Science

Math

Humanities

... and beyond

Socratic Meta
Featured Answers

Topics

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

1 Answer

Feb 26, 2016

$C_{2} H_{4} O$ $C_2H_4O$

Explanation:

The ratio of no of atoms of C,H&O= $(\frac{54.53}{12}) : (\frac{9.15}{1}) : (\frac{36.32}{16})$ $(54.53/12):(9.15/1):(36.32/16)$

= $4.54 : 9.15 : 2.27$ $4.54:9.15:2.27$

= $\frac{4.54}{2.27} : \frac{9.15}{2.27} : \frac{2.27}{2.27}$ $4.54/2.27:9.15/2.27:2.27/2.27$

= $2 : 4 : 1$ $2:4:1$

Empirical formula $C_{2} H_{4} O$ $C_2H_4O$

Related questions

See all questions in Empirical and Molecular Formulas

Impact of this question

38222 views around the world

You can reuse this answer
Creative Commons License