1) If 0.5 moles of a mono alkanoic acid weighs 44g,determine the molecular formular. 2) molecular formular =emprical * n. 3) molecular formular = 44*0.5

Answer 1

The molecular formula is not 44*0.5. If 0.5 mol weighs 44 g then a whole mole must weigh 88 g.
An acid has a -COOH group which is 45 so the rest of the molecule must be 88-45 = 43. A -CH3 group is 15 so 43-15 = 28 and CH2 groups are 14 each so the molecular formula must be
CH3CH2CH2COOH. Check it out and see that it adds up to 88.

Answer 2

C3H7COOH