EQUATIONS AND OTHER INFORMATION FOR CHM152L FINAL EXAM

Experiment A:

MM:    NaOH=40.00 g/mol

Heat Capacity in kJ/oC: Cp = ((55.90 kJ/mol)n)/(change in T)

Heat of Reaction in kJ/mol = -Cp(change in T)/n

Where n = moles and T equals temperature

Experiment B:

General Reaction: E + S = ES = E + P

where E is the enzyme (papain), S is substrate (BAPNA),

P is products (p-nitroaniline), and ES is the enzyme-substrate complex.

p-nitroaniline is yellow in solution so its concentration can be directly related to the absorbance at the wavelength for maximum absorbance.

T=I/Io    A=-log10(T)

Experiment C: Synthesis of the iron salt

Percent by Mass = (part/whole)*100%

%yield = (Actual Yield/Theoretical Yield)/100%

Experiment D:   MM:    NaSCN = 81.07 g/mol

Fe3+ + SCN- = FeSCN2+

KC = [FeSCN2+]/([Fe3+][SCN-])

Wavelenght for Maximum absorbance for FeSCN2+ is 447 nm

Beer's Law:  A = €lC where A is absorbance, € is molar absorptivity, l is path length, and C is concentration

Experiment E:     AW:   Fe=55.847g/mol     K=39.098g/mol

pH = pKa + log([A-]/[HA])

Equivalent formula weight of the weak acid:

= (grams of weak acid)/(moles NaOH to reach equivalence point)

Moles NaOH = (liters NaOH)(molarity NaOH)

H+ + OH- = H2O        Fe3+ + 3OH- = Fe(OH)3 (1 mol Fe3+ reacts with 3 mol of OH-)

Experiment F:    MM g/mol   Na2C2O4=134.0   C2O42-=88.00    KMnO4=158.04

2MnO4-+5C2O4-2+16H+ = 2Mn+2+10CO2+8H2O

Conversion factors (obtained from the balanced reaction)

2 mole MnO4- reacts with 5 mole C2O4-2

Percent by Mass Na2C204 = (g Na2C204)/g Unknown) x 100%