4:10

All you have to do is apply the rule that distinguishes molecular from empirical formulae to your set A,B,C,D. Empirical formulae have the subscripts in their lowest terms, i.e, their greatest common divisor is one.

A) $\ce{C6H5COOH}$ can be rewritten as $\ce{C7H6O2}$ which is an empirical formula.

B) $\ce{C6H4(COOH)2}$ can be rewritten as $\ce{C8H6O4}$ which is not an empirical formula. The empirical formula for this compound is $\ce{C4H3O2}$: this is the formula you'd get from an elemental analysis.

Best of luck with the other two!

-
$\ce{CH2O}$ is the empirical formula for the last one? – Frank Jul 2 '12 at 2:56

The best way to know it is to convert it is easy form like:
Molecular Formulas :-

1.$\ce{C6H5COOH}$ --> $\ce{C7H6O2}$

2.$\ce{C6H4(COOH)2}$ ---> $\ce{C8H6O4}$

3.$\ce{HOOCCOOH}$ ----> $\ce{C2H2O4}$

4.$\ce{CH3COOH}$ ---> $\ce{C2H4O2}$

So, the Empirical formulas will be:

1.$\ce{C7H6O2}$ #The H.C.F between 7,6,2 is 1

2.$\ce{C4H3O2}$ #The H.C.F between 8,6,4 is 2

3.$\ce{CHO2}$ #The H.C.F between 2,2,4 is 2

4.$\ce{CH2O}$ #The H.C.F between 2,4,2 is 2

-