Question

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 .

Answer 1

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

Step-by-step explanation:

QUESTION-1

Given,

Mass of boron and chlorine is 9.224% and 90.74%

Then, Boron- 9.224×
`(1)/(10.811)`=0.85 mol (Mass value of boron=10.811)

Chlorine- 90.74×
`(1)/(35.453)`=2.559 mol.(Mass value of chlorine=35.453)

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

Boron-
`(0.85)/(0.68)`=1

Chlorine-
`(2.559)/(0.68)`=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=
`(molar mass)/(empirical formula)`

=
`(117.17)/(117)`≈1

The molecular formula is same with the empirical formula.