Chapter 1 · Some Basic Concepts of Chemistry

1.1 Matter, Measurement and Significant Figures

What matter is made of, how we sort it, and how to measure it honestly: SI units, density, temperature scales, significant figures, precision versus accuracy and dimensional analysis.

In September 1999 NASA lost the Mars Climate Orbiter, a spacecraft that had cost more than 300 million US dollars. It flew far too low into the Martian atmosphere because one piece of software reported thruster impulses in pound-force seconds while the navigation software expected newton seconds — a factor of $4.45$. Nobody had made a chemistry mistake or a physics mistake. They had made a measurement mistake: a number was passed on without its unit.

Chemistry is a quantitative science. A doctor deciding a dose, a pharmaceutical company checking the purity of a tablet and a JEE examiner marking your answer all want the same three things from a number: the right unit, an honest statement of how certain it is, and a value that can be trusted. This section builds those habits. First we sort matter into its kinds; then we learn the SI system of units, the meaning of the digits we write down (significant figures), the difference between precision and accuracy, and a fool-proof method for converting units.

Chemistry and the Nature of Matter

Chemistry is the study of the composition, structure and properties of matter and of the changes it undergoes. It sits at the centre of modern life: fertilisers and pesticides that feed more than eight billion people, medicines such as the anti-cancer drug cisplatin and the AIDS drug AZT, new materials for batteries and solar cells, clean-water treatment, and the understanding of air pollution and climate. Whatever career you choose — engineering, medicine, environmental science — you will use the ideas in this course.

📘 Definition 1.1 — Matter, pure substances and mixtures

Matter is anything that has mass and occupies space. It exists in three common physical states — solid (definite shape and volume), liquid (definite volume, takes the shape of its container) and gas (neither definite shape nor definite volume).

A pure substance has a fixed composition and properties that are the same throughout. It is either an element (made of only one kind of atom, e.g. $\mathrm{Na}$, $\mathrm{Cu}$, $\mathrm{O_{2}}$) or a compound (two or more elements combined in a fixed ratio by mass, e.g. $\mathrm{H_{2}O}$, $\mathrm{CO_{2}}$, $\mathrm{NaCl}$). A compound can be split into simpler substances only by chemical methods, and its properties differ from those of its elements.

A mixture contains two or more substances in any ratio, each keeping its own properties. In a homogeneous mixture the composition is uniform throughout (air, sugar solution, brass); in a heterogeneous mixture it is not, and different parts can often be seen (sand in water, a mixture of salt and iron filings). The components of a mixture can be separated by physical methods such as filtration, crystallisation and distillation.

✏️ Example 1 — Classifying matter
Classify each as an element, a compound, a homogeneous mixture or a heterogeneous mixture: (a) mercury; (b) sodium chloride, $\mathrm{NaCl}$; (c) air filtered free of dust; (d) sand stirred into water; (e) brass (an alloy of copper and zinc); (f) ice floating in water.
Test 1 Is there only one substance with a fixed composition? (a) mercury, (b) $\mathrm{NaCl}$ and (f) ice + water ($\mathrm{H_{2}O}$ in two states) are pure substances.
Test 2 Can the pure substance be broken down chemically? Mercury contains only $\mathrm{Hg}$ atoms: element. $\mathrm{NaCl}$ contains Na and Cl in a fixed 1 : 1 ratio: compound. Ice in water is the compound $\mathrm{H_{2}O}$ — two physical states, but still one substance.
Test 3 For the mixtures, is the composition uniform? Clean air (mainly $\mathrm{N_{2}}$ and $\mathrm{O_{2}}$) and brass are homogeneous mixtures; sand in water is a heterogeneous mixture — the sand settles and can be seen.
Watch out "Two visible layers" does not automatically mean a mixture: (f) shows that a single compound can be present in two states at once.

Properties and Their Measurement: the SI System

Every substance has characteristic properties. Physical properties can be measured without changing the identity of the substance — colour, odour, melting point, boiling point, density. Chemical properties describe how a substance changes into new substances — burning, rusting, reacting with an acid — and can be observed only through a chemical change. Most properties are quantitative: they are a number times a unit, such as a boiling point of $78\ ^{\circ}\mathrm{C}$ or a mass of $2.5\ \mathrm{kg}$.

📘 Definition 1.2 — SI base units

The International System of Units (SI, from the French Le Système International d'Unités) has seven base units; all other units (derived units) are products and quotients of them. Since 2019 every base unit is defined by fixing the exact value of a constant of nature (for example the kilogram via the Planck constant $h$, and the mole via the Avogadro constant $N_{A}$).

Physical quantitySymbolSI unitUnit symbol
length$l$metrem
mass$m$kilogramkg
time$t$seconds
electric current$I$ampereA
thermodynamic temperature$T$kelvinK
amount of substance$n$molemol
luminous intensity$I_{v}$candelacd

Very large and very small quantities are written with prefixes, each meaning a power of ten. Chemists use the small ones most: bond lengths in picometres, wavelengths of light in nanometres, drug doses in milligrams.

Multiple$10^{-15}$$10^{-12}$$10^{-9}$$10^{-6}$$10^{-3}$$10^{-2}$$10^{-1}$$10^{3}$$10^{6}$$10^{9}$
Prefixfemtopiconanomicromillicentidecikilomegagiga
SymbolfpnµmcdkMG

Mass and weight

Mass is the amount of matter in an object; it is the same everywhere. Weight is the gravitational force on the object, $W=mg$, measured in newtons, and it changes with $g$ (an astronaut weighs about one-sixth as much on the Moon, but has the same mass). In the laboratory mass is found with an analytical balance, which compares masses and so is not fooled by $g$. The SI unit of mass is the kilogram, but the gram ($1\ \mathrm{g}=10^{-3}\ \mathrm{kg}$) is more convenient for chemical quantities.

Volume and density

Volume has units of $(\text{length})^{3}$; the SI unit $\mathrm{m^{3}}$ is huge for laboratory work, so we use $\mathrm{cm^{3}}$ and $\mathrm{dm^{3}}$. The litre is not an SI unit but is accepted for use: $1\ \mathrm{L}=1\ \mathrm{dm^{3}}=1000\ \mathrm{cm^{3}}=10^{-3}\ \mathrm{m^{3}}$ and $1\ \mathrm{mL}=1\ \mathrm{cm^{3}}$. Liquids are measured with a measuring cylinder (rough), a burette or pipette (accurate, delivers a volume) and a volumetric flask (accurate, contains a fixed volume).

📘 Definition 1.3 — Density
The density of a substance is its mass per unit volume, $$\rho=\frac{m}{V}.$$ The SI unit is $\mathrm{kg\,m^{-3}}$; chemists usually quote $\mathrm{g\,cm^{-3}}$ (or $\mathrm{g\,mL^{-1}}$), and $1\ \mathrm{g\,cm^{-3}}=10^{3}\ \mathrm{kg\,m^{-3}}$. Density is a physical property that helps identify a substance; it depends on temperature, because volume changes when a substance is heated.
✏️ Example 2 — Identifying a metal by its density
A rectangular metal block measures $2.50\ \mathrm{cm}\times3.00\ \mathrm{cm}\times1.20\ \mathrm{cm}$ and has mass $24.2\ \mathrm{g}$. (a) Find its density in $\mathrm{g\,cm^{-3}}$ and in $\mathrm{kg\,m^{-3}}$. (b) Which metal is it likely to be: aluminium ($2.70$), iron ($7.87$) or copper ($8.96\ \mathrm{g\,cm^{-3}}$)? (c) What mass of this metal occupies $50.0\ \mathrm{cm^{3}}$?
Volume $V=2.50\times3.00\times1.20=9.00\ \mathrm{cm^{3}}$ (three significant figures, like each length).
Density $$\rho=\frac{m}{V}=\frac{24.2\ \mathrm{g}}{9.00\ \mathrm{cm^{3}}}=2.688\ldots\ \mathrm{g\,cm^{-3}}\approx2.69\ \mathrm{g\,cm^{-3}}.$$ In SI units: $2.69\ \mathrm{g\,cm^{-3}}\times\dfrac{1\ \mathrm{kg}}{10^{3}\ \mathrm{g}}\times\dfrac{10^{6}\ \mathrm{cm^{3}}}{1\ \mathrm{m^{3}}}=2.69\times10^{3}\ \mathrm{kg\,m^{-3}}$.
Identify $2.69$ is within $0.4\%$ of aluminium's $2.70\ \mathrm{g\,cm^{-3}}$ and far from iron or copper: the block is most likely aluminium.
Mass $m=\rho V=2.688\ \mathrm{g\,cm^{-3}}\times50.0\ \mathrm{cm^{3}}=134\ \mathrm{g}$ (keep the unrounded density in the calculator; round only the final answer).

Temperature

Three scales are in common use. On the Celsius scale ($^{\circ}\mathrm{C}$) water freezes at $0$ and boils at $100$ (at 1 atm); on the Fahrenheit scale ($^{\circ}\mathrm{F}$) the same points are $32$ and $212$, so $100$ Celsius degrees span $180$ Fahrenheit degrees. The SI scale is the kelvin scale, which starts at absolute zero, the lowest temperature possible; a kelvin is the same size as a Celsius degree.

📐 Rule 1.1 — Converting between temperature scales
$$T(\mathrm{K})=t(^{\circ}\mathrm{C})+273.15,\qquad t(^{\circ}\mathrm{F})=\frac{9}{5}\,t(^{\circ}\mathrm{C})+32 .$$ Temperatures below $0\ ^{\circ}\mathrm{C}$ (and below $0\ ^{\circ}\mathrm{F}$) are negative, but a kelvin temperature is never negative: $0\ \mathrm{K}=-273.15\ ^{\circ}\mathrm{C}$. A temperature difference of $1\ \mathrm{K}$ equals $1\ ^{\circ}\mathrm{C}$ and $\tfrac95\ ^{\circ}\mathrm{F}$.
✏️ Example 3 — Three temperature scales
(a) Normal body temperature is $98.6\ ^{\circ}\mathrm{F}$. Express it in $^{\circ}\mathrm{C}$ and in K. (b) At what temperature do the Celsius and Fahrenheit scales give the same reading? (c) Liquid nitrogen boils at $77\ \mathrm{K}$. Express this in $^{\circ}\mathrm{C}$.
(a) Invert Rule 1.1: $t(^{\circ}\mathrm{C})=\tfrac59\left(t(^{\circ}\mathrm{F})-32\right)=\tfrac59(98.6-32)=\tfrac59(66.6)=37.0\ ^{\circ}\mathrm{C}$. Then $T=37.0+273.15=310.15$, which we report to one decimal place (like $37.0$): $T=310.2\ \mathrm{K}$.
(b) Put $t(^{\circ}\mathrm{F})=t(^{\circ}\mathrm{C})=x$: $x=\tfrac95x+32\Rightarrow-\tfrac45x=32\Rightarrow x=-40$. So $-40\ ^{\circ}\mathrm{C}=-40\ ^{\circ}\mathrm{F}$.
(c) $t(^{\circ}\mathrm{C})=T-273.15=77-273.15=-196.15$; since $77$ has no decimal places, $t=-196\ ^{\circ}\mathrm{C}$.
Check (a) is a little above room temperature ($\approx298\ \mathrm{K}$) — sensible. In (c) the kelvin value is positive even though the Celsius value is very negative, as it must be.

Uncertainty in Measurement: Significant Figures

Every measurement has some uncertainty. If a burette is marked every $0.1\ \mathrm{mL}$, you can read it to about $\pm0.01\ \mathrm{mL}$ by estimating between the lines: a reading such as $24.36\ \mathrm{mL}$ has three certain digits ($2$, $4$, $3$) and one estimated digit ($6$). Writing $24.360$ would claim a precision you do not have; writing $24$ would throw away information. The digits we are entitled to write are called significant figures.

Scientific notation

Chemistry spans enormous ranges — $602\,200\,000\,000\,000\,000\,000\,000$ particles in a mole, $0.000\,000\,000\,074\ \mathrm{m}$ for the length of an $\mathrm{H{-}H}$ bond. We write every number as $N\times10^{n}$ with $1\le N\lt10$ and $n$ an integer: $6.022\times10^{23}$ and $7.4\times10^{-11}\ \mathrm{m}$. To multiply, multiply the $N$'s and add the exponents; to divide, subtract the exponents; to add or subtract, first rewrite both numbers with the same exponent. In scientific notation every digit of $N$ is significant, which removes all ambiguity about zeros.

📘 Definition 1.4 — Significant figures
The significant figures of a measured value are all the digits known with certainty plus the first uncertain (estimated) digit. By convention the uncertainty is $\pm1$ in the last significant digit: $11.2\ \mathrm{mL}$ means $11.2\pm0.1\ \mathrm{mL}$.
📐 Rule 1.2 — Counting significant figures
1 · Non-zero digitsare always significant: $285$ has 3, $0.25$ has 2.
2 · Leading zeros(before the first non-zero digit) are never significant — they only place the decimal point: $0.0034$ has 2.
3 · Captive zeros(between non-zero digits) are significant: $2.005$ has 4, $3006$ has 4.
4 · Trailing zerosare significant if the number has a decimal point: $0.200$ has 3, $100.$ has 3, $100.0$ has 4. Without a decimal point they are not counted: $100$ has 1.
5 · Exact numbers(counted objects, defined conversions such as $1\ \mathrm{m}=100\ \mathrm{cm}$) have infinitely many significant figures.
6 · Scientific notationremoves ambiguity: $1.0\times10^{2}$ has 2, $1.00\times10^{2}$ has 3.
📐 Rule 1.3 — Significant figures in calculations and rounding

Addition and subtraction: the result has the same number of decimal places as the quantity with the fewest decimal places. Multiplication and division: the result has the same number of significant figures as the quantity with the fewest significant figures. Exact numbers never limit the result.

Rounding off (the convention used in the Class 11 syllabus and in Board marking): if the first digit to be dropped is greater than $5$ (or is $5$ followed by non-zero digits), increase the preceding digit by one; if it is less than $5$, leave the preceding digit unchanged; if it is exactly $5$, leave the preceding digit unchanged when it is even and increase it by one when it is odd. So $7.45\to7.4$ but $7.55\to7.6$ (two significant figures).

✏️ Example 4 — Counting and rounding
(a) How many significant figures are there in $0.002050$, $3.400\times10^{3}$, $5000$, $5000.$, $2.0300$ and "12 eggs"? (b) Round $2.718$ to three, $0.04449$ to three, and $7.45$ and $7.55$ to two significant figures.
$0.002050$ The three leading zeros are not significant; $2$, the captive $0$, $5$ and the trailing $0$ after the decimal point are: 4.
$3.400\times10^{3}$ Every digit of the coefficient counts: 4.
$5000$ / $5000.$ With no decimal point the trailing zeros are not counted: 1. With the decimal point they are: 4. To avoid the ambiguity, write $5\times10^{3}$ or $5.000\times10^{3}$.
$2.0300$, 12 eggs $2.0300$ has 5 (captive zero plus two trailing zeros after the point). "12 eggs" is a count — an exact number with unlimited significant figures.
(b) $2.718\to2.72$ (dropped $8\gt5$). $0.04449\to0.0445$ (keep $4,4,4$ and drop $9\gt5$, so the last kept $4$ becomes $5$). $7.45\to7.4$ (dropped digit exactly $5$, preceding $4$ is even: unchanged). $7.55\to7.6$ (preceding $5$ is odd: increased).
⚠️ Common mistakes with significant figures

1. Rounding too early. Keep at least one extra digit in intermediate steps and round only the final answer; rounding at every step lets errors pile up.

2. Calculators and spreadsheets round a final $5$ upwards. The syllabus convention (Rule 1.3) rounds to the even digit when the dropped part is exactly $5$, and Board answer keys follow it.

3. Mixing the two rules. Decimal places for $+$ and $-$; significant figures for $\times$ and $\div$. In a mixed calculation apply the rules step by step — Example 5(c) shows how a subtraction can destroy significant figures.

✏️ Example 5 — Calculating with significant figures
Give each result with the correct number of significant figures: (a) $12.11+18.0+1.013$; (b) $4.512\times3.6$; (c) $\dfrac{12.5-10.21}{0.050}$; (d) $\left(6.022\times10^{23}\right)\times0.0500$.
(a) The raw sum is $31.123$. The least precise term, $18.0$, has one decimal place, so the answer is $31.1$.
(b) $4.512\times3.6=16.2432$. $3.6$ has two significant figures, so the answer is $16$.
(c) Numerator: $12.5-10.21=2.29$, good only to one decimal place ($2.3$, two significant figures) — the subtraction has cost us a figure. Then $2.29/0.050=45.8$; both numerator and divisor have two significant figures, so the answer is $46$.
(d) $6.022\times0.0500=0.3011$, so the product is $0.3011\times10^{23}=3.011\times10^{22}$. $0.0500$ has three significant figures: $3.01\times10^{22}$.
Interactive 1 — Significant-Figures Trainer

Choose a mode and answer with the buttons. After you answer, every digit is coloured: green = significant, grey = only a place-holder. In the add/subtract mode the shaded column is the last decimal place that every number knows; in the multiply/divide mode the factor with the fewest significant figures sets the answer. Press ▶ Play to watch worked demonstrations.

problems answered0
correct0
current streak0
rule used—

Precision, Accuracy and Dimensional Analysis

Two words that are used loosely in everyday speech have exact meanings in science.

📘 Definition 1.5 — Precision and accuracy
Precision is the closeness of repeated measurements of the same quantity to one another. Accuracy is the agreement of a measured value (usually the average of the repeats) with the true value. Random errors (small, unpredictable variations in reading and handling) spoil precision; systematic errors (a mis-calibrated balance, a parallax error made the same way every time) spoil accuracy, and repeating the measurement does not remove them.

Three students each weigh a certified $2.000\ \mathrm{g}$ standard mass three times:

StudentTrial 1 / gTrial 2 / gTrial 3 / gMean / gSpread (max − min) / gVerdict
A1.951.931.941.940.02precise, not accurate
B1.942.052.012.000.11accurate on average, not precise
C2.011.992.002.000.02precise and accurate

Student A probably has a systematic error (perhaps the balance was not zeroed); student B's results scatter, so any single reading is unreliable even though the mean happens to be right. Interactive 2 lets you create these situations yourself.

Interactive 2 — Precision versus Accuracy at the Balance

A balance repeatedly weighs a certified $2.000\ \mathrm{g}$ mass. The systematic error shifts every reading the same way; the random error scatters them. Left: the dart-board picture (centre = true value). Right: each reading on a number line, with the mean (▲) and the acceptable band $2.000\pm0.020\ \mathrm{g}$ shaded. Precise means a small standard deviation; accurate means the mean lies inside the band.

06
readings $n$—
last reading (g)—
mean (g)—
standard deviation (g)—
mean − true (g)—

Dimensional analysis

Converting units is where many marks — and one Mars spacecraft — are lost. The safe method, called dimensional analysis (or the factor-label method), treats units like algebraic symbols. Every true statement such as $1\ \mathrm{kg}=1000\ \mathrm{g}$ gives two unit factors, $\dfrac{1000\ \mathrm{g}}{1\ \mathrm{kg}}$ and $\dfrac{1\ \mathrm{kg}}{1000\ \mathrm{g}}$, each equal to $1$. Multiplying by a unit factor changes the unit without changing the quantity.

🧭 Problem-solving strategy — dimensional analysis
  1. Write the given quantity with its unit, and the unit you want.
  2. Multiply by unit factors chosen so that the unwanted unit is in the denominator of the factor and cancels.
  3. For squared or cubed units, raise the whole factor to the same power: $\left(\dfrac{100\ \mathrm{cm}}{1\ \mathrm{m}}\right)^{3}=\dfrac{10^{6}\ \mathrm{cm^{3}}}{1\ \mathrm{m^{3}}}$.
  4. Cancel units, compute, and check that the units left are the ones you wanted and that the size of the answer is sensible. Defined conversion factors are exact, so they do not limit significant figures.
✏️ Example 6 — Dimensional analysis
(a) Express $2.5\ \mathrm{nm}$ in metres and in picometres. (b) Mercury has density $13.6\ \mathrm{g\,cm^{-3}}$. Express this in $\mathrm{kg\,m^{-3}}$. (c) A cubical water tank has inner side $1.20\ \mathrm{m}$. How many litres does it hold, and what is the mass of the water when it is full (density $1.00\ \mathrm{g\,mL^{-1}}$)?
(a) $2.5\ \mathrm{nm}\times\dfrac{10^{-9}\ \mathrm{m}}{1\ \mathrm{nm}}=2.5\times10^{-9}\ \mathrm{m}$, and $2.5\times10^{-9}\ \mathrm{m}\times\dfrac{1\ \mathrm{pm}}{10^{-12}\ \mathrm{m}}=2.5\times10^{3}\ \mathrm{pm}$.
(b) $$13.6\ \frac{\mathrm{g}}{\mathrm{cm^{3}}}\times\frac{1\ \mathrm{kg}}{1000\ \mathrm{g}}\times\left(\frac{100\ \mathrm{cm}}{1\ \mathrm{m}}\right)^{3}=13.6\times\frac{10^{6}}{10^{3}}\ \mathrm{kg\,m^{-3}}=1.36\times10^{4}\ \mathrm{kg\,m^{-3}}.$$
(c) Volume $V=(1.20\ \mathrm{m})^{3}=1.728\ \mathrm{m^{3}}$, and $1.728\ \mathrm{m^{3}}\times\dfrac{1000\ \mathrm{L}}{1\ \mathrm{m^{3}}}=1728\ \mathrm{L}\approx1.73\times10^{3}\ \mathrm{L}$ (three significant figures).
(c) Mass $1728\ \mathrm{L}\times\dfrac{1000\ \mathrm{mL}}{1\ \mathrm{L}}\times\dfrac{1.00\ \mathrm{g}}{1\ \mathrm{mL}}\times\dfrac{1\ \mathrm{kg}}{1000\ \mathrm{g}}=1.73\times10^{3}\ \mathrm{kg}$ — almost two tonnes, which is why water tanks sit on strong frames.

Exercises

I
Classify each material as an element, a compound, a homogeneous mixture or a heterogeneous mixture (parts 1–6); then say whether each property is physical or chemical (parts 7–10).
  • 1 solid iodine
  • 2 glucose, $\mathrm{C_{6}H_{12}O_{6}}$
  • 3 clear tap water
  • 4 granite
  • 5 stainless steel
  • 6 dry ice, solid $\mathrm{CO_{2}}$
  • 7 Iron rusts in moist air.
  • 8 Ethanol boils at $78\ ^{\circ}\mathrm{C}$.
  • 9 Magnesium ribbon burns with a dazzling white flame.
  • 10 Copper is a good conductor of electricity.
1 Element — only iodine atoms (as $\mathrm{I_{2}}$ molecules).
2 Compound — C, H and O in a fixed ratio.
3 Homogeneous mixture — water with dissolved salts and gases, uniform throughout.
4 Heterogeneous mixture — grains of different minerals are visible.
5 Homogeneous mixture — an alloy of iron with chromium, nickel and carbon, uniform in composition.
6 Compound — $\mathrm{CO_{2}}$ in the solid state.
7 Chemical — a new substance (hydrated iron oxide) forms.
8 Physical — boiling changes the state, not the identity, of ethanol.
9 Chemical — magnesium becomes magnesium oxide.
10 Physical — measured without changing the copper.
II
SI units, conversions, density and temperature. Give each answer with a unit and a sensible number of significant figures.
  • 1 Express $5.6\ \mathrm{\mu m}$ in metres and in nanometres.
  • 2 Express $250\ \mathrm{mL}$ in $\mathrm{L}$, in $\mathrm{dm^{3}}$ and in $\mathrm{m^{3}}$ (treat $250$ as three significant figures).
  • 3 Express a density of $1.5\ \mathrm{g\,cm^{-3}}$ in $\mathrm{kg\,m^{-3}}$.
  • 4 A $25.0\ \mathrm{mL}$ sample of a liquid has mass $19.7\ \mathrm{g}$. Find its density.
  • 5 What volume does $100.0\ \mathrm{g}$ of mercury occupy? ($\rho=13.6\ \mathrm{g\,cm^{-3}}$)
  • 6 Convert $25\ ^{\circ}\mathrm{C}$ to kelvin.
  • 7 Convert $-196\ ^{\circ}\mathrm{C}$ to kelvin and to $^{\circ}\mathrm{F}$.
  • 8 A recipe asks for an oven at $350\ ^{\circ}\mathrm{F}$. What is this in $^{\circ}\mathrm{C}$?
  • 9 An astronaut has mass $70.0\ \mathrm{kg}$. Find her weight on Earth ($g=9.81\ \mathrm{m\,s^{-2}}$) and on the Moon ($g=1.62\ \mathrm{m\,s^{-2}}$). What is her mass on the Moon?
  • 10 Applied. A $2.00\ \mathrm{L}$ bottle of a soft drink has density $1.04\ \mathrm{g\,mL^{-1}}$. What is the mass of the drink in kilograms?
1 $5.6\ \mathrm{\mu m}=5.6\times10^{-6}\ \mathrm{m}=5.6\times10^{-6}\ \mathrm{m}\times\dfrac{1\ \mathrm{nm}}{10^{-9}\ \mathrm{m}}=5.6\times10^{3}\ \mathrm{nm}$.
2 $250\ \mathrm{mL}=0.250\ \mathrm{L}=0.250\ \mathrm{dm^{3}}=0.250\times10^{-3}\ \mathrm{m^{3}}=2.50\times10^{-4}\ \mathrm{m^{3}}$.
3 $1.5\ \mathrm{g\,cm^{-3}}\times\dfrac{1\ \mathrm{kg}}{10^{3}\ \mathrm{g}}\times\dfrac{10^{6}\ \mathrm{cm^{3}}}{1\ \mathrm{m^{3}}}=1.5\times10^{3}\ \mathrm{kg\,m^{-3}}$.
4 $\rho=\dfrac{19.7\ \mathrm{g}}{25.0\ \mathrm{mL}}=0.788\ \mathrm{g\,mL^{-1}}$ (close to ethanol, $0.789\ \mathrm{g\,mL^{-1}}$).
5 $V=\dfrac{m}{\rho}=\dfrac{100.0\ \mathrm{g}}{13.6\ \mathrm{g\,cm^{-3}}}=7.35\ \mathrm{cm^{3}}$ (three significant figures, set by $13.6$).
6 $T=25+273.15=298.15$; $25$ has no decimal places, so $T=298\ \mathrm{K}$.
7 $T=-196+273.15=77.15\to77\ \mathrm{K}$. $t(^{\circ}\mathrm{F})=\tfrac95(-196)+32=-352.8+32=-320.8\to-321\ ^{\circ}\mathrm{F}$.
8 $t(^{\circ}\mathrm{C})=\tfrac59(350-32)=\tfrac59(318)=176.7\approx177\ ^{\circ}\mathrm{C}$.
9 Earth: $W=mg=70.0\times9.81=686.7\approx687\ \mathrm{N}$. Moon: $W=70.0\times1.62=113.4\approx113\ \mathrm{N}$. Her mass is still $70.0\ \mathrm{kg}$ — mass does not depend on $g$.
10 $2.00\ \mathrm{L}\times\dfrac{1000\ \mathrm{mL}}{1\ \mathrm{L}}\times\dfrac{1.04\ \mathrm{g}}{1\ \mathrm{mL}}\times\dfrac{1\ \mathrm{kg}}{1000\ \mathrm{g}}=2.08\ \mathrm{kg}$.
III
Significant figures, scientific notation and rounding (use the rounding convention of Rule 1.3).
  • 1 How many significant figures in $0.0300$?
  • 2 How many significant figures in $4.050\times10^{-2}$?
  • 3 How many significant figures in $7000$, and how would you write it to show four?
  • 4 How many significant figures in $10.020$?
  • 5 Write $0.0000450$ in scientific notation.
  • 6 Round $2.3456$ to three significant figures.
  • 7 Round $0.008765$ to two significant figures.
  • 8 Round $0.0625$ to two significant figures, and $0.0635$ to two significant figures.
  • 9 $12.06+3.1+0.425$
  • 10 $\left(3.26\times10^{-5}\right)\times1.78$
  • 11 $\dfrac{5.0\times10^{-3}}{2.50\times10^{2}}$
  • 12 $(8.71-8.6)\times4.002$
  • 13 The mean of three titre readings $24.31$, $24.35$ and $24.33\ \mathrm{mL}$.
Check: type any number to see its significant figures
e.g. 0.0300, 7000, 7.000e3, 10.020
significant figures—
scientific notation—
digits that count—
1 3: the two leading zeros do not count; $3$ and the two trailing zeros after the point do.
2 4 ($4$, $0$, $5$, $0$).
3 1 by convention (no decimal point). Four significant figures: $7.000\times10^{3}$ (or $7000.$).
4 5: captive zeros and the trailing zero after the point all count.
5 $4.50\times10^{-5}$ (three significant figures, keeping the trailing zero).
6 $2.35$ (the dropped part $56$ is more than half a unit).
7 $0.0088$ (keep $8,7$; dropped $65$ is more than half, so $7\to8$).
8 $0.0625\to0.062$ (exactly $5$ dropped, preceding $2$ even); $0.0635\to0.064$ (preceding $3$ odd, increased).
9 Raw sum $15.585$; fewest decimal places is one ($3.1$): $15.6$.
10 $3.26\times1.78=5.8028$: $5.80\times10^{-5}$ (three significant figures).
11 $\dfrac{5.0}{2.50}\times10^{-3-2}=2.0\times10^{-5}$ (two significant figures, from $5.0$).
12 $8.71-8.6=0.11$, known only to one decimal place: $0.1$ (one significant figure). Then $0.11\times4.002=0.440$, which must be reported to one significant figure: $0.4$. The subtraction destroyed almost all the information.
13 Sum $=72.99\ \mathrm{mL}$; dividing by the exact number $3$ gives $24.33\ \mathrm{mL}$ (two decimal places, like the readings).
IV
Precision, accuracy and dimensional analysis.
  • 1 A pipette is certified to deliver $50.0\ \mathrm{mL}$. A student's three deliveries measure $49.2$, $49.3$ and $49.1\ \mathrm{mL}$. Comment on precision and accuracy.
  • 2 Another student's four deliveries measure $50.6$, $49.1$, $50.3$ and $49.9\ \mathrm{mL}$. Find the mean and comment.
  • 3 How many seconds are there in a year of exactly $365$ days? How many significant figures does the answer have?
  • 4 Medicine. A child of mass $62\ \mathrm{kg}$ is prescribed $5.0\ \mathrm{mg}$ of a drug per kilogram of body mass per day. The syrup contains $125\ \mathrm{mg}$ of drug in every $5.0\ \mathrm{mL}$. What volume of syrup should be given each day?
  • 5 Engineering. $1$ pound-force second $=4.448$ newton seconds. A program reported an impulse of $100$ (meaning $\mathrm{lbf\,s}$) but the receiving program read it as $100\ \mathrm{N\,s}$. What was the true impulse in $\mathrm{N\,s}$, and by what factor was it underestimated?
  • 6 A human hair is about $70\ \mathrm{\mu m}$ thick and a carbon atom is about $0.15\ \mathrm{nm}$ across. Roughly how many carbon atoms would fit side by side across the hair?
  • 7 Challenge. A rectangle of aluminium foil $30.0\ \mathrm{cm}\times20.0\ \mathrm{cm}$ has mass $1.62\ \mathrm{g}$. The density of aluminium is $2.70\ \mathrm{g\,cm^{-3}}$. Find the thickness of the foil in micrometres.
1 Mean $=49.2\ \mathrm{mL}$, spread only $0.2\ \mathrm{mL}$: precise but not accurate — every delivery is about $0.8\ \mathrm{mL}$ short, a systematic error (perhaps the pipette was not drained or filled to the mark correctly).
2 Mean $=\dfrac{199.9}{4}=49.975\to50.0\ \mathrm{mL}$, but the spread is $1.5\ \mathrm{mL}$: accurate on average but not precise; no single delivery can be trusted.
3 $365\ \mathrm{d}\times\dfrac{24\ \mathrm{h}}{1\ \mathrm{d}}\times\dfrac{60\ \mathrm{min}}{1\ \mathrm{h}}\times\dfrac{60\ \mathrm{s}}{1\ \mathrm{min}}=31\,536\,000\ \mathrm{s}=3.1536\times10^{7}\ \mathrm{s}$. Every factor is exact, so the answer is exact (unlimited significant figures).
4 Dose $=62\ \mathrm{kg}\times\dfrac{5.0\ \mathrm{mg}}{1\ \mathrm{kg}}=310\ \mathrm{mg}$ per day. Volume $=310\ \mathrm{mg}\times\dfrac{5.0\ \mathrm{mL}}{125\ \mathrm{mg}}=12.4\ \mathrm{mL}\approx12\ \mathrm{mL}$ per day (two significant figures).
5 True impulse $=100\ \mathrm{lbf\,s}\times\dfrac{4.448\ \mathrm{N\,s}}{1\ \mathrm{lbf\,s}}=444.8\approx445\ \mathrm{N\,s}$. It was read as $100\ \mathrm{N\,s}$, an underestimate by a factor of $4.448\approx4.45$.
6 $\dfrac{70\times10^{-6}\ \mathrm{m}}{0.15\times10^{-9}\ \mathrm{m}}=4.7\times10^{5}$ atoms — about half a million.
7 Volume $V=\dfrac{m}{\rho}=\dfrac{1.62}{2.70}=0.600\ \mathrm{cm^{3}}$; area $=600.\ \mathrm{cm^{2}}$; thickness $=\dfrac{0.600\ \mathrm{cm^{3}}}{600.\ \mathrm{cm^{2}}}=1.00\times10^{-3}\ \mathrm{cm}=1.00\times10^{-5}\ \mathrm{m}=10.0\ \mathrm{\mu m}$.
🎯 Exam-style problems BoardJEE MainJEE AdvancedNEET

E1. JEE Main NEET Which of the following has the largest number of significant figures? (A) $0.002500$ (B) $25.0$ (C) $2.5\times10^{-3}$ (D) $0.0250$

Count (A) leading zeros do not count; $2,5,0,0$ do: 4. (B) 3. (C) 2. (D) $2,5,0$: 3.
Answer (A), with four significant figures. Trap: the many zeros in (A) and (D) tempt you to count them all; only zeros after the first non-zero digit can count.

E2. JEE Main The value of $\dfrac{(29.2-20.2)\left(1.79\times10^{5}\right)}{1.37}$, reported to the correct number of significant figures, is (A) $1.176\times10^{6}$ (B) $1.18\times10^{6}$ (C) $1.2\times10^{6}$ (D) $1\times10^{6}$

Subtract first $29.2-20.2=9.0$: one decimal place, so two significant figures.
Multiply/divide $\dfrac{9.0\times1.79\times10^{5}}{1.37}=\dfrac{1.611\times10^{6}}{1.37}=1.1759\times10^{6}$. The least number of significant figures among $9.0$, $1.79\times10^{5}$ and $1.37$ is two.
Answer (C) $1.2\times10^{6}$. Option (B) forgets that the subtraction reduced $29.2$ and $20.2$ (three figures each) to a two-figure result.

E3. Board (i) Write $0.000307$ and $60\,700$ in scientific notation and state the number of significant figures in each. (ii) Round each to two significant figures. (iii) Express $0.000307\ \mathrm{kg}$ in milligrams. (3 marks)

(i) $0.000307=3.07\times10^{-4}$ (3 significant figures); $60\,700=6.07\times10^{4}$ (3 significant figures — the trailing zeros of $60\,700$ have no decimal point, so they are not counted).
(ii) $3.07\times10^{-4}\to3.1\times10^{-4}$ and $6.07\times10^{4}\to6.1\times10^{4}$ (dropped digit $7\gt5$).
(iii) $3.07\times10^{-4}\ \mathrm{kg}\times\dfrac{10^{3}\ \mathrm{g}}{1\ \mathrm{kg}}\times\dfrac{10^{3}\ \mathrm{mg}}{1\ \mathrm{g}}=3.07\times10^{2}\ \mathrm{mg}=307\ \mathrm{mg}$.

E4. NEET At what temperature is the reading on the Fahrenheit scale exactly twice the reading on the Celsius scale? (A) $80\ ^{\circ}\mathrm{C}$ (B) $160\ ^{\circ}\mathrm{C}$ (C) $-40\ ^{\circ}\mathrm{C}$ (D) $320\ ^{\circ}\mathrm{C}$

Equation Let the Celsius reading be $x$. Then $\tfrac95x+32=2x\Rightarrow32=\tfrac15x\Rightarrow x=160$.
Check $\tfrac95(160)+32=288+32=320\ ^{\circ}\mathrm{F}=2\times160$. ✓
Answer (B) $160\ ^{\circ}\mathrm{C}$ (option (C) is where the readings are equal; (D) is the Fahrenheit reading itself).

E5. JEE Advanced One or more options correct. Which statements are correct? (A) $1.0\times10^{2}$ and $100.0$ have the same number of significant figures. (B) The number of students in a class, $40$, has an unlimited number of significant figures. (C) The sum $3.0\ \mathrm{g}+0.0045\ \mathrm{g}$ is correctly reported as $3.0\ \mathrm{g}$. (D) A set of readings can be precise without being accurate.

(A) False: $1.0\times10^{2}$ has 2 significant figures, $100.0$ has 4.
(B) True: a count is an exact number.
(C) True: the raw sum $3.0045$ is limited to one decimal place by $3.0$, giving $3.0\ \mathrm{g}$ — the tiny mass is lost in the uncertainty of the larger one.
(D) True: a systematic error gives close (precise) readings that are all wrong by the same amount (student A in the table above).
Answer (B), (C), (D).

E6. JEE Main Board Numerical answer. Gold (density $19.3\ \mathrm{g\,cm^{-3}}$) can be beaten into leaf $0.100\ \mathrm{\mu m}$ thick. What area, in $\mathrm{m^{2}}$, can be covered by $1.00\ \mathrm{g}$ of gold? Give the answer to three significant figures.

Volume $V=\dfrac{m}{\rho}=\dfrac{1.00\ \mathrm{g}}{19.3\ \mathrm{g\,cm^{-3}}}=0.05181\ \mathrm{cm^{3}}$.
Thickness $0.100\ \mathrm{\mu m}=0.100\times10^{-6}\ \mathrm{m}=1.00\times10^{-5}\ \mathrm{cm}$.
Area $A=\dfrac{V}{\text{thickness}}=\dfrac{0.05181\ \mathrm{cm^{3}}}{1.00\times10^{-5}\ \mathrm{cm}}=5181\ \mathrm{cm^{2}}$, and $5181\ \mathrm{cm^{2}}\times\left(\dfrac{1\ \mathrm{m}}{100\ \mathrm{cm}}\right)^{2}=0.518\ \mathrm{m^{2}}$.
Answer $0.518\ \mathrm{m^{2}}$ — one gram covers about half a square metre.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. How accurate and how precise must a home blood-glucose meter be before a person with diabetes can trust it to decide an insulin dose?

Why it matters. More than 500 million people worldwide live with diabetes, and many adjust their medicine using a finger-prick meter. The international standard ISO 15197:2013 requires that at least $95\%$ of a meter's results lie within $\pm15\ \mathrm{mg\,dL^{-1}}$ of the laboratory value when glucose is below $100\ \mathrm{mg\,dL^{-1}}$, and within $\pm15\%$ at or above $100\ \mathrm{mg\,dL^{-1}}$. That single sentence contains accuracy, precision, units and significant figures.

What to do.

  1. Sample case: a laboratory analyser gives a reference value of $120\ \mathrm{mg\,dL^{-1}}$. One meter tested eight times on the same sample reads $126,\ 131,\ 118,\ 128,\ 135,\ 124,\ 129,\ 122\ \mathrm{mg\,dL^{-1}}$.
  2. Find the mean, the bias (mean $-$ reference, also as a percentage) and the standard deviation. Is the meter more limited by accuracy or by precision?
  3. Find the $\pm15\%$ window around $120$ and check what fraction of the readings fall inside it.
  4. Many countries (including India and the UK) report glucose in $\mathrm{mmol\,L^{-1}}$. Using dimensional analysis with $1\ \mathrm{mmol\,L^{-1}}$ of glucose $=18.0\ \mathrm{mg\,dL^{-1}}$, convert the reference value. Then show where the factor $18.0$ comes from, using the molar mass of glucose, $180\ \mathrm{g\,mol^{-1}}$ (Section 1.2).

Tools: Calculator or spreadsheet; Interactive 2 to visualise the readings.

Hand in: A half-page note with your statistics, a verdict on whether this meter passes at this glucose level, and one source of systematic error a patient could introduce (for example unwashed fingers or expired strips).

Going further. Look up how sports doping laboratories or food-safety agencies state the uncertainty of a result (an "expanded uncertainty" such as $\pm0.05\ \mathrm{mg\,kg^{-1}}$). Why must a result exceed the legal limit by more than the uncertainty before an athlete or a food batch is penalised?

Instructor note: mean $126.6\ \mathrm{mg\,dL^{-1}}$, bias $+6.6\ \mathrm{mg\,dL^{-1}}$ ($+5.5\%$), standard deviation $\approx5.3\ \mathrm{mg\,dL^{-1}}$; the window is $102$–$138\ \mathrm{mg\,dL^{-1}}$ and all eight readings lie inside it. $120\ \mathrm{mg\,dL^{-1}}=6.67\ \mathrm{mmol\,L^{-1}}$; $1\ \mathrm{mmol\,L^{-1}}=180\ \mathrm{mg\,L^{-1}}=18.0\ \mathrm{mg\,dL^{-1}}$. Bias and spread are of similar size, so both matter.

🧪 Self-check quiz

0/5 correct

Q1. Which of these is a homogeneous mixture?

Q2. The SI base unit of temperature is

Q3. How many significant figures are there in $0.04060$?

Q4. A rectangle measures $2.0\ \mathrm{cm}\times3.145\ \mathrm{cm}$. Its area, correctly reported, is

Q5. Repeated readings are very close to each other but all far from the true value. They are

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