The situation is as follows:

A sample of 19.40 g of a compound of K, Cr and O receives a treatment and this results into 7.6 g. of Cr_2O_3 and 9.70 g. of the compound is transformed into 13.85 g of KClO_4.

Given this situation, what would be the most adequate way to to find the empirical formula of such compound?.

I've attempted to solve this riddle using the usual procedure for a combustion analysis but this doesn't seem to be the case, as no information was given to assume such situation.

First I did found the grams of Chromium in the given 7.6 g.

Assuming FW for Cr_2O_3 = 2(52)+3(16)=152 g/mol

$7.6 g Cr_2O_3 x 2x 52 g Cr / 152 g Cr_2O_3=5.2 g Cr

FW for KClO_4 = 39+35.5+4(16)= 138.5 g/mol

Then:

13.85 g KClO_4 x 39 g K / 138.5 g KClO_4=3.9 g K

Then the rest of the mass should be the oxygen, subtracted from the sample:

$19.40-(3.9+5.2)=10.3$

Therefor the moles for each would be as follows:

5.2 g Cr x 1 mol Cr / 52 g Cr =0.1 mol Cr

3.9 g K x 1 mol K/ 39 g K=0.1 mol K

10.3 g O x 1 mol O/ 16 g O=0.64375 mol O

Dividing between the smallest yields:

0.1 mol Cr / 0.1 =1

0.1 mol K / 0.1 =1

0.64375 mol K / 0.1 =6.4375

Which would give:

K_1Cr_1O_6.4375

However this isn't very convincing to me. However since the chromium and potassium are in the same proportion it looks very suspiciously to be potassium dichromate. But does it exist a justification for this. Can somebody help me to clear out the basis for the right calculation?.

Btw this forum doesnt support latex so I had to resort to the limitations of html.

Your calculations are right and moles of K ,Cr and O calculated are correct.May be there some error in information provided in the question.

No it was two experiments. 19.4 g result to 5,2 g Cr and  9,7 g result to 3,9 g mol K

19,4 g - 5,2 g = 14,2 g KxOy
9,7 g -3,9 g = 5,8 g CrzOy |* 2

19,4 g -7,8 g = 11,6 g CrzOy

It means the compound should have 2 K and  Cr

The Differenz of 14,2 g - 7,8 g = 6,4 g for Oy
And 11,6 g- 5,2 g  = 6,4 g Oy

What means it 4 x for oxygen = 0,4 mol * 16 g/mol

The compound is K2CrO4 potassium chromate.