Question

In 3 trials in a Bonding and Acidity experiment, a student titrated a compound with a molecular formula C₆H₈O₇ with a 0.0964 M NaOH solution.

Trial 1) 0.53 g C₆H₈O₇ titrated with 85.6 mL of 0.0964 M NaOH

Trial 2) 0.22 g C₆H₈O₇ titrated with 35.69 mL of 0.0964 M NaOH

Trial 3) 0.27 g C₆H₈O₇ titrated with 43.51 mL of 0.0964 M NaOH

1. Determine the average ratio of moles of hydrogen ions to moles of acid. This value will give you the number of ionizable hydrogens in each acid molecule.

2. Rewrite the molecular formula from the vial (C₆H₈O₇) to reflect the number of ionizable hydrogens.

3. Draw a Lewis-dot or line sketch consistent with the modified formula for your acid.

4. Develop a hypothesis to account for your experimental results.

Answer 1

MWofC6H8O7   = 192.124 g/mol

so..

Trial1.

mol of C6H8O7 = mass/MW = 0.53/192.124 = 0.002758

mol of NaOH = MV = 0.0964*85.6/1000 = 0.00825184

ratio is 0.00825184/ 0.002758 = 2.99 so 3 H+ per acid

Trial2.

mol of C6H8O7 = mass/MW = 0.22/192.124 = 0.0011450

mol of NaOH = MV = 0.0964*35.69/1000 = 0.003440

ratio is 0.003440/ 0.0011450= 3.00 so 3 H+ per acid

Trial3.

mol of C6H8O7 = mass/MW = 0.27/192.124 = 0.00140534

mol of NaOH = MV = 0.0964*43.51/1000 = 0.00419

ratio is 0.00419/ 0.00140534= 2.9814 so 3 H+ per acid

Q2.

C3H5O(COOH)3

Q3.

Lewis dot