Some Basic Concepts in Chemistry

A sealed vessel contains equal masses of hydrogen and oxygen at the same temperature. What fraction of the total pressure does the hydrogen exert?

Not half. About 94%.

Pressure depends on the number of molecules, not on their mass. For a mass $m$ of each gas,

$n_{{\text{H}}_2}=\frac{m}{2},n_{{\text{O}}_2}=\frac{m}{32}\Rightarrow \frac{n_{{\text{H}}_2}}{n_{{\text{O}}_2}}=16$

so by Dalton's law $p_{{\text{H}}_2}=\frac{16}{17}{p}_{total}=94.1\%$.

Every question in this chapter punishes mass intuition. Main asks you to convert grams to moles and balance an equation. Advanced asks you to work in equivalents, to analyse a titration whose end point depends on the indicator, and to extract two unknown concentrations from two volume readings.

1. Average masses, and where isotopes matter

The atomic mass on the periodic table is a weighted average over isotopes, not the mass of any actual atom:

$\bar{A}=\sum f_i{A}_i$

Chlorine's $35.5$ is $75.77\%$ of mass $35$ and $24.23\%$ of mass $37$. No chlorine atom weighs $35.5$.

The distinction matters whenever a question mixes counting with weighing. One mole of ${\text{Cl}}_2$ weighs $71$ g, but its molecules come in three kinds — ${}^{35}{\text{Cl}}_2$, ${}^{35}{\text{Cl}}^{37}\text{Cl}$ and ${}^{37}{\text{Cl}}_2$ — with a mass spectrum showing peaks at $70$, $72$ and $74$.

Illustration 1

Boron has two isotopes of masses $10.01$ and $11.01$, with an average atomic mass of $10.81$. Find the percentage abundance of each.

Let the fraction of ${}^{10}\text{B}$ be $x$:

$10.01x+11.01\left(1-x\right)=10.81$

$11.01-x=10.81\Rightarrow x=0.20$

So $20\%$ ${}^{10}\text{B}$ and $80\%$ ${}^{11}\text{B}$.

The average always sits closer to the more abundant isotope. Since $10.81$ is much nearer $11.01$, the heavier isotope had to dominate — a check worth doing before the algebra.

2. Limiting reagent through several steps

With one reaction, divide each reagent's moles by its coefficient and take the smallest. With a sequence of reactions, the product of one step is the reagent of the next, and yields multiply:

$\text{overall yield}=y_1\times y_2\times y_3\times \cdots$

Illustration 2

Nitrogen and hydrogen are mixed in the mass ratio $28:9$ and reacted to give ammonia. Identify the limiting reagent and the mass of ammonia from $28$ g of nitrogen.

$n_{{\text{N}}_2}=\frac{28}{28}=1$ mol; $n_{{\text{H}}_2}=\frac{9}{2}=4.5$ mol

The reaction needs $1:3$, so dividing by coefficients gives $1$ and $1.5$. Nitrogen is limiting.

Ammonia formed $=2$ mol $=34$ g, with $1.5$ mol of hydrogen left over.

The reagent present in greater mass is limiting here, which is exactly why the test must be done on moles divided by coefficients and never on masses.

Illustration 3

A three-step synthesis runs at $80\%$, $75\%$ and $60\%$ yield. Find the overall yield and the mass of starting material needed for $50$ g of product, assuming a $1:1$ mole ratio and equal molar masses.

Overall $=0.80\times 0.75\times 0.60=0.36$, that is $36\%$.

Starting material $=\frac{50}{0.36}=139$ g

Yields multiply, they do not average. Three respectable steps compound into a poor overall figure, which is why synthetic chemists count steps as carefully as they count yields.

3. Formulas from combustion analysis

Burn a compound of carbon, hydrogen and oxygen; all carbon becomes ${\text{CO}}_2$ and all hydrogen becomes ${\text{H}}_2\text{O}$. Oxygen is found by difference, never measured directly.

$m_{\text{C}}=\frac{12}{44}{m}_{{\text{CO}}_2},m_{\text{H}}=\frac{2}{18}{m}_{{\text{H}}_2\text{O}},m_{\text{O}}=m_{sample}-m_{\text{C}}-m_{\text{H}}$

Divide each mass by its atomic mass, divide through by the smallest, and clear fractions to get the empirical formula. The molecular formula then needs an independent molar mass, from a vapour density or a colligative measurement.

Illustration 4

A $0.90$ g sample of a compound of C, H and O gives $1.32$ g of carbon dioxide and $0.54$ g of water on combustion. Its vapour density is $30$. Find the molecular formula.

$m_{\text{C}}=\frac{12}{44}\times 1.32=0.36$ g; $m_{\text{H}}=\frac{2}{18}\times 0.54=0.06$ g; $m_{\text{O}}=0.90-0.42=0.48$ g

Moles: $\frac{0.36}{12}=0.03$, $\frac{0.06}{1}=0.06$, $\frac{0.48}{16}=0.03$, giving a ratio $1:2:1$ and the empirical formula ${\text{CH}}_2\text{O}$.

Molar mass $=2\times 30=60$, and the empirical mass is $30$, so the molecular formula is ${\text{C}}_2{\text{H}}_4{\text{O}}_2$.

Vapour density is half the molar mass, a definition that trips people every year. It is the density relative to hydrogen, and hydrogen's molar mass is $2$.

4. Concentration terms, and converting between them

Converting between molarity and molality needs the density of the solution, and this is the step most often botched:

$m=\frac{1000M}{1000d-M{M}_{solute}}$

where $d$ is in g mL${}^{-1}$. The denominator is the mass of solvent in one litre — total mass minus solute mass — and forgetting to subtract the solute is the standard error.

Illustration 5

A $2.0$ M aqueous solution of a solute of molar mass $60$ g mol${}^{-1}$ has density $1.05$ g mL${}^{-1}$. Find its molality and mole fraction.

One litre contains $2$ mol $=120$ g of solute and weighs $1050$ g.

Solvent mass $=1050-120=930$ g $=0.930$ kg

$m=\frac{2}{0.930}=2.15$ mol kg${}^{-1}$

$n_{water}=\frac{930}{18}=51.7$ mol, so $x_{solute}=\frac{2}{53.7}=0.0373$

Molality always exceeds molarity for a solution denser than water only if the solute is light. The reliable route is the one used here: take exactly one litre, compute both masses, and never memorise the conversion formula.

5. Equivalents and the n-factor

This is the machinery Advanced assumes and Main never introduces. The equivalent is defined so that one equivalent of any species reacts with exactly one equivalent of any other:

$\text{equivalents}=n\times \left(\text{n-factor}\right),N=M\times \left(\text{n-factor}\right)$

$\boxed{N_1{V}_1=N_2{V}_2}$

The n-factor depends on what the species does, not on what it is:

The last row is where the difficulty lives, because the same reagent has different n-factors in different media.

Permanganate is reduced to ${\text{Mn}}^{2+}$ in acid ($n=5$), to ${\text{MnO}}_2$ in neutral or weakly basic solution ($n=3$), and to manganate in strong base ($n=1$). Dichromate goes to $2{\text{Cr}}^{3+}$ and always has $n=6$.

Illustration 6

Find the equivalent mass of potassium permanganate in acidic and in neutral medium. Its molar mass is $158$ g mol${}^{-1}$.

Acidic: $n=5$, so equivalent mass $=\frac{158}{5}=31.6$ g

Neutral: $n=3$, so equivalent mass $=\frac{158}{3}=52.7$ g

A single substance has no single equivalent mass. Quoting one without stating the medium is meaningless, which is why the reaction must be written before any equivalent calculation is attempted.

Illustration 7

$20$ mL of a ferrous sulphate solution requires $25$ mL of $0.02$ M potassium permanganate in acidic medium. Find the molarity of the ferrous solution.

${\text{Fe}}^{2+}\to {\text{Fe}}^{3+}$ has $n=1$; permanganate in acid has $n=5$.

$N_{{\text{KMnO}}_4}=0.02\times 5=0.1$ N

$N_1{V}_1=N_2{V}_2$: $N_1\times 20=0.1\times 25\Rightarrow N_1=0.125$ N

Since $n=1$ for iron(II), its molarity is also $0.125$ M.

The whole point of normality is that no balanced equation is needed once the n-factors are known. Working in molarity here would require the $1:5$ stoichiometry to be derived first.

6. Double titration and back titration

Double titration analyses a mixture of sodium hydroxide and sodium carbonate, using the fact that the two indicators change colour at different stages.

Phenolphthalein turns at the point where all the hydroxide has been neutralised and the carbonate has gone half way, to bicarbonate. Methyl orange turns only when everything is done. Writing the two readings as $P$ and $M$ (both measured from the start),

$\text{NaOH}\equiv 2P-M,{\text{Na}}_2{\text{CO}}_3\equiv 2\left(M-P\right)$

Back titration is used when the analyte reacts too slowly or is insoluble. Add a known excess of reagent, let it finish, then titrate what is left:

$\text{reacted}=\text{added}-\text{left over}$

Illustration 8

A mixture of sodium hydroxide and sodium carbonate is titrated with $0.1$ N acid. Phenolphthalein gives an end point at $22$ mL and methyl orange at $30$ mL. Find the equivalents of each present.

$P=22$ mL, $M=30$ mL

NaOH $\equiv 2\left(22\right)-30=14$ mL of $0.1$ N acid $=1.4$ meq

${\text{Na}}_2{\text{CO}}_3\equiv 2\left(30-22\right)=16$ mL $=1.6$ meq

Check the consistency: $M>P$ confirms carbonate is present, and $2P>M$ confirms hydroxide is. If $2P=M$ the sample is pure carbonate; if $M=P$ it is pure hydroxide.

Illustration 9

$1.0$ g of impure calcium carbonate is treated with $50$ mL of $0.5$ N hydrochloric acid. The excess acid needs $12$ mL of $0.5$ N sodium hydroxide. Find the purity.

Acid added $=50\times 0.5=25$ meq

Acid left $=12\times 0.5=6$ meq

Acid reacted $=19$ meq, so equivalents of carbonate $=19$ meq.

Equivalent mass of ${\text{CaCO}}_3=\frac{100}{2}=50$, so mass $=19\times 50=950$ mg.

Purity $=95\%$

7. Iodometry and iodimetry

Iodine sits conveniently in the middle of the redox range, so it can be used in two opposite ways — and the two names are constantly confused.

Iodimetry is the direct titration of a reducing agent against a standard iodine solution. Iodometry is indirect: an oxidising agent liberates iodine from excess potassium iodide, and the liberated iodine is then titrated against standard sodium thiosulphate:

${\text{I}}_2+2{\text{S}}_2{\text{O}}_3^{2-}\to 2{\text{I}}^{-}+{\text{S}}_4{\text{O}}_6^{2-}$

Thiosulphate has $n=1$, since two of them together lose only two electrons in forming the tetrathionate link. Because the chain from oxidant to thiosulphate conserves equivalents at every step,

$\text{equivalents of oxidant}=\text{equivalents of thiosulphate}$

so the original oxidant is found without ever isolating the iodine.

Starch is added only near the end point, when the solution has faded to pale straw. Added early, it traps iodine in a complex that releases it too slowly and the end point overshoots.

Illustration 10

$25$ mL of a copper sulphate solution is treated with excess potassium iodide, and the liberated iodine needs $20$ mL of $0.1$ N sodium thiosulphate. Find the molarity of the copper solution.

The reaction is $2{\text{Cu}}^{2+}+4{\text{I}}^{-}\to 2\text{CuI}+{\text{I}}_2$, so each copper(II) ion gains one electron and $n=1$.

Equivalents of thiosulphate $=20\times 0.1=2$ meq $=$ equivalents of ${\text{Cu}}^{2+}$

$M=\frac{2}{25}=0.08$ M

The iodine is never weighed or isolated. It is simply a carrier of equivalents from the copper to the thiosulphate, which is what makes the indirect method so much more accurate than titrating copper directly.

Illustration 11

$2.0$ g of bleaching powder is dissolved and made up to $250$ mL. A $25$ mL portion treated with potassium iodide liberates iodine needing $20$ mL of $0.1$ N thiosulphate. Find the percentage of available chlorine.

Equivalents in $25$ mL $=2$ meq, so in $250$ mL $=20$ meq.

Available chlorine has an equivalent mass of $35.5$:

mass $=20\times 10^{-3}\times 35.5=0.71$ g

Percentage $=\frac{0.71}{2.0}\times 100=35.5\%$

Available chlorine is reported as an equivalent, not as a real chemical species. It is the mass of chlorine that would deliver the same oxidising power, which is how bleaching powders of different composition are compared.

8. Volume strength, hardness and eudiometry

Volume strength of hydrogen peroxide is the volume of oxygen at STP that one volume of solution releases:

$\text{volume strength}=5.6N=11.2M,{\text{strength in g L}}^{-1}=17N$

Hardness of water is quoted as parts per million of calcium carbonate equivalent, so any calcium or magnesium salt present is converted to the mass of carbonate that would supply the same number of equivalents.

Eudiometry analyses gas mixtures by burning them and measuring the contraction. All volumes are taken at the same temperature and pressure, so volume ratios are mole ratios directly. Water condenses on cooling, so it contributes nothing to the final volume.

Illustration 12

A sample of hydrogen peroxide is labelled "$20$ volume". Find its normality, molarity and strength in grams per litre.

$N=\frac{20}{5.6}=3.57$ N

$M=\frac{N}{2}=1.79$ M

Strength $=17\times 3.57=60.7$ g L${}^{-1}$

The factor $5.6$ is half of $11.2$, the molar volume divided by two, because each mole of peroxide releases only half a mole of oxygen.

Illustration 13

$10$ mL of a gaseous hydrocarbon is burnt with excess oxygen. After cooling, the volume contracts by $25$ mL, and adding potassium hydroxide removes a further $30$ mL. Identify the hydrocarbon.

Potassium hydroxide absorbs carbon dioxide, so $30$ mL of ${\text{CO}}_2$ came from $10$ mL of hydrocarbon: three carbons.

For ${\text{C}}_3{\text{H}}_y$, oxygen used $=10\left(3+\frac{y}{4}\right)$ mL and ${\text{CO}}_2$ formed $=30$ mL.

Contraction on cooling $=$ (hydrocarbon $+$ oxygen used) $-$ carbon dioxide $=10+30+2.5y-30=25$, giving $y=6$.

The hydrocarbon is ${\text{C}}_3{\text{H}}_6$, propene.

Water is invisible in eudiometry because the measurement is made after cooling, which is exactly what makes the contraction carry the hydrogen count.

Illustration 14

A water sample contains $120$ mg of magnesium sulphate per litre. Express its hardness in ppm of calcium carbonate.

Molar mass of ${\text{MgSO}}_4$ is $120$, so the sample holds $1$ mmol per litre.

The same number of millimoles of ${\text{CaCO}}_3$ weighs $1\times 100=100$ mg.

Hardness $=100$ ppm

Hardness is always reported as an equivalent, never as the actual salt, so that samples containing different salts can be compared on one scale.

Summary

• Pressure and reaction stoichiometry count molecules, so equal masses of different gases are never equal amounts.
• Atomic masses on the table are weighted averages over isotopes; no single atom has that mass.
• Limiting reagent: divide moles by coefficients, take the smallest. Never compare masses.
• Across a sequence of reactions the fractional yields multiply.
• Combustion: $m_{\text{C}}=\frac{12}{44}{m}_{{\text{CO}}_2}$, $m_{\text{H}}=\frac{2}{18}{m}_{{\text{H}}_2\text{O}}$, oxygen by difference.
• Vapour density is half the molar mass.
• Molarity depends on temperature; molality, mole fraction and ppm do not. Converting needs density and the solvent mass, not the solution mass.
• Equivalents: $N=M\times n$, and $N_1{V}_1=N_2{V}_2$ needs no balanced equation.
• n-factor is set by what the species does: basicity, acidity, charge, or electrons transferred.
• ${\text{MnO}}_4^{-}$ has $n=5$ in acid, $3$ in neutral, $1$ in strong base; ${\text{Cr}}_2{\text{O}}_7^{2-}$ always has $n=6$.
• Double titration: $\text{NaOH}\equiv 2P-M$ and ${\text{Na}}_2{\text{CO}}_3\equiv 2\left(M-P\right)$.
• Back titration: reacted equals added minus left over, used when the analyte is slow or insoluble.
• Iodimetry is direct against standard iodine; iodometry liberates iodine from KI and titrates it with thiosulphate ($n=1$).
• Add starch only near the end point, or the iodine-starch complex releases too slowly and the reading overshoots.
• ${\text{H}}_2{\text{O}}_2$: volume strength $=5.6N=11.2M$; strength $=17N$ g L${}^{-1}$.
• In eudiometry, volume ratios are mole ratios, and water condenses out before the final reading.