A compound is 40.0 % C , 6.70 % H , and 53.3 % O by mass. Assume that we have a

Question

A compound is 40.0% C, 6.70% H, and 53.3% O by mass. Assume that we have a 100.-g sample of this compound.

What are the subscripts in the empirical formula of this compound? Enter the subscripts for C, H, and O, respectively, separated by commas (e.g., 5,6,7).
A compound is 40.0% C, 6.70% H, and 53.3% O by mass. Assume that we have a 100.-g sample of this compound. A compound is 40.0% C, 6.70% H, and 53.3% O by mass. Assume that we have a 100.-g sample of this compound.

What are the subscripts in the empirical formula of this compound? Enter the subscripts for C, H, and O, respectively, separated by commas (e.g., 5,6,7).


What are the subscripts in the empirical formula of this compound? Enter the subscripts for C, H, and O, respectively, separated by commas (e.g., 5,6,7).

Explanation / Answer

40% C = 40 grams = 40/12 = 3.33 mol C

6.7 % H = 6.7 grams = 6.7/1 = 6.7 mol H

53.3% O = 53.3 grams = 53.3 /16 = 3.33 mol O

Divide all ratios by 3.33 we get

C = 1

H = 2

O = 1

Empirical formula = CH2O

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