g A laboratory analysis of an unknown compound found the following composition: C 75.68% ; H 8.80% ; O 15.52%. What is the empirical formula of the compound?

THE EMPIRICAL FORMULA FOR THE UNKNOWN COMPOUND IS C7H9O

Explanation:

The empirical formula for the unknown compound can be obtained by following the processes below:

1 . Write out the percentage composition of the individual elements in the compound

C = 75.68 %

H = 8.80 %

O = 15.52 %

2. Divide the percentage composition by the atomic masses of the elements

C = 75 .68 / 12 = 6.3066

H = 8.80 / 1 = 8.8000

O = 15.52 / 16 = 0.9700

3. Divide the individual results by the lowest values

C = 6.3066 / 0.9700 = 6.5016

H = 8.8000 / 0.9700 = 9.0722

O = 0.9700 / 0.9700 = 1

4. Round up the values to the whole number

C = 7

H = 9

O = 1

5 Write out the empirical formula for the compound

C7H90

In conclusion, the empirical formula for the unknown compound is therefore C7H9O

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.