Stoichiometry
An unknown acid contains carbon, hydrogen, and oxygen. When a 10.46mg sample
was burned in oxygen it gave 22.17mg of carbon dioxide and 3.40mg of water. The
molecular mass was determined to be 166g/mol. When a 0.1680g sample of this acid
was titrated with 0.1250 M NaOH, the end point was reached after 16.18mL of base
had been added. What is the empirical formula? What is the molecular formula? Is the
acid mono-,di-, or triprotic?
y z⎞
y
⎛
CxHyOz + ⎜ x + − ⎟ O2  x CO2 + H2O
⎝
4 2⎠
2
Number of moles of acid burned = 10.46 x 10–3 /166 = 6.3012 x 10–5 mol
Number of moles CO2 produced = 22.17 x 10–3 / 44 = 5.0386 x 10–4 mol
Number of moles H2O produced = 3.40 x 10–3 / 18 = 1.8889 x 10–4 mol
C x H y O z ≡ x CO 2
By comparing CO2 and the acid,
n ( CO 2 )
(
n C x H yO z
)
=
5.0386 × 10 −4
6.3012 × 10 −5
⇒x≈8
C x H yO z ≡
By comparing H2O and the acid,
n ( H 2O )
(
n C x H yO z
)
y
H 2O
2
1.8889 × 10 −4
=
6.3012 × 10 −5
y
≈3
2
⇒y≈6
⇒
We know that the Mr of the acid is 166, so
12x + y + 16z = 166
Plugging in x = 8 and y = 6 yields
12 ( 8 ) + 6 + 16z = 166
⇒ 16z = 64
⇒z=4
Therefore, the molecular formula of the acid is C8H6O4.
The empirical formula of the acid is C4H3O2.
Number of moles of acid used for titration = 0.1680 / 166 = 1.0120 x 10–3 mol
Number of moles of NaOH neutralized = 0.1250 x 16.18/1000 = 0.0020225 mol
n ( NaOH )
0.0020225
=
≈ 1.998 ≈ 2
n ( acid )
1.0120 × 10 −3
Therefore, the acid is a diprotic acid.