A major textile dye manufacturer developed a new yellow dye. The dye has a percent composition of 75.95% C, 17.72% N, and 6.33% H by mass with a molar mass of about 240 g/mol. Determine the molecular formula of the dye.

C₁₅N₃H₁₅

Explanation:

Follow the sequence ...

%/100wt => grams/100wt => moles => mole ratio => reduce mole ratio (divide by smaller mole value) => empirical ratio.

Note: If the above calculation comes out with a fraction (for most textbook problems) do the following:

--.25 => multiply empirical ratio by 4

--.50 => multiply empirical ratio by 2

--.75 => multiply empirical ratio by 4

Or, if initial data is given in grams, convert to %/100wt. (example at end of problem)

This Problem:  

         %/100wt.   grams         moles =mass/formula wt.

%C = 75.95% => 75.95g => 75.95g/12.0107g·mol⁻¹ = 6.325 mole OS

%N = 17.72% =>  17.72g => 17.72g/14.0067g·mol⁻¹ = 1.2651 mole N  

%H =  6.33% =>   6.33g =>  6.33g/1.00794g·mol⁻¹ = 6.28 mol H              

(∑  =   100%    | assume 100g cpd  => ( grams => 100g compound would contain 75.95g C, 17.72g N & 6.33g H )

=> mole ratio =>  C : N : H => 6.3235 : 1.2651 : 6.28  

=> reduce mole ratio ( divide by smaller mole value ) ...

=> 6.3235/1.2651 : 1.2651/1.2651 : 6.28/1.2651 =>  empirical ratio is => 5:1:5

empirical formula => C₅NH₅

empirical formula wt. = 5C + 1N + 5H = 5(12) + 1(14) + 5(1) = 79

=> molecular ratio (n) => molecular wt (given) = N · empirical formula wt

=> N =  molecular wt / empirical wt = 240/79 = 3

∴ Molecular Formula is =>  C₅NH₅ X 3 => C₁₅N₃H₁₅

EXAMPLE: Empirical Formula Problem with initial data in grams:

Assume an oxide of Osmium (compd wt) = 2.89 grams contains 2.16 gram Osmium, what is the empirical formula?

Use given mass values to calculate %/100wt then use the sequence

% => grams => moles => reduce => emp. ratio => emp. formula

%Os = 2.16g/2.89g x 100% = 74.74%

%Oxy = [(2.89g -2.16g)/2.89] x 100% = 25.26% or use 100% - 74.74% = 25.26%

%  => g => moles => reduce => emp formula

%Os   = 74.74% => 74.74g => 74.74g/190.23g·mol⁻¹ = 0.3929 mol Os

%Oxy = 25.26% => 25.26g => 25.26g/16g·mol⁻¹ = 1.5788 mole Oxy

Reduce mole ratio (divide by smaller mole value) => empirical ratioo

=> Os:Oxy => 0.3929/0.3929 : 1.5788/0.3929 = 1 : 4

=> Empirical Formula => OsO₄