What is the empirical formula of a compound that contains 29% Na, 41% S, and 30% O by mass? ..My answer is NaSO and I am not sure... can anyone help me for it?????????

Answer 1

I'm relatively certain that it isn't NaSO.
Take 100 g sample which will give you
29 g Na
41 g S
30 g O

Now convert grams to moles. Remember moles = gram/molar mass.
moles Na = 29/23 =
moles S = 41/32 =
moles O = 30/16 =
You may want to look up the atomic masses of the elements Na, S, and O. I'm just using from my memory.
Now convert moles to the ratio of small whole numbers. The easy way to do that is to divide the smallest number by itself. That gives you 1.00 for it. Then divide all of the other numbers by the same small number. Round to whole numbers and that should be the subscripts that go with the formula. A word of caution: Don't round 0.5 or 0.3 or 0.25 to a whole number since the values usually are more accurate than that. That is, if you get 1.00 to 1.25, then the ratio would be 4:5 and not 1:1. Post your work if you get stuck.