Why is the empirical formula of A pheromone- C(19)H(38)O simply C(19)H(38)O and why not CH2O? An explanation would be greatly appreciated.. Created by eaglesfan1717. science-mathematics-en - chemistry-en

Hi there!The empirical formula does involve cancelling the ratio of elements down to the lowest possible, but if we look at the formula C(19)H(38)O, we can see that there is only one O present, and so the ratio of these elements cannot get any lower (you can't have 1/2 of an oxygen). If the formula were C(19)H(38)O(19), then you would be left with the empirical formula CH(2)O.Hope this helps!