If 0.5mole of monoalkanioc acid weigh 44g.Determine the molecular formula and the name of the acid
moles = grams/molar mass or
molar mass = grams/mol = 44/0.5 = 88
I would guess something like CH3(CH2)nCOOH so COOH is 45 and the end CH3 is 15 which adds to 45 + 15 = 60 and that leaves 88 -60 = 28. Each CH2 is 14; 28/14 = 2 so the best guess would be CH3(CH2)2COOH of C4H8O2.