A compound containing chromium, Cr; chlorine, Cl; and oxygen, O, is analyzed and found to be 33.6% chromium, 45.8% chlorine, and 20.6% oxygen by mass. What is the empirical formula of the compound? The molar mass of chromium, Cr, is 51.996 gmol; the molar mass of chlorine, Cl, is 35.45 gmol; and the molar mass of oxygen, O, is 15.999 gmol.

Answer

The empirical formula is CrO₂Cl₂

Explanation:

Empirical formula is the simplest whole number ratio of an atom present in a compound.

The compound contain, Chromium=33.6%

                                         Chlorine=45.8%

                                          Oxygen=20.6%

And the molar mass of Chromium(Cr)=51.996 g mol.

                 Chlorine containing molar mass (Cl)= 35.45    g mol.

                 Oxygen containing molar mass (O)=15.999  g mol.

Step-1

 Then,we will get,

Cr= mol

Cl= mol.

O= mol.

Step-2

Divide the mole value with the smallest number of mole, we will get,

Cr=

Cl=

O=

Then, the empirical formula of the compound is CrO₂Cl₂ (Chromyl chloride)

Explanation:

Calculation,

Given,       Cr = 33.6% and molar mass of Cr = 51.996%

                 Cl = 45.8% and molar mass of Cl = 35.45%

                 O = 20.6% and molar mass of O = 15.999%

Element   %    Molar mass    relative number   divide by    simples

                                                 of atoms               least            ratio

Cr       33.6     51.996                                    1

Cl       45.8     35.45                         2

O       20.6     15.999                     2

So,  Empirical formula =

your answer would be “dry”

thats how its supposed to be