This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Cory CHE 201 M1/8 Wednesday 5:00pm TA: Gameli Agbeleze Experiment 1: A. The Effect of pH on a Food Preservative B. Measuring the Density of an Unknown Liquid. Introduction: The purpose of part a in the experiment was to place sodium benzoate into a medium that simulated stomach acid (pH 1-3) and see whether or not a new substance was formed. The purpose of part b in the experiment was to measure the density of an unknown liquid by graphical analysis. My working hypothesis for part a is that if sodium benzoate is exposed to an acid, it will undergo a reaction resulting in a new substance; this substance will most likely be benzoic acid. This is shown in the reaction below: The net reaction for part a was: C 6 H 5 COONa + HCl C 6 H 5 COOH + NaCl This was an acid-base reaction. Procedure/Results: For part a of the experiment, 0.400g of C 6 H 5 COONa was weighed and dissolved in 3.0mL of water until all of the sold was gone. Next, about 0.8mL of 3M HCl was added to the aqueous C 6 H 5 COONa solution until the solution reached a pH of 2 or lower. This yielded a thick milky white precipitate. The vial containing the reaction was set in a cold water bath for approx. 5 minutes. Since a new substance was formed it had to be separated and measured, this was accomplished by transferring the solution to a Hirsch funnel and vacuum filtration was carried out until the white precipitate was dried to a consistent mass. The substance was then weighed and recorded. The reaction from above has a 1:1 mole ratio so the moles of product should be equal to the moles of reactant. Since 0.400g of C 6 H 5 COONa is 0.002776moles (mass/gfm) then the theoretical yield for C 6 H 5 COOH is 0.002776moles as well. Moles x Molar mass = 0.002776 x 122.1 = 0.339g of C 6 H 5 COOH. So I would want a product as close to 0.339g as possible. After the experiment was carried out the mass of C 6 H 5 COOH was weighed and a mass of 0.33 grams was produced. The %yield was calculated as (mass of actual yield / theoretical yield) x 100 = (0.33/0.339) x 100 = 97.3%yield....
View Full Document