Determining the Atomic Mass of an Unknown Diprotic Acid – Lab Report

Zach Gordon

Determining the Atomic Mass of an Unknown Diprotic Acid  Lab Report  Introduction The goal of this experiment was to calculate the mass of an unknown diprotic acid. In the experiment, 3 samples of a specific diprotic acid were dissolved d issolved in roughly 10 mL of water. Titrations were then preformed with each sample by adding phenolphthalein to the dissolved acid, and then slowly adding add ing 1.0853 molar Sodium Hydroxide (NaOH) until the solution reaches the end point and turns pink. Stoichiometry was then used to find the molar mass of each acid sample and then the average was taken. The average atomic mass that I found was 123.8 123 .8 g/mol.

Data Table Titration # 1 2 3

Mass of  Acid 0.158 g 0.178 g 0.195 g

Initial Volume 34.90 mL 37.29 mL 39.91 mL

Final Volume 37.29 mL 39.91 mL 42.80 mL

Net Volume of NaOH 2.39 mL 2.62 mL 2.89 mL

Calculations H2 A + 2NaOH Na2 A + 2H2O Titration 1: 1.0853mol/L NaOH × 0.00239 L = 0.002594 mol NaOH 0.002594 mol NaOH ÷ 2 = 0.001297 mol H2A 0.158g H2A ÷ 0.001297 mol = 121.8 g/mol Titration 2: 1.0853mol/L NaOH × 0.00262 L = 0.00284 mol NaOH 0.00284 mol NaOH ÷ 2 = 0.00142 mol H2A 0.178g H2A ÷ 0.00142 mol = 125.35 g/mol Titration 3: 1.0853mol/L NaOH × 0.00289 L = 0.0031365 mol NaOH 0.0031365 mol NaOH ÷ 2 = 0.001568 mol H2A 0.195g H2A ÷ 0.001568 mol = 124.36 g/mol  Average: 123.8 g/mol

Conclusion The molar mass of the unknown u nknown diprotic acid is about 123.8 g/mol.

Quality of  End Point  Light Pink  Dark Pink  Dark Pink 

Zach Gordon Post Lab Questions

1. If the actual molar mass of H2A is 126 g/mol then my percent error is (126g-123.8g)/126g * 100% = 2.2g/126g*100% = 0.01746 * 100% = 1.75% error 2. If all of the solid acid was not transferred to the Erlenmeyer flask used in the titration, then it would take less base to reach the end point and neutralize the acid (since theres less acid it would take less base to neutralize), and then that would make the calculated molar mass heavier because we would be dividing the weight of the pre-weighed acid by less moles meaning theres more mass per mol.