Question

A brown dye has a percent composition of 62.41% C, 5.24% H, and 32.36% N by mass with a molar mass of 346.40 g/mol. Determine the molecular formula of the dye

Answer 1

The molecular formula : C₁₈H₁₈N₈

Further explanation

Given

62.41% C, 5.24% H, and 32.36% N

Required

The molecular formula

Solution

mol ratio

C : 62.41/12.0096 = 5.1967

H : 5.24/1.00784 = 5.1992

N : 32.36/14.0067 = 2.310

Divide by 2.310(smallest)

C : 5.1967/2.31=2.25

H : 5.1992/2.31 = 2.25

N : 2.31/2.31 = 1

Multiplied by 4

C : H : N = 9 : 9 : 4

The empirical formula : C₉H₉N₄

(C₉H₉N₄)n=346.40 g/mol

(12.0096 x 9 + 1.00784 x 9 + 14.0067 x 4)n=346.4

(108.0864+9.07056+56.0268)n=346.4

(173.184)n=346.4

n=2

The molecular formula : C₁₈H₁₈N₈