Extra Credit 28

Q11.5.39

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 K_b 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 T_b= i x k_b x m which you can change to get

moles Sugar fructose = [delta T_b/k_b] 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

K_c = [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 \(\ce{[HgCl4^2- ]}\) in a solution prepared by adding 0.0200 mol of NaCl to 0.250 L of a 0.100-M HgCl₂ solution.

[HgCl₄^2-]=

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 → S^2- + H₃O⁺ K₂= 1.0 x 10^-19

	HS^-	H2O	H3O^+	S^2-
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

[S^2-]= 2.63 x 10^-12 M

Answer: The concentration for [HS-], [H₃O+],[H₂S],[S^2-] 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.

1. \(\ce{C}(s,\, \ce{graphite})+\ce{O2}(g)⟶\ce{CO2}(g)\)
2. \(\ce{O2}(g)+\ce{N2}(g)⟶\ce{2NO}(g)\)
3. \(\ce{2Cu}(s)+\ce{S}(g)⟶\ce{Cu2S}(s)\)
4. \(\ce{CaO}(s)+\ce{H2O}(l)⟶\ce{Ca(OH)2}(s)\)
5. \(\ce{Fe2O3}(s)+\ce{3CO}(g)⟶\ce{2Fe}(s)+\ce{3CO2}(g)\)
6. \(\ce{CaSO4⋅2H2O}(s)⟶\ce{CaSO4}(s)+\ce{2H2O}(g)\)

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 ΔG^o₂₉₈ you use the equation ΔG^o₂₉₈ = ΔH- TΔS

If ΔG^o₂₉₈ is negative, then the reaction is spontaneous, but if it is positive, then it is non spontaneous.

a. ΔG^o₂₉₈= (-393.51) - 298 x (0.00286) = -394.36 kJ/mole (Spontaneous)

b. ΔG^o₂₉₈= (180.5) - 298 x (0.0248) = 173.11 kJ/mol (Non Spontaneous)

c. ΔG^o₂₉₈= (-358.31) - 298 x (-0.11322) = -324.57 kJ/mol (Spontaneous)

d. ΔG^o₂₉₈= (-64.47) - 298 x (-0.0247) = -57.1094 kJ/mol (Spontaneous)

e. ΔG^o₂₉₈= (-24.72) - 298 x (0.00155) = -29.339 kJ/mol (Spontaneous)

f. ΔG^o₂₉₈= (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