38.7% C, 9.7% H, and 51.6% O. Its molecular mass is 62.0 g/mole. A) Determine the empirical formula of the compound.?

B)

(62.0 g/mol) x (38.7 / 100) / (12.01078 g C/mol) = 1.998

(62.0 g/mol) x (9.7 / 100) / (1.007947 g H/mol) = 5.967

(62.0 g/mol) x (51.6 / 100) / (15.99943 g O/mol) = 1.9996

Round to the nearest whole numbers to find the molecular formula:

C2H6O2

[You didn't ask, but the engine coolant with a molecular formula like that is

ethylene glycol (1,2-ethanediol).]

A)

Since the molecular formula has coefficients with the common factor of 2, divide the molecular formula by 2 to find the empirical formula:

CH3O

A)

Relative atomic masses: C = 12.0, H = 1.0, O = 16.0

In the compound, mole ratio C : H : O

= (38.7/12.0) : (9.7/1.0) : (51.6/16.0)

= 3.23 : 9.7 : 3.23

= (3.23/3.23) : (9.7/3.23) : (3.23/3.23) …… Divided by the smallest number

= 1 : 3 : 1

Hence, empirical formula = CH₃O

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B)

Let molecular formula = (CH₃O)ₙ

Molecular mass:

(12.0 + 1.0×3 + 16.0) × n = 62.0

31.0n = 62.0

Hence, molecular formula = C₂H₆O₂