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?

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.

How did you get 23,52, 16?

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.

To determine the empirical formula of the compound, we need to find the ratio of the elements present in the compound.

Let's assume we have 100 grams of the compound. From the given information, we know that it contains 28.4 grams of sodium, 32.1 grams of chromium, and the rest is oxygen.

To find the amount of oxygen present, we subtract the mass of sodium and chromium from the total mass:

Mass of oxygen = Total mass - Mass of sodium - Mass of chromium
Mass of oxygen = 100 g - 28.4 g - 32.1 g
Mass of oxygen = 39.5 g

Now, we need to convert these masses into moles. To do this, we use the molar mass of each element:

Molar mass of sodium (Na) = 22.99 g/mol
Molar mass of chromium (Cr) = 52.00 g/mol
Molar mass of oxygen (O) = 16.00 g/mol

Moles of sodium = Mass of sodium / Molar mass of sodium
Moles of sodium = 28.4 g / 22.99 g/mol
Moles of sodium ≈ 1.234 mol

Moles of chromium = Mass of chromium / Molar mass of chromium
Moles of chromium = 32.1 g / 52.00 g/mol
Moles of chromium ≈ 0.617 mol

Moles of oxygen = Mass of oxygen / Molar mass of oxygen
Moles of oxygen = 39.5 g / 16.00 g/mol
Moles of oxygen ≈ 2.469 mol

Now we divide the number of moles of each element by the smallest number of moles to find the simplest whole number mole ratios:

Sodium: 1.234 mol / 0.617 mol = 2
Chromium: 0.617 mol / 0.617 mol = 1
Oxygen: 2.469 mol / 0.617 mol = 4

The empirical formula represents the ratio of the elements in the compound in its simplest form. Therefore, the empirical formula of the compound is Na2CrO4.