A 0.380 g sample of an unknown monoprotic acid is titrated with 0.300 M KOH to the equivalence point. It requires 20.1 mL of KOH to reach this point. Elemental analysis is also performed, showing that the acid contains 1.6% H, 22.2% N, and 76.2% O. Use this information to determine the molecular formula and the molar mass of the unknown acid.

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 .

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.