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Sec 2: Synthesis of 4-Hydroxy-phthalazinone

Calculation: ⸪ One mole of any substance = Molecular weight in grams
Amount of phthalic Amount of hydrazine Amount of phthalazinone (g)
anhydride (g) (ml) (Theoretical Yield)
1 mole → 148.11 g 1 mole → 50.06 g 1 mole →162.14 g

0.005 mol → ? g 0.005 mol → ? g 0.005 mol → ? g

⸫ phthalic anhydride Amount = 0.25 g ⸫ Phthalazinone amount
(theoretical yield) = 0.81 g
amount= 0.74 ≅ As hydrazine is liquid,

. ⸪ Density of phenylhydrazine
m (grams) %Yield =
◼ Phthalic = ρ = ( ) ( )
anhydride is the ( ) ×

limiting reactant ⸫ ( ) = m (g) = 0.25 = ⋯ %
used up first. ρ (ml) 1.032 ( )

⸫ . ( ) ×
= . ≅ . = ⋯ %

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