Question

What is the empirical formula of a compound containing 24.56% potassium, 34.81% manganese, and 40.50% oxygen?


KMnO
KMnO2
K0.63Mn0.63O2.53
KMnO4
K3MnO2

Answer 1

Assume a 100g sample to make things easier; so 24.56% of 100g is 24.56g, for example. The empirical formula is simply the ratio of the moles of each substance. So convert each to moles so that you can compare them:

24.56g K *(1 mol/39.10g)= .6281 mol K
34.81g Mn *(1mol/54.94)= .6336 mol Mn
40.50g O *(1mol/16.00g)= 2.531 mol O

Remember that we are trying to find the ratio between them. The easiest way to do that is to divide each by the smallest number:

.6281 mol K /.6281 mol = 1
.6336 mol Mn /.6281mol = 1.0
2.531 mol O /.6281 mol =~ 4.0

So you have a ratio of 1:1:4 (there is some degree of approximation that you have to make)
So the empirical formula is:
KMnO4

(the 4 is a subscript)

Answer 2

%'s are % by weight,
so use them as grams,

& use their molar masses,
i norder to find moles:
(24.7 grams of K) / (39.1 grams per mol K) = 0.63 mol K
(34.8grams of Mn) / (54.9 grams per mol Mn) = 0.63 mol Mn
(40.5 grams of O) / (16 grams per mol O) = 2.53 mol O


divide by the smallest number of moles,
in order to ratio the moles:
(0.63 mol K) / (0.63) = 1 mol K
(0.63 mol Mn) / (0.63) = 1 mol Mn
(2.53 mol O) / (0.63) = 4 mol O


your answer is
KMnO4