A compound having an approximate molar mass of 165 - 170 g has the following... A compound having an approximate molar mass of 165 - 170 g has the following percentage composition by mass: carbon, 42.87%; hydrogen, 3.598%; oxygen,
28.55%; nitrogen, 25.00%. Determine the empirical and molecular formulas of
the compound.

1. The empirical formula of the compound is C₂H₂ON

2. The molecular formula of the compound is C₆H₆O₃N₃

From the question given above, the following data were obtained:

Carbon (C) = 42.87%

Hydrogen (H) = 3.598%

Oxygen (O) = 28.55%

Nitrogen (N) = 25.00%

C = 42.87%

H = 3.598%

O = 28.55%

N = 25.00%

Divide by their molar mass

C = 42.87 / 12 = 3.5725

H = 3.598 / 1 = 3.598

O = 28.55 / 16 = 1.7844

N = 25.00 / 14 = 1.7857

Divide by the smallest

C = 3.5725 / 1.7844 = 2

H = 3.598 / 1.7844 = 2

O = 1.7844 / 1.7844 = 1

N = 1.7857 / 1.7844 = 1

Therefore, the empirical formula of the compound is C₂H₂ON

Molar mass of compound = (165 + 170)/2

Molar mass of compound = 335/2

Molar mass of compound = 167.5 g/mol

Empirical formula = C₂H₂ON

Molecular formula = Empirical formula × n = molar mass of compound

[C₂H₂ON]n = 167.5

[(12×2) + (2×1) + 16 + 14]n = 167.5

[24 + 2 + 16 + 14]n = 167.5

56n = 167.5

Divide both side by 56

n = 167.5 / 56

n = 3

Molecular formula = [C₂H₂ON]n

Molecular formula = [C₂H₂ON]₃

Therefore, the Molecular formula of the compound is C₆H₆O₃N₃

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Find ratio of atoms for empirical formula, divide mass by atomic mass

C=42.87/12 = 3.57

H= 3.598/1 = 3.598

O= 28.55/16 = 1.784

N= 25.00/14 =1.1785

to see ratio, divide by smallest no

Gives C2H2ON for empirical formula

Empirical mass = 24 +2 +16 + 14 = 56

165 to 170 /56 = 3

so molecular formula = empirical formula x 3= C6H6O3N3