What you can do is organize them by color, what matter they are in room temperature, their molecular structure, or what kind of conductor in electricity and heat it is. I'm not sure what the format is supposed to look like but first just organize them all in different categories.
Since there is no weight, I would assume that this is a 100g of pure compound.
Okay so I would be changing the percentage to gram to solve for the mole.
So
40.0g C (1 mol C/12.01 g C) = 3.33 mol C
6.73g H (1 mol H/1.01 g H ) = 6.66 mol H
53.3g O (1 mol O/16.00 g O) = 3.33 mol O
With that, two of our moles is 3.33, so we consider that are our 1, as it is also the lowest. Therefore the empirical formula is CH2O
Na 2Co3
1*23=23. 2*12=24. 6*16=96
23+24+96=143
(23*100)/143. (24*100)/143. (96*100)/143
=16%. =16.7%. =67.1%
B. Precipitates are formed when two solutions combine to form another solution and a solid.
