An unknown gas X effuses 0.507 times as fast as C4H8 . What is the molecular mass of gas X

Another difficult problem...my brain is dead!

Graham's Law:

r1 / r2 = √(MM2 / MM1)

The C4H8 effuses at a rate of 1.000 and gas X effuses at 0.507. We assign C4H8 to the r1 and MM1 combo. That puts gas X as r2 and MM2, putting its molecular weight in the numerator.

1.000 / 0.507 = √(x / 56.1072)

square both sides:

3.8903 = x / 56.1072

x = 218 g/mol (to three sig figs)