Answer:
21.63 %
Step-by-step explanation:
The molar mass of Mn₃[Mn(CO)₄]₃ is 665.64 g/mol.
Let's assume we have 1 mol of Mn₃[Mn(CO)₄]₃, if that were the case then we would have 665.64 grams.
There are 12 C moles per Mn₃[Mn(CO)₄]₃, with that in mind we calculate the weight of 12 C moles:
- 12 mol C * 12 g/mol = 144 g
Finally we calculate the mass percent of carbon:
- 144 g / 665.64 g * 100% = 21.63 %