Chemical reactions, stoichiometry, molarity and the limiting reactant

Stoichiometry means “recipe calculation”. If a cake recipe says 2 eggs and 3 cups of flour, and you have 6 eggs, you know how many cups of flour you need. A chemical equation is also a recipe; the only difference is that you have to count the ingredients in moles. This week we combine the gram-mole bridge from week 1 with the equation.

Golden rules:

1. Balance first. Every calculation done with an unbalanced equation is wrong.
2. Coefficients are mole ratios. Not gram ratios. Always convert grams to moles first.
3. In a solution, moles = M × V. The volume must be in liters.
4. If two reactants are given, find the limiting one first. The product is computed from the limiting reactant.

Writing and balancing chemical equations

A chemical equation shows the reactants on the left and the products on the right. The physical state is in parentheses: (s) solid, (l) liquid, (g) gas, (aq) dissolved in water.

Because of conservation of mass, the number of atoms of each element must be equal on both sides. You achieve this with coefficients, and you never change the subscripts. Turning $\mathrm{H_2O}$ into $\mathrm{H_2O_2}$ means writing a different substance.

A practical order:

1. Start with elements that appear in only one reactant and one product (usually C or a metal).
2. Balance H.
3. Do O last; if it stands alone as $\mathrm{O_2}$, you can adjust it with any fraction you want.
4. If a fraction remains, multiply all the coefficients by the denominator.

Example 1. Ethanol combustion: $\mathrm{C_2H_5OH + O_2 \rightarrow CO_2 + H_2O}$

C: 2 on the left      → 2 CO2
H: 6 on the left      → 3 H2O
O: right side 2·2 + 3 = 7;  ethanol has 1 on the left, 6 more are needed from O2 → 3 O2
C2H5OH + 3 O2 → 2 CO2 + 3 H2O     check: C 2=2, H 6=6, O 7=7

Stoichiometry with mole ratios

The coefficients in a balanced equation give you conversion factors. From the ethanol equation:

$\frac{3\ \text{mol O}_2}{1\ \text{mol C}_2\text{H}_5\text{OH}} \qquad \frac{2\ \text{mol CO}_2}{1\ \text{mol C}_2\text{H}_5\text{OH}} \qquad \frac{3\ \text{mol H}_2\text{O}}{3\ \text{mol O}_2}$

The standard chain:

$\text{g A} \xrightarrow{\div M_A} \text{mol A} \xrightarrow{\times \frac{b}{a}} \text{mol B} \xrightarrow{\times M_B} \text{g B}$

Example 2. How many grams of water form when 10.0 g of ethanol burns? ($M = 46.069$ g/mol)

$10.0\ \text{g} \times \frac{1\ \text{mol C}_2\text{H}_5\text{OH}}{46.069\ \text{g}} \times \frac{3\ \text{mol H}_2\text{O}}{1\ \text{mol C}_2\text{H}_5\text{OH}} \times \frac{18.015\ \text{g}}{1\ \text{mol H}_2\text{O}} = 11.7\ \text{g H}_2\text{O}$

The oxygen needed, in the same way: $10.0/46.069 \times 3 \times 31.998 = 20.8$ g $\mathrm{O_2}$.

PITFALL · setting up a ratio with grams. Saying “1 g of ethanol reacts with 3 g of oxygen” is wrong. Coefficients count particles (moles). To set up a gram ratio you must first convert to moles.

Solutions and concentration

A solution is made of a solute and a solvent. The concentration tells how “strong” it is. Two units this week:

Percent by mass

$\%\ \text{mass} = \frac{m_{\text{solute}}}{m_{\text{solution}}} \times 100 \qquad m_{\text{solution}} = m_{\text{solute}} + m_{\text{solvent}}$

If 20.0 g of sugar is dissolved in 80.0 g of water, the solution is 100.0 g and the sugar is 20.0%.

Molarity

$M = \frac{n_{\text{solute}}\ (\text{mol})}{V_{\text{solution}}\ (\text{L})}$

Its unit is mol/L, M for short. The denominator is the total volume of the solution, not the volume of the added water.

Example 3. 7.45 g of KCl ($M = 74.55$ g/mol) is dissolved and made up to 500.0 mL.

$n = \frac{7.45}{74.55} = 0.09993\ \text{mol} \qquad M = \frac{0.09993\ \text{mol}}{0.5000\ \text{L}} = 0.200\ \text{M}$

Molarity is really a conversion factor: 0.200 M means “0.200 mol in 1 L of solution”, or $\frac{0.200\ \text{mol}}{1\ \text{L}}$. If a volume is given and moles are wanted, you multiply: $n = M \times V$.

Dilution

When you add water to a solution the number of moles of solute does not change, only the volume grows:

$M_1 V_1 = M_2 V_2$

It is enough to write the $V$ values in the same unit (both in mL or both in L).

Example 4. To prepare 250.0 mL of 0.100 M solution from a 2.00 M stock solution:

$V_1 = \frac{M_2 V_2}{M_1} = \frac{0.100 \times 250.0}{2.00} = 12.5\ \text{mL}$

12.5 mL of stock is taken and made up to 250.0 mL in a volumetric flask. You do not say “add 237.5 mL of water”, because liquid volumes do not always add exactly.

Stoichiometry in solutions

If a reactant is given as a solution, the start of the chain is $M \times V$ instead of grams:

$V_A \times M_A \rightarrow \text{mol A} \xrightarrow{\times \frac{b}{a}} \text{mol B} \rightarrow \text{g B or } V_B = \frac{\text{mol B}}{M_B}$

Example 5. $\mathrm{Pb(NO_3)_2(aq) + 2KI(aq) \rightarrow PbI_2(s) + 2KNO_3(aq)}$

The volume of 0.200 M KI that will react completely with 25.0 mL of 0.100 M $\mathrm{Pb(NO_3)_2}$, and the mass of $\mathrm{PbI_2}$ formed:

mol Pb(NO3)2 = 0.0250 L × 0.100 mol/L = 2.50×10^-3 mol
mol KI       = 2.50×10^-3 × 2          = 5.00×10^-3 mol
V(KI)        = 5.00×10^-3 / 0.200      = 0.0250 L = 25.0 mL
m(PbI2)      = 2.50×10^-3 × 461.0      = 1.15 g

The same volume was needed even though the molarity is twice as large, because the coefficient of KI is also 2.

Molarity from percent and density

Commercial acids are given like “37% HCl by mass, density 1.18 g/mL”. The method: assume 1 L of solution.

$1000\ \text{mL} \xrightarrow{\times d} \text{g solution} \xrightarrow{\times \%} \text{g solute} \xrightarrow{\div M} \text{mol} \Rightarrow M$

The limiting and the excess reactant

In reality reactants are rarely mixed in exactly the ratio of the equation. The reactant that runs out first stops the reaction: the limiting reactant. The other is left over: the excess reactant.

The method of finding it:

1. Convert each reactant to moles.
2. Divide each number of moles by its coefficient in the equation.
3. The one that gives the smallest result is limiting.
4. Compute the product from the limiting reactant.

Example 6. 4.00 g $\mathrm{H_2}$ reacts with 24.0 g $\mathrm{O_2}$: $\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$.

H2: 4.00 / 2.016  = 1.984 mol    ÷ 2 = 0.992
O2: 24.0 / 31.998 = 0.7501 mol   ÷ 1 = 0.750   ← smaller, O2 is limiting
H2O = 0.7501 × 2 × 18.015 = 27.0 g
H2 used = 0.7501 × 2 = 1.500 mol;  left = 1.984 − 1.500 = 0.484 mol = 0.976 g
check: 4.00 + 24.0 = 28.0 g = 27.0 + 0.98

In grams, the oxygen looked 6 times larger but it was the limiting one. Do not decide by looking at the grams.

Example 7 (molarity from percent and density, then dilution). Commercial nitric acid is 70.0% $\mathrm{HNO_3}$ by mass, with a density of 1.42 g/mL. How many mL are needed for 500.0 mL of 1.00 M solution?

1 L solution:  1000 mL × 1.42 g/mL = 1420 g solution
mass of HNO3:  1420 × 0.700        = 994 g
mol HNO3:      994 / 63.012         = 15.77 mol  → 15.8 M
dilution:      V1 = (1.00 × 500.0) / 15.77 = 31.7 mL

Consecutive reactions

In industry the product is often obtained in several steps. The product of each step is the reactant of the next. You do not need to weigh the intermediates one by one; you multiply the mole ratios in a chain.

Sulfuric acid production: $\mathrm{S + O_2 \rightarrow SO_2}$, $\mathrm{2SO_2 + O_2 \rightarrow 2SO_3}$, $\mathrm{SO_3 + H_2O \rightarrow H_2SO_4}$. Because the S atom is conserved at each step, the overall ratio is 1 mol S : 1 mol $\mathrm{H_2SO_4}$.

$100.0\ \text{g S} \times \frac{1\ \text{mol S}}{32.06\ \text{g}} \times \frac{1\ \text{mol SO}_2}{1\ \text{mol S}} \times \frac{2\ \text{mol SO}_3}{2\ \text{mol SO}_2} \times \frac{1\ \text{mol H}_2\text{SO}_4}{1\ \text{mol SO}_3} \times \frac{98.07\ \text{g}}{1\ \text{mol}} = 306\ \text{g}$

TIP · follow the atom. In consecutive reactions, tracking the key atom (here S) is often the shortest way: if all the S at the start goes into the final product, the ratio can be read from the number of S in the formulas.

Stoichiometry strategy in the exam

1. Balance the equation and check it once more by eye.
2. Convert everything given to moles (grams → $\div M$, solution → $M \times V$, gas → $PV/RT$).
3. If there are two reactants, find the limiting one.
4. Use the mole ratio to get to the moles of the wanted substance.
5. Convert to the wanted unit, round to the right significant figures, write the unit.

TIP · check in two ways. The alternative method: for each reactant separately, compute “how much product if it is all used up”. The one that gives less product is limiting, and that amount is the real product. In the exam use one way and check with the other.

Common mistakes