Question

1. What is the theoretical yeild od asprin (C9H12O4) if you started with 2.0156g of salicylic acid (C7H6O3) and excess acetic anhydride (C4H6O2)?

C4H6O2 + C7H6O3 --> C9H8O4 + C2H4O2

2. Caculate the moles of KHP (204.233 g/mol) if you took 0.5123g of KHP to standardize your NaOH.

3. A student used 0.5123g KHP to standarize the NaOH during day 1 of the titration experiment. The inital buret reading was 0.56mL and the buret reading after the solution turned a faint pink color was 25.67mL. What was the concentration of the NaOH solution to 4 sig figs?

NaOH + KHP --> NaKP + H2O

4. In day 2, a student used 4.00mL of vinegar solution of unknown concentration and titrated with 0.1058 M MaOH. The inital and final volume reading of the buret was 0.25mL and 28.54mL. What is the concentration of the vinegar solution?

NaOH + HC2H3O2 --> NaC2H3O2 + H2O

5. A student comes upto the instructor and says her solution did not change color even after adding all the solution from the buret. What could be the mistake made by the student? Explain.

Answer 1

1)

Molar mass of C7H6O3,

MM = 7*MM(C) + 6*MM(H) + 3*MM(O)

= 7*12.01 + 6*1.008 + 3*16.0

= 138.118 g/mol

mass of C7H6O3 = 2.016 g

mol of C7H6O3 = (mass)/(molar mass)

= 2.016/1.381*10^2

= 1.459*10^-2 mol

According to balanced equation

mol of C9H12O4 formed = moles of C7H6O3

= 1.459*10^-2 mol

Molar mass of C9H12O4,

MM = 9*MM(C) + 12*MM(H) + 4*MM(O)

= 9*12.01 + 12*1.008 + 4*16.0

= 184.186 g/mol

mass of C9H12O4 = number of mol * molar mass

= 1.459*10^-2*1.842*10^2

= 2.688 g

Answer: 2.688 g

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