Answer:
Explanation:
20.1 mL of .3 M KOH reacts with .38 g of sample monoprotic acid
20.1 mL of .3 M KOH = .021 x .3 mole of KOH
= .0063 mole of KOH
.0063 mole of KOH will react with .0063 mole of monoprotic acid
.0063 mole of monoprotic acid = .38 g
1 mole of monoprotic acid = 60.3 or 60 g
molecular weight of monoprotic acid = 60
Acid contains 1.6 % H , 22.2 % N , 76.2 % O
Ratio of weight of H , N , O in sample
= 1.6 : 22.2 : 76.2
Ratio of moles of H , N , O
1.6 / 1 : 22.2 / 14 : 76.2 / 16
1.6 : 1.585 : 4.76
= 1.6 / 1.585 : 1 : 4.76 / 1.585
= 1 : 1 : 3
Empirical formula
= HNO₃
Mol formula = ( let )
Mol weight = n ( 1 + 14 + 3 x 16 )
= 63n .
Mol weight calculated above = 60
63n = 60
n = 1 ( approx )
Mol formula = HNO₃
Mol mass = 63 .