What is the empirical formula of a compound composed of 72.36% Fe and 27.64% O by mass? empirical formula:

Empirical formula = Fe₃O₄

Explanation:

Given data:

Percentage of Fe = 72.36%

Percentage of oxygen = 27.64%

Empirical formula = ?

Solution:

Number of gram atoms of O = 27.64 / 16 = 1.7

Number of gram atoms of Fe = 72.36 / 55.8 = 1.3

Atomic ratio:

                          Fe                    :         O

                           1.3/1.3              :     1.7/1.3

                              1                    :       1.3            

Fe : O = 1 : 1.3

Empirical formula = 3( 1: 1.3)

Empirical formula = Fe₃O₄

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

58750.08068 is the answer i think