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.

V = 10 m³

m = 70 grams

density = ?

density = m/v

= 70/10

= 7 g/cm³ (B)