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

2 Answers
Apr 4, 2016

Answer:

#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#

Apr 4, 2016

Answer:

#"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....