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

`CrCl_(2)O_(2)`

Step-by-step 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
`(33.6)/(51.996) = 0.646`
`(0.646)/(0.646) = 1` 1

Cl 45.8 35.45
`(45.8)/(35.45) = 1.291`
`(1.291)/(0.646) = 1.998 =2` 2

O 20.6 15.999
`(20.6)/(15.999)= 1.287`
`(1.287)/(0.646) =1.992=2` 2

So, Empirical formula =
`CrCl_(2)O_(2)`

Answer 2

The empirical formula is CrO₂Cl₂

Step-by-step 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=
`(1)/(51.996) *33.6=0.64` mol

Cl=
`(1)/(35.45) *45.8=1.29` mol.

O=
`(1)/(15.99) *=1.28` mol.

Step-2

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

Cr=
`(0.64)/(0.64) =1`

Cl=
`(1.29)/(0.64) =2`

O=
`(1.28)/(0.64) =2`

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