Dr. Collins PCAT practice problem help

[soonerfan989]

A sample of a compound contains 0.90 g of element A and 0.60 g of element B. If the atomic weight of A is 60 g/ mole and that of B is 40 g/mole, then what is the empirical formula for this compound?
I can't seem to figure this question out, can someone please help me out? Thanks

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[Skrumpy]

A sample of a compound contains 0.90 g of element A and 0.60 g of element B. If the atomic weight of A is 60 g/ mole and that of B is 40 g/mole, then what is the empirical formula for this compound?
I can't seem to figure this question out, can someone please help me out? Thanks

I believe it should just be: AB

The question tests your knowledge of what an empirical formula is and how to find it. Basically it is just the simplest (reduced form of the molecular formula), therefore for example cyclononane would be C9H18 while the empirical formula for it would be CH2 (showing that the ratio of C to H is 1:2).

Basically just convert to moles:

0.90g x 1mol/60g = 0.015 moles of element A in the compound
0.60g x 1mol/40g = 0.015 moles of element B in the compound

Then divide both moles by the smallest # of moles (0.015) to get the ratio. Since they are both 0.015 moles 0.015/0.015 obviously = 1 so the ratio is 1:1 and the empirical formula should be AB. If element A had 0.030 moles then 0.030/0.015 = 2 and the ratio would then be 2:1 or A2B, and so on. Someone can of course correct me if I'm off.. been awhile since I've done any problems like this .

 

[Sugoi Travis]

I believe it should just be: AB

The question tests your knowledge of what an empirical formula is and how to find it. Basically it is just the simplest (reduced form of the molecular formula), therefore for example cyclononane would be C9H18 while the empirical formula for it would be CH2 (showing that the ratio of C to H is 1:2).

Basically just convert to moles:

0.90g x 1mol/60g = 0.015 moles of element A in the compound
0.60g x 1mol/40g = 0.015 moles of element B in the compound

Then divide both moles by the smallest # of moles (0.015) to get the ratio. Since they are both 0.015 moles 0.015/0.015 obviously = 1 so the ratio is 1:1 and the empirical formula should be AB. If element A had 0.030 moles then 0.030/0.015 = 2 and the ratio would then be 2:1 or A2B, and so on. Someone can of course correct me if I'm off.. been awhile since I've done any problems like this .

 

[soonerfan989]

I believe it should just be: AB

The question tests your knowledge of what an empirical formula is and how to find it. Basically it is just the simplest (reduced form of the molecular formula), therefore for example cyclononane would be C9H18 while the empirical formula for it would be CH2 (showing that the ratio of C to H is 1:2).

Basically just convert to moles:

0.90g x 1mol/60g = 0.015 moles of element A in the compound
0.60g x 1mol/40g = 0.015 moles of element B in the compound

Then divide both moles by the smallest # of moles (0.015) to get the ratio. Since they are both 0.015 moles 0.015/0.015 obviously = 1 so the ratio is 1:1 and the empirical formula should be AB. If element A had 0.030 moles then 0.030/0.015 = 2 and the ratio would then be 2:1 or A2B, and so on. Someone can of course correct me if I'm off.. been awhile since I've done any problems like this .

It's been awhile for me too, thanks for the help.