Extra Credit 28
- Page ID
- 96912
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The sugar fructose contains 40.0% C, 6.7% H, and 53.3% O by mass. A solution of 11.7 g of fructose in 325 g of ethanol has a boiling point of 78.59 °C. The boiling point of ethanol is 78.35 °C, and Kb for ethanol is 1.20 °C/m. What is the molecular formula of fructose?
Solution:
First find the empirical formula by finding moles of C, H, and O in the sugar fructose by assuming there is 100 grams total of the three elements.
moles C= 40 g C x 1 mole C/ 12 g C = 3.33 moles C
moles H= 6.7 g H x 1 mole H/ 1.008 g H = 6.66 moles H
moles O=53.3 g O x 1 mole O/16 g O = 3.33 moles
This means the empirical formula of the sugar fructose is CH2O. In order to calculate moles of sugar fructose you use the formula Delta Tb= i x kb x m which you can change to get
moles Sugar fructose = [delta Tb/kb] x kg solvent
Plugging in the information provided, you get
moles fructose = [(78.59 - 78.35) / 1.20] x 0.325 kg ethanol
moles fructose = 0.065 moles
Once you find moles of fructose, you can use that information to find the molar mass of the fructose by dividing grams of fructose by moles of fructose.
MM= 11.7 g / 0.065 moles = 180 g/mole
Then using that molar mass, you can find the molecular formula by multiplying the molar mass by moles of element divided by 100.
C= 3.33 moles C / 100% x 180 g/mole = 5.99
H= 6.66 moles H / 100% x 180 g/mole = 11.98
O= 3.33 moles O / 100% x 180 g/mole = 5.99
Rounding to the nearest whole number gives you a molecular formula of C6H12O6.
Answer:C6H12O6
Q13.4.23
Calculate the equilibrium concentrations that result when 0.25 M O2 and 1.0 M HCl react and come to equilibrium.
\[\ce{4HCl}(g)+\ce{O2}(g)⇌\ce{2Cl2}(g)+\ce{2H2O}(g) \hspace{20px} K_c=3.1×10^{13}\]
Solution:
To find the equilibrium concentrations, an ICE table needs to be made first.
4HCl | O2 | 2 Cl2 | 2 H2O | |
Initial | 1.0 M | 0.25 M | 0 | 0 |
Change | -4x | -x | + 2x | +2x |
Equilibrium | 1.0 - 4x | 0.25 -x | 2x | 2x |
Kc = [Cl2]2[H2O]2 / [O2][HCl]4
3.1 x 1013 = 16x2 / (1-4x)4 x (0.25 -x)
Solving this equation for x, you get x=0.25M
This means that [HCl]≈ 0M, [O2]≈ 0M, and [Cl2] and [H2O] both equal 0.5M.
Answer: [HCl] = 0.0016M
[O2] = 0.0004M
[Cl2] = 0.4992M
Q15.2.X
Calculate
[HgCl42-]=
1.2 x 10-6
Q15.5.2
Calculate the concentration of each species present in a 0.050-M solution of H2S.
Solution:
Since H2S is a diprotic acid, you will need to construct two ICE tables to find the concentrations of all of the species in the solution. The first reactions is
H2S + H2O → H3O+ + HS- K1= 9.5 x 10-8
H2S | H2O | H3O+ | HS- | |
Initial | 0.050 M | --- | 0 | 0 |
Change | - x | --- | + x | + x |
Equilibrium | 0.050 - x | --- | x | x |
K = x2 / (0.050 - x)
Solving for x gives you x= 6.892 x 10-5 which means that
[H2S]=0.050 M
[H3O+] = 6.892 x 10-5 M
[HS-] = 6.892 x 10-5 M.
The second reaction is
HS- + H2O → S2- + H3O+ K2= 1.0 x 10-19
HS- | H2O | H3O+ | S2- | |
Initial | 6.892 x 10-5 | --- | 0 | 0 |
Change | - x | --- | + x | + x |
Equilibrium | 6.892 x 10-5 - x | --- | x | x |
K2= x2 / (6.892 x 10-5 - x )
Solving for x gives you x= 2.63 x 10-12 . This means that
[HS-)= 6.892 x 10-5 M
[H3O+] =2.63 x 10-12 M
[S2-]= 2.63 x 10-12 M
Answer: The concentration for [HS-], [H3O+],[H2S],[S2-] and [OH-] are
6.67*10-5M
6.67*10-5M
0.050M
1.0*10-19M
1.0*10-19M
Respectively
Q16.4.4
Use the standard free energy data in Appendix G to determine the free energy change for each of the following reactions, which are run under standard state conditions and 25 °C. Identify each as either spontaneous or nonspontaneous at these conditions.
\(\ce{C}(s,\, \ce{graphite})+\ce{O2}(g)⟶\ce{CO2}(g)\) \(\ce{O2}(g)+\ce{N2}(g)⟶\ce{2NO}(g)\) \(\ce{2Cu}(s)+\ce{S}(g)⟶\ce{Cu2S}(s)\) \(\ce{CaO}(s)+\ce{H2O}(l)⟶\ce{Ca(OH)2}(s)\) \(\ce{Fe2O3}(s)+\ce{3CO}(g)⟶\ce{2Fe}(s)+\ce{3CO2}(g)\) \(\ce{CaSO4⋅2H2O}(s)⟶\ce{CaSO4}(s)+\ce{2H2O}(g)\)
Given:
\[\ce{P4}(s)+\ce{5O2}(g)⟶\ce{P4O10}(s) \hspace{20px} ΔG^\circ_{298}=\mathrm{−2697.0\: kJ/mol}\]
\[\ce{2H2}(g)+\ce{O2}(g)⟶\ce{2H2O}(g) \hspace{20px} ΔG^\circ_{298}=\mathrm{−457.18\: kJ/mol}\]
\[\ce{6H2O}(g)+\ce{P4O10}(g)⟶\ce{4H3PO4}(l) \hspace{20px} ΔG^\circ_{298}=\mathrm{−428.66\: kJ/mol}\]
Solution:
To find ΔGo298 you use the equation ΔGo298 = ΔH- TΔS
If ΔGo298 is negative, then the reaction is spontaneous, but if it is positive, then it is non spontaneous.
a. ΔGo298= (-393.51) - 298 x (0.00286) = -394.36 kJ/mole (Spontaneous)
b. ΔGo298= (180.5) - 298 x (0.0248) = 173.11 kJ/mol (Non Spontaneous)
c. ΔGo298= (-358.31) - 298 x (-0.11322) = -324.57 kJ/mol (Spontaneous)
d. ΔGo298= (-64.47) - 298 x (-0.0247) = -57.1094 kJ/mol (Spontaneous)
e. ΔGo298= (-24.72) - 298 x (0.00155) = -29.339 kJ/mol (Spontaneous)
f. ΔGo298= (104.49) - 298 x (0.28996) = 18.082 kJ/mol (Non Spontaneous)
a. G = -394.39 kJ/mol; Spontaneous
b. G = 87.60 kJ/mol; Nonspontaneous
c. G= -115 kJ/mol; Spontaneous
d. G= -351 kJ/mol; Spontaneous
e. G= -333.17 kJ/mol;Spontaneous
f. G=-228.59 KJ/mol;Spontaneous