What is the standard entropy change for the formation of butyric acid at 25 degrees C

I need help solving this little thermodynamics issue.

Standard Heat of Absolute
Formation, DHf°, Entropy, S°,
Substance in kJ mol-1 in J mol-1 K-1
------------------ ----------------------------- ------------------------
C(s) 0.00 5.69
CO2(g) -393.5 213.6
H2(g) 0.00 130.6
H2O(l) -285.85 69.91
O2(g) 0.00 205.0
C3H7COOH(l) ? 226.3

The enthalpy change for the combustion of butyric acid at 25°C, DH°comb, is -2,183.5 kilojoules per mole. The combustion reaction is

C3H7COOH(l) + 5 O2(g) ® 4 CO2(g) + 4 H2O(l)

(a) From the above data, calculate the standard heat of formation, DHf°, for butyric acid.

(b) Write a correctly balanced equation for the formation of butyric acid from its elements.

(c) Calculate the standard entropy change, DSf°, for the formation of butyric acid at 25°C. The entropy change, DS°, for the combustion reaction above is -117.1 J K-1 at 25°C.

(d) Calculate the standard free energy of formation, DG°f, for butyric acid at 25°C.

I already have the answers, and am mostly interested solely in parts C and D, but i need to see the process and how it is done. Please show the steps taken to calculate entropy and free energy of formation, any help is appreciated.

c) Calculate the standard entropy change, ∆Sf°, for the formation of butyric acid at 25° C. The entropy change, ∆S°, for the combustion reaction above is - 117.1 J K-1 at 25° C.

∆S°f (butyric acid) = S°(butyric acid) - [4 S°C + 4 S°H2 + S°O2]
226.3 - [4(5.69) + 4(130.6) + 205] = - 523.9 J/K

d) Calculate the standard free energy of formation, ∆Gf°, for butyric acid at 25° C.

∆G°f = ∆H°f - T∆S°f = 533.8 - (298)(-0.5239) kJ = - 377.7 kJ