1. An oxide of chromium is found to have the following % composition: 68.4% Cr and 31.6% 0. Determine this compound's empirical formula.

Empirical formula is Cr₂O₃.

Explanation:

Given data:

Percentage of Cr = 68.4%

Percentage of O = 31.6%

Empirical formula = ?

Solution:

Number of gram atoms of Cr = 68.4 / 52 = 1.3 2

Number of gram atoms of O = 31.6 / 16 = 1.98

Atomic ratio:

                            Cr               :         O

                           1.32/1.32     :       1.98/1.32

                               1              :        1.5

Cr : O = 1 :  1.5

Cr : O = 2(1 : 1.5)

Empirical formula is Cr₂O₃.

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.