A compound is found to contain 9.224 % boron and 90.74 % chlorine by mass. To answer the question, enter the elements in the order presented above. QUESTION 1: The empirical formula for this compound is . QUESTION 2: The molar mass for this compound is 117.2 g/mol. The molecular formula for this compound is .

Empirical formula of a compound means that it provides simplest ratio of whole number.

Explanation:

QUESTION-1

Given,

Mass of boron and chlorine is 9.224% and 90.74%

Then, Boron- 9.224×=0.85 mol   (Mass value of boron=10.811)

      Chlorine- 90.74×=2.559 mol.(Mass value of chlorine=35.453)

By dividing the mole value with smallest number, then we will get-

Boron-=1

Chlorine-=3

The empirical formula of the compound is -BCl₃

QUESTION-2

BCl₃⇄B+ 3.Cl  (Putting the actual mass number of boron and chlorine we will get-)

           1×(10.811)+ 3×(35.453)=117.17g/mol

Molecular formula=

                               = ≈1

  The molecular formula is same with the empirical formula.

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.