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A compound has a molar mass of 100 g/mol and the percent composition (by mass) of 65.45% C, 5.45% H, and 29.09% 0. Determine the empirical formula and the molecular formula.

A) CHO and C6H6O6

B) C3H3O and C6H6O2

C) C3HO and C6H2O2

D) CH2O and C4H8O4

E) CH4O and C3H12O3

Sagot :

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Answer:

B) C3H3O and C6H6O2

Explanation:

Given data:

Molar mass of compound = 100 g/mol

Percentage of hydrogen = 5.45%

Percentage of carbon = 65.45%

Percentage of oxygen = 29.09%

Empirical formula = ?

Molecular formula = ?

Solution:

Number of gram atoms of H = 5.45 / 1.01 = 5.4

Number of gram atoms of O = 29.09/ 16 = 1.8

Number of gram atoms of C = 65.45 / 12 = 5.5

Atomic ratio:

            C                      :      H            :         O

           5.5/1.8              :     5.4/1.8     :        1.8/1.8

            3                      :        3          :        1

C : H : O = 3 : 3 : 1

Empirical formula is C₃H₃O.

Molecular formula:

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

Empirical formula mass = 12×3 + 1.01 ×3 + 16×1 = 55.03  

n = 100 / 5503

n = 2

Molecular formula = n (empirical formula)

Molecular formula = 2 (C₃H₃O)

Molecular formula = C₆H₆O₂

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