Corrected exercises: stoichiometry, limiting reactant and yield (202-SN1-RE)

$M=6(12.01)+12(1.01)+6(16.00)=72.06+12.12+96.00=180.18$ g/mol.

$n=\dfrac{m}{M}=\dfrac{45.0}{180.18}=0.250$ mol.

Number of molecules $=n\times N_{A}=0.250\times 6.022\times 10^{23}=1.50\times 10^{23}$ molecules.

Each glucose molecule contains 6 carbon atoms, so $6\times 1.50\times 10^{23}=9.02\times 10^{23}$ carbon atoms.

$M(NH_{4}NO_{3})=2(14.01)+4(1.01)+3(16.00)=28.02+4.04+48.00=80.06$ g/mol. Mass percents: $\%N=\dfrac{28.02}{80.06}\times 100=35.0\%$; $\%H=\dfrac{4.04}{80.06}\times 100=5.05\%$; $\%O=\dfrac{48.00}{80.06}\times 100=59.96\%$. Sum: $35.0+5.05+59.96=100.0\%$. Watch for the classic pitfall: nitrogen appears twice in the formula, once in $NH_{4}^{+}$ and once in $NO_{3}^{-}$, so you must count 2 atoms.

Urea: $M=12.01+16.00+2(14.01+2\times 1.01)=60.07$ g/mol, so $\%N=\dfrac{28.02}{60.07}\times 100=46.6\%$. Ammonium sulfate: $M=2(14.01+4\times 1.01)+32.07+4(16.00)=132.17$ g/mol, so $\%N=\dfrac{28.02}{132.17}\times 100=21.2\%$. Ranking by decreasing nitrogen content: urea (46.6 %), ammonium nitrate (35.0 %), ammonium sulfate (21.2 %).

Each kilogram of urea supplies 0.466 kg of nitrogen. So you need $m=\dfrac{25.0}{0.466}=53.6$ kg of urea. The reasoning is exactly that of a purity: the total mass to buy is always larger than the mass of the element you are after.

Urea: one kilogram costs 0.85 dollars and supplies 0.466 kg of nitrogen, that is $\dfrac{0.85}{0.466}=1.82$ dollars per kilogram of nitrogen. Ammonium nitrate: $\dfrac{0.60}{0.350}=1.71$ dollars per kilogram of nitrogen. Ammonium nitrate is therefore cheaper, although it is clearly less rich in nitrogen. Conclusion: nitrogen content alone decides nothing. What counts is the cost per unit of useful element, that is, the ratio of the price to the mass percent. The same reasoning governs the choice of an ore in metallurgy.

In 100 g: 40.0 g of C, 6.71 g of H, 53.3 g of O. $n_{C}=\dfrac{40.0}{12.01}=3.33$ mol; $n_{H}=\dfrac{6.71}{1.01}=6.64$ mol; $n_{O}=\dfrac{53.3}{16.00}=3.33$ mol.

Divide by the smallest number of moles (3.33): $C : 1$; $H : \dfrac{6.64}{3.33}=1.99\approx 2$; $O : 1$. The empirical formula is $CH_{2}O$.

Empirical molar mass: $M(CH_{2}O)=12.01+2(1.01)+16.00=30.03$ g/mol. The factor is $\dfrac{180}{30.03}=6.0$, so the molecular formula is $C_{6}H_{12}O_{6}$: it is most likely glucose.

Balance carbon, then hydrogen, and oxygen last because it appears in two products. With 1 $C_{4}H_{10}$: 4 $CO_{2}$ and 5 $H_{2}O$, which requires $4\times 2+5=13$ oxygen atoms, that is 6.5 $O_{2}$. To get whole-number coefficients, double everything: $2\,C_{4}H_{10}+13\,O_{2}\rightarrow 8\,CO_{2}+10\,H_{2}O$. Check: C $8=8$, H $20=20$, O $26=16+10=26$.

Iron and carbon first, oxygen next: $Fe_{2}O_{3}+3\,CO\rightarrow 2\,Fe+3\,CO_{2}$. Check: Fe $2=2$, C $3=3$, O $3+3=6$ on the left and $3\times 2=6$ on the right.

Start with the element that appears only once on each side and in the heaviest formula. Phosphorus exists only as $P_{4}$ on the right, so you need 2 $Ca_{3}(PO_{4})_{2}$ to supply 4 P atoms. That forces 6 Ca, hence 6 $CaSiO_{3}$, hence 6 $SiO_{2}$. Carbon remains, and it all goes into CO, so count the oxygens. On the left: $2\times 8=16$ (from the phosphates) $+\ 6\times 2=12$ (from the silica), that is 28. On the right: $6\times 3=18$ (in $CaSiO_{3}$), which leaves 10 for the CO, so 10 CO and 10 C on the left. The balanced equation is $2\,Ca_{3}(PO_{4})_{2}+6\,SiO_{2}+10\,C\rightarrow 6\,CaSiO_{3}+10\,CO+P_{4}$.

The coefficients read as amounts of substance: 2 mol of butane react with 13 mol of dioxygen to give 8 mol of carbon dioxide and 10 mol of water. Since a mole always contains $6.022\times 10^{23}$ entities, the same ratio holds for molecules: 2 molecules for 13. And since, at the same temperature and pressure, equal volumes of gas contain the same number of molecules (Avogadro's law), the ratio also holds in volumes for the gaseous species: 2 L of butane consume 13 L of dioxygen and produce 8 L of carbon dioxide. On the other hand, it would be wrong to speak of 2 g for 13 g: the coefficients count entities, not masses, and the molar masses of butane (58.14 g/mol) and dioxygen (32.00 g/mol) are different. By mass, 2 mol of butane weigh 116 g and 13 mol of dioxygen weigh 416 g. This is precisely why you always convert to moles before applying a stoichiometric ratio.

On the left: 3 C, 8 H, 10 O. On the right: $3\,CO_{2}$ gives 3 C and 6 O; $4\,H_{2}O$ gives 8 H and 4 O, that is $6+4=10$ O. All three elements are balanced (3 C, 8 H, 10 O on each side).

$n(C_{3}H_{8})=\dfrac{22.0}{44.11}=0.499$ mol.

According to the equation, 1 mol of propane gives 3 mol of $CO_{2}$: $n(CO_{2})=3\times 0.499=1.50$ mol. So $m(CO_{2})=1.50\times 44.01=65.9$ g. This ratio of coefficients is the only bridge between two different substances: you never go directly from one mass to another, you first step down to moles.

At SATP, $V(CO_{2})=n\times 24.5=1.50\times 24.5=36.7$ L.

$n(N_{2})=\dfrac{28.0}{28.02}=1.00$ mol; $n(H_{2})=\dfrac{5.00}{2.02}=2.48$ mol.

For $N_{2}$: $\dfrac{1.00}{1}=1.00$. For $H_{2}$: $\dfrac{2.48}{3}=0.825$. The smallest ratio is that of dihydrogen, so $H_{2}$ is the limiting reactant: it runs out before the dinitrogen. Once the limiting reactant is used up, adding more excess reactant produces nothing more: the amount of product levels off on a plateau.

The product is calculated from the limiting reactant: $n(NH_{3})=n(H_{2})\times\dfrac{2}{3}=2.48\times\dfrac{2}{3}=1.65$ mol. Theoretical mass: $m(NH_{3})=1.65\times 17.04=28.1$ g.

Percent yield $=\dfrac{22.0}{28.1}\times 100=78.2\%$.

Each mole of $CO_{2}$ comes from one mole of carbon: $n_{C}=\dfrac{1.998}{44.01}=0.04540$ mol, so $m_{C}=0.04540\times 12.01=0.5452$ g. Each mole of water comes from two moles of hydrogen: $n_{H}=2\times\dfrac{0.818}{18.02}=2\times 0.04540=0.09079$ mol, so $m_{H}=0.09079\times 1.01=0.0917$ g.

Oxygen is obtained by difference: $m_{O}=1.000-0.5452-0.0917=0.3631$ g, so $n_{O}=\dfrac{0.3631}{16.00}=0.02269$ mol. You cannot go through the dioxygen consumed: the oxygen in the products comes both from the compound and from the $O_{2}$ of the combustion, and nothing allows the two contributions to be separated. This is why oxygen is always the one element determined by difference in a combustion analysis, and also why the method requires the exact mass of the starting sample.

Divide by the smallest number of moles, 0.02269: $C:\dfrac{0.04540}{0.02269}=2.00$; $H:\dfrac{0.09079}{0.02269}=4.00$; $O:1$. The empirical formula is therefore $C_{2}H_{4}O$, with a mass of $2(12.01)+4(1.01)+16.00=44.06$ g/mol. The factor is $\dfrac{88.1}{44.06}=2.00$: the molecular formula is $C_{4}H_{8}O_{2}$, that of ethyl acetate or butanoic acid, for example.

Combustion analysis never gives more than the empirical formula: the proportions are identical, so the empirical formula is still $C_{2}H_{4}O$. With $M=44.1$ g/mol, the factor is $\dfrac{44.1}{44.06}=1.00$ and the molecular formula is $C_{2}H_{4}O$ itself, that is acetaldehyde (ethanal). The two compounds have exactly the same percent composition and give the same analysis, but molar masses in the ratio 2. This is why an elemental analysis alone is never enough: you always need an independent measurement of the molar mass, by mass spectrometry or by a colligative property, to decide among $C_{2}H_{4}O$, $C_{4}H_{8}O_{2}$ and $C_{6}H_{12}O_{3}$.

$n(NH_{3})=\dfrac{34.0}{17.04}=1.995$ mol and $n(O_{2})=\dfrac{96.0}{32.00}=3.000$ mol. Divide by the coefficients: $\dfrac{1.995}{4}=0.4988$ for ammonia and $\dfrac{3.000}{5}=0.600$ for dioxygen. The smallest ratio is that of ammonia: $NH_{3}$ is the limiting reactant. Here the masses (34.0 g against 96.0 g) happen to point to the same reagent, but only by coincidence: a larger mass means nothing until you have divided by the molar mass, then by the coefficient. With other masses, the comparison of grams would name the wrong reactant.

The dioxygen consumed is $n(O_{2})_{cons}=1.995\times\dfrac{5}{4}=2.494$ mol. So $3.000-2.494=0.506$ mol remains, that is $m=0.506\times 32.00=16.2$ g of unconsumed dioxygen.

Theoretical amount of $NO$: the ratio is 4 to 4, so $n(NO)_{theor}=1.995$ mol. With a yield of 90.0 %: $n(NO)_{actual}=0.900\times 1.995=1.796$ mol. Second step, ratio 2 to 2: $n(NO_{2})_{theor}=1.796$ mol, and with an 85.0 % yield, $n(NO_{2})_{actual}=0.850\times 1.796=1.526$ mol. The mass obtained is $m=1.526\times 46.01=70.2$ g.

If both steps were perfect, each mole of ammonia would give one mole of $NO_{2}$, that is $1.995\times 46.01=91.8$ g. The overall yield is therefore $\dfrac{70.2}{91.8}\times 100=76.5\%$. We get back exactly the product of the two yields: $0.900\times 0.850=0.765$. This is general as soon as each step processes all of the product of the previous one: yields multiply, they are neither added nor averaged. An important industrial consequence: a third step at 80 % would bring the overall yield down to $0.765\times 0.80=0.612$, that is 61.2 %. Every extra step can only lower the overall yield, which is why a short synthesis is almost always preferable to a long one, even if each step taken alone looks good.

$NaCl+AgNO_{3}\rightarrow AgCl\downarrow+NaNO_{3}$ and $KCl+AgNO_{3}\rightarrow AgCl\downarrow+KNO_{3}$. The excess silver nitrate guarantees that all the chloride ions are precipitated. If there were too little, silver would be the limiting reactant, the mass of precipitate would measure the amount of silver added and not the amount of chloride present, and the whole analysis would collapse. This is the very principle of gravimetric analysis: the precipitating reagent is always in excess so that the weighing reflects only the species being sought.

Each mole of $AgCl$ contains one mole of chloride ion: $n(Cl^{-})=n(AgCl)=\dfrac{11.65}{143.32}=0.08129$ mol.

Each mole of $NaCl$, like each mole of $KCl$, releases one chloride ion, hence the first equation: $x+y=0.08129$. The total mass gives the second: $58.44x+74.55y=5.00$. Substitute $x=0.08129-y$ into the second: $58.44(0.08129-y)+74.55y=5.00$, that is $4.750+16.11y=5.00$, so $y=\dfrac{0.250}{16.11}=0.01549$ mol of $KCl$ and $x=0.08129-0.01549=0.06579$ mol of $NaCl$. Masses: $m(KCl)=0.01549\times 74.55=1.155$ g and $m(NaCl)=0.06579\times 58.44=3.845$ g, whose sum does give 5.00 g again. Mass percents: 76.9 % $NaCl$ and 23.1 % $KCl$.

If everything were $NaCl$: $n=\dfrac{5.00}{58.44}=0.08556$ mol of chloride, that is $0.08556\times 143.32=12.26$ g of precipitate. If everything were $KCl$: $n=\dfrac{5.00}{74.55}=0.06707$ mol, that is 9.61 g of precipitate. The measured value, 11.65 g, does fall between these two bounds, which validates the result. It is also clearly closer to the upper bound, so the mixture must be largely dominated by $NaCl$, as the 76.9 % found confirms. The physical reason is simple: for equal mass, $NaCl$ contains more chloride because its molar mass is lower. This test of the bounds costs only two divisions and immediately detects a sign error or an inversion in the solution of the system.

For pure $CaCO_{3}$: $n=\dfrac{50.0}{100.09}=0.500$ mol, and since 1 mol of $CaCO_{3}$ gives 1 mol of $CaO$, $m(CaO)=0.500\times 56.08=28.0$ g.

Start from the actual mass of $CaO$: $n(CaO)=\dfrac{22.0}{56.08}=0.392$ mol. This amount comes from as many moles of $CaCO_{3}$: $m(CaCO_{3})=0.392\times 100.09=39.3$ g. Purity $=\dfrac{39.3}{50.0}\times 100=78.5\%$.

$n(CO_{2})=n(CaO)=0.392$ mol. Mass: $m(CO_{2})=0.392\times 44.01=17.3$ g. Volume at SATP: $V=0.392\times 24.5=9.61$ L.

$m(CaO)+m(CO_{2})=22.0+17.3=39.3$ g, which does equal the mass of $CaCO_{3}$ decomposed: mass is conserved in the reaction. The difference $50.0-39.3=10.7$ g did not vanish: it is the inert impurities, which never reacted and remain in the solid residue with the $CaO$. This is the key distinction of the problem: conservation of mass applies only to the part that reacts, and purity separates that part from the rest.