## Chemistry and Chemical Reactivity (9th Edition)

$Fe_2(CO)_9$
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 $Fe_2(CO)_9$