Question

A sample of phosphorus of mass 3.654 g reacts with dichlorine to form 16.20 g of a molecular compound. What is the empirical formula of this compound?

Answer 1

Compound of P and Cl = mass 16.20 g
Started with 3.654 g P = -3.654
Amount chlorine added = 16.20-3.654 = 12.546.

moles P = 3.546/molar mass P
moles Cl = 12.546/atomic mass Cl

Then find the molar ratios of the two elements with the smallest being 1.00. The easy way to do that is to divide the smallest number by itself thereby producing 1.00 for that element. Then divide the other number by the same small number. The numbers you get will be the x and y of PxCly.
Post your work if you get stuck.

Answer 2

I don't understand when you say divide by the smallest numbers. For instnace, am I supposed to divide 3.546 (P) by 3.546 again?

Answer 3

No. Did you convert g P to moles. Then convert g chlorine to moles. You want to find the molar ratios of THOSE two numbers (the moles).

Answer 4

P:
3.654 g P x (1 mol/30.97 g P) = .12 mol

Cl:
12.546 g Cl x (1 mol/35.45 g Cl) = .35 mol


Therefore, the empirical formula will be PCl3? Is this correct?

Answer 5

Yes, but you rounded one of your numbers before you should have rounded.
3.546/30.97 = 0.1180 (don't round to 0.12--to two significant figures when you are allowed 4.
12.546/35.45 = 0.3539.

Your ratios are 0.12/0.12 = 1.00
0.35/0.12 = 2.92 so you rounded to P1Cl3 (PCl3) which is correct. Using my numbers, you would have obtained,
0.1180/0.1180 = 1.000
0.3539/0.1180 = 2.999
The final numbers are closer to whole numbers if you round correctly. In some cases it would make a difference in the formula obtained; in this case it did not.