Question

A component in a sleeping pill is a weak monoprotic acid (HX). A titration required 12.00 mL of 0.200 M NaOH to titrate 0.4421 g of this acid.

(a) What is the molar mass of this compound?

(b)If this acid has the percent composition of 52.162 % C, 6.567 % H, 15.212 % N, and 26.059 % O, what is the empirical formula and molecular formula of this aicd?

Answer 1

a)
for titration,
number of mol of acid = number of moles of base
mass of acid/molar mass of acid = M(NaOH) *V (NaOH)
0.4421/molar mass of acid = 0.200 M * 0.012 L
molar mass of acid = 184.2 g/mol

b)
consider 100 g of acid.

C : 52.162 g
H : 6.567 g
N : 15.212 g
O : 26.059 g

divide each by molar mass:
C : 52.162/12 = 4.35 mol
H : 6.567/1 = 6.567 mol
N : 15.212/14 = 1.0866 mol
O : 26.059/16 = 1.629 mol

divide by smallest numer:
C : 4.35/1.0866 = 4
H : 6.567/1.0866 = 6
N : 1.0866 / 1.0866 = 1
O : 1.629/1.0866 = 1.5

multiply each by 2 to get whole number:
C : 8
H : 12
N : 2
O : 3
empirical formula = C8H12N2O3
empirical formula mass = 8*12 + 12*1 + 2*14 + 3*16 = 184 g
molar mass is also same
so, empirical formula is same as molecular formula

Molecular formula will be:
C₈H₁₂N₂O₃