Question

What is the empirical formula of a compound that is 27.3% C and 72.7% O?

Answer 1

@vanessanne

Answer 2

this should be easy stoichiometry :)

Answer 3

@nincompoop

Answer 4

that it what u were trying to said hopefully something instead leave that comment

Answer 5

Assume that there is 100 gr of the compound so we got 27.3 gr C and 72.7 gr O
\[?mol C = 27.3 gr C \times \frac{ 1 mol C }{12 gr C } = 2.275 mol C\]
\[?mol O = 72.2 gr O \times \frac{ 1mol O }{16 gr O } = 4.5125 mol O\]
\[4.5125 \div 2.275 \approx 2\]
so the empirical formula would be CO2