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Chemistry 19 Online
OpenStudy (anonymous):

In which pair would both compounds have the same empirical formula? A. H2O and H2O2 B. BaSO4 and BaSO3 C. FeO and Fe2O3 D. C6H12O6 and HC2H3O2

OpenStudy (anonymous):

basically i'm confused as to how to figure it out. i think the answer is A but not sure how to get there.

thomaster (thomaster):

The empirical formula is the simplest formula for a compound. To figure it out you would need to divide the numbers in the molecules by 2 until it can no longer. For example: C2H2 and C6H6 you can both divide them by 2 untill you get CH as empirical formula for both Next example C2H6 and C4H10 You can divide C2H6 by 2 untill you have CH3 For C4H10 it would stop at C2H5 So in your case: A. H2O and H2O2 = H2O and HO B. BaSO4 and BaSO3 = BaSO4 and BaSO3 (you can't divide Barium by 2) C. FeO and Fe2O3 = FeO and Fe2O3 D. C6H12O6 and HC2H3O2 =CH2 and CH2 (HC2H3O2 = same as C2H4O2) Hope that helps

thomaster (thomaster):

Sorry forgot oxygen at D should be: D. C6H12O6 and HC2H3O2 =CH2O and CH2O (HC2H3O2 = same as C2H4O2)

OpenStudy (anonymous):

so the answer is D? i understand dividing by 2 until you get the lowest possible number but none seem to work out that way

OpenStudy (anonymous):

or is it A?

OpenStudy (anonymous):

????

thomaster (thomaster):

Its D ofcourse

thomaster (thomaster):

In A, the pairs do not have the same empirical formula

OpenStudy (anonymous):

ok..i'm still just a little confused..in D i divided by 2 so for

OpenStudy (anonymous):

c6h12o6 i got 333 dividing byy 2...and hc2h3o2 i got 121

OpenStudy (anonymous):

so how are they the same?

thomaster (thomaster):

Eehh I didn't explain the division part very well I see. its not only 2 You can divide C6H12P6 by 6 so you get CH2O The other one is C2H4O2, you can divide that one bio 2 so you get CH2O The point is that you divide untill the smallest number in the formula can't be divided anymore. the number you choose must be so that you van divide all numbers in the formulas with it. So for 6 12 and 15 you can use 3

OpenStudy (anonymous):

ooohhhh ok...ok that makes much more sense to me, thank you very much :-)

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