Sodium Benzoate Reaction with Hcl

Lauren Correa

CHM2210C-Organic Chemistry I

“Sodium Benzoate Reaction with HCl”

Purpose:

The purpose of this lab is to prove that sodium benzoate, when combined with HCl, yields benzoic acid.

Procedure and Apparatus:

2.00g of sodium benzoate was weighed out and transferred to a 100mL beaker. 5.0mL of DI water was added to the beaker and stirred until the sodium benzoate dissolved. 4.4mL of 3.0M HCL was added to the aqueous sodium benzoate solution. The new solution was placed into an ice water bath for approximately 1 minute until the desired temperature of 10 degrees centigrade was reached. The newly formed precipitate was transferred from the beaker to a vacuum filtration device. The beaker was rinsed with DI water to ensure the least amount of product possible was left behind. After vacuum filtration, the leftover substance was transferred to a watch glass and placed in a drying oven at 90 degrees centigrade until the excess water was removed.

Data:

The sodium benzoate, in its solid form, was a white, powdery substance. When mixed with the 10.0 mL of water, the new solution became a cloudy white mix. After adding the HCl, the pH of the solution dropped to between 1-2. As the HCl was being added, bubble formation was noted as well as a white opaque new substance. The final weight of the new product after being left in the drying oven was

Calculations:

NaC7H5O2 + HCl ↔ C6H5COOH + NaCl

To find the theoretical yield, the equation above had to be verified as balanced. Once balanced, stoichiometry could be performed.

To perform the stoichiometry, the molar mass of both NaC7H5O2 and C6H5COOH had to be computed. They were found to be 144.11 mol/g and 122.12 mol/g respectively.

The math for the theoretical yield of NaC7H5O2 is found below;

2.00g NaC7H5O2 ∙   1mol      =  0.0139 mol NaC7H5O2[pic 1]

                                   144.11g

0.0139 mol NaC7H5O2 ∗ 1mol C6H5COOH  ∗ 1.22.12g =.  1.70g NaC7H5O2 [pic 2][pic 3]

                                          1mol NaC7H5O2               1mol

To find the %yield, the actual yield was divided by the theoretical yield, providing the answer below:

Discussion:

Error Analysis:

The new product was difficult to remove from the beaker as it was a thick substance. Even after trying to rinse it out with DI water, some amount remained on the beaked and on the stirring rod.

Some of the Product was lost while stirring after adding the HCl into the solution.

The product was extremely wet, so even after vacuum filtration and longer than 30 minutes in the oven drying, the weight of the new product was still significantly higher than the theoretical yield. The increased weight was a sign that the product was still too wet and needed longer in the over. The product had to be left over the week, and lack of access during the week sets up potential for outside sources affecting the outcome.