Answer:
C₁₅N₃H₁₅
Step-by-step 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
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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