Question

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 1

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=$\frac{1}{51.996} \times33.6=0.64$ mol

Cl= $\frac{1}{35.45} \times45.8=1.29$ mol.

O=$\frac{1}{15.99} \times=1.28$ mol.

Step-2

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

Cr= $\frac{0.64}{0.64} =1$

Cl= $\frac{1.29}{0.64} =2$

O= $\frac{1.28}{0.64} =2$

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

Answer 2

Answer:

$CrCl_{2}O_{2}$

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         $\frac{33.6}{51.996} = 0.646$           $\frac{0.646}{0.646} = 1$                 1

Cl       45.8     35.45         $\frac{45.8}{35.45} = 1.291$         $\frac{1.291}{0.646} = 1.998 =2$        2

O       20.6     15.999       $\frac{20.6}{15.999}= 1.287$         $\frac{1.287}{0.646} =1.992=2$       2

So,  Empirical formula = $CrCl_{2}O_{2}$