A 0.5130 g aspirin tablet which was dissolved in 95 percent ethanol required 27.98 mL of 0.1000 M NaOH for neutralization. Then an additional 42.78 mL of 0.1000 M NaOH was added and the soln was heated to hydrolyze the aceytlsalicyclic acid. After the reaction mixture was cooled, the excess base was back-titrated with 14.29 mL of 0.1056 M HCL. How many grams of acetylsalicylic acid are in the tablet? What % w/w of the tablet is acetylsalicyclic acid?
The first titration neutralizes the ASA. The hydrolysis and back titration determines the amount of acetic acid produced in the hydrolysis and indirectly, the amount of ASA initially. Each can be used to determine the percent ASA. Therefore, I can obtain two answers for the ASA. Do you have any instructions as to which is more accurate? Or to take the average of the two results? Anything to help here. I can show you how to do the two but they differ by about 1%.
Why did the acetylsalicylic acid go to school? Because it wanted to be a tablet!
To find the grams of acetylsalicylic acid in the tablet, we need to use stoichiometry. First, we'll find the moles of NaOH used in the neutralization:
moles NaOH = Molarity NaOH × Volume NaOH
moles NaOH = 0.1000 M × 0.02798 L
moles NaOH = 0.002798 mol
Since the ratio of NaOH to acetylsalicylic acid is 1:1, we can say that the moles of acetylsalicylic acid in the tablet is also 0.002798 mol.
Now, let's find the grams of acetylsalicylic acid:
grams acetylsalicylic acid = moles acetylsalicylic acid × molar mass acetylsalicylic acid
grams acetylsalicylic acid = 0.002798 mol × 180.16 g/mol
grams acetylsalicylic acid = 0.5033 g
Therefore, there are approximately 0.5033 grams of acetylsalicylic acid in the tablet.
Now let's calculate the percentage w/w of the tablet that is acetylsalicylic acid:
% w/w acetylsalicylic acid = (grams acetylsalicylic acid / grams tablet) × 100%
% w/w acetylsalicylic acid = (0.5033 g / 0.5130 g) × 100%
% w/w acetylsalicylic acid = 98.08%
So, approximately 98.08% of the tablet is acetylsalicylic acid. Hope this answers your question with a dose of laughter!
To calculate the grams of acetylsalicylic acid in the tablet, we need to follow these steps:
Step 1: Calculate the number of moles of NaOH used for neutralization.
Given:
Volume of NaOH used for neutralization = 27.98 mL = 0.02798 L
Concentration of NaOH used for neutralization = 0.1000 M
Number of moles of NaOH used for neutralization = volume x concentration
= 0.02798 L x 0.1000 mol/L
Step 2: Calculate the number of moles of acetylsalicylic acid (ASA) in the tablet.
From the balanced chemical equation, the molar ratio between NaOH and ASA is 1:1. This means that the number of moles of ASA is equal to the number of moles of NaOH used for neutralization.
Number of moles of ASA = Number of moles of NaOH used for neutralization
Step 3: Calculate the mass of acetylsalicylic acid in the tablet.
Given:
Mass of aspirin tablet = 0.5130 g
Mass of ASA = Number of moles of ASA x molar mass of ASA
To find the molar mass of ASA, we refer to the periodic table and add up the atomic masses of the elements carbon (C), hydrogen (H), and oxygen (O) present in ASA.
The molar mass of C = 12.01 g/mol
The molar mass of H = 1.01 g/mol
The molar mass of O = 16.00 g/mol
The molecular formula of ASA is C9H8O4.
Hence, the molar mass of ASA = (9 * 12.01 g/mol) + (8 * 1.01 g/mol) + (4 * 16.00 g/mol)
Step 4: Calculate the % w/w of acetylsalicylic acid in the tablet.
The % w/w of ASA in the tablet is given by:
% w/w = (mass of ASA / mass of tablet) x 100
Now, we can calculate the grams of acetylsalicylic acid (ASA) in the tablet and the % w/w of the tablet that is ASA.
To determine the number of grams of acetylsalicylic acid (ASA) in the tablet and the percentage w/w of ASA in the tablet, we need to follow these steps:
Step 1: Calculate the number of moles of NaOH used for neutralization.
Given:
Volume of NaOH used for neutralization = 27.98 mL
Molarity of NaOH = 0.1000 M
Using the formula:
moles of NaOH = (volume of NaOH used / 1000) x Molarity of NaOH
moles of NaOH = (27.98 mL / 1000 mL) x 0.1000 M
moles of NaOH = 0.002798 mol
Step 2: Calculate the number of moles of ASA in the tablet using the neutralization reaction.
The balanced neutralization reaction between NaOH and ASA is:
ASA + NaOH → Sodium Acetylsalicylate + H2O
From the chemical equation, we can see that the stoichiometric ratio between ASA and NaOH is 1:1. This means that every mole of NaOH used for neutralization is equal to one mole of ASA in the tablet.
Therefore, the number of moles of ASA in the tablet is also 0.002798 mol.
Step 3: Calculate the molar mass of ASA.
The molar mass of ASA can be calculated using the atomic masses of the elements in the compound. The molar mass of ASA is:
ASA = C9H8O4
C = 12.01 g/mol
H = 1.008 g/mol (x8)
O = 16.00 g/mol (x4)
Molar mass of ASA = (12.01 x 9) + (1.008 x 8) + (16.00 x 4) = 180.16 g/mol
Step 4: Calculate the mass of ASA in the tablet.
mass of ASA = moles of ASA x molar mass of ASA
mass of ASA = 0.002798 mol x 180.16 g/mol
mass of ASA = 0.5003 g
Therefore, there are 0.5003 grams of acetylsalicylic acid (ASA) in the tablet.
Step 5: Calculate the percentage w/w of ASA in the tablet.
The percentage w/w (weight/weight) of ASA in the tablet can be calculated using the formula:
% w/w ASA = (mass of ASA / total mass of tablet) x 100
Given:
mass of tablet = 0.5130 g
% w/w ASA = (0.5003 g / 0.5130 g) x 100
% w/w ASA = 97.51%
Therefore, the tablet is approximately 97.51% w/w acetylsalicylic acid.