Some Basic Concepts of Chemistry

This chapter lays the measurement and counting foundation for the entire Class 11–12 chemistry syllabus. The guided map below follows the logical order the NCERT text itself builds in.

• Nature and classification of matter — the three states, and the mixture/pure-substance/element/compound hierarchy.
• Measurement in chemistry — the SI base units, mass versus weight, volume, density and the three temperature scales.
• Handling uncertain numbers — scientific notation, the rules for counting significant figures, precision versus accuracy, and dimensional analysis (the unit-factor method).
• The historical laws of chemical combination and how they add up to Dalton's atomic theory.
• Atomic mass, average atomic mass, molecular mass and formula mass — how "the" atomic mass of an element is actually a weighted average over isotopes.
• The mole concept and molar mass — the bridge between the microscopic (atoms, molecules) and the macroscopic (grams you can weigh).
• Percentage composition and empirical/molecular formula — working backward from experimental mass data to a compound's identity.
• Stoichiometry and the limiting reagent — using a balanced equation to calculate exact masses of reactants consumed and products formed.
• Concentration of solutions — mass percent, mole fraction, molarity and molality, and why only some of these change with temperature.

Matter: states and classification

Matter is anything that has mass and occupies space, and it exists in three familiar physical states — solid, liquid and gas — distinguished purely by how tightly and how orderly the constituent particles are packed and how freely they can move. In a solid the particles sit very close together in an orderly arrangement with little freedom of movement, which is why solids have a definite shape and a definite volume. In a liquid the particles are still close but can move past one another, so a liquid has a definite volume but takes the shape of its container. In a gas the particles are far apart and move fast and freely, so a gas has neither a definite shape nor a definite volume — it expands to fill whatever container holds it. Heating drives a substance solid to liquid to gas; cooling reverses the sequence.

At the macroscopic level, matter is either a mixture or a pure substance. A mixture contains two or more pure substances in variable proportion; it can be separated back into its components by purely physical methods such as filtration, distillation or crystallisation. A mixture is homogeneous if the composition is uniform throughout (sugar solution, air) or heterogeneous if it is not (salt mixed with sand). A pure substance has a fixed composition throughout and splits further into elements, whose particles (atoms or, for gases like hydrogen and oxygen, molecules made of one kind of atom) are all of one type, and compounds, in which atoms of two or more different elements combine in a fixed, definite ratio — and which can only be broken back into elements by chemical, not physical, means.

Measurement: SI units, mass, volume, density and temperature

Any physical property worth reporting is a number followed by a unit — "$6\ \text{m}$" means nothing without the metre. Chemistry has settled on the International System of Units (SI), established by the General Conference on Weights and Measures, which fixes seven base quantities: length (metre, m), mass (kilogram, kg), time (second, s), electric current (ampere, A), thermodynamic temperature (kelvin, K), amount of substance (mole, mol) and luminous intensity (candela, cd). Every other unit — speed, volume, density — is derived from these seven. SI also allows standard prefixes (kilo- for $10^3$, milli- for $10^{-3}$, micro- for $10^{-6}$, nano- for $10^{-9}$, and so on) so that a quantity is never written with an unwieldy string of zeros. A distinction the chapter insists on: mass is the amount of matter in an object and is constant wherever the object goes, while weight is the force gravity exerts on that mass and therefore varies from place to place. In the laboratory mass is measured with an analytical balance and reported in grams, a convenient fraction of the SI kilogram. Volume, the space a substance occupies, has SI units of $\text{m}^3$ but is more conveniently expressed in a chemistry lab as $\text{cm}^3$ or $\text{dm}^3$, or using the non-SI litre, where $1\ \text{L} = 1000\ \text{mL} = 1000\ \text{cm}^3 = 1\ \text{dm}^3$. Density links mass and volume:

where: density carries SI units of $\text{kg m}^{-3}$, but a chemist almost always reports it in $\text{g cm}^{-3}$, since laboratory masses and volumes are small. A denser substance packs more mass into the same volume — its particles are more closely packed. Three temperature scales are in everyday use — Celsius ($^\circ\text{C}$), Fahrenheit ($^\circ\text{F}$) and the SI unit, kelvin (K) — related by

where: the Celsius scale is pinned so that water freezes at $0^\circ\text{C}$ and boils at $100^\circ\text{C}$; the Fahrenheit scale runs from $32^\circ$ to $212^\circ$ over the same physical span. Negative temperatures are perfectly ordinary in Celsius, but the kelvin scale starts at absolute zero, so a negative kelvin temperature is never possible — this is the reason thermodynamic calculations later in the course are always done in kelvin.

Uncertainty in measurement: scientific notation, significant figures, precision and accuracy

Because chemistry deals with numbers as large as $6.022\times10^{23}$ molecules and as small as $1.66\times10^{-24}$ g, every quantity is written in scientific notation, $N \times 10^{n}$, where $N$ is a digit term between $1.000\ldots$ and $9.999\ldots$ and $n$ is a positive or negative integer exponent. Thus $232.508$ becomes $2.32508 \times 10^{2}$ (decimal moved two places left, exponent $+2$) and $0.00016$ becomes $1.6 \times 10^{-4}$ (decimal moved four places right, exponent $-4$). Multiplication and division of numbers in this form multiply or divide the digit terms and add or subtract the exponents; addition and subtraction require both numbers to first share the same exponent, after which only the digit terms are combined. Every real measurement carries some uncertainty from the limits of the instrument and the observer's skill, and significant figures are how that honesty is recorded: they are all the digits known with certainty plus one final estimated digit. The standard rules are that every non-zero digit is significant; zeros before the first non-zero digit are never significant (they only fix the decimal point, so $0.0052$ has two significant figures); zeros sandwiched between non-zero digits are always significant ($2.005$ has four); trailing zeros after a decimal point are significant ($0.200$ has three), but trailing zeros in a number with no decimal point are ambiguous and best avoided by writing in scientific notation ($1.00\times10^{2}$ unambiguously carries three significant figures); and counted (exact) numbers, like $20$ eggs, have effectively infinite significant figures. When combining measured numbers, addition and subtraction are limited by the operand with the fewest digits after the decimal point, while multiplication and division are limited by the operand with the fewest significant figures overall — these are different rules, and mixing them up is one of the most common marks lost in this chapter. Rounding follows the standard "round half to even" convention: if the digit being dropped is $5$ exactly, the preceding digit is left unchanged if it is already even and increased by one if it is odd (so $6.25 \to 6.2$ but $6.35 \to 6.4$).

Precision and accuracy answer different questions. Precision is how close repeated measurements of the same quantity are to each other; accuracy is how close a measurement (or its average) is to the true value. A student who measures $1.95\ \text{g}$ and $1.93\ \text{g}$ against a true value of $2.00\ \text{g}$ is precise (the two readings agree closely) but not accurate (both miss the true value); a student who gets $1.94\ \text{g}$ and $2.05\ \text{g}$ is neither precise nor accurate; a student who gets $2.01\ \text{g}$ and $1.99\ \text{g}$ is both. High precision never guarantees high accuracy — a systematically miscalibrated instrument can give beautifully consistent, uniformly wrong readings. Finally, dimensional analysis (also called the factor-label or unit-factor method) is the systematic technique for converting a quantity from one system of units to another. It rests on constructing a unit factor — a ratio built from an equivalence such as $1\ \text{in} = 2.54\ \text{cm}$, so that $\frac{2.54\ \text{cm}}{1\ \text{in}} = 1$ — and multiplying the original quantity by whichever orientation of that ratio cancels the unwanted unit and leaves the desired one. Because the unit factor equals exactly $1$, multiplying by it never changes the physical quantity, only how it is expressed. This method is developed fully, and generalised to a chain of several conversions at once, in the derivation below.

Laws of chemical combination and Dalton's atomic theory

Long before atoms could be seen, five carefully measured experimental regularities forced chemists to conclude that matter is discrete rather than continuous. The Law of Conservation of Mass (Antoine Lavoisier, 1789) states that in any physical or chemical change, the total mass of the system stays exactly the same — matter is neither created nor destroyed, only rearranged. The Law of Definite Proportions (Joseph Proust) states that a given compound always contains the same elements combined in exactly the same proportion by mass, regardless of the source or method of preparation — Proust proved this by showing natural and synthetic cupric carbonate both contained $51.35\%$ copper, $9.74\%$ carbon and $38.91\%$ oxygen. The Law of Multiple Proportions (John Dalton, 1803) states that when two elements combine to form more than one compound, the different masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers — hydrogen and oxygen form both water ($2\ \text{g H}$ with $16\ \text{g O}$) and hydrogen peroxide ($2\ \text{g H}$ with $32\ \text{g O}$), and the two oxygen masses, $16\ \text{g}$ and $32\ \text{g}$, stand in the small whole-number ratio $1:2$. Gay Lussac's Law of Gaseous Volumes (1808) states that gases combine, or are produced, in a simple whole-number ratio by volume at the same temperature and pressure — $100\ \text{mL}$ hydrogen reacts with $50\ \text{mL}$ oxygen to give $100\ \text{mL}$ water vapour, a $2:1:2$ volume ratio. Avogadro's Law (1811) then supplied the explanation: equal volumes of any gas, at the same temperature and pressure, contain equal numbers of molecules — a claim that only made sense once elements like hydrogen and oxygen were accepted as existing as diatomic molecules, $\text{H}_2$ and $\text{O}_2$, rather than single atoms.

Dalton's atomic theory (1808) is the single framework that accounts for all five laws at once: matter consists of indivisible atoms; all atoms of a given element are identical, including in mass, and atoms of different elements have different masses; compounds form when atoms of different elements combine in a fixed ratio; and chemical reactions merely reorganise atoms — none are created or destroyed. Conservation of mass follows because atoms are neither created nor destroyed; definite proportions follows because a compound always has the same atom ratio; multiple proportions follows because atoms combine in small whole numbers. What Dalton's original theory could not explain was Gay Lussac's law of gaseous volumes — that gap was closed only by Avogadro's distinction between atoms and molecules.

Atomic mass, average atomic mass, molecular mass and formula mass

A single atom is far too light to weigh directly, so atomic masses are reported relative to a standard. The modern standard, agreed in 1961, is the carbon-12 isotope: one atomic mass unit (u), also written amu, is defined as exactly $\dfrac{1}{12}$ the mass of one $^{12}\text{C}$ atom, so that $^{12}\text{C}$ itself has a mass of exactly $12\ \text{u}$, and $1\ \text{u} = 1.66\times10^{-24}\ \text{g}$. On this scale a hydrogen atom, whose true mass is $1.6736\times10^{-24}\ \text{g}$, works out to $1.008\ \text{u}$. Most elements occur naturally as a mixture of isotopes — atoms with the same number of protons but different numbers of neutrons, hence different masses — so the atomic mass quoted in a periodic table is really the average atomic mass, the abundance-weighted mean over all naturally occurring isotopes of that element. Carbon, for example, occurs as $^{12}\text{C}$ ($98.892\%$ abundant, mass $12\ \text{u}$ exactly), $^{13}\text{C}$ ($1.108\%$, mass $13.00335\ \text{u}$) and a trace of $^{14}\text{C}$, giving the familiar $12.011\ \text{u}$ used throughout the syllabus — never $12$ exactly. Molecular mass is the sum of the atomic masses of every atom in one molecule of a compound, so methane's molecular mass is $12.011 + 4(1.008) = 16.043\ \text{u}$ and water's is $2(1.008) + 16.00 = 18.02\ \text{u}$ (NCERT Problem 1.1 works the same idea for glucose, $\text{C}_6\text{H}_{12}\text{O}_6$, below). Ionic solids such as sodium chloride do not exist as discrete NaCl molecules — Na⁺ and Cl⁻ ions instead pack into an extended three-dimensional lattice, with each Na⁺ surrounded by six Cl⁻ and vice versa — so chemists instead quote a formula mass, computed exactly like a molecular mass from the formula unit: NaCl's formula mass is $23.0 + 35.5 = 58.5\ \text{u}$.

The mole concept and molar mass

Just as a dozen always means $12$ items and a gross always means $144$, chemists needed one fixed number to count atoms, molecules and ions at the microscopic scale, because the actual numbers involved in any weighable sample are always astronomically large. That number is the mole, the seventh SI base quantity: one mole contains exactly $6.02214076\times10^{23}$ elementary entities, a constant known as the Avogadro constant, $N_A \approx 6.022\times10^{23}\ \text{mol}^{-1}$. It was fixed historically by dividing the mass of one mole of $^{12}\text{C}$ ($12\ \text{g}$, by definition) by the mass of a single $^{12}\text{C}$ atom measured with a mass spectrometer ($1.992648\times10^{-23}\ \text{g}$).

where: $n$ is the amount of substance in moles, $m$ is the sample's mass in grams, $M$ is its molar mass in $\text{g mol}^{-1}$, and $N$ is the number of elementary entities (atoms, molecules, ions or formula units, whichever the substance is made of). The molar mass, the mass of one mole of a substance, is numerically equal in grams to the atomic, molecular or formula mass already computed in u — so water's molar mass is $18.02\ \text{g mol}^{-1}$ and sodium chloride's is $58.5\ \text{g mol}^{-1}$. Equation (3) is the single most-used relation in the whole chapter: it is the two-way bridge between what a balance can weigh (mass) and what a chemical equation actually counts (moles of particles).

Percentage composition and empirical/molecular formula

Given an unknown compound, the first practical question is what elements it contains and in what ratio. This is answered by the mass percentage of each element:

where: the numerator is the total mass contributed by all atoms of that element in the formula (for water's hydrogen: $2\times1.008 = 2.016\ \text{g}$ out of a $18.02\ \text{g}$ molar mass, giving $11.18\%$ hydrogen and, by difference, $88.79\%$ oxygen). The empirical formula is the simplest whole-number ratio of atoms in a compound, while the molecular formula gives the actual number of each type of atom in one molecule; the molecular formula is always a positive-integer multiple $n$ of the empirical formula, $n = \dfrac{\text{molar mass}}{\text{empirical formula mass}}$, and the two coincide exactly when $n = 1$ (water and methane are their own empirical formulas). The full working, from raw mass-percent data to a final molecular formula, is carried out step by step in the derivation below, reusing NCERT's own Problem 1.2.

Stoichiometry, balanced equations and the limiting reagent

A balanced chemical equation, obeying the law of conservation of mass, has the same number of atoms of every element on both sides; its coefficients (the stoichiometric coefficients) give the mole ratio — and therefore, through molar mass, the mass ratio — in which reactants combine and products form. For the combustion of methane,

where: the coefficients read that $1\ \text{mol CH}_4$ reacts with $2\ \text{mol O}_2$ to give $1\ \text{mol CO}_2$ and $2\ \text{mol H}_2\text{O}$ — equivalently, $16\ \text{g CH}_4$ reacts with $64\ \text{g O}_2$ to give $44\ \text{g CO}_2$ and $36\ \text{g H}_2\text{O}$ (mass is conserved: $16+64=80=44+36$). Balancing itself is done by inspection: write the correct formulas first (subscripts, once fixed by the actual compound, are never altered to balance an equation — only the coefficients in front of a formula may change), then balance one element at a time, usually finishing with oxygen, and finally re-check every element on both sides. In practice, reactants are rarely supplied in the exact mole ratio a reaction demands. Whichever reactant would be completely consumed first, if the reaction ran to completion, is the limiting reagent — it caps how much product can possibly form, no matter how much excess of the other reactant is present. Identifying it, and computing the resulting product mass, is the subject of the stoichiometry derivation below, worked with NCERT's own ammonia-synthesis data.

Concentration of solutions

Most reactions in a laboratory happen in solution, so chemists need a compact way to state how much solute is dissolved in how much solvent or solution. Four standard measures cover almost every situation.

where: molarity ($M$, $\text{mol L}^{-1}$) is moles of solute per litre of solution (not solvent), while molality ($m$, $\text{mol kg}^{-1}$) is moles of solute per kilogram of solvent. Because volume expands and contracts with temperature, molarity is temperature-dependent; because mass does not change with temperature, molality is not — this is exactly why molality, not molarity, is the concentration unit used in precise, temperature-sensitive work such as colligative-property calculations later in the syllabus. Diluting a stock solution obeys the simple identity $M_1V_1 = M_2V_2$, since the moles of solute present are unchanged by adding solvent — only the volume, and hence the concentration, changes.

where: $x_A$ is dimensionless and the mole fractions of all components of a solution always sum to $1$ ($x_A + x_B = 1$ for a two-component solution). Mass percent (or weight percent, w/w %) is simply $\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 100$, structurally identical to equation (4) but applied to a solute-in-solvent mixture rather than to a compound's own atoms. As a quick illustration of why the molarity/molality distinction matters: a $3\ \text{M}$ NaCl solution of density $1.25\ \text{g mL}^{-1}$ contains $175.5\ \text{g}$ NaCl and $1074.5\ \text{g}$ water per litre, giving a molality of $\dfrac{3}{1.0745} = 2.79\ \text{m}$ — a different number from the molarity, and one that would stay $2.79\ \text{m}$ even if the same solution were warmed until its volume, and hence its molarity, changed.

Board level - Scientific notation, significant figures and molecular mass of glucose (reusing NCERT Problem 1.1)

Express $602{,}200{,}000{,}000{,}000{,}000{,}000{,}000$ in scientific notation to four significant figures, and calculate the molecular mass of glucose, $\text{C}_6\text{H}_{12}\text{O}_6$, using $A_{\text{C}} = 12.011\ \text{u}$, $A_{\text{H}} = 1.008\ \text{u}$, $A_{\text{O}} = 16.00\ \text{u}$.

Step 1. Convert the large number to scientific notation: moving the decimal point $23$ places to the left gives $6.022\times10^{23}$ — this is simply the Avogadro constant, $N_A$, written out in full, and it already has exactly four significant figures.

Step 2. Multiply each atomic mass by the number of atoms of that element in one glucose molecule: carbon contributes $6\times12.011 = 72.066\ \text{u}$; hydrogen contributes $12\times1.008 = 12.096\ \text{u}$; oxygen contributes $6\times16.00 = 96.00\ \text{u}$.

Step 3. Add the three contributions: $72.066 + 12.096 + 96.00 = 180.162\ \text{u}$. Answer: $6.022\times10^{23}$ (Avogadro's number in scientific notation), and the molecular mass of glucose is $180.162\ \text{u}$, so one mole of glucose weighs $180.162\ \text{g}$.

Board level - Percentage composition of ethanol and mass of oxygen in a sample

Ethanol has molecular formula $\text{C}_2\text{H}_5\text{OH}$, molar mass $46.068\ \text{g mol}^{-1}$. Find its percentage composition, then find the mass of oxygen contained in a $23.034\ \text{g}$ sample of pure ethanol.

Step 1. Compute each element's mass in one mole of ethanol: carbon $= 2\times12.01 = 24.02\ \text{g}$; hydrogen $= 6\times1.008 = 6.048\ \text{g}$; oxygen $= 16.00\ \text{g}$.

Step 2. Apply equation (4) to each: mass \% carbon $= \dfrac{24.02}{46.068}\times100 = 52.14\%$; mass \% hydrogen $= \dfrac{6.048}{46.068}\times100 = 13.13\%$; mass \% oxygen $= \dfrac{16.00}{46.068}\times100 = 34.73\%$ (the three percentages sum to $100.00\%$, a useful self-check).

Step 3. The $23.034\ \text{g}$ sample is exactly half a mole of ethanol ($23.034/46.068 = 0.500\ \text{mol}$), so it contains exactly half of one mole's worth of oxygen mass: $0.500\times16.00\ \text{g} = 8.00\ \text{g}$. Equivalently, using the percentage directly: $23.034\times0.3473 = 8.00\ \text{g}$. Answer: ethanol is $52.14\%$ C, $13.13\%$ H, $34.73\%$ O by mass, and a $23.034\ \text{g}$ sample contains $8.00\ \text{g}$ of oxygen.

Competitive level (JEE / NEET) - Empirical and molecular formula from mass percentage (reusing NCERT Problem 1.2)

A compound is found by analysis to contain $4.07\%$ hydrogen, $24.27\%$ carbon and $71.65\%$ chlorine by mass, with a measured molar mass of $98.96\ \text{g mol}^{-1}$. Determine its empirical and molecular formulas.

Step 1. Take $100\ \text{g}$ of compound as the basis, giving $4.07\ \text{g}$ H, $24.27\ \text{g}$ C and $71.65\ \text{g}$ Cl, and convert each to moles: $n_{\text{H}} = 4.07/1.008 = 4.04$, $n_{\text{C}} = 24.27/12.01 = 2.021$, $n_{\text{Cl}} = 71.65/35.453 = 2.021$.

Step 2. Divide by the smallest value, $2.021$: $\text{H} : \text{C} : \text{Cl} = 2 : 1 : 1$, giving empirical formula $\text{CH}_2\text{Cl}$ with empirical formula mass $12.01 + 2.016 + 35.453 = 49.48\ \text{g mol}^{-1}$.

Step 3. Find $n = \dfrac{98.96}{49.48} = 2.0$ and scale every subscript in the empirical formula by $2$. Answer: empirical formula $\text{CH}_2\text{Cl}$; molecular formula $\text{C}_2\text{H}_4\text{Cl}_2$ (this is 1,2-dichloroethane). Timing note: under two minutes if the moles are computed in one pass. The single biggest speed trap is stopping at the empirical formula and forgetting the question asked for the molecular formula too — always check whether a molar mass is given in the problem statement, because its presence is the signal that a Step 3 (find $n$) is expected. Keep atomic masses to four significant figures throughout so the final $n$ lands cleanly on a whole number; sloppy intermediate rounding is what makes $n$ come out as $1.97$ or $2.04$ instead of a clean $2.0$.

Competitive level (JEE / NEET) - Limiting reagent and mass of ammonia formed (reusing NCERT Problem 1.5)

$50.0\ \text{kg}$ of $\text{N}_2(g)$ and $10.0\ \text{kg}$ of $\text{H}_2(g)$ are mixed and allowed to react completely by $\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)$. Calculate the mass of $\text{NH}_3$ formed, identify the limiting reagent, and state which reactant remains and how much.

Step 1. Convert both masses to moles: $n_{\text{N}_2} = 50{,}000/28.0 = 1786\ \text{mol}$; $n_{\text{H}_2} = 10{,}000/2.016 = 4960\ \text{mol}$.

Step 2. Test the limiting reagent: fully consuming $1786\ \text{mol}\ \text{N}_2$ would need $1786\times3 = 5358\ \text{mol}\ \text{H}_2$, but only $4960\ \text{mol}$ is available — so $\text{H}_2$ is exhausted first and is the limiting reagent.

Step 3. Base the product calculation entirely on the limiting reagent: $4960\ \text{mol}\ \text{H}_2 \times \dfrac{2\ \text{mol NH}_3}{3\ \text{mol H}_2} = 3307\ \text{mol NH}_3$, which is $3307\times17.0 = 56{,}200\ \text{g} = 56.2\ \text{kg NH}_3$.

Step 4. Find the leftover reactant: $\text{N}_2$ actually consumed is $4960/3 = 1653\ \text{mol}$, so unreacted $\text{N}_2$ is $1786 - 1653 = 133\ \text{mol} \times 28.0\ \text{g mol}^{-1} = 3.7\ \text{kg}$. Answer: $56.2\ \text{kg}\ \text{NH}_3$ forms; $\text{H}_2$ is the limiting reagent; about $3.7\ \text{kg}$ of $\text{N}_2$ remains unreacted. Timing note: about two to three minutes. JEE/NEET routinely disguise a limiting-reagent question as a plain "calculate the mass of product" problem with two reactant masses given instead of one — seeing two masses for two different reactants is itself the cue to run the limiting-reagent test before touching the product calculation. Never average, add or otherwise combine the two reactant masses; always convert both independently to moles first, then compare via the stoichiometric ratio as in Step 2.

Confusing the significant-figure rule for addition/subtraction with the rule for multiplication/division. Addition and subtraction are limited by the fewest decimal places among the operands; multiplication and division are limited by the fewest significant figures overall. Applying the wrong rule silently drops or invents precision in the final answer.

Treating trailing zeros in a whole number as automatically significant. $100$ has one significant figure unless a decimal point or scientific notation says otherwise; write $1.00\times10^{2}$ if all three digits are meant to be significant. This ambiguity is exactly why scientific notation exists.

Mixing up empirical formula and molecular formula. Percentage composition data alone, no matter how carefully worked, can only ever deliver the empirical (simplest-ratio) formula. A separately measured molar mass is always required to compute $n$ and recover the true molecular formula — a formula question that gives only mass percentages, with no molar mass, is asking for the empirical formula and nothing more.

Forgetting that molarity depends on temperature while molality does not. Molarity is defined per litre of solution, and solution volume expands on heating; molality is defined per kilogram of solvent mass, which heating does not change. Using molarity in a context that assumes a temperature-independent concentration is a conceptual, not just an arithmetic, error.

Skipping the limiting-reagent test whenever two reactant masses are given. If a problem states the amounts of both reactants, it is testing whether the student checks which one runs out first before computing the product mass — computing product yield directly from just one reactant's moles, without comparing, is the single most common stoichiometry error.

Changing subscripts to balance an equation. Only the coefficients in front of a formula may be changed when balancing; the subscripts inside a formula are fixed by the compound's actual composition and altering them describes an entirely different (and usually non-existent) substance.

Assuming average atomic mass is the mass of a real atom. No chlorine atom actually has a mass of $35.45\ \text{u}$ — every real atom is close to $35$ or $37$ u. The average is a weighted statistical property of a natural sample, valid for bulk calculations but not for any single atom.

• Classify any given sample of matter as a homogeneous mixture, a heterogeneous mixture, an element or a compound, and justify the classification.
• Convert a measured physical quantity between SI base units and their common prefixed multiples using the chained unit-factor method.
• State the correct number of significant figures in a given measurement and apply the correct rounding rule when the operations are addition/subtraction versus multiplication/division.
• Distinguish precision from accuracy and identify which, both, or neither a given data set demonstrates.
• State all five laws of chemical combination in your own words and connect each one to a postulate of Dalton's atomic theory.
• Calculate the average atomic mass of an element from isotopic abundance data.
• Calculate molecular mass or formula mass for any given chemical formula, and convert freely between mass, moles and number of particles using $n = m/M$ and $N = nN_A$.
• Determine a compound's empirical formula from percentage composition and its molecular formula given the molar mass.
• Balance a chemical equation by inspection and use its coefficients to calculate reactant or product masses, correctly identifying the limiting reagent when two reactant amounts are given.
• Compute and convert between mass percent, mole fraction, molarity and molality for a solution.

Ask the tutor: Can you walk me through a limiting reagent problem where three reactants are mixed together instead of two?