A chromium compound 28.4% by mass of sodium 32.1% by mass of chromium, the rest being oxygen. Determine the empirical formula of the compound?...Can any one help me for that?

Answer 1

Take a 100 g sample. In that 100 g sample there will be
28.4 g Na
32.1 g Cr
100-(32.1+28.4) = 39.5 g oxygen.

Now convert grams to moles.
moles Na = 28.4/approx 23 = 1.23
moles Cr = 32.1/approx 52 = 0.617
moles O = 39.5/approx 16 = 2.47

You need to look these the atomic masses in the periodic table and use the exact number. I'm just using these from memory.

Now you want to convert these moles to a ratio of small whole numbers. The easiest way to do that is to divide each by the smallest of the group, namely 0.617.
Therefore, ratios
Na = 1.23/0.617 = 1.99
Cr = 0.617/0.617 = 1
O = 2.47/0.617 = 4.00

Now round each to a whole number which gives Na = 2, Cr = 1, O = 4 so the empirical formula is
Na2CrO4.

Answer 2

How did you get 23,52, 16?

Answer 3

moles = grams/atomic mass.
The atomic mass of Na is 22.9897 which I rounded to 23.
The atomic mass of Cr is 51.99 which I rounded to 52.
The atomic mass of O is 15.999 which I rounded to 16.
Those are the numbers I told you that I used from memory and you should use the exact value as found in the periodic table.