Question

An oxide of iron has 69.9% iron and 30.1% oxygen by mass. What would be its empirical formula

Answer 1

Answer:

Fe2O3

Explanation:

Step 1: divide the composition of both elements by their RFM

Iron= 69.9/56.                         Oxygen= 30.1/16      

      = 1.25                                              =  1.88

Step 2: divide the answers from step 1 by the smallest answer.

Iron= 1.25/1.25.                          Oxygen= 1.88/1.25

      = 1                                                     = 1.50

Step 3: since the formula cannot be decimals, the ratio needs to be multiplied in order to get whole numbers

Iron= 1 X 2.                                  Oxygen= 1.50 X 2

      = 2.                                                     = 3

Therefore , since ratio is 2:3, the empirical formula is Fe2O3