Question

What is the empirical formula for a compound containing 26.57 g potassium, 35.36 g chromium, and 38.07 g oxygen?

Answer 1

`KCrO_3`

Explanation:

Find the number of moles for each element by dividing the mass present with the relative mass of the atom.

`K -> (26.57g)/(39.098gmol^-1) = 0.68molK`
`Cr -> (35.36g)/(51.996gmol^-1) = 0.68molCr`
`O -> (38.07g)/(15.999gmol^-1) = 2.34molO`

Now divide each number of moles by the lowest one, in this case `0.68`.

`(0.68molK)/(0.68) = 1molK`
`(0.68molCr)/(0.68) = 1molCr`
`(2.34molO)/(0.68) = 3.44molO`

Round to the nearest whole number

`K_1Cr_1O_3 = KCrO_3`

Answer 2

`"Empirical formula"` `-=` `K_2Cr_2O_7`

Explanation:

We got `100 *g` of stuff, and we calculate the number of atoms element by element:

`K` `=` `(26.57*g)/(39.1*g*mol^-1)` `=` `0.680*mol` `K`

`Cr` `=` `(35.36*g)/(52.00*g*mol^-1)` `=` `0.680*mol` `Cr`

`O` `=` `(38.07*g)/(16.0*g*mol^-1)` `=` `2.38*mol` `O`

Note that NORMALLY you would NEVER be given the percentage oxygen. Why not? Because there are very few ways to measure the proportion of this gas. At a 1st year undergrad level this question would have proposed an oxide of chromium that contained `26.57%` potassium, and `35.36%` chromium, and expected the student to twig that the missing percentage was due to the oxygen.

When we divide thru by the lowest number of moles (`0.68*mol`) we get an `"empirical formula"` of....

`K_((0.680*mol)/(0.680*mol))Cr_((0.680*mol)/(0.680*mol))O_((2.38*mol)/(0.680*mol))`, i.e. `KCrO_(3.5)`.

But by definition, the empirical formula is the simplest WHOLE number ratio defining constituent atoms in a species, so we must double this provisional formula to give:

`K_2Cr_2O_7`, `"i.e. potassium dichromate"`.

And this is a crystalline, orange powder, that is widely used in oxidations of organic materials....