suppose a 22.092 g sample of 1 1 mixture of acetylferrocene and ferrocene was sepeerated by column chromatography and the recovered fractions weighed 9.017 g acetylferrocene and 8.075 g ferrocene what was the eprcent recovery of acetyl ferrocene

Percentage recovery of acetylferrocene = 81.6%

Explanation:

Mass of the sample mixture = 22.092 g

Ratio of mixture of acetylferrocene and ferrocene = 1 : 1

This means that the sample conatains equal amounts of acetylferrocene and ferrocene.

Therefore the mass of each sample in the mixture = 22.092 g / 2 = 11.046 g

Mass of acetylferrocene recovered = 9.017 g

Percentage recovery of acetylferrocene = (mass of recovered/ mass in sample) * 100%

Percentage recovery of acetylferrocene = (9.017 g / 11.046 g) *100%

Percentage recovery of acetylferrocene = 81.6%

2b2t

explanation:

2b2t