Corrected practice final exam in general chemistry (202-SN1-RE)

Sulfur: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{4}$, that is 6 valence electrons in the $n=3$ shell. Sulfur can exceed the octet because its valence shell contains vacant $3d$ orbitals, energetically accessible, which allow up to 12 electrons around it. Oxygen is in period 2: its valence shell has only the $2s$ and $2p$ orbitals, at most four boxes, so eight electrons and not one more. The octet is not an arbitrary rule, it is a count of the available boxes.

$SF_{4}$: $6+4(7)=34$ valence electrons, that is 17 pairs. Four S-F bonds use up 4 pairs, each fluorine keeps 3, that is 12, and 1 LONE pair is left on sulfur. Sulfur therefore carries 5 domains: TRIGONAL BIPYRAMIDAL arrangement, $sp^{3}d$ hybridization. The lone pair sits in an equatorial position, where it is least crowded, and the molecular geometry is SEESAW.

$XeF_{4}$: $8+4(7)=36$ electrons, that is 18 pairs. Four bonds, 12 pairs on the fluorines, which leaves 2 lone pairs on xenon, so 6 domains: OCTAHEDRAL arrangement, $sp^{3}d^{2}$ hybridization. The two lone pairs take opposite positions, 180° apart, to minimize their repulsion, and the four fluorines end up in one plane: the geometry is SQUARE PLANAR.

Polarity does not depend on the presence of lone pairs but on the SYMMETRY of the whole. In square planar $XeF_{4}$, the four Xe-F bond dipoles cancel in pairs and the two opposite lone pairs cancel as well: the molecule is NONPOLAR despite four very polar bonds. In seesaw $SF_{4}$ the arrangement is asymmetric: neither the bonds nor the lone pair find an opposite partner, the dipoles add up partially and the molecule is POLAR. Two $AF_{4}$ molecules with lone pairs, two opposite answers: only the geometry decides.

$CF_{4}$ is tetrahedral and therefore nonpolar, $XeF_{4}$ is nonpolar too, $SF_{4}$ is polar. Between the two nonpolar molecules only dispersion forces act, and they grow with molar mass and number of electrons: $XeF_{4}$ (207 g/mol) far outweighs $CF_{4}$ (88 g/mol). $SF_{4}$ (108 g/mol) adds dipole-dipole interactions to its dispersion. The increasing order is therefore $CF_{4}$, then $SF_{4}$, then $XeF_{4}$, and the observed values are indeed $-128$ °C, $-38$ °C and $117$ °C. Note that the heaviest molecule beats the polar one: dispersion, often presented as the weakest force, becomes dominant as soon as the electron cloud is large.

In 100 g: $n_{S}=\frac{40.05}{32.07}=1.249$ mol and $n_{O}=\frac{59.95}{16.00}=3.747$ mol. Dividing by the smaller: $\frac{1.249}{1.249}=1.000$ and $\frac{3.747}{1.249}=3.000$. The empirical formula is therefore $SO_{3}$.

The mass of the empirical formula is $32.07+3(16.00)=80.07$ g/mol, exactly the measured molar mass. The multiplying factor is 1 and the molecular formula is also $SO_{3}$. Recall why both steps are necessary: elemental analysis gives ONLY ratios, it cannot tell $SO_{3}$ from $S_{2}O_{6}$; it is the molar mass, obtained by a completely independent measurement, that removes the ambiguity.

$6+3(6)=24$ valence electrons, that is 12 pairs. Whether we place one S=O double bond and two single bonds with formal charges, or three double bonds with an expanded octet, we get in both conventions THREE equivalent resonance forms, obtained by cyclic permutation of the double bond. Sulfur carries 3 domains and no lone pair: the geometry is TRIGONAL PLANAR, the hybridization $sp^{2}$, and the three bond dipoles, at 120° from one another, cancel exactly: the molecule is NONPOLAR. The real molecule is perfectly symmetric, with three strictly identical bonds intermediate between single and double, which no single Lewis structure can represent.

$\Delta EN=3.44-2.58=0.86$: the bond is POLAR COVALENT, well below the threshold of 1.7 often used for ionic bonding. Yes, the acidic character was predictable: sulfur is a NONMETAL, and nonmetal oxides are acidic oxides, which react with water to give an acid, here $SO_{3}+H_{2}O\rightarrow H_{2}SO_{4}$. This is the general rule of periodicity: the acid-base character of an oxide follows the metallic character of the element, and it shifts gradually from basic on the left of the table to acidic on the right.

Sodium is an alkali metal, with electronegativity 0.93: $\Delta EN=3.44-0.93=2.51$, far above 1.7, so the bond is IONIC. $Na_{2}O$ is therefore an ionic solid, made of a lattice of $Na^{+}$ and $O^{2-}$ ions held together by powerful electrostatic forces in all three directions: its melting point is very high, 1132 °C, whereas $SO_{3}$, a molecular solid held together only by weak intermolecular forces, melts at 17 °C. And as the oxide of a metal, it is BASIC: $Na_{2}O+H_{2}O\rightarrow 2NaOH$. The same element, oxygen, thus gives two compounds with opposite properties depending on its partner, and everything can be read from a single quantity, the electronegativity difference.

$\Delta EN(H-F)=3.98-2.20=1.78$ and $\Delta EN(H-Cl)=3.16-2.20=0.96$. The H-F bond is therefore much more polar, and above all fluorine is one of only three elements, with nitrogen and oxygen, small enough and electronegative enough to take part in a HYDROGEN BOND. $HF$ therefore forms between its molecules a network of hydrogen bonds of some twenty kJ per mole, whereas $HCl$ has only ordinary dipole-dipole interactions of a few kJ per mole. This network has to be broken to make $HF$ boil, hence the 105-degree gap.

Because dispersion forces, although real and indeed stronger in $HCl$, remain an order of magnitude below hydrogen bonding. The gain brought by the extra molar mass, a few kJ per mole, does not make up for the loss of the hydrogen-bond network, several tens. The same logic makes water a liquid at room temperature while $H_{2}S$, twice as heavy, is a gas: within a group, the anomaly of the first member always signals hydrogen bonding.

The two compounds simply do not have the same architecture. Carbon dioxide is a MOLECULAR solid: small, discrete $CO_{2}$ molecules, held to one another by weak dispersion forces since they are nonpolar. Silicon dioxide is a NETWORK COVALENT solid: there is no $SiO_{2}$ molecule, only an infinite three-dimensional network in which each silicon is bonded to four oxygens by covalent bonds. The reason is that silicon, being larger, forms pi bonds poorly and prefers four single bonds to two double bonds. Subliming the first therefore only requires pulling apart molecules that already exist; melting the second requires BREAKING covalent bonds.

$\frac{452}{25}\approx 18$: about eighteen times more energy is needed, and the calculation still underestimates the gap since melting quartz breaks several of them per formula unit. The ratio says the essential thing: subliming carbon dioxide is a PHYSICAL phenomenon, in which no chemical bond is touched and the molecules stay intact, whereas melting silica is closer to a transformation in which the chemical bond itself is attacked. Confusing the two energy scales, a few tens against a few hundred kJ per mole, is the error that leads to absurd boiling-point predictions.

$SiC$: network covalent solid, insulator as a solid and very high melting point, about 2700 °C, which makes it an abrasive and a furnace material. $I_{2}$: nonpolar molecular solid, insulator, low melting point at 114 °C, and it sublimes easily. $NaCl$: ionic solid, an INSULATOR as a solid since the ions are immobilized, but a conductor once molten or dissolved, and a high melting point at 801 °C. $Cu$: metallic solid, a CONDUCTOR as a solid thanks to its delocalized electrons, high melting point at 1085 °C, and malleable besides, unlike the ionic solid, which cleaves. Remember the question that sorts everything out: what is broken when you melt, forces between molecules or bonds within a network?

$CH_{4}+2O_{2}\rightarrow CO_{2}+2H_{2}O$ and $2C_{2}H_{6}+7O_{2}\rightarrow 4CO_{2}+6H_{2}O$, which is better written $C_{2}H_{6}+\frac{7}{2}O_{2}\rightarrow 2CO_{2}+3H_{2}O$ so as to reason per mole of hydrocarbon.

Let $x$ be the moles of methane and $y$ those of ethane. The total gives $x+y=0.500$. Each methane supplies one carbon and each ethane supplies two, hence $x+2y=0.700$. Subtracting: $y=0.200$ mol of ethane and so $x=0.300$ mol of methane. Note how neat the setup is: the second equation is a balance of carbon ATOMS, the first a balance of molecules, and it is their difference that isolates the ethane.

In moles: $\frac{0.300}{0.500}=60.0$% methane and 40.0% ethane. By mass: $m_{CH_{4}}=(0.300)(16.04)=4.81$ g and $m_{C_{2}H_{6}}=(0.200)(30.07)=6.01$ g, for a total of $10.83$ g, that is $\frac{4.81}{10.83}=44.4$% methane by mass. The two values differ because one mole of ethane weighs almost twice as much as one mole of methane: ethane is a minority by NUMBER but a majority by MASS. This is the most common confusion when reading a composition, and it is cleared up only if the statement specifies what it is counting.

Water: $2x+3y=0.600+0.600=1.200$ mol. Dioxygen: $2x+\frac{7}{2}y=0.600+0.700=1.300$ mol, that is a volume of $(1.300)(22.4)\approx 29.1$ L at STP. Note that more than two and a half times the volume of the mixture is needed in oxygen, which is why a poorly ventilated gas appliance chokes and produces carbon monoxide.

Dalton's law: in a mixture of ideal gases, each component exerts the pressure it would exert if it alone occupied the whole volume, and the partial pressure is the MOLE fraction times the total pressure. $P_{CH_{4}}=(0.600)(2.00)=1.20$ atm and $P_{C_{2}H_{6}}=(0.400)(2.00)=0.800$ atm, whose sum indeed gives back 2.00 atm. The law is justified by the kinetic model: the molecules of an ideal gas do not interact, each one strikes the walls independently of the others, and the pressure is merely a count of collisions. That is why the partial pressure follows the mole fraction and never the mass fraction.

Because the reaction CONSUMES four moles of gas to produce only two: the total number of molecules decreases. At constant volume and temperature, the ideal gas law gives $P=\frac{nRT}{V}$, where everything is fixed except $n$: the pressure is therefore directly PROPORTIONAL to the total amount of gas. Here, $\frac{8.00}{4.00}=2.00$ atm per mole, and this constant will be used throughout the problem. The pressure gauge thus becomes a molecule counter.

$n_{final}=\frac{6.80}{2.00}=3.40$ mol, against 4.00 at the start: $0.600$ mol of gas has disappeared. Now each unit of extent $\xi$ makes $1+3-2=2$ net moles disappear, so $2\xi=0.600$ and $\xi=0.300$ mol.

Table: $N_{2}$ goes from 1.00 to $1.00-0.300=0.700$ mol; $H_{2}$ goes from 3.00 to $3.00-3(0.300)=2.10$ mol; $NH_{3}$ goes from 0 to $2(0.300)=0.600$ mol. Total: $0.700+2.10+0.600=3.40$ mol, consistent with b), which validates the table. Mass of ammonia: $(0.600)(17.03)\approx 10.2$ g.

The two reactants are introduced in exactly the stoichiometric proportions, $1$ to $3$: neither is in shortage, they would be used up together. The maximum extent would therefore be $\xi_{max}=1.00$ mol, which would give $2.00$ mol of ammonia. Yield: $\frac{0.600}{2.00}=30.0$%. This mediocre figure is not due to an impurity or a leak: the reaction is an EQUILIBRIUM, and it stops by itself long before everything has been consumed.

Multiplying each amount by $2.00$ atm/mol: $P_{N_{2}}=1.40$ atm, $P_{H_{2}}=4.20$ atm and $P_{NH_{3}}=1.20$ atm, whose sum is $6.80$ atm, the measured pressure. The engineer's proposal is SOUND, for a reason that can be read from the stoichiometry: compressing the mixture shifts the equilibrium toward the side with the FEWEST gas molecules, that is toward ammonia, four moles becoming two. This is exactly the principle exploited by the industrial Haber-Bosch process, run at around 200 atm. The quantitative treatment of this shift belongs to the chapter on chemical equilibrium, in 202-SN2-RE, but the stoichiometry alone is already enough to predict its direction.

$CaCO_{3}+2HCl\rightarrow CaCl_{2}+H_{2}O+CO_{2}$. At STP, one mole of gas occupies 22.4 L, so $n_{CO_{2}}=\frac{0.112}{22.4}=5.00\times 10^{-3}$ mol.

The equation requires two moles of acid per mole of carbon dioxide: $n_{HCl}=2(5.00\times 10^{-3})=1.00\times 10^{-2}$ mol. Since the carbonate is in excess, all the acid has reacted, and $c=\frac{1.00\times 10^{-2}}{0.0250}=0.400$ mol/L. Note the principle of the method: a liquid is titrated by measuring a VOLUME OF GAS, which avoids any color indicator and any burette.

$n_{CaCO_{3}}=n_{CO_{2}}=5.00\times 10^{-3}$ mol, so $m=(5.00\times 10^{-3})(100.09)\approx 0.500$ g. Only this mass reacted; the rest of the solid, in excess, stayed at the bottom.

$c'=c\frac{V}{V'}=(0.400)\frac{25.0}{250.0}=4.00\times 10^{-2}$ mol/L, ten times less concentrated. What was conserved during the dilution is the AMOUNT OF SOLUTE: only solvent was added, so $n=cV$ stays constant, and that is where the relation $cV=c'V'$ comes from. The volume changed, and the concentration with it.

A gas collected over water becomes saturated with water vapor, which contributes to the measured pressure. Dalton's law therefore requires $P_{CO_{2}}=752-19.8=732.2$ torr, that is $\frac{732.2}{760}=0.9634$ atm. The volume becomes $V=\frac{nRT}{P}=\frac{(5.00\times 10^{-3})(0.08206)(295.15)}{0.9634}\approx 0.126$ L, about 126 mL against 112 mL at STP. The measured volume is therefore LARGER, for two cumulative reasons: the temperature is higher than at 0 °C, and the presence of water vapor lowers the partial pressure of the carbon dioxide. Forgetting the water-vapor correction would overestimate the amount of gas by nearly 3%, which is enough to spoil a titration.

The bonds BROKEN absorb energy: four C-H bonds in methane and two O=O bonds in dioxygen, that is $4(413)+2(498)=1652+996=2648$ kJ. The bonds FORMED give it back: two C=O bonds in carbon dioxide and four O-H bonds in the two water molecules, that is $2(799)+4(467)=1598+1868=3466$ kJ. Balance: $\Delta H=2648-3466=-818$ kJ/mol. The negative sign confirms an exothermic reaction, as expected for a combustion.

$\Delta H=\sum\Delta H_{f}(\text{products})-\sum\Delta H_{f}(\text{reactants})=\left[-393.5+2(-241.8)\right]-\left[-74.6+2(0)\right]=-877.1+74.6=-802.5$ kJ/mol. Recall that the enthalpy of formation of an element in its standard state, here dioxygen, is zero by convention.

Gap: $818-802.5=15.5$ kJ/mol, that is $\frac{15.5}{802.5}\approx 1.9$%. The origin lies in the very nature of bond energies: they are AVERAGES established over a large number of different molecules. A C-H bond is not worth exactly the same in methane, in ethane or in benzene, and the table gives a compromise value. Enthalpies of formation, on the other hand, are measured on the PRECISE compound in question, by calorimetry: they are therefore far more reliable. Bond energies keep all their value when no enthalpy of formation is tabulated, notably for an exotic compound or a reaction intermediate, and an estimate within 2% is still excellent for choosing between two synthetic routes.

$\Delta H=\left[-393.5+2(-285.8)\right]+74.6=-965.1+74.6=-890.5$ kJ/mol, that is $88.0$ kJ more than the value with gaseous water. This difference is exactly $2\times 44.0$ kJ/mol, that is twice the enthalpy of VAPORIZATION of water, for the two moles produced. It makes physical sense: if the water stays vapor, the reaction has not recovered the latent heat of condensation; if it liquefies, that heat is given back on top. The physical state of the products is therefore never a mere detail of notation in a thermochemical equation.

The two numbers refer to different bases. The efficiency of 92% is calculated on the HIGHER heating value, that of d), which assumes the water in the flue gas is condensed. The efficiency of 102% is calculated on the LOWER heating value, that of b), which assumes the water leaves as vapor. A condensing boiler recovers precisely part of those 88 kJ that the lower convention has already removed from the denominator: it therefore gives back more than the reference credits it with, and the quotient exceeds 100%. No principle is violated, only the convention is badly chosen, and advertising is quite happy with that. It is an excellent example of an exact figure that misleads.

If the solution cools, it has GIVEN heat to the dissolving process: this is therefore ENDOTHERMIC and $\Delta H_{\text{solution}}>0$. Mass of solution: $200.0+11.7=211.7$ g. Heat absorbed: $Q=mc\Delta T=(211.7)(4.18)(0.94)\approx 832$ J. Per amount dissolved: $\Delta H=\frac{+832}{0.200}\approx +4.2\times 10^{3}$ J/mol, that is about $+4.2$ kJ/mol. Note that the heat is related to the TOTAL MASS of solution, solid included, and not to that of the solvent alone.

Dissolving breaks down into two conceptual steps. First the ions must be TORN out of their crystal lattice, which costs the lattice energy, $+787$ kJ/mol: this is the price of destroying a very stable electrostatic stacking. The free ions are then SURROUNDED by water molecules, whose poles turn toward them, which gives back the enthalpy of hydration, $-783$ kJ/mol. Sum: $+787-783=+4$ kJ/mol, in excellent agreement with the measurement of a). Two enormous numbers, almost equal, whose difference is tiny: this is exactly why the dissolution of a salt can be slightly exothermic or slightly endothermic, depending on whether hydration wins by a little or not.

Because enthalpy is not the only driving force of a change. Going from a perfectly ordered crystal to ions scattered at random in the solvent considerably increases the DISORDER of the system, that is its entropy, and it is this gain that largely makes up for the enthalpy cost of 4 kJ/mol. An endothermic change can therefore be spontaneous as soon as the entropy gain is large enough, which the full criterion, the free energy, will let you quantify in the next course. It is also the principle of commercial instant cold packs, which dissolve a salt chosen for its clearly positive enthalpy of solution.

Molality: $\frac{0.200\text{ mol}}{0.2000\text{ kg}}=1.00$ mol/kg. With complete dissociation, each $NaCl$ unit gives two particles, so $i=2.00$ and $\Delta T_{f}=iK_{f}m=(2.00)(1.86)(1.00)=3.72$ °C. The solution freezes at $-3.72$ °C. This is the principle of road salting, and its limit is immediately visible: a considerable molality would be needed to go much lower, and in practice salt becomes ineffective below about $-10$ °C.

$\Delta T_{f}=(1.87)(1.86)(1.00)\approx 3.48$ °C, so freezing at $-3.48$ °C, a quarter of a degree less than predicted. The discrepancy comes from the dissociation not being quite complete from the OSMOTIC point of view: in solution, some of the $Na^{+}$ and $Cl^{-}$ ions momentarily move in pairs, because of their mutual electrostatic attraction, so that the number of really independent particles is slightly below twice the number of units dissolved. The effect grows as the concentration increases, since the ions meet more often, and it is more pronounced the higher the charges: for $MgSO_{4}$, where two doubly charged ions attract each other four times as much, the measured factor falls well below 2. Colligative properties count particles, and they count what is really free.

$P=\frac{nRT}{V}=\frac{(1.00)(0.08206)(300)}{0.500}\approx 49.2$ atm.

$P=\frac{nRT}{V-nb}-\frac{an^{2}}{V^{2}}=\frac{24.618}{0.500-0.0427}-\frac{3.59}{0.250}=53.8-14.4\approx 39.5$ atm. The term $nb$ subtracted from the volume stands for the OWN VOLUME of the molecules: they are not point-like, the space really available for moving about is smaller than the container, which INCREASES the pressure. The term $\frac{an^{2}}{V^{2}}$ stands for the ATTRACTIVE FORCES: a molecule about to strike the wall is held back by its neighbors left behind, so it hits less hard, which DECREASES the pressure. The two corrections act in opposite directions, and it is their competition that makes the equation interesting.

$Z=\frac{PV}{nRT}=\frac{(39.47)(0.500)}{24.618}\approx 0.802$. Since $Z<1$, it is the ATTRACTIVE forces that win under these conditions: the gas is more compressible than an ideal gas. The ideal gas law predicts 49.2 atm where reality gives 39.5: it OVERESTIMATES the pressure by $\frac{49.24-39.47}{39.47}\approx 25$%, which is considerable and disqualifies the ideal model as soon as one approaches a hundred atmospheres or approaches liquefaction. At very high pressure, on the other hand, the volume of the molecules eventually dominates and $Z$ climbs back above 1.

The parameter $a$ directly measures the strength of the intermolecular attractions: $3.59$ for carbon dioxide, $2.25$ for methane, $0.0341$ for helium, a ratio of 105 between the extremes. The order of the critical temperatures follows exactly the same ranking, 304 K, 191 K and then 5.2 K, and this is no coincidence: the critical temperature is the one above which thermal agitation is too violent for the attractions to still condense the gas. The stronger the forces, the higher one must go to overcome them. A single parameter of the equation of state thus predicts a property that seemed unrelated.

Because helium is the least polarizable atom there is: only two electrons, tightly held, in a tiny and perfectly spherical cloud. Its only intermolecular forces are dispersion forces of extreme weakness, which is reflected in its $a$ a hundred times smaller than that of carbon dioxide. Its critical temperature of $5.2$ K means that NO pressure, however high, can liquefy it above that temperature: it was therefore necessary first to learn how to go below 5 K, which technology allowed only in 1908, with Kamerlingh Onnes. This same work led him, three years later, to discover superconductivity, and liquid helium remains today the coolant of medical imaging magnets.

All the carbon of the compound ends up in the carbon dioxide: $n_{C}=\frac{4.40}{44.01}=0.1000$ mol, that is $m_{C}=(0.1000)(12.01)=1.20$ g. All the hydrogen ends up in the water, at two atoms per molecule: $n_{H}=2\frac{2.70}{18.02}=0.2997$ mol, that is $m_{H}=0.302$ g. The oxygen, however, can NOT be deduced from the products, because they contain two indistinguishable sources of it: that of the compound and that of the dioxygen used in the combustion, supplied in excess. It is therefore obtained by DIFFERENCE: $m_{O}=2.30-1.20-0.302=0.797$ g, that is $n_{O}=\frac{0.797}{16.00}=0.0498$ mol.

The three amounts are divided by the smallest, $0.0498$: carbon $\frac{0.1000}{0.0498}\approx 2.01$, hydrogen $\frac{0.2997}{0.0498}\approx 6.02$, oxygen $1.00$. The empirical formula is therefore $C_{2}H_{6}O$.

$M=\frac{dRT}{P}=\frac{(1.50)(0.08206)(373.15)}{1.00}\approx 45.9$ g/mol. Now the mass of the empirical formula is $2(12.01)+6(1.008)+16.00=46.07$ g/mol: the ratio is 1.00, so the molecular formula is identical to the empirical formula, $C_{2}H_{6}O$. Note that the two measurements are completely independent, one chemical and the other physical, which makes the conclusion robust.

The two compounds are ETHANOL, $CH_{3}-CH_{2}-OH$, which boils at 78 °C, and METHOXYMETHANE, or dimethyl ether, $CH_{3}-O-CH_{3}$, which boils at $-24$ °C. They are CONSTITUTIONAL ISOMERS: same molecular formula, same molar mass, hence practically identical dispersion forces. The whole difference comes down to one detail of connectivity: ethanol carries a hydrogen bonded directly to oxygen, so it can form HYDROGEN BONDS between its molecules, whereas in the ether the two hydrogens next to the oxygen are carried by carbons, far too weakly electronegative for that. A hydrogen bond is worth about twenty kJ/mol against a few kJ/mol for dispersion, hence more than a hundred degrees of gap. It is the clearest illustration of the fact that in chemistry, structure takes precedence over molecular formula.

It is ethanol, because its $-OH$ group lets it form hydrogen bonds WITH water and not only with its own kind; the ether is only moderately soluble in it. Around the oxygen there are two bonds and two lone pairs, that is four domains: the hybridization is $sp^{3}$ and the expected angle is that of the tetrahedron, $109.5°$. The real angle is a little smaller, about $105°$, because LONE pairs take up more room than a bond: held by only one nucleus, they spread out more and squeeze the angle between the two bonds. It is the same correction that takes water from 109.5 to 104.5 degrees.