Question

A compound contains carbon, hydrogen, and nitrogen as its only elements. a 9.353 g sample of the compound contains 5.217 g of carbon and 1.095 g of hydrogen; the remainder is nitrogen. what is the empirical formula for the compound?

Answer 1

9,353 - 1,095 - 5,217 = mN
3,041 = mN

H:C:N= 1,095/1 : 5,217/12 : 3,041/14
H:C:N = 1,095 : 0,43475 : 0,2172
H:C:N = 5,037 : 1,99985 : 0,99912 ≈ 5 : 2 : 1

H₅C₂N (C₂H₅N)

Answer 2

Answer:

Empirical formular is given as ; C₂H₅N

Explanation:

Information given;

Mass of sample = 9.353g

Mass of Carbon = 5.217g

Mass of Hydrogen = 1.095g

Remainder is Nitrogen.

Mass of Nitrogen = Mass of Sample - (Mass of carbon + Hydrogen)

Mass of Nitrogen = 9.353 - (5.217 + 1.095) = 3.041g

Step 1: Determine the masses. We just calculated the value for this.

Step 2: Determine the number of moles by dividing the grams by the atomic mass.

Carbon = 5.217/12 = 0.4348

Hydrogen = 1.095/1 = 1.095

Nitrogen = 3.041/14 = 0.2172

Step 3: Divide the number of moles of each element by the smallest number of moles.

The smallest in this case is 0.2172

Carbon = 0.4348/ 0.2172 = 2.044

Hydrogen = 1.095/ 0.2172 = 5.14

Nitrogen = 0.2172/ 0.2172 = 1

Step 4: Convert numbers to whole numbers

Carbon = 2.044 = 2

Hydrogen = 5.14 = 5

Nitrogen = 1 = 1

Empirical formular is given as ; C₂H₅N