Question

Using your data, calculate the moles and mass of acetylsalicylic acid in each tablet (molar mass = 180.16 g/mol).

Data Concentration of NaOH 107 M Solutions Mass of Aspirin Tablet #11 Mass of Aspirin Tablet #21 Volume of NaOH mixed with Tablet #1 | Volume of NaOH mixed with tablet #2 Value .37 .36 50 mL 50 mL Standardization of HCl Trial #1 Value Mass of Na2CO3 .070 a Initial Volume of HCI in burettel 2.20 mL Volume of HCl in burette at equivalence point8.13 mL Standardization of HCl Trial #2 Value Mass of Na2CO3.069 a Initial Volume of HCl in burette3.19 mL Volume of HCl in burette at equivalence point 10.20 mL Titration of NaOH Tablet solutions Value Titration #1 - Initial Volume of HCl in burette| 8.13 |mL Titration #1 -Final Volume of HCl in burette 25 Titration #2-Initial Volume of HCl in burette 10.20 mL Titration #2-Final Volume of HCl in burette| 21.04 |mL .58 mL Analysis Analysis #1 Value Concentration of HCl from trial #1 .223 M Correct Concentration of HCl from trial #2 .185 M Correct Average Concentration of HCI Solution .204M Correct Analysis #2 Value Trial 1 -Moles of Acetylsalicylic acid in tablet Trial 1 Masss of Acetylsalicvlic acid in tablet161 Trial 2 Moles of Acetvlsalicvlic acid in tablet Trial 2- Mass of Acetvlsalicylic acid in tablet.282 mol Incorrect g Correct mol Incorrect g Correct

Answer 1

Molar mass of Acetyl salicylic acid given as 180.16g/mol

That means 180.16 gm of acetylsalicylic acid is = 1 mol

                     1 gm of acetylsalicylic acid is = 1/180.16 mol

Trial 1 mass 0.161 gm of acetylsalicylic acid is = (1/180.16) $\times$ 0.161 mol = 8.94 $\times$ 10^-4 mol

For trial 2 mass 0.282 gm of acetylsalicylic acid is = (1/180.16) $\times$ 0.282 mol = 1.57 $\times$ 10^-3 mol