Answer
$Fe_2(CO)_9$
Work Step by Step
Atomic weight $Fe: 55.845\ g/mol$
Molar mass of the compound, assuming x = 1: $55.845\ \div \frac{30.70}{100}=181.9\ g/mol$
Molar mass of CO $28.01\ g/mol$
$55.845+y\times28.01=181.9\rightarrow y=4.50$
Multiplying by 2 to make x and y integers:
Empirical formula $Fe_2(CO)_9$
