Answer:
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
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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