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A 1.400 g sample of dry, ground barley was digested and the resulting ammonia absorbed into 50.0 cm3 of 0.0200 M sulfuric acid. The remaining acid produced a 9.05 cm3 titre with 0.0400M sodium hydroxide solution. Would this sample pass the brewery specification of a maximum of 1.70% nitrogen by mass?
Throughout the calculation we shall use the equation, number of moles = concentration × volume in order to work out number of moles
Number of moles of sodium hydroxide = 9.05 ×
0.0400/1000 = 3.62 × 10-4
This is
the number of moles of sodium hydroxide used in the
titration
|
2NaOH(aq) + H2SO4(aq) ⇒ Na2SO4(aq) + 2H2O(l) |
\Number of
moles of H2SO4 titrated = ½ × 3.62 ×
10-4 = 1.81 × 10-4
This
is the number of moles of sulfuric acid left over after the
reaction with the ammonia gas. The equation shows that there is
twice as much sodium hydroxide compared to the sulfuric acid. This
means there must be half as much sulfuric acid.
Number of moles of H2SO4
initially used = 0.0200 × 50.0/1000 = 1.00 ×
10-3
This is the number of moles
of sulfuric acid in the 50 cm3 that we started
with
\Number of
moles of H2SO4 reacted with ammonia = 1.00 ×
10-3 - 1.81 × 10-4 = 8.19 ×
10-4
If we take away the number of
moles of sulfuric acid that were left after the ammonia was
absorbed (1.8 × 10-4) from the number of moles we
started with (1 × 10-3), we have the number of moles
which must have reacted with the absorbed ammonia
|
2NH3(g) + H2SO4(aq) ⇒ (NH4)2SO4(aq) |
\Number of
moles of ammonia = 2 × 8.19 × 10-4 = 1.64 ×
10-3
This is the ammonia from the
barley which was absorbed into, and reacted with, the sulfuric
acid. From the equation we see there is twice as much ammonia
compared to the sulfuric acid
\Mass of
nitrogen = 1.64 × 10-3 × 14 = 0.0230 g
This is the number of moles of nitrogen (which is
the same as the number of moles of ammonia because one ammonia
molecule contains one nitrogen atom) multiplied by the atomic mass
of nitrogen (14)
\Percentage
of nitrogen = 0.0230/1.400 × 100 = 1.64%
The mass of nitrogen
(0.0230 g) divided by the original mass of the sample (1.400 g)
times by 100 to give the percentage
This passes the brewery specification.
That is, it is less than 1.70%