Question
what is the empirical formular of the following compound: 0.142 g phosphorus with 0.436 g of fluorine.
Answers
DrBob222
mols P = 0.142/31 = 0.0046 approx
mols F = 0.436/19 = 0.023
Now find the radio of these to each other with the smallest being no less than 1.00. The easy way to do that is divide the smallest number by itself then divide the other number by the same small numer.
mols F = 0.436/19 = 0.023
Now find the radio of these to each other with the smallest being no less than 1.00. The easy way to do that is divide the smallest number by itself then divide the other number by the same small numer.