A compound is 54.53 % C , 9.15 % H , and 36.32 % O by mass.

Question

A compound is 54.53 % C , 9.15 % H , and 36.32 % O by mass. What is its empirical formula? Insert subscripts as needed.

The molecular mass of the compound is 132 amu. What is its molecular formula? Insert subscripts as needed.

Answer

Divide the percent mass by the atomic mall of each element to have 4.54 moles C to 9.08 moles H to 2.27 moles O. Divide these by the smallest to have the ratoi of 2 C to 4 H to 1 O.

Thus the empirical formula is C2H4O. Multiply this by the mass of the elements with each subscript for a total of 44.02. Divide 132 by 44.02 to have 3 times the empirical formula for the moleular formula or C6H12O3.

Answered by: MS in Chemistry and over 25 years industrial experience.

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