Respuesta :
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=[tex]\frac{1}{51.996} \times33.6=0.64[/tex] mol
Cl= [tex]\frac{1}{35.45} \times45.8=1.29[/tex] mol.
O=[tex]\frac{1}{15.99} \times=1.28[/tex] mol.
Step-2
Divide the mole value with the smallest number of mole, we will get,
Cr= [tex]\frac{0.64}{0.64} =1[/tex]
Cl= [tex]\frac{1.29}{0.64} =2[/tex]
O= [tex]\frac{1.28}{0.64} =2[/tex]
Then, the empirical formula of the compound is CrO₂Cl₂ (Chromyl chloride)
Answer:
[tex]CrCl_{2}O_{2}[/tex]
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 [tex]\frac{33.6}{51.996} = 0.646[/tex] [tex]\frac{0.646}{0.646} = 1[/tex] 1
Cl 45.8 35.45 [tex]\frac{45.8}{35.45} = 1.291[/tex] [tex]\frac{1.291}{0.646} = 1.998 =2[/tex] 2
O 20.6 15.999 [tex]\frac{20.6}{15.999}= 1.287[/tex] [tex]\frac{1.287}{0.646} =1.992=2[/tex] 2
So, Empirical formula = [tex]CrCl_{2}O_{2}[/tex]
