Archive for June 20th, 2008
Stoichiometry calculation of acetyl salicylic acid is easier we solve if we know the complete reaction that involve. The synthesis of acetyl salicylic acid (ASA) is categorized as an esterification of salicylic acid and acetic anhydride that yielding acetyl salicylic acid and acetic acid as by product, a few drops of sulfuric acid often added as a catalyst. This technique is usually employed in undergraduate teaching labs.
Now examine this problem on how you can calculate the theoretical yield of acetylsalicylic synthesis.
“Chemistry student reacts 4.81 g of salicylic acid with an excess of acetic anhydride in the presence of a few drops of sulfuric acid to compose acetylsalicylic acid (aspirin), which the student then purifies. (1) Calculate the theoretical yield mass of product (2) if the student only get aspirin 5,781 g calculate the %yield of the student result ”
Free Chemistry Tutor
In the synthesis of aspirin we usually used excess acetic anhydride; this should be done to ensure that the salicylic acid will be completely reacted and we get the maximum result. But unfortunately there are not chemical reactions that 100% complete, many factors influence them. As the result, we have the theoretical result and % yield of the reaction. Let’s we solve the problem.
The balanced reaction :
+ (CH3CO)2O -> 
+ CH3COOH
Firstly calculate the molecular mass of the salicylic acid and the acetylsalicylic acid (aspirin)
Mr salicylic acid (C7H6O3)
= (7 x 12.01) + (6 x 1.008) + (3 x 16.00)
= 138.118 g/mol
Mr acetylsalicylic acid (C9H8O4)
= (9 x 12.01) + (8 x 1.008) + (4 x 16.00)
= 180.154 g/mol
Secondly calculate the moles of salicylic acid
Moles salicylic acid (C7H6O3)
= mass / molar mass
= 4.81 / 138.118
= 0.035 mol
the mole ratio of salicylic acid and acetylsalicylic acid (aspirin) is 1:1 thus from this stoichiometry we can calculate the moles of acetylsalicylic acid (aspirin) that produced.
Moles of acetylsalicylic acid (aspirin)
= 1/1 x moles of salicylic acid (C7H6O3)
= 1/1 x 0.035 mol
= 0.035 mol
The theoretical mass of acetic salicylic acid (aspirin) obtains by multiplying the moles of aspirin with its molecular mass.
Theoretical mass of acetylsalicylic acid (C9H8O4)
= moles x molar mass
= 0.035 mol x 180.154 g/mol
= 6.274 g
The mass of percentage yield calculated by dividing actual mass of the aspirin that the student got with the theoretical that we solve above in the percent, and we have
The % yield of the aspirin
= actual mass / theoretical yield x 100%
= 5,781 / 6.274 x 100%
= 92.143 %
Stoichiometry calculation of theoretical yield and mass percentage yield of acetylsalicylic synthesis from the calculation above we have 6.274 g and 92.143 % respectively.
See also :
Calculate the actual mass of aspirin
June 20th, 2008
Standardization denotes a process to determine the concentration of a solution using a standard solution. Base usually has a hydroscopic property because of this we can not use base like this for a primer standard. We use another substance to standardize this solution. using this technique we have a base standard solution that can be reacted to determine acid solution. take a look for the example
A student is asked to standardize a solution of potassium hydroxide. He weighs out 1.01 g potassium hydrogen phthalate (KHC8H4O4, treat this as a monoprotic acid). It requires 36.5 mL of potassium hydroxide to reach the endpoint. What is the molarity of the potassium hydroxide solution? Same solution used to titrate an unknown solution of hydrochloric acid, if 22.1 mL of the potassium hydroxide solution is required to neutralize 21.9 mL of HCl, find molarity of HCl?
First calculate the moles of potassium hydrogen phthalate KHC8H4O4
= mass/ molar mass
= 1.01 g / 204.2
= 4.946 x 10-3 mol
= 4.946 x 10-3 equivalents
= 4.946 milli-equivalent
to calculate the normality of KOH we used titration formula
V x N acid = V x N base
eq acid = V x N base
4.946 = 36.5 mL x N
N = 0.136 N
Potassium hydroxide has one ion hydroxide thus 0.136 N of KOH should be 0.136 M of KOH. By the same way we can calculate the molarity of HCl,
V x N acid = V x N base
21.9 x N = 22.1 x 0.136 N
N HCl = 0.137 N
As KOH, HCl has one hydrogen ion thus 0.137 N of HCl = 0.137 M of HCl.
June 20th, 2008
Back titration can be use to determine the concentration of unknown acid or base. The unknown concentration of the base solution firstly reacted with the excess acid and then the excess can be titrated with the standard base solution, from the difference of the moles of the acid we can determine the concentration of the unknown base. Here is the example:
38.7 mL of 0.514 M nitric acid is added to 20.7 mL of calcium hydroxide, and the resulting solution is found to be acidic. 27.0 mL of 0.411 M potassium hydroxide is required to reach neutrality. What is the molarity of the original calcium hydroxide solution?
Answer:
moles of HNO3 original = 38.7 mL x 0.514 = 19.892 mmol
mmol of HNO3 that reacted with KOH is calculated using titration formula, because KOH has one hydroxide ion thus 0.411 M KOH would be equal to 0.4111 N of KOH
meq acid = meq base
meq acid = V x N
meq acid = 27.0 mL x 0.411 N
meq acid = 11.097 milli-equivalent
because HNO3 has one ion hydrogen thus 11.097 milli-equivalent of HNO3 are equal to 11.0971 mmol HNO3. The moles of HNO3 that reacted with Ca(OH)2 can be calculated by subtracting the origin moles of HNO3 with the moles of HNO3 that reacted with KOH, and these moles are represent the moles of Ca(OH)2 too.
moles HNO3 that reacted with Ca(OH)2 are
= moles of HNO3 original - moles HNO3 reacted with KOH
= 19.892 mmol - 11.0971 mmol
= 8.795 mmol
Molarity of Ca(OH)2 solution are
M = mmol / mL
M = 8.795 mmol / 20.7 mL
M = 0.425 M
The molarity oh calcium hydroxide Ca(OH)2 that calculated using back titration from above calculation are 0.425 M.
June 20th, 2008
We can calculate the mass percentage of a substance using titration method. Titration formula is used to determine the mole-equivalent of the substance, if the mole equivalent of the substance is known we convert it into moles and then multiplying with molar mass to get the mass of the species. Dividing the actual mass of the substance with its original weight we have its mass percentage. Here the example
A 12.1 g sample of an aqueous solution of perchloric acid contains an unknown amount of the acid. If 20.4 mL of 0.450 M barium hydroxide is required to neutralize the perchloric acid, what is the percent by weight of perchloric acid in the mixture?
First we calculate the mole of the perchloric acid (HClO4) using titration formula. Remember V x N (volume multiply with normality produce mole-equivalent or meq), and we convert molarity of Ba(OH)2 solution into normality because Ba(OH)2 has two hydroxide ions thus 0.450 M of barium hydroxide is equal to 0.9 N.
V x N acid = V x N base
meqacid = V x N base
meq acid = 20.4 mL x 0.9 N
meq scid = 18.36 milli-equivalent
because HClO4 has one ion hydrogen thus one equivlent of HClO4 is equal to one mole of HClO4, then 18.36 milli-equivalents of HClO4 = 18.36 millimoles of HClO4. The mass of HClO4 can be calculate as
mass of acid (HClO4)
= moles x molar mass
= 18.36 mmol x 100.458
= 1844.409 mgr
= 1.844409
then the mass percentage of HClO4 are
= mass of HClO4 / mass of sampel x 100%
= 1.844409/12.1x 100%
= 15.243 %
from these calculation we get that the mass percentage of perchloric acid in the sample using titration calculation are 15.243%
June 20th, 2008