Grade 11 ChemistryStudy Guide

The mole is chemistry's counting unit: a fixed number of particles large enough to weigh on a bench balance. Dividing a measured mass by the molar mass is the bridge between the two worlds: weigh out 36.0 g of water, divide by its molar mass of 18.02 g/mol, and you know you have 2.00 mol, about 1.2 × 10²⁴ molecules. Every stoichiometry problem starts or ends with this conversion, because balanced equations speak in moles while balances speak in grams.

The word mole was coined by Wilhelm Ostwald in the 1890s, but the idea goes back to Avogadro's 1811 hypothesis that equal gas volumes hold equal numbers of molecules. Since the 2019 SI redefinition, the mole is defined by an exact count, 6.02214076 × 10²³ particles, so molar masses in g/mol are now measured quantities rather than definitions.

A mole is a count and nothing more mysterious: the chemist's dozen, scaled up to something useful. Multiply an amount in moles by the Avogadro constant and you have the literal number of particles in front of you. The only real question is why the constant is that particular size, and the answer is that it was chosen to make the bridge between two worlds land cleanly: $N_A$ is the number of atoms that makes a mole of carbon-12 weigh exactly 12 grams. That choice is what lets a mass in grams read off a balance be converted into a count of atoms, which is the single most useful trick in chemistry.

Scale is the thing worth feeling here. A 250 g glass of water is 250/18.015 = 13.9 mol, so it holds $13.9 \times 6.022\times10^{23} = 8.4\times10^{24}$ molecules. Now count the other way: all the water on Earth, about $1.4\times10^{21}$ litres, divided into 250 g glasses, comes to roughly $5.4\times10^{21}$ glasses. There are about fifteen hundred times more molecules in one glass of water than there are glasses of water in every ocean on the planet. That ratio is why chemists never think about individual molecules and why statistical behaviour is so reliable at this scale.

The constant carries Avogadro's name but not his arithmetic: he proposed in 1811 that equal gas volumes hold equal numbers of particles, and never estimated the number. Jean Perrin did, from Brownian motion, and named it for Avogadro in 1909, work that took him the 1926 Nobel Prize. The 2019 SI redefinition then reversed the logic entirely. $N_A$ is now exact by decree at 6.02214076 × 10²³ per mole, and the mole is defined as that many entities. Carbon-12's role is retired: a mole of it now weighs 12 grams only to within experimental uncertainty, rather than by definition. So 2.00 mol contains exactly 1.204428152 × 10²⁴ particles, with no uncertainty in the constant at all, though your measured 2.00 mol still has its own.

The error that swallows this page whole is leaving "particles of what" unstated. A mole of O₂ is 6.022 × 10²³ molecules but 1.204 × 10²⁴ atoms. A mole of NaCl is 6.022 × 10²³ formula units, which is 6.022 × 10²³ sodium ions plus the same number of chloride ions: 1.204 × 10²⁴ ions in total. A mole of Al₂(SO₄)₃ contains three moles of sulfate. Almost every wrong answer here is a correct calculation attached to the wrong noun, so write the noun down before you multiply: molecules, atoms, ions, or formula units.

Two smaller slips. Dividing a molar mass by $N_A$ gives the mass of one particle in grams, water comes out at $2.99\times10^{-23}$ g, and people routinely lose a factor of 1000 by mixing kilograms into that step, or confuse it with the mass in unified atomic mass units, which is just the molar mass number again with different units. And because $N_A$ is exact, it never limits your significant figures; only the amount you measured does.

A chemical formula is a recipe by count, and this equation translates it into a recipe by mass. Molar mass is additive (the compound weighs exactly what its atoms weigh, summed) so an element's share of the total is its atoms' contribution divided by the whole. In water, hydrogen brings 2 × 1.008 = 2.016 g/mol out of 18.02, which is 11.2% by mass; oxygen carries the other 88.8%. Two atoms out of three, and barely a ninth of the weight.

A case with real stakes. Ammonium sulfate, (NH₄)₂SO₄, has a molar mass of 132.14 g/mol. Nitrogen appears twice, so $\%N = (2 \times 14.007)/132.14 \times 100 = 21.2\%$. That is not a coincidence of the classroom: it is why the fertilizer is sold as 21-0-0, and why a 25 kg bag delivers 5.3 kg of actual nitrogen to a field.

Run the arithmetic backwards and it becomes analysis rather than description. Burn an unknown compound, weigh the CO₂ and H₂O produced, convert those to masses of carbon and hydrogen, take oxygen by difference, then divide each element's mass percent by its atomic mass and normalise to the smallest result. What comes out is the empirical formula. This is combustion analysis, and it was the central technique of organic chemistry from Liebig's refinement of it in the 1830s until spectroscopy arrived. Behind all of it sits Proust's law of definite proportions: a given compound always holds the same elements in the same mass ratio, which is exactly the claim this equation encodes.

The subscript $a$ is the number of atoms of that element per formula unit, and parentheses are where people lose it. In (NH₄)₂SO₄ the hydrogen count is 8, not 4: the subscript outside the bracket multiplies everything inside. Hydrates catch people the same way: copper(II) sulfate pentahydrate, CuSO₄·5H₂O, has a molar mass of 249.7 g/mol including its water, so the copper is 25.4%, not the 39.8% you would get from the anhydrous salt. If the bottle says pentahydrate, the water is part of what you weighed.

Two further traps. Percent composition can only ever give you an empirical formula, never a molecular one: formaldehyde CH₂O, acetic acid C₂H₄O₂ and glucose C₆H₁₂O₆ all analyse to identical percentages, and separating them needs an independent molar mass. And on fertilizer bags, only the first number is what it appears to be. The N in N-P-K is genuinely percent nitrogen, but P is reported as P₂O₅ equivalent and K as K₂O: a nineteenth-century convention that never died. A bag marked 0-46-0 is 46% P₂O₅, which works out to about 20% actual phosphorus. Reading it as elemental phosphorus overstates the dose by well over a factor of two.

This is the hinge every stoichiometry problem turns on, and it is one line long. The coefficients in a balanced equation are a ratio of moles: never of masses, never of volumes, never of anything you can weigh directly. So the trip from one substance to another has to pass through moles, and the mole ratio b/a is the bridge.

Worked: the Haber process, N₂ + 3H₂ → 2NH₃. Feed 6.00 mol of hydrogen and ask how much ammonia it can make. The coefficient on hydrogen is 3, on ammonia 2, so n(NH₃) = 6.00 × 2/3 = 4.00 mol. Not 6.00, not 9.00: the equation says three hydrogens buy two ammonias, and no amount of arithmetic elsewhere changes that exchange rate.

The failure mode is reading coefficients off an unbalanced equation, and nothing downstream will catch it. Write N₂ + H₂ → NH₃ and the ratio looks like 1:1; every mass, every volume, every yield you compute afterwards is then wrong by a clean factor, and every one of them will look perfectly reasonable. Balance first, count atoms on both sides, and only then read the numbers.

Two smaller traps. First, the ratio is not symmetric: going from A to B multiplies by b/a, and going back multiplies by a/b, so the direction of the arrow matters as much as the numbers. Second, a balanced equation can legitimately be written with fractional coefficients, as in H₂ + ½O₂ → H₂O. That is fine and the ratio still works; it just means the ΔH quoted alongside belongs to that version of the equation and not to the doubled one.

Everything else in a mass-to-mass calculation is unit conversion. Molar mass turns grams into moles on the way in, this ratio does the chemistry in the middle, and molar mass turns moles back into grams on the way out. Students who find stoichiometry hard almost always find the middle step easy and the surrounding conversions hard; the ratio itself is the part with the chemistry in it.

Three conversions collapsed into one line: grams of A to moles of A, moles of A to moles of B through the balanced coefficients, then moles of B back to grams. Written out it is m_B = (m_A/M_A)(b/a)M_B, and it is the single most-used calculation in a teaching lab, because a balance is the only instrument on the bench that measures anything directly.

Worked: burning hydrogen, 2H₂ + O₂ → 2H₂O. Take 4.032 g of hydrogen, molar mass 2.016 g/mol, so 2.000 mol. The coefficients on H₂ and H₂O are both 2, so 2.000 mol of water. Times 18.015 g/mol gives 36.03 g. Note that the mass grew ninefold while the moles did not change at all: the mole ratio was 1:1 and every gram of the difference came from the oxygen.

This answer is the theoretical yield, and it is only true if A is the limiting reagent. The equation as written assumes there is enough of everything else. In a real flask there rarely is. The test is to run this calculation from each reactant in turn and take the smallest answer: that reactant is the limiter, and its figure is the real ceiling. Doing it from whichever reactant the question mentions first is how a calculation comes out generously and confidently wrong.

And limiting is decided by moles, not by mass. Given 100 g of hydrogen and 100 g of oxygen, the masses match and the amounts do not: 49.6 mol against 3.13 mol, with the reaction needing two hydrogens per oxygen. Oxygen limits by a factor of eight. Hydrogen simply looks generous because it is light.

What comes out of the flask is always less than this. Percent yield compares the two, and a figure above 100% means the product is still carrying solvent it has not finished drying out of: stoichiometry itself never returns more than it predicts.

Fix the temperature and pressure and a mole of any ideal gas occupies the same volume, so converting between moles and litres needs one constant and no information about the substance. This is Avogadro's principle, and it deserves a moment of surprise before it becomes routine. A mole of hydrogen weighs 2 g and a mole of sulfur hexafluoride weighs 146 g, seventy-three times more, yet at the same conditions they fill the same flask. The reason is that pressure comes from the number of impacts and their momentum, and at a common temperature the heavier molecules move proportionally slower. The mass cancels out of everything the container can feel.

This page uses classic STP, 0 °C and 1 atm, where $V_m = 22.414$ L/mol. Burning one mole of methane, $\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$, consumes 2 mol of oxygen and produces 1 mol of carbon dioxide, which is 44.83 L of O₂ in and 22.41 L of CO₂ out at STP. Note that the two moles of water do not appear in that volume tally, because at 0 °C the water is a liquid and the equation only counts gases.

The constant is not independent; it is $PV = nRT$ evaluated once. $V_m = RT/P = (8.314 \times 273.15)/101\,325 = 0.022414$ m³/mol, and every other molar volume in circulation is the same calculation at different conditions. Historically the logic ran the other way. Gay-Lussac reported in 1808 that gases combine in simple whole-number volume ratios, two volumes of hydrogen to one of oxygen, and Avogadro's 1811 hypothesis explained why: equal volumes hold equal counts, so the volume ratios are the mole ratios of the balanced equation. That insight was ignored for half a century before Cannizzaro revived it at Karlsruhe in 1860.

The error that dominates this page is using 22.4 L/mol for conditions that are not STP. Room temperature is not 0 °C. At 25 °C and 1 atm the molar volume is 24.47 L/mol, so applying 22.4 to a bench-top measurement understates the volume by 9%. And "standard conditions" is not one definition but several: classic STP gives 22.414 L/mol, IUPAC's post-1982 STP (0 °C, 100 kPa) gives 22.711, and SATP (25 °C, 100 kPa) gives 24.79. The spread between the smallest and largest is over 10%, which is far more than the precision most people think they are carrying. Check which convention a source assumes before you borrow its number, and when the conditions are anything other than a listed standard, abandon the shortcut and use the ideal gas law directly.

Two smaller cautions. This is an ideal-gas result, so it degrades for gases near their condensation point: ammonia and sulfur dioxide at STP already deviate by a percent or two, and water vapour at 0 °C is not a gas at all. And when you use volume ratios as mole ratios in a reaction, they apply only to the species that are actually gaseous at the stated conditions. A dissolved or condensed product contributes no volume, and counting it is the quiet way to get a stoichiometry problem wrong.

Two reactants go into a flask, and one of them runs out first. That one, the limiting reactant, decides how much product you get; the other is left over. The test is to divide each reactant's moles by its coefficient in the balanced equation. Each quotient counts how many times that reactant could run the equation as written, and the smaller count wins.

Worked: 2H₂ + O₂ → 2H₂O, with 3.00 mol of hydrogen and 2.00 mol of oxygen. Hydrogen supplies 3.00/2 = 1.50 batches and oxygen supplies 2.00/1 = 2.00. Hydrogen runs out first, so the water is 2 × 1.50 = 3.00 mol, and 2.00 − 1.50 = 0.50 mol of oxygen is left in the flask.

The reactant with fewer moles is not always the limiting one. In the example, oxygen had fewer moles and was still in excess, because the equation asks for twice as much hydrogen. Compare moles divided by coefficient, never raw moles, and never grams: comparing masses ignores both the molar masses and the coefficients.

The answer is a theoretical yield, a ceiling on paper. What the balance shows afterwards is the actual yield, and the ratio of the two is the percent yield. Working backwards only goes one branch at a time: given the yield, the calculator can tell you the least A that makes it, but not how much A was really there, because once B runs out any extra A changes nothing.

Once you know which reactant ran out, the interesting question is what is left of the other one. Subtract what the limiting reagent consumed from what you charged: n_excess = n_B − (b/a)n_A. It is the second half of a limiting-reagent problem and the half that gets rushed, even though in industry it is the half that matters: leftover reagent is cost, and it is usually also a separation problem.

Worked: N₂ + 3H₂ → 2NH₃, charged with 5.00 mol of nitrogen and 20.0 mol of hydrogen. Nitrogen runs out first, so it is the limiter. Hydrogen consumed = (3/1) × 5.00 = 15.0 mol, leaving 20.0 − 15.0 = 5.00 mol of hydrogen sitting unreacted. In an ammonia plant that leftover is not thrown away: it is separated from the ammonia and recycled, which is why the real recycle loop runs a large excess of both gases past the catalyst.

A negative answer here is not an error; it is a diagnosis. It means B ran out before A did, so you labelled the wrong reactant as limiting. Swap the two, run it again, and the question you actually wanted answered is how much of A is left over. The solver says so rather than blocking the calculation, because the negative number carries the information.

The limiting-reagent test comes first and has its own page: divide each reactant's moles by its coefficient, n_A/a against n_B/b, and the smaller quotient limits. It only runs backwards one branch at a time, because the forward direction throws away the losing reactant's number entirely.

To turn the leftover moles back into something you can weigh, multiply by molar mass. Chemists usually run the cheap or easily removed reactant in excess on purpose, to push a reversible reaction forward: Le Chatelier's principle applied with a purchase order.

A balanced equation promises a certain mass of product; the flask rarely delivers all of it. Side reactions consume reactant, some product stays dissolved in the mother liquor, and a little is always lost on filter paper and glassware. Percent yield is the honest scorecard: the mass you actually isolated divided by the stoichiometric maximum, times 100. In a classic teaching lab, synthesizing aspirin from salicylic acid might predict 5.00 g of product; if 4.21 g of dry crystals come off the funnel, the yield is 84.2%.

The number matters far beyond the classroom. Process chemists judge manufacturing routes largely by yield, because in a multi-step synthesis losses multiply: five steps at 80% each deliver only 33% overall. A reported yield above 100% is a red flag, not a triumph: it usually means the product is still wet or carries impurities.

Enthalpy is an extensive quantity: run a reaction twice and you get twice the heat. That is the entire justification for $q = n\Delta H$, and it is why tabulating a single per-mole figure is enough to describe a reaction at any scale from a test tube to a boiler. The sign convention runs from the system's point of view: ΔH negative for an exothermic reaction, because the system's enthalpy falls as heat leaves it. A negative $q$ is not an error message; it is the answer telling you the heat came out.

Something concrete. Heating 200 L of water from 10 °C to 60 °C takes $200 \times 4.186 \times 50 = 41\,860$ kJ. Methane burns at ΔH = −890.3 kJ/mol, so the amount required is $n = 41\,860/890.3 = 47.0$ mol: about 754 g of methane, or 1.05 m³ at STP, before any consideration of how much of that heat actually reaches the water. Chain those three pages together and you have most of a combustion calculation.

The relation's real power is that enthalpy is a state function, which is Hess's law: the heat of a reaction depends only on where it starts and ends, not on the route. So you can add reactions like algebra, and you can build any ΔH you need from tabulated standard enthalpies of formation, products minus reactants, without ever running the reaction. One caution on what is being measured: ΔH is the heat at constant pressure, which is what an open vessel or a flowing burner delivers. A bomb calorimeter holds volume constant and measures ΔU instead, and the two differ by the work done pushing the atmosphere aside, $\Delta H = \Delta U + \Delta n_{gas}RT$.

The question this page most needs you to ask is "per mole of what?" ΔH belongs to the balanced equation as written, not to any one substance in it. $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ has ΔH = −571.6 kJ, but $\text{H}_2 + \tfrac{1}{2}\text{O}_2 \rightarrow \text{H}_2\text{O}$ has ΔH = −285.8 kJ. Same chemistry, same physical world, different bookkeeping, and a factor of two waiting for anyone who reads a table without reading the equation above it. Write the equation first, decide which species your $n$ counts, and make the two agree.

Then the trap that costs more marks than any other in stoichiometry. The limiting reagent is found by moles, never by mass. Given 100 g of hydrogen and 100 g of oxygen, the masses are equal and the amounts are not remotely: 49.6 mol of H₂ against 3.13 mol of O₂, and the reaction needs two hydrogens per oxygen. Oxygen limits by a factor of eight, and 100 g of hydrogen looks generous only because hydrogen is light. Convert everything to moles, divide each by its coefficient in the balanced equation, and the smallest quotient is the limiter. The $n$ that goes into $q = n\Delta H$ is then the extent of reaction that limiter permits, not the amount of whatever you happened to weigh out.

One last note for anyone comparing fuel figures. Combustion enthalpies come in two flavours depending on whether the product water is counted as liquid or as vapour, differing by 44 kJ per mole of water. That is the higher-heating-value and lower-heating-value distinction, and for methane it is about a 10% gap. Two sources can disagree by that much while both being correct.

Molarity answers the practical question "how much stuff is in this bottle?" by counting moles of solute per liter of solution. Dissolve 58.44 g of table salt, exactly one mole of NaCl, in water and top up to the 1.00 L mark of a volumetric flask, and you have a 1.00 M solution. Note the fine print: it is per liter of solution, not per liter of water added, which is why chemists fill to a calibrated mark instead of adding a measured liter of solvent.

Molarity is the workhorse concentration unit because reactions are mole-to-mole affairs: multiplying C by a dispensed volume immediately gives the moles delivered, which is exactly what a titration calculation needs. Physiological saline is about 0.154 M NaCl, ocean water roughly 0.5 M, and concentrated hydrochloric acid around 12 M: a span that dilution calculations cross daily in every lab.

Mass percent is the concentration measure that needs nothing but a balance: the solute's mass divided by the mass of the whole solution, solute plus solvent, times 100. Dissolve 25 g of salt in 225 g of water and the solution is 25/250 × 100 = 10% salt by mass. The classic mistake is dividing by the solvent mass alone, which would wrongly give 11.1%.

Because it is temperature-proof and instrument-free, mass percent dominates industrial and household labels: household vinegar is about 5% acetic acid, physiological saline 0.9% NaCl, seawater roughly 3.5% dissolved salts, and concentrated sulfuric acid ships at 98%. Converting to molarity requires the solution's density, which is why bottle labels for concentrated acids list both figures side by side.

Salt dissolving in water does not stay as salt: the crystal comes apart into its ions. Sodium chloride gives one Na⁺ and one Cl⁻ for every formula unit, so a 0.10 mol/L NaCl solution holds 0.10 mol/L of each. Calcium chloride gives one Ca²⁺ and two Cl⁻, so a 0.10 mol/L CaCl₂ solution holds 0.10 mol/L of calcium and 0.20 mol/L of chloride.

The count ν comes straight off the dissociation equation, and the dissociation equation comes straight off the formula: the subscript on each ion becomes its coefficient. CaCl₂(s) → Ca²⁺(aq) + 2Cl⁻(aq). Na₂SO₄(s) → 2Na⁺(aq) + SO₄²⁻(aq). Al₂(SO₄)₃(s) → 2Al³⁺(aq) + 3SO₄²⁻(aq). The sulfate stays in one piece: a polyatomic ion is one ion, and the subscripts inside it are not counted.

Check the charges. A dissociation equation balances charge as well as atoms: 2(+3) + 3(−2) = 0 for aluminium sulfate. If the charges do not cancel, an ion count is wrong.

The rule assumes the compound dissociates completely, which holds for soluble ionic compounds. Sugar dissolves without making any ions at all, and a weak acid such as the acetic acid in vinegar releases only a small fraction of its hydrogen. When two solutions carrying the same ion are mixed, find the moles of that ion from each, add them, and divide by the total volume.

Adding solvent to a solution spreads the same solute through a larger volume: the moles do not change, only their crowding. Since moles equal concentration times volume, $C_1V_1$ must equal $C_2V_2$ before and after any dilution. To prepare 250 mL of 1.0 M hydrochloric acid from a 12.1 M concentrated stock, solve for $V_1$: (1.0 × 250)/12.1 ≈ 20.7 mL of stock, made up to the 250 mL mark with water. (And always add acid to water, never the reverse.)

The law is not limited to molarity: any concentration measure proportional to moles per volume works, as long as both sides use the same one. Water-treatment operators lean on it daily when dosing inhibitor or biocide from concentrated drums into recirculating loops, and biologists use the identical arithmetic for serial dilutions, where each step divides concentration by a fixed factor.

$PV = nRT$ says that for a gas, four quantities are not independent: fix any three and the fourth is decided. Squeeze it and the pressure rises; warm it and it pushes harder or swells; add more of it and both go up. What makes the equation remarkable is not that those things are true, anyone with a bicycle pump knows them, but that one constant serves every gas. Helium, nitrogen and steam all obey it with the same $R = 8.314$ J/(mol·K), which is a strong hint that pressure has nothing to do with what the molecules are and everything to do with how many there are and how fast they are moving.

A worked case in units you would actually read off a gauge. A 20 L cylinder sits at 150 kPa absolute on a 20 °C morning. Rearranged for amount, $n = PV/RT = (150\,000 \times 0.020)/(8.314 \times 293.15) = 3000/2437 \approx 1.23\ \text{mol}$. Note what had to happen before the arithmetic: pascals not kilopascals, cubic metres not litres, and kelvin not Celsius. This page converts your entries for you, but the discipline is worth keeping in your head, because a scrap of paper will not.

The law arrived in pieces. Boyle established $PV$ constant at fixed temperature in 1662; Charles and Gay-Lussac tied volume and pressure to temperature around 1800; Avogadro proposed in 1811 that equal volumes of gases hold equal numbers of particles. Émile Clapeyron folded them into a single expression in 1834. Kinetic theory later derived the whole thing from mechanics: treat molecules as point masses that bounce elastically and never attract one another, average over their collisions with the walls, and $PV = nRT$ falls out, with $RT$ revealed as a measure of the average kinetic energy per mole.

Those two assumptions are also the fine print, and here a common textbook line deserves correcting. The ideal gas law is a limit, not a fact about gases. Real molecules do occupy volume and do attract each other, so the equation is exact only as pressure approaches zero and the gas gets out of its own way. Near condensation it fails plainly: at 100 atm, or anywhere close to the boiling point, the error runs to tens of percent and you want van der Waals or a compressibility factor. Under ordinary room conditions the error is well under 1%, which is why the approximation earns its keep.

Two errors account for most wrong answers on this page, and both are unit errors rather than physics errors. The first is feeding in Celsius. Doubling a gas from 20 °C to 40 °C does not double anything: in kelvin that is 293 to 313, a rise of 7%, and a calculation that used 20 and 40 would be wrong by a factor of nearly two. The second is feeding in a gauge pressure. A tire gauge reading 220 kPa means 321 kPa absolute; $P$ here is absolute pressure, measured from vacuum, because the equation counts molecular impacts and vacuum is where there are none. A third, quieter trap: the familiar 22.4 L per mole belongs to 0 °C and 1 atm. IUPAC redefined standard pressure to 100 kPa in 1982, and at that pressure the molar volume is 22.71 L. Both numbers circulate, and quoting one against the other's conditions is a 1.3% error hiding inside a memorised constant.

Robert Boyle published his gas law in 1662, making it one of the oldest quantitative laws in physics: at constant temperature, the pressure and volume of a trapped gas are inversely proportional, so their product never changes. Cap a syringe and squeeze: compress 60 mL of air at 100 kPa down to 20 mL and the pressure climbs to 300 kPa. Molecularly, shrinking the space raises how often molecules hammer the walls, and pressure rises in exact proportion.

The law matters wherever gas gets squeezed. Scuba divers learn it first: air breathed at depth expands as they ascend, which is why the cardinal rule is never to hold your breath on the way up. Boyle's law assumes the temperature and the amount of gas stay fixed; change either and you need Charles's law or the combined gas law instead.

Hold the pressure on a gas constant and its volume follows its absolute temperature in strict proportion: warm it by 10% and it swells by 10%. The reason is worth having rather than memorising. Temperature is a measure of how fast the molecules are moving; pressure is how hard their impacts push on each square metre of wall. If the gas is to keep pushing with the same force while its molecules move faster, the only thing it can do is spread out, so that each patch of wall is struck less often. Volume rises exactly as fast as temperature to keep that balance, and $V_1/T_1 = V_2/T_2$ is that sentence in symbols.

A 2.5 L balloon leaves a 22 °C room and goes into a −18 °C freezer. In kelvin those are 295.15 and 255.15, so $V_2 = 2.5 \times 255.15/295.15 = 2.16\ \text{L}$: it loses about 14% of its volume and visibly puckers. Bring it back out and it recovers. Nothing left the balloon; the same molecules simply stopped needing as much room.

Jacques Charles found this with hydrogen balloons around 1787 and never published it. Joseph Louis Gay-Lussac did the careful work and published in 1802, and generously named the result after Charles. The interesting part is what came out of extending the straight line. Plot volume against Celsius temperature and you get a line that, extrapolated backwards, hits zero volume at about −273 °C. No one in 1802 could reach anywhere near that temperature, yet the graph pointed straight at it. That extrapolation is how absolute zero was first located, and it is why William Thomson could propose an absolute scale in 1848 by simply moving the origin to where the gases were pointing. The modern value, −273.15 °C, is the zero of the kelvin scale.

Which is exactly why a ratio in Celsius is meaningless here. Going from 20 °C to 40 °C does not double the volume; in kelvin that is 293.15 to 313.15, a rise of under 7%. Worse, a Celsius ratio breaks outright when the temperature crosses zero: 0 °C in the denominator gives infinity, and a negative Celsius temperature gives a negative volume. This page converts your °C or °F entries to kelvin before it does anything, but the trap is worth recognising when you meet the equation off-screen.

Two smaller conditions do real work. The pressure must actually be constant: a gas sealed in a rigid tank obeys Gay-Lussac's law instead, where pressure rises and volume does not move at all. And the gas must stay a gas. Cool steam through 100 °C and the relation does not merely become inaccurate, it stops applying: the vapour condenses and the volume collapses by a factor of about 1600. A balloon is also only approximately constant-pressure, since the stretched rubber adds a little tension of its own, which is why the real shrinkage runs slightly under what the arithmetic predicts.

Seal a gas in a container that cannot change size and there is only one thing left for it to do when you heat it: push harder. Pressure then tracks absolute temperature in direct proportion, $P_1/T_1 = P_2/T_2$. The mechanism is the same one behind Charles's law with the outcome swapped. Faster molecules strike the walls both more often and with more momentum each time, and since the walls will not move aside, all of that arrives as pressure.

Here is the case everyone actually meets, worked carefully, because the careless version is the standard error. A tire is set to 220 kPa gauge on a 5 °C morning and warms to 45 °C after an hour on the highway. Convert to absolute pressure first: 220 + 101 = 321 kPa absolute. Then $P_2 = 321 \times 318.15/278.15 = 367\ \text{kPa}$ absolute, which is 266 kPa on the gauge: a rise of 46 kPa, about 6.6 psi. Run the same calculation on the gauge reading alone and you get 252 kPa, understating the rise by a third. This is why tire pressures are specified cold, and why topping up a hot tire leaves it soft in the morning.

The relation is usually credited to Gay-Lussac's 1802 paper, though Guillaume Amontons had it a century earlier: around 1702 he built an air thermometer that worked on precisely this principle, and noticed that the pressure line extrapolated toward a temperature below which it could not go. Some texts call it Amontons's law for that reason. Combine it with Boyle's law and Charles's law and you have the combined gas law; add Avogadro and you have $PV = nRT$, of which this is the constant-volume slice.

Absolute pressure is the trap here, more than absolute temperature. The ratio $P_1/P_2$ is only meaningful when both pressures are measured from vacuum, because the equation is counting molecular impacts and a gauge has quietly subtracted an atmosphere from the count. Temperature has the same requirement for the same reason (kelvin, not Celsius) and this page converts your entries on both fronts. But when you meet the equation on paper, ask twice whether the pressure in your hand is gauge or absolute. It usually is gauge; almost every instrument in a mechanical room reads that way.

The other honest limit is that constant volume is an idealisation. A tire is not rigid: it grows a little as it warms, which relieves some of the pressure rise, so the measured increase runs slightly below the calculation. A steel cylinder is much closer to the ideal, which is what makes this law genuinely dangerous rather than merely academic. An aerosol can left on a dashboard, a propane cylinder in a closed vehicle, or a sealed pressure vessel in a fire all follow this line with nothing to relieve them, and the pressure keeps climbing until something gives. Relief valves exist because the equation has no upper bound.

The combined gas law merges Boyle's, Charles's, and Gay-Lussac's laws into a single statement: for a fixed amount of gas, PV/T is constant. A weather balloon shows all three variables moving at once. Launched with 2.0 m³ of helium at 101 kPa and 288 K, it rises to where the pressure is 30 kPa and the temperature 228 K; its new volume is V₂ = P₁V₁T₂ ÷ (P₂T₁) = 2.0 × 101 × 228 ÷ (30 × 288) ≈ 5.3 m³: more than double.

Hold any one variable constant and the named laws drop out: fix T for Boyle's law, fix P for Charles's, fix V for Gay-Lussac's. Temperatures must be absolute (the ratio of 20 °C to 40 °C is not 1:2 but 293:313), and Celsius or Fahrenheit inputs convert to kelvin automatically. Add Avogadro's insight about the amount of gas n and this law becomes the full ideal gas law, PV = nRT.

Start from $PV = nRT$, substitute $n = m/M$, and rearrange for mass over volume: $\rho = PM/RT$. What the result says is that a gas has no density of its own. Unlike a solid or a liquid, whose density is close enough to a fixed property to tabulate, a gas takes whatever density its pressure and temperature impose, and the only thing the substance itself contributes is $M$. Squeeze it and it gets denser in exact proportion; warm it and it thins in inverse proportion.

Air at 101.325 kPa and 20 °C, using $M = 0.028964$ kg/mol: $\rho = (101\,325 \times 0.028964)/(8.314 \times 293.15) = 1.204\ \text{kg/m}^3$, the figure every ventilation calculation starts from. Helium at the same conditions, with $M = 0.0040026$, comes to 0.166 kg/m³. Subtract, and a cubic metre of helium lifts about 1.04 kg. That is the entire physics of a party balloon, and it explains why a balloon large enough to lift a person has to be the size of a house.

Rearranged for molar mass the equation becomes a measurement rather than a prediction, and a historically important one. Weigh a bulb of known volume empty, fill it with a vapour at measured temperature and pressure, weigh it again, and $M = \rho RT/P$ hands you the molar mass of an unknown. This is the Dumas method, and through the middle of the nineteenth century it was one of the few routes to a molecular formula. It also settled arguments: measured vapour densities are what showed that many elemental gases travel as diatomic molecules rather than lone atoms.

The dominant error here is the molar mass unit, and it is a clean factor of a thousand. With $R = 8.314$ J/(mol·K) the equation demands $M$ in kilograms per mole. Air is 0.029 kg/mol. Enter 29 and the answer comes back as 1204 kg/m³, air denser than water, which at least announces itself. Enter 0.029 when the calculation wanted grams and you get the mirror error. The usual companions apply too: $P$ must be absolute, not a gauge reading, and $T$ must be in kelvin.

Two conceptual notes. There is no such thing as "the molar mass of air" in the strict sense: air is a mixture, and 28.96 g/mol is a mole-weighted average of nitrogen, oxygen and argon. It works precisely because an ideal gas is indifferent to what its neighbours are; only the total count matters. That same indifference produces a result most people find backwards: humid air is lighter than dry air. Water is 18 g/mol against air's 29, so at a given pressure and temperature every water molecule that joins the mixture has displaced a heavier one. Muggy days are low-density days, which is why aircraft performance charts include humidity and why a hot, humid runway is a long takeoff.

This is the arithmetic behind every hydroponic feed chart, and doing it yourself is what makes a nutrient program legible rather than a matter of following the bottle. The concentration of a nutrient element is the fertiliser weight times the element's share of that fertiliser, divided by the water volume.

One convenient fact makes the mental version easy: because a litre of water weighs a kilogram, one gram in one hundred litres is ten ppm. So 10 g of calcium nitrate at 15.5% nitrogen into 100 L gives 1.55 g of N in 100 L, which is 15.5 ppm.

The trap is the same one that catches field growers, and it is worth stating plainly: fertiliser labels report phosphorus and potassium as oxides, not elements. The P in an N-P-K analysis is P₂O₅ and the K is K₂O. To get elemental phosphorus multiply by 0.436; for elemental potassium multiply by 0.830. Hydroponic recipes are almost always written in elemental ppm, so feeding to a label's P number without converting overfeeds phosphorus by a factor of 2.3.

Reference targets for a general vegetative feed, in elemental ppm: N 150–200, P 50, K 200, Ca 150–200, Mg 50, S 60, with iron near 3 and the other micronutrients below 1. Those numbers are why calcium and magnesium deserve their own attention: they are needed at concentrations comparable to phosphorus, and a base nutrient designed for soft water will not supply them.

Evaporating a litre of water to dryness and weighing the residue is the only true TDS measurement, and it takes a drying oven and several hours. A conductivity meter gives an answer in two seconds, and because dissolved salts are what carry the current the two track each other closely enough for daily work: TDS in mg/L ≈ k × conductivity in µS/cm. The factor k is not a constant of nature: it depends on which ions dominate. Chloride-rich waters run near 0.55, typical fresh surface and well waters around 0.65, and sulphate- or bicarbonate-heavy waters climb toward 0.70. A tower reading 500 µS/cm at k = 0.65 is carrying roughly 325 mg/L of dissolved solids.

Two things will bite you. First, conductivity is strongly temperature-dependent, about 2% per °C, so every meaningful reading is temperature-compensated to 25 °C; a probe without compensation reading warm blowdown will overstate TDS by 15% or more. Second, non-ionic dissolved solids are invisible to the meter. Silica, dissolved organics and sugars contribute real gravimetric TDS but carry almost no current, which is why a high-silica cooling water or a food-plant effluent can gravimetrically test far above what its conductivity predicts. Calibrate k against one oven-dried sample of your own water and it becomes a genuinely reliable daily tool.

Hardness is a property, not a substance: it is the combined effect of every divalent cation in the water, and in practice calcium and magnesium account for essentially all of it. Because they have different atomic weights, the trade converts both to a single reference salt, calcium carbonate, so that numbers from different labs, different waters and different treatment steps can be added and subtracted. The factors come straight from equivalent weights: 50.04/20.04 = 2.497 for calcium and 50.04/12.15 = 4.118 for magnesium. A well water reporting 80 mg/L Ca and 25 mg/L Mg therefore has a total hardness of 2.497 × 80 + 4.118 × 25 = 199.8 + 103.0 = 303 mg/L as CaCO₃, or about 17.7 grains per gallon.

The single most common field error is reading a lab sheet that already says "as CaCO₃" and multiplying by 2.497 anyway: instant 2.5× overstatement, an oversized softener and a customer paying for salt they do not need. Check the units column before you touch the calculator. The second trap is the magnesium factor: because Mg²⁺ is light, a modest 25 mg/L of it contributes as much hardness as 41 mg/L of calcium, which is why waters from dolomitic aquifers feel far harder than their calcium number suggests and why magnesium-heavy water scales heat exchangers with a stubborn magnesium silicate rather than easily-acid-cleaned calcite.

The grain is a survivor from the old apothecaries' system (1/7000 of a pound, originally the weight of a single barleycorn) and North American water treatment never let it go. One grain of CaCO₃ dissolved in one US gallon is 64.79891 mg in 3.785412 L, which is 17.118 mg/L. That is the whole conversion: 10 gpg is 171 mg/L, and a residential softener rated "30,000 grains" will remove 30,000 grains of hardness before it needs salt. Softener control valves, resin capacity tables and salt-dose settings are all published in grains, while every laboratory report and every regulatory limit is in mg/L, so the technician lives in both worlds at once.

Watch out for the imperial gallon. A British or Canadian "grain per gallon" historically meant grains per imperial gallon (4.546 L), which works out to 14.25 mg/L: a 20% difference that will undersize a softener if you take a UK spec sheet at face value. And remember the informal hardness bands are quoted in grains: under 1 gpg is soft, 1–3.5 slightly hard, 3.5–7 moderately hard, 7–10.5 hard, and anything over 10.5 gpg (180 mg/L) is very hard and will scale a water heater within a few years untreated.

A water analysis lists a dozen ions with a dozen different atomic weights, and you cannot add or subtract them directly: 20 mg/L of calcium and 20 mg/L of sodium are not the same amount of chemistry. Expressing everything "as CaCO₃" fixes that by converting each ion to the mass of calcium carbonate that carries the same number of charge equivalents. Calcium carbonate has a molar mass of 100.09 and a valence of 2, so its equivalent weight is 50.04 g/eq; divide that by the ion's own equivalent weight and you have the factor. Calcium (EW 20.04) gets 2.497, magnesium (12.15) gets 4.118, sodium (23.0) gets 2.18, bicarbonate (61.0) gets 0.82, and sulphate (48.03) gets 1.04. Forty mg/L of Ca²⁺ becomes 40 × 2.497 = 99.9 mg/L as CaCO₃.

Once every ion is on the CaCO₃ scale you can do the things that make an analysis useful: check that cations balance anions to within a few percent (the classic sanity test on any lab report), split total hardness into carbonate and non-carbonate fractions by comparing hardness to alkalinity, and size a softener or a dealkalizer directly from the numbers. The trap is the word "equivalent": the divisor is the equivalent weight, molar mass divided by charge, not the molar mass. Using 40.08 for calcium instead of 20.04 halves every hardness figure you produce.

Reactions in water happen equivalent-for-equivalent, not gram-for-gram: one charge neutralizes one charge, one H⁺ neutralizes one OH⁻. The equivalent weight is simply the mass of a species that carries one mole of that reacting capacity: molar mass divided by valence. Calcium carbonate, M = 100.087 with a divalent cation, has EW = 50.04 g/eq, which is why 50.04 turns up in every "as CaCO₃" conversion in this trade. Sulphuric acid, M = 98.08 with two replaceable protons, has EW = 49.04 g/eq, so a pound of pure H₂SO₄ neutralizes almost exactly a pound of alkalinity expressed as CaCO₃: the near-1:1 coincidence that lets operators do acid-feed arithmetic in their heads.

The valence you divide by is the one that applies to the reaction in question, and that is where people go wrong. Sodium carbonate reacting as a base uses z = 2 (EW 53.0), but if you are counting sodium ions it behaves as z = 2 for a different reason. Bicarbonate is z = 1 (EW 61.0) in the alkalinity titration but the carbonate it came from is z = 2 (EW 30.0). Write down which reaction you mean before you pick z, and the rest of the water chemistry falls into line.

Chlorine added to water does not stay chlorine. It is consumed by iron, manganese, sulphide, ammonia, and every scrap of organic matter it meets, that consumption is the demand, and only what survives shows up on a DPD test as residual. The bookkeeping is a straight sum: dose equals demand plus residual. Feed 3.3 mg/L into a water with 2.5 mg/L of demand and you keep 0.8 mg/L residual, which is a typical distribution target. There is no way around the demand; it has to be satisfied before a single tenth of a milligram of residual appears, which is why a well with high iron can swallow 4 mg/L and still test zero at the tap.

The practical procedure is a chlorine demand study: dose a set of bottles of the actual water at increasing rates, wait the real contact time (usually 30 minutes, but overnight for a distribution main) and test each. The plot of residual against dose is flat until demand is satisfied, then rises with unit slope. Two field cautions. Demand is not a fixed property: it climbs in warm weather, after a main break, and with every degree of surface-water turbidity, so the study needs repeating seasonally. And test at the far end of the system, not at the plant: the residual that matters legally and microbiologically is the one still standing at the last service connection.

Every North American operator learns this one as a chant: pounds per day equals milligrams per litre times million gallons per day times 8.34. The 8.34 is nothing mysterious: it is the weight of a US gallon of water, 8.345 lb at 60 °F, rounded. Dosing 2 mg/L of chlorine into a 1 MGD plant needs 2 × 1 × 8.34 = 16.7 lb/day. Underneath the trade constant it is simply concentration times flow, mass balance in its plainest form, and this page will do it in kilograms per hour and cubic metres per hour just as happily.

Three warnings from the field. First, this gives pounds of active chemical; if you are feeding 12.5% sodium hypochlorite you still have to divide by the strength to get pounds of product, and again by the solution's density to get gallons the pump must deliver. Second, the 8.34 quietly assumes the water is water: feeding into a brine, a slurry or a glycol loop needs the actual density. Third, calibrate the pump against a drawdown cylinder rather than trusting the dial: a diaphragm pump's stroke setting drifts with discharge pressure, and the difference between a calculated feed rate and a measured one is where half of all chemical overspend hides.

Hydrogen ion concentrations in water span more than ten orders of magnitude, so in 1909 the Danish chemist Søren Sørensen, working at the Carlsberg brewery laboratory, compressed them onto a logarithmic scale. Each pH unit is a factor of ten: lemon juice at pH 2 carries a hundred times the hydrogen ion concentration of tomato juice at pH 4. Pure water at 25 °C sits at pH 7, where [H⁺] = 1.0 × 10⁻⁷ mol/L. The logarithm takes the concentration in mol/L, and the inverse direction is exact: [H⁺] = $10^{-\text{pH}}$.

In water treatment pH is the master variable. Boiler water is typically held between 10.5 and 11.5 to suppress corrosion, cooling towers near 7–9 to balance scale against corrosion, and municipal drinking water around 7.0–8.5. A pH swing of a single unit, a tenfold chemistry change, is often the first sign that a chemical feed pump has failed.

Water is never merely water. A small fraction of it is always pulled apart into H⁺ and OH⁻, and the equilibrium between them fixes the product of their concentrations at a constant: $K_w = [\text{H}^+][\text{OH}^-] = 1.0\times10^{-14}$ at 25 °C. Because it is a product that is constant, the two concentrations are locked in a see-saw: push one up and the other must come down by the same factor. Take negative logarithms of both sides and the product becomes a sum: $\mathrm{pH} + \mathrm{pOH} = 14$. So pH and pOH are not two independent measurements. They are one number written two ways, and knowing either gives the other for free.

A worked case. Dissolve enough sodium hydroxide to make a 0.010 M solution: it dissociates completely, so $[\text{OH}^-] = 1.0\times10^{-2}$ and pOH = 2.00. Then pH = 14 − 2.00 = 12.00. Check it the long way, $[\text{H}^+] = K_w/[\text{OH}^-] = 10^{-14}/10^{-2} = 10^{-12}$, giving pH 12.00, and the two routes agree, as they must.

The "14" is not a magic number, it is $\mathrm{p}K_w$, and the exact value at 25 °C is 13.995. The whole logarithmic apparatus came from Søren Sørensen at the Carlsberg laboratory in 1909, who needed a compact way to talk about hydrogen ion concentrations spanning more than ten orders of magnitude in brewing. The p-prefix now attaches to anything worth compressing that way: pOH, pKa, pKw, pCa.

The trap is that 14 belongs to 25 °C and nothing else. Self-ionisation is endothermic, so heating water drives it forward and $K_w$ rises: pKw is about 14.9 at 0 °C, 13.0 at 60 °C, and 12.0 at 100 °C. Which means neutral water at 100 °C has a pH of 6.0, and it is still perfectly neutral, not acidic, because pH still equals pOH. This matters wherever hot water is measured: boiler water sampled at temperature and boiler water sampled after cooling do not read the same, and the difference is the equation moving, not the chemistry changing. Anyone reading pH on a hot sample needs to know whether the meter is compensating and what reference temperature it is compensating to.

Which leads to the deeper misconception. Neutral does not mean pH 7; neutral means pH equals pOH. The two coincide at 25 °C by arithmetic accident, and part company at every other temperature. Two smaller notes: the relation is aqueous only (it says nothing about a non-aqueous solvent, which has its own autoionisation constant if it has one at all) and the pH scale is not fenced between 0 and 14. Those bounds are just where 1 M solutions land. Concentrated hydrochloric acid at 12 M has a negative pH, and concentrated sodium hydroxide runs above 14, with the sum still holding.

At the equivalence point the moles of titrant delivered, $C_b V_b$, exactly match the moles of analyte present, $C_a V_a$, scaled by the balanced equation's mole ratio n. Everything else in a titration (the burette, the indicator, the swirling) exists only to find that point precisely. Titrate a 25.00 mL aliquot of hydrochloric acid with 0.1000 M sodium hydroxide and take 23.45 mL to reach the endpoint: with n = 1, $C_a$ = (1 × 0.1000 × 23.45)/25.00 = 0.09380 M, good to four figures from nothing but glassware.

The mole ratio is where marks are lost. For a diprotic acid such as H₂SO₄ titrated with NaOH, one mole of acid consumes two of base, so n = 0.5: a 25.00 mL aliquot needing 30.00 mL of 0.100 M NaOH is 0.5 × 0.100 × 30.00/25.00 = 0.0600 M. Karl Friedrich Mohr systematised the whole technique in his 1855 Lehrbuch der chemisch-analytischen Titrirmethode, introducing the burette clamp and the pinchcock that made reproducible volumetric analysis possible; his methods still underpin water-hardness and chlorine testing today. Note the difference between the endpoint (where the indicator changes) and the equivalence point (where the stoichiometry balances): the gap between them is the indicator error, which is why the indicator is chosen to change colour on the steep part of the titration curve.