SI units in chemistry — kg, mol, K, Pa; derived units (J, L): Worked examples

The scenario matrix

Below is every class of situation this topic can throw at you. Each of the 8 examples that follow is tagged with the cell it covers, so the whole grid gets filled.

Cell	Case class	What makes it tricky	Example
A	Straight forward conversion (mass → mol → particles)	keeping mass units consistent	E1
B	Reverse direction (particles → mol → mass)	dividing by $N_{A}$ , not multiplying	E2
C	Unit trap: kg vs g in molar mass	the "molar mass is kg" myth	E3
D	Temperature: negative Celsius, absolute-zero limit	Celsius can be negative; kelvin cannot go below 0	E4
E	Volume scaling: L ↔ m³ ↔ cm³	cube of a length ratio	E5
F	Derived unit built from scratch (pressure, units only)	reading Pa off $F / A$	E6
G	Everything at once — ideal gas in pure SI (word problem)	convert L, °C, kPa all before plugging in	E7
H	Degenerate / limiting inputs (zero moles, zero kelvin, huge $N$ )	what the formulas say at the edges	E8

The examples

Worked example E3 — Cell C: the kg-vs-g molar-mass trap

Compute $n$ for $0.036 kg$ of water two ways — once keeping everything in grams, once in kg — and show they agree. ( $M_{H_{2} O} = 18 g/mol = 0.018 kg/mol$ .)

Forecast: The two methods must give the same $n$ — moles is a count, it cannot depend on which mass unit we chose. Guess: same answer both times.

Grams route: $m = 0.036 kg = 36 g$ , $M = 18 g/mol$ . $n = \frac{36}{18} = 2 mol$ . Why this step? Both $m$ and $M$ are now in grams, so grams cancel cleanly.
Kilograms route: $m = 0.036 kg$ , $M = 0.018 kg/mol$ . $n = \frac{0.036}{0.018} = 2 mol$ . Why this step? Both $m$ and $M$ are now in kg, so kg cancel cleanly. The rule is not "use kg" — it's "use the SAME mass unit on top and bottom."

Verify: Both give $2 mol$ ✓. The myth " $M$ must be kg" mixes units ( $36 g \div 0.018 kg/mol$ ) and gives $2000$ — wrong by $1000 \times$ , exactly the g↔kg factor.

Worked example E4 — Cell D: negative Celsius & the absolute-zero limit

(a) Convert $- 40^{\circ} C$ to kelvin. (b) Convert $- 273.15^{\circ} C$ to kelvin — what does the result mean? (c) Can a temperature of $- 300^{\circ} C$ exist?

Forecast: Kelvin is Celsius shifted up by 273.15. So a negative Celsius can still become a positive kelvin. And there should be a floor below which nothing exists.

$T_{K} = - 40 + 273.15 = 233.15 K$ . Why this step? Same-size steps, shifted zero — we just add the offset. Note the answer is positive, even though °C was negative.
$T_{K} = - 273.15 + 273.15 = 0 K$ . Why this step? This is absolute zero — zero thermal motion. It's the physical origin the kelvin scale was built around, which is exactly why the gas law uses kelvin.
$- 300^{\circ} C \Rightarrow - 300 + 273.15 = - 26.85 K$ — a negative kelvin. Why this step? Negative kelvin means less-than-zero thermal motion, which is impossible. So $- 300^{\circ} C$ cannot exist; the coldest possible is $- 273.15^{\circ} C$ .

Verify: (a) $233.15 > 0$ ✓; (b) $= 0$ exactly ✓; (c) $< 0$ , flagged impossible ✓. See Kinetic Theory of Gases for why motion stops at $0 K$ .

Worked example E5 — Cell E: volume scaling L ↔ m³ ↔ cm³

A flask holds $250 mL$ . Express this in (a) litres, (b) cubic metres, (c) cubic centimetres, and confirm it is a cube of side ≈ 6.3 cm. See figure.

Forecast: $250 mL$ is a quarter-litre. In m³ it should be a tiny number (because $1 m^{3} = 1000 L$ ), around $1 0^{- 4}$ .

$250 mL = 0.250 L$ . Why this step? "milli" = $1 0^{- 3}$ , so $250 mL = 250 \times 1 0^{- 3} L$ .
$0.250 L = 0.250 \times 1 0^{- 3} m^{3} = 2.5 \times 1 0^{- 4} m^{3}$ . Why this step? $1 L = 1 0^{- 3} m^{3}$ (a 10 cm cube inside a 1 m cube). Look at the red cube in the figure — it's a tiny corner of the black metre-cube.
$0.250 L = 250 cm^{3}$ (since $1 L = 1000 cm^{3}$ ). Side of a cube of this volume: $3250 = 6.30 cm$ . Why this step? Volume is length cubed, so to get the side we take the cube root — the geometric inverse of cubing.

Verify: $6.299 6^{3} = 250.0 cm^{3}$ ✓, and $2.5 \times 1 0^{- 4} m^{3} \times 1000 = 0.250 L$ ✓. Tiny in m³ as forecast.

Worked example E6 — Cell F: build the pascal from base units alone

A block of ice pushes down with a force of $150 N$ on a table contact patch of $30 cm^{2}$ . Find the pressure in Pa, and show the answer's units are kg·m⁻¹·s⁻² directly.

Forecast: Pressure = force ÷ area. Watch the area: $30 cm^{2}$ is not $30 m^{2}$ — converting will make the pressure bigger than a naive guess.

Convert area: $30 cm^{2} = 30 \times (1 0^{- 2} m)^{2} = 30 \times 1 0^{- 4} m^{2} = 3.0 \times 1 0^{- 3} m^{2}$ . Why this step? $1 cm = 1 0^{- 2} m$ , and area is length squared, so the factor is $(1 0^{- 2})^{2} = 1 0^{- 4}$ .
$p = \frac{F}{A} = \frac{150}{3.0 \times 1 0 ^{- 3}} = 5.0 \times 1 0^{4} Pa$ . Why this step? Straight from the definition of pressure — the same push over a smaller patch means more pressure.
Units: $\frac{N}{m ^{2}} = \frac{kg m s ^{- 2}}{m ^{2}} = kg m^{- 1} s^{- 2}$ . Why this step? Substituting the newton's base-unit spelling ( $N = kg m s^{- 2}$ ) and simplifying $m / m^{2} = m^{- 1}$ reads the pascal off the formula — no memorisation.

Verify: $5.0 \times 1 0^{4} Pa = 0.5 bar$ ✓ (divide by $1 0^{5}$ ). Had we forgotten the cm² conversion we'd get $150/30 = 5 Pa$ — off by $1 0^{4}$ , the area-scaling factor.

Worked example E7 — Cell G: everything at once (word problem, ideal gas in pure SI)

A weather balloon holds $12.0 L$ of helium at $27^{\circ} C$ and a pressure of $95 kPa$ . How many moles of helium are inside? ( $R = 8.314 J mol^{- 1} K^{- 1}$ .) This is the meeting point of all four units — see Ideal Gas Law.

Forecast: Roughly room conditions, about half a mole of gas per 11 L is typical, and 12 L is near a half-mole scale. Guess: around 0.4–0.5 mol.

Convert every quantity to SI before touching the formula:
- Volume: $V = 12.0 L = 12.0 \times 1 0^{- 3} m^{3}$ .
- Temperature: $T = 27 + 273.15 = 300.15 K$ .
- Pressure: $p = 95 kPa = 95 \times 1 0^{3} Pa = 9.5 \times 1 0^{4} Pa$ . Why this step? $R$ is written in joules, and a joule is $Pa \cdot m^{3}$ . If any of $p, V, T$ is in the "wrong" unit (L, °C, kPa), the joules won't cancel and the answer is garbage.
Rearrange $p V = n R T$ for the unknown: $n = \frac{p V}{R T}$ . Why this step? We want $n$ alone, so divide both sides by $R T$ .
$n = \frac{( 9.5 \times 1 0 ^{4} ) ( 12.0 \times 1 0 ^{- 3} )}{( 8.314 ) ( 300.15 )} = \frac{1140}{2495.4} = 0.457 mol$ . Why this step? Numerator $= p V = 1140 Pa\cdotm^{3} = 1140 J$ ; denominator $= R T = 2495.4 J/mol$ ; joules cancel, leaving mol.

Verify: $0.457 mol$ lands inside the forecast 0.4–0.5 ✓. Units: $\frac{J}{J/mol} = mol$ ✓.

Worked example E8 — Cell H: degenerate and limiting inputs

Read each edge case and predict what the formula says. (a) $m = 0 g$ of any substance — how many moles and particles? (b) A rigid $2.0 L$ sealed box at $0 K$ — what pressure does $p V = n R T$ predict? (c) If you had $N = 6.022 \times 1 0^{25}$ atoms of neon ( $M = 20 g/mol$ ), what mass is that?

Forecast: Zero mass → zero of everything. Zero kelvin → the formula gives zero pressure (a limit, not a real gas). $1 0^{25}$ is 100× $N_{A}$ , so expect ~100 mol = ~2 kg.

(a) $n = \frac{0}{M} = 0 mol$ , so $N = 0 \times N_{A} = 0$ particles. Why this step? No matter of substance means no moles and no particles — the formulas degrade gracefully to zero, never divide-by-zero (we divide by $M$ , not by $m$ ).
(b) $p = \frac{n R T}{V}$ with $T = 0 K$ gives $p = 0 Pa$ . Why this step? At absolute zero the model has no thermal motion, so no molecular impacts on the walls — pressure vanishes. (Real gases liquefy long before this; the ideal-gas limit is what's shown.)
(c) $n = \frac{6.022 \times 1 0 ^{25}}{6.022 \times 1 0 ^{23}} = 100 mol$ , so $m = 100 \times 20 = 2000 g = 2 kg$ . Why this step? $N$ is exactly $100 \times N_{A}$ , so it's 100 mol; multiply by molar mass for grams.

Verify: (a) $0$ and $0$ ✓; (b) $0 Pa$ ✓; (c) $2000 g = 2 kg$ ✓, matching the ~2 kg forecast.

Recall Which formula, which direction?

Going mass → count you multiply by $N_{A}$ ; going count → mass you divide by $N_{A}$ ::: divide when going from a big raw count to moles; multiply when going from moles to a count Before plugging into $p V = n R T$ , what three conversions must you always check? ::: L → m³, °C → K, kPa/bar → Pa Why did E3's two routes agree? ::: moles is a count; it cannot depend on whether mass was written in g or kg, as long as $m$ and $M$ share a unit What does $p V = n R T$ predict at $T = 0 K$ ? ::: $p = 0$ — zero thermal motion, zero wall impacts

Connections

The Mole Concept — the $n = m / M$ and $n = N / N_{A}$ engine used in E1–E3, E8
Molar Mass and Molecular Mass — where the $M$ values come from
Ideal Gas Law — E7's home turf; all four units meet
Dimensional Analysis — the unit-cancelling checks in E6, E7
Significant Figures and Measurement Uncertainty — how many digits E7's $0.457$ deserves
Kinetic Theory of Gases — why $p \to 0$ as $T \to 0 K$ (E8)