If the molecular weight is 146 amu. what is the molecular formula?

a certain organic acid is used in the manufacturing of nylon. the composition of the acid 49.3% C, 6.9% H, and 43.8% O (dry mass), find the empirical formula of this compund.

Answer:
49.3 g C (1 mol C/12.0 g C) = 4.11 mol C

6.9 g H (1 mol H/1.0 g H) = 6.9 mol H

43.9 g O (1 mol O/16.0 g O) = 2.74 mol O

C4.11/2.74 H6.9/2.74 O2.74/2.74 ==> C1.5H2.5O

Multiply by 2 to get rid of the 0.5 ==> C3H5O2 (empirical formula)

Mass of empirical formula = (3 X 12) + (5 X 1) + (2 X 16) = 73 amu.

146 amu/73 amu = 2, so the molecular formula is C6H10O4

HOOC-CH2CH2CH2CH2-COOH (1, 6 -hexanedioic acid, common name, adipic acid)
C6O4H10

Most likely, it would be C6H6(OH)4
Empirical formula method-

Convert % to grams by assuming a 100 gram sample.

C=49.3 g, H=6.9 g, O=43.8g

Divide each element weight by its atomic mass-

C: 49.3/12.01=4.10
H: 6.9/1.008= 6.84
O: 43.8/16=2.73

divide each number by the smallest number (2.73)
C: 4.10/2.73=1.5
H: 6.84/2.73=2.5
O: 2.73/2.73=1.0

Since we want whole numbers, multiply each number by 2

C: 1.5x2=3
H: 2.5x2=5
O: 1.0x2=2

The empirical formula is C3H5O2 which would have a MW of 73. Since the molecular weight is 146 the molecular formula must be twice the empirical formula-

C6H10O4

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