How many grams of acid and alcohol must be taken to obtain 1 g of isobutyl acetate with an ether

How many grams of acid and alcohol must be taken to obtain 1 g of isobutyl acetate with an ether yield of 60% of the theoretically possible

Given:
m pract. (C6H12O2) = 1 g
η (C6H12O2) = 60%

To find:
m (CH3COOH) -?
m (C4H9OH) -?

Solution:
1) CH3COOH + C4H9OH => C6H12O2 + H2O;
2) m theor. (C6H12O2) = m practical. * 100% / η = 1 * 100% / 60% = 1.667 g;
3) n theory. (C6H12O2) = m theory. / M = 1.667 / 116 = 0.014 mol;
4) n (CH3COOH) = n theory. (C6H12O2) = 0.014 mol;
5) m (CH3COOH) = n * M = 0.014 * 60 = 0.84 g;
6) n (C4H9OH) = n theory. (C6H12O2) = 0.014 mol;
5) m (C4H9OH) = n * M = 0.014 * 74 = 1.04 g.

Answer: The mass of CH3COOH is 0.84 g; C4H9OH – 1.04 g.