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# Stoichiometry and empirical formulae

Reaction stoichiometry gives us the tools in chemistry to figure out the relative amounts of reactants and products in a chemical reaction. We can also use stoichiometric tools to figure out the number of atoms present in a compound or amount of substance or solute in a solution, respectively called composition and solution stoichiometry. We will focus our discussion on reaction stoichiometry.
There are, however, tools common to all types of stoichiometry, which are dimensional analysis, Avogadro’s number, and molecular weight. Dimensional analysis allows conversion from unit to unit, molecular weight between mass and moles, and Avogadro’s number between number of atoms, molecules, or ions and moles. Molecular weight is sometimes called molar mass, and for our purposes, we will consider them synonymous.

## Calculating mole fraction

Suppose we were asked to figure out the mole fraction of cadmium nitrate in a 12% (wt-wt) aqueous solution. 12% (wt-wt) tells us that there is 12% in weight of the solute in 100 g of the solution. For a 12% (wt-wt) aqueous Cd(NO${}_{3}$)${}_{2}$ solution, let’s suppose we have 100 grams of the substance. There would be 12 g of the solute and 88 g of water.
$\begin{array}{rl}Cd\left(N{O}_{3}{\right)}_{2}:& 112+\left(14×2\right)+\left(16×6\right)\\ \\ =& 112+28+96\\ \\ =& 236g/mol\end{array}$
${H}_{2}O:16+\left(1×2\right)=18g/mol$
Once we have figured the molecular weights, we should divide each of the masses with its respective molecular weight to obtain an amount of moles.
$\frac{12g}{236g/mol}$ $\approx$ $\frac{12}{240}$ $=$ $\frac{1}{2x{10}^{1}}$ $=$ $0.05molCd\left(N{O}_{3}{\right)}_{2}$
$\frac{88g}{18g/mol}$ $\approx$ $\frac{90}{18}$ $=$ $5mol{H}_{2}O$
Next, to find out the mole fraction, we can put the moles of substance of interest in the numerator and the sum of the moles of the two substances in the denominator:
$\frac{0.05molCd\left(N{O}_{3}{\right)}_{2}}{5mol{H}_{2}O+0.05molCd\left(N{O}_{3}{\right)}_{2}}$ $=$ $\frac{0.05}{5.05}$ $\approx$ $\frac{0.05}{5.00}$ $=$ $0.01\mathrm{%}$
We may encounter answer choices that differ in power of 10, like 0.01, 0.1, 0.05, and 5.0, but we will select the answer choice closest to our calculated answer. Since there was very little rounding, we should expect to see the answer of 0.01%.

## Calculating the number of particles

Suppose we are asked to figure out the approximate number of ammonium ions in 6 mL of a (NH${}_{4}$)${}_{3}$PO${}_{4}$ ∙3H${}_{2}$O solution prepared by dissolving 4.0 g of ammonium phosphate trihydrate in enough water to make 25 mL of the solution. To convert between moles and particles, we must use Avogadro’s number, which has the value of 6.023 x 10${}^{2}$${}^{3}$ particles/mole.
The goal is always to convert into moles, which is the language of stoichiometry. First, we should figure out the molar mass of ammonium phosphate trihydrate:
${H}_{1}$${}_{8}$${N}_{3}$${O}_{7}$$P:$ $\left(18×1\right)$ $+$ $\left(14×3\right)$ $+$ $\left(16×7\right)$ $+$ $31$ $=$ $18$ $+$ $42$ $+$ $112$ $+$ $31$ $=$ $203$ $g/mol$
Next, we should figure out the moles of ammonium phosphate trihydrate we have on hand:
$mole$ $=$ $\frac{mass}{molecularweight}$ $=$ $\frac{4.0g}{203g/mol}$ $=$ $\frac{4.0}{20.3×{10}^{1}}$ $=$ $\frac{1}{5}$ X ${10}^{-}$${}^{1}$ $=$ $0.02$ $moles$
Since we are asked to find out how many are in 6 mL out of 25 mL:
$\frac{0.02moles}{25mL}$ $=$ $\frac{x}{6mL}$ $=>$ $25$ $x$ $=$ $6\left(0.02\right)$ $=>$ $4x$ $=$ $0.02$ $=>$ $x$ $=$ $0.005$ $moles$
Next we must multiply the moles by Avogadro’s number to find the number of molecules or ions:
$molecules$ $=$ $moles$ $\left(Avogadr{o}^{\prime }s$ $number\right)$ $=$ $0.005$ $mol×6.023×{10}^{2}$${}^{3}$ $molecules/mol$ $=$ $5×{10}^{-}$${}^{3}$ $×6.0×{10}^{2}$${}^{3}$ $=$ $30.0×{10}^{2}$${}^{0}$ $=$ $3×{10}^{2}$${}^{1}$ $molecules$
Finally, we must consider that there are three ammonium ions that dissociate into solution when dissolved:
$3×{10}^{2}$${}^{1}$ $molecules$ $×$ $3$ $ions/molecule$ $=$ $9×{10}^{2}$${}^{1}$ $ions$ $N{H}_{4}+$
The actual answer with 2 significant figures is 8.5 x 10${}^{2}$${}^{1}$, which rounds up to 9 x 10${}^{2}$${}^{1}$. We may encounter answer choices like 3 x 10${}^{2}$${}^{1}$, 9 x 10${}^{2}$${}^{2}$, 3 x 10${}^{2}$${}^{0}$, and 9 x 10${}^{2}$${}^{1}$. Since there is once again little rounding, our calculated answer should closely match up with the answer choices.

## Solving across the equation using stoichiometric coefficients

Suppose we are given the following unbalanced equation:
$A{s}_{\left(}$${}_{s}$${}_{\right)}$ $+$ $NaO{H}_{\left(}$${}_{a}$${}_{q}$${}_{\right)}$ $->$ $N{a}_{3}$$As{O}_{3}$${}_{\left(}$${}_{a}$${}_{q}$${}_{\right)}$ $+$ ${H}_{2}$${}_{\left(}$${}_{g}$${}_{\right)}$
and we are asked to calculate how many milliliters (mL) of a 4 molar (M) solution of sodium hydroxide are required to produce 4.5 liters (L) of hydrogen gas.
The first step is to balance the equation, and there are multiple methods to solve this, but we ultimately get the following:
$2$ $A{s}_{\left(}$${}_{s}$${}_{\right)}$ $+$ $6$ $NaOH$${}_{\left(}$${}_{a}$${}_{q}$${}_{\right)}$ $->$ $2N{a}_{3}$$As{O}_{3}$${}_{\left(}$${}_{a}$${}_{q}$${}_{\right)}$ $+$ $3$ ${H}_{2}$${}_{\left(}$${}_{g}$${}_{\right)}$
Let’s recall that 1 mole of an ideal gas at standard temperature and pressure (STP) has a volume of 22.4 L:
$4.5$ $L$ ${H}_{2}$ $\left($ $\frac{1mol{H}_{2}}{22.4L{H}_{2}}$ $\right)$ $\left($ $\frac{6molNaOH}{3mol{H}_{2}}$$\right)$ $\left($ $\frac{1LNaOH}{4molNaOH}$ $\right)$ $=$ $0.10$ $L$ $or$ $100$ $mL$ $NaOH$
We start with 4.5 L divided by 22.4 L, which is ⅕ mol H${}_{2}$. The reaction stoichiometry indicates that for every 3 moles of hydrogen gas we need 6 moles of sodium hydroxide, so we need mol NaOH. However, we need to put that in terms of our 4 molar solution, so we divide by 4 to get 2/20, which is 0.10 L or 100 mL of the NaOH solution.

## Empirical formulae

Empirical formulae represent the simplest ratio between the elements in a compound, while the actual number of atoms in a compound represent the molecular formula. The molecular formula may be the empirical formula or some multiple of the empirical formula. For instance, formaldehyde and glucose share the same empirical formula, but have different molecular formula, where formaldehyde is CH${}_{2}$O and glucose is C${}_{6}$H${}_{1}$${}_{2}$O${}_{6}$. To convert from empirical to molecular formula, we need the molecular weight of the compound. To find the empirical formula, we need the mass composition of a compound given as either grams or mass percentages.
Supposed we were asked to find the empirical formula of a compound that has a mass composition of 44.7% carbon, 7.5% hydrogen, and 47.8% oxygen.
$C$ $:$ $\frac{44.7g}{12g}$ $\approx$ $\frac{45}{12}$ $=$ $3$ $\frac{9}{12}$ $=$ $3.75$ $or$ $3$ $\frac{3}{4}$
$H$ $:$ $\frac{7.5g}{1g}$ $=$ $7.5$ $7$ $\frac{1}{2}$
$O$ $:$ $\frac{47.8g}{16g}$ $\approx$ $\frac{48}{16}$ $=$ $3$
Here we can be more exact in our calculation. We can round up 44.7 to 45 and 47.8 to 48, but the rest should be calculable in our heads. To get whole number ratios, multiply by 4 to get rid of the denominator for the following values:
$C$ $:$ $3.75×4$ $=$ $15$
$H$ $:$ $7.5×4$ $=$ $30$
$O$ $:$ $3×4$ $=$ $12$
We can obtain the empirical (and molecular) formula by dividing each of the values by 3 to obtain C5H10O4, which is the formula for the following compound deoxyribose:

## Transaminase reaction

Suppose we encounter a passage about the transaminase reaction (Reaction 1) that is catalyzed by aspartate transaminase (AST) in our liver.
${C}_{4}$${H}_{7}$$N{O}_{4}$ $+$ ${C}_{5}$${H}_{6}$${O}_{5}$ $=>$ ${C}_{5}$${H}_{9}$$N{O}_{4}$ $+$ ${C}_{4}$${H}_{4}$${O}_{5}$
Reaction 1
A researcher performs an experiment preparing the enzyme at a concentration of one mg/ml in 0.1 M potassium phosphate, pH 7.4. Immediately prior to use, she dilutes further in this buffer to a concentration of 0.05-0.25 u/ml. 3.3 g of aspartate (C${}_{4}$H${}_{7}$NO${}_{4}$) and 7.3 g of α-ketoglutarate (C${}_{5}$H${}_{6}$O${}_{5}$) were added to the mix. After incubation, she was able to isolate 2.4 g of pure glutamate (C${}_{5}$H${}_{9}$NO${}_{4}$) from the reaction. The other product in the reaction is the α-keto acid oxaloacetate (C${}_{4}$H${}_{4}$O${}_{5}$).

## Calculating mass percent

Suppose we were asked to find the compound with the greatest mass percent of oxygen. The first step in tackling such a question would be to calculate the molar masses of the four compounds.
C${}_{4}$H${}_{7}$NO${}_{4}$: (12 x 4) + (1 x 7) + 14 + (16 x 4) = 48 + 7 + 64+ 14 = 133 g/mol
C${}_{5}$H${}_{6}$O${}_{5}$: (12 x 5) + (1 x 6) + (16 x 5) = 60 + 6 + 80 = 146 g/mol
C${}_{4}$H${}_{4}$O${}_{5}$: (12 x 4) + (1 x 4) + (16 x 5) = 48 + 4 + 80 = 132 g/mol
C${}_{5}$H${}_{9}$NO${}_{4}$: (12 x 5) + (1 x 9) + 14 + (16 x 4) = 60 + 9 + 14 + 64 = 147 g/mol
For a problem with such a set of difficult numbers and where the passage does not demand an exact value, we must determine which fraction is the biggest without actually calculating the actual value. Try to approximate the numerator and the denominator by substituting numbers that are easier to handle with in your head but remain as close as possible to the original numbers:
C${}_{4}$H${}_{7}$NO${}_{4}$: 64 g O/133 g total $\approx$ 65/130 (about ½, actual value 0.48)
C${}_{5}$H${}_{6}$O${}_{5}$: 80 g O/146 g total $\approx$ 80/150 (slightly greater than ½, actual value 0.55)
C${}_{4}$H${}_{4}$O${}_{5}$: 80 g O/132 g total $\approx$ 80/130 (greater than ½, actual value 0.61)
C${}_{5}$H${}_{9}$NO${}_{4}$: 64 g O/147 g total $\approx$ 64/150 (less than ½, actual value 0.44)
For aspartate, we have 64/133. We can approximate that to 65/130, and those are appropriate since 65 is exact one-half of 130. For α-ketoglutarate and glutamate, we can round up to the denominator to 150. Our value for glutamate is clearly below ½ and since we are looking for the greatest value, we can eliminate this answer choice. For the remaining two compounds, 80/150 is close to ½ since 75/150 equals ½, and 80/130 must be larger since only the denominator changes between the two, i.e. 80/130 > 80/150.
We would select oxaloacetate (C${}_{4}$H${}_{4}$O${}_{5}$) as our answer choice.

## Calculating limiting reactant

Suppose then we are asked to determine the limiting reactant of the transaminase reaction as performed above:
$\frac{3.3g}{133g/mol}$ $=$ $\frac{33×{10}^{-}1}{133}$ $=$ $\frac{1}{4}$ $×{10}^{-}$${}^{1}$ $=$ $0.025$ $mol$
$\frac{7.4g}{146g/mol}$ $=$ $\frac{7.4}{14.6×{10}^{1}}$ $=$ $\frac{1}{2}$ $×{10}^{-}$${}^{1}$ $=$ $0.05$ $mol$
Since the reaction tells us the stoichiometry of the reaction is 1:1, this is somewhat straightforward, and the compound with the smallest moles is the limiting reagent since it will run out first. Aspartate as the limiting reagent will run out before α-ketoglutarate, and there will be α-ketoglutarate remaining in the solution. A more surefire way to determine the limiting reagent is to determine which has the smallest resultant value only after dividing each reagent by its stoichoimetric coefficient.
Let’s try this method with a trickier stoichiometric problem. Phosphorus pentoxide is a powerful desiccant, and can be used in organic chemistry to obtain anhydrides.
$10$ $KCl{O}_{3}$ $+$ $3$ ${P}_{4}$ $=>$ $3$ ${P}_{4}$${O}_{1}$${}_{0}$ $+$ $10$ $KCl$
Suppose we are asked to figure out the limiting reagent when there is 12 g of potassium chlorate and 41 g of white phosphorus in the reaction tube and how much of the other reactant will remain unreacted. Let’s start off by finding the molar mass of the two reactants so that we can convert to moles:
$KCl{O}_{3}$ $:$ $39$ $+$ $35.5$ $+$ $\left(16×3\right)$ $=$ $122.5$ $g/mol$
${P}_{4}$ $:$ $31×4$ $=$ $124$ $g/mol$
$KCl{O}_{3}$ $:$ $\frac{12g}{122.5g/mol}$ $0.10$ $mol$
${P}_{4}$ $:$ $\frac{41g}{124g/mol}$ $+$ $0.33$ $mol$
We need to divide each by its stoichiometric coefficient to find the limiting reagent:
$KCl{O}_{3}$ $:$ $\frac{0.10mol}{10}$ $=$ $0.01$
${P}_{4}$ $:$ $\frac{0.33mol}{3}$ $=$ $0.11$
Since 0.01 < 0.11 and hence the smaller value, potassium chlorate is the limiting reagent.
To figure out the remaining amount of phosphorus, when 0.01 mol of potassium chlorate reacts with white phosphorus, it reacts in a ratio of 10:3. So for every 10 mol of potassium chlorate that reacts, 3 mol of phosphorus is also consumed.
$\frac{0.10mol}{10}$ $=$ $\frac{xmol}{3}$
$0.30$ $=$ $10x$
$x=0.03$ $mol$ ${P}_{4}$ $reacted$
We can subtract 0.03 mol from the original amount of 0.33 mol to obtain 0.30 mol phosphorus remaining.

## Theoretical yield

Coming back to the transaminase experiment mentioned above, suppose we were asked to find the percent yield of glutamate. From earlier, we were told that the researcher isolated 2.4 g of glutamate from the experiment. From our calculations, we determined that aspartate was our limiting reagent which was 3.3 g or 0.025 mol. The stoichiometry is 1:1 for aspartate and glutamate; for every mole of aspartate consumed, a corresponding mole of glutamate is produced.
$0.025$ $mol$ $Glu×147$ $\frac{gGlu}{molGlu}$ $\approx$ $150×2.5×{10}^{-}$${}^{2}$ $=$ $\left(300$ $+$ $75\right)$ $×$ $\left({10}^{-}$${}^{2}$$\right)$ $=$ $3.75$ $g$ $Glu$
We multiply the moles by the molecular weight to find the mass of glutamate. Let’s approximate 147 as 150, and to calculate (150 x 2.5), split it up into (150 x 2) and (150 x 0.5) to get 375. We then multiply 375 by 10${}^{-}$${}^{2}$ to get the mass, which is 3.75 grams. That is the theoretical yield, while our actual yield is the 2.4 grams obtained by the researcher.
$\frac{2.4g}{3.75g}$ $\approx$ $\frac{2.4g}{3.6g}$ $=$ $0.66$ $or$ $66\mathrm{%}$
To find percent yield, we divide the actual by the theoretical yield and round 3.75 to 3.6 to simplify the fraction to 2.4/3.6 or ⅔, which is 0.66. By multiplying 0.66 by 100, we convert it into a percentage for a final value of 66%.

## Handling complex math in answer choices

Suppose we are asked to figure out the mole fraction of the solute for a saturated solution of hydrogen sulfide, which is prepared using 0.385 grams of H${}_{2}$S gas (density = 1.36 kg/m${}^{3}$, solubility in water = 4 g/dm${}^{3}$ , 4 g/L) dissolving in 100 grams of water at 20oC and 1 atm.
We are given more information that is necessary to figure out the answer. Instead of having to complete such grueling calculations, we may encounter answer choices that have the following format:
Let’s try to spend a little time figuring out where these numbers come from. 0.385 represents the mass of hydrogen sulfide, and 100 represents the mass of the water. With a quick look at the periodic table, we can deduce that 18 is the molar mass of water and 34 hydrogen sulfide. 4 is the solubility of hydrogen sulfide in water, but dividing solubility with the molar mass will not give us moles. 1.36 is the density of hydrogen sulfide, but mass divided by density gives us volume not moles. Since we want the mole fraction of hydrogen sulfide, we would select the first fraction.

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