Question

Phosphorus combines with oxygen to form two oxides. Find the empirical formula for each oxide of phosphorus if the percentage composition is 43.6% oxygen.

Answer 1

I keep getting the wrong answer: PO instead of the correct answer: P2O3.

Answer 2

I tried oxygen as O2 and I still get the wrong answer...

Answer 3

We know from the question that there are two elements in the compound: P and O, in an unknown ratio.

If 43.6% of the compound (by weight) is oxygen, then 56.4% is phosphorous.

The molecular weight of O is 16, of P is 31.

We can convert the percentages (which are "by weight") into ratio values which correspond to the amount (in terms of atoms, as a ratio - remember that different elements' atoms weigh different amounts)

O: 43.6/16 = 2.73
P: 56.4/31 = 1.82

So we have a O:P ratio of 2.73:1.82

This suggests the formula \[P_2O_3\]

Though the question you wrote asks for the identities of two oxides of phosphorous. Is more data given?

Answer 4

It had another question b) 56.6% but I wanted to know how to solve the first one. And don't I divide by 1.82 to obtain the lowest ratio?

Answer 5

Yep - that'd give you O:P 1.5:1
Which is equal to O:P 3:2 - which corresponds to the same formula. I'm perhaps just a bit lazy for rounding :)

Answer 6

how does O:P equal to 3:2 when it's 1.5:1? O.O

Answer 7

Well - since it's a ratio, as long as you do the same operation to both sides, it remains "the same".

So from O:P 1.5:1 we multiply both sides by 2 to get:
O:P 3:2.

In words - "for every one P we have 1.5 O" is the same as saying "for every 2 P we have 3 O".
Hope this helps

Answer 8

ohhhh.. okay last question: why did you multiply both side by 2? Why not 3, 4, 5? Is it to get the lowest ratio?

Answer 9

exactly! We were looking for the lowest integer ratio.

Cheers