This is a practice page for Density, molar mass, molar volume . We will not re-derive the bridges here — we will use them until every possible type of problem feels familiar . First we map out what kinds of questions even exist , then we walk one worked example for each kind.
Intuition The three bridges, on one pictureEvery problem on this page is a journey between three "islands": mass (grams, what a balance reads), volume (litres or cm³, what a flask reads), and moles (how many particles — what chemistry actually cares about). The three bridges connect them:
density ρ = V m molar mass n = M m molar volume n = V m V
Every example below is just "which island am I on, which island do I want, which bridges cross the gap."
Before solving anything, let's list every distinct kind of situation this topic can hand you. If we cover each row, nothing on an exam can surprise us.
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Case class
What makes it different
Covered by
A
One bridge, forward given mass, want moles (or vice-versa)
Ex 1
B
Two bridges chained mass→moles→volume across an island
Ex 2
C
Gas at STP use V m = 22.7 L mol − 1
Ex 3
D
Gas NOT at STP must use P V = n R T
Ex 4
E
Unit-trap (mL/L, cm³/m³) numbers right, units wrong
Ex 5
F
Degenerate / limiting V → 0 Ex 6
G
Solid/liquid density packing matters, 22.7 forbidden
Ex 7
H
Reverse: find M from density measure ρ
Ex 8
I
Word problem (real world) strip the story to numbers
Ex 9
J
Exam twist (find %, comparison) multi-part reasoning
Ex 10
Definition Symbols we will reuse (earned once, here)
m ( g ) V ( L ) ( cm 3 ) n ( mol ) count of 6.022 × 1 0 23 M ( g mol − 1 ) ρ ( g cm − 3 g L − 1 ) V m ( L mol − 1 ) P , V , T , R P V = n R T Ideal Gas Law PV=nRT ).
Worked example Example 1 — grams to molesHow many moles are in 8.0 g M H e = 4.0 g mol − 1
Forecast: one mole of He weighs 4 g. We have 8 g — that's two full bags. Guess: 2 mol .
Pick the bridge. We stand on the mass island, want the moles island → molar-mass bridge n = M m Why this step? Only this bridge connects grams to moles.
Plug in: n = 4.0 g mol − 1 8.0 g = 2.0 mol Why this step? Dividing grams by grams-per-mole cancels grams and leaves mol.
Verify: units g / mol g = mol
Worked example Example 2 — mass of solid → volume of its gas at STP6.6 g C O 2 M = 44 g mol − 1 V m = 22.7 L mol − 1
Forecast: 6.6 g is much less than 44 g, so less than one mole — expect less than 22.7 L. Guess a few litres.
Cross bridge 1 (mass → moles): n = 44 6.6 = 0.15 mol Why this step? We start on the mass island; volume of a gas is found through moles, not directly.
Cross bridge 2 (moles → volume): V = n V m = 0.15 × 22.7 = 3.405 L Why this step? Molar-volume bridge turns a count of moles into litres of gas.
Verify: 0.15 mol < 1 mol 3.4 L < 22.7 L
Worked example Example 3 — density of a gas at STPFind the density of nitrogen gas N 2 M = 28 g mol − 1
Forecast: one mole (28 g) fills 22.7 L, so density = 28 ÷ 22.7, a bit over 1 g/L. Guess ~1.2.
Use the gas density link ρ = V m M Why this step? One mole is simultaneously M V m Avogadro's Law .
ρ = 22.7 28 = 1.234 g L − 1
Verify: air is ~1.2 g/L and is mostly N₂ — perfectly sensible ✓.
Common mistake The 22.7 shortcut only works at STPV m = 22.7 L mol − 1 P R T at 273.15 K and 1 bar . Change T P P V = n R T
Worked example Example 4 — gas away from standard conditionsWhat volume does 0.50 mol P = 2.0 × 1 0 5 Pa T = 300 K R = 8.314 J mol − 1 K − 1
Forecast: double the pressure and slightly higher temperature than STP — pressure squeezes volume down, so expect well under the 0.5 × 22.7 = 11.35
Rearrange the ideal gas law for volume: V = P n R T Why this step? We can't use V m P , T
Plug in SI units: V = 2.0 × 1 0 5 0.50 × 8.314 × 300 = 6.24 × 1 0 − 3 m 3 Why this step? Joules and pascals give cubic metres.
Convert: 6.24 × 1 0 − 3 m 3 × 1000 = 6.24 L
Verify: 6.24 L < 11.35 L
Worked example Example 5 — mL vs L / cm³ vs m³A 500 mL 13.6 g cm − 3
Forecast: mercury is famously heavy; half a litre should be a hefty ~7 kg. Guess several kilograms.
Match the volume unit to the density unit. Density is per cm³ , so express volume in cm³: 500 mL = 500 cm 3 Why this step? ρ = m / V V same unit density uses.
m = ρ V = 13.6 × 500 = 6800 g = 6.8 kg
Verify: the number 6800 g ≈ 6.8 kg matches the "several kilograms" forecast ✓. Had we sloppily left volume as "0.5 L" and multiplied by 13.6, we'd get 6.8 g — a 1000 ×
Intuition What happens at the edges?Good problems test the boundaries: what if the volume is zero? What if you don't know the gas's identity? These "degenerate" cases reveal whether you actually understand the formulas.
Worked example Example 6 — the identity-independent limit(a) As V → 0 m ρ = m / V (b) You have 1 mol unknown ideal gas at STP. Can you state its volume without knowing what it is?
Forecast: (a) shrinking the box while keeping the mass should make density blow up. (b) Avogadro says gas volume ignores identity, so yes.
(a) In ρ = V m m V → 0 + ρ → + ∞ Why this step? Cramming any mass into no space means infinite density — a black-hole-like limit; physically it's why solids resist compression.(b) By Avogadro's law every ideal gas has V m = 22.7 L mol − 1 M V = 1 × 22.7 = 22.7 L Why this step? Molar volume is set by spacing (T , P STP and Standard Conditions .
Verify: (a) sign check — as V ρ mass would differ per gas (via ρ = M / V m volume is fixed at 22.7 L ✓.
Worked example Example 7 — molar volume of liquid water (why it is NOT 22.7 L)Liquid water has ρ = 1.00 g cm − 3 M = 18 g mol − 1
Forecast: water is packed tight, not gas-spaced, so expect a tiny molar volume — nowhere near 22.7 L. Guess a few tens of mL.
Get the volume of one mole's worth of mass. One mole weighs M = 18 g Why this step? Molar volume V m = V / n n = 1 M
Convert mass to volume using density: V = ρ m = 1.00 g cm − 3 18 g = 18 cm 3 = 18 mL
Verify: 18 mL 22 700 mL touch .
Worked example Example 8 — identify a mystery gasAn unknown ideal gas at STP has density ρ = 1.25 g L − 1
Forecast: its density is close to nitrogen's (1.23 from Ex 3), so expect M ≈ 28 N 2 C O
Invert the gas density link. From ρ = V m M M M = ρ V m Why this step? We know ρ V m M
M = 1.25 × 22.7 = 28.375 g mol − 1 ≈ 28 g mol − 1 Why this step? Multiplying grams-per-litre by litres-per-mole cancels litres, leaving grams-per-mole.
Verify: M ≈ 28 N 2 M Relative Atomic Mass & Atomic Mass Unit .
Worked example Example 9 — the propane tankA cook uses propane C 3 H 8 M = 44 g mol − 1 11 kg as a gas at STP?
Forecast: 11 kg = 11000 g; divide by 44 ≈ 250 mol; times 22.7 L ≈ 5700 L. That huge gas volume is why it's stored as a compressed liquid!
Strip the story to a mass: m = 11 kg = 11000 g Why this step? Formulas need grams; "a tank" is just m
Bridge to moles: n = 44 11000 = 250 mol
Bridge to gas volume: V = n V m = 250 × 22.7 = 5675 L Why this step? Same chain as Example 2 (mass→moles→volume), just bigger numbers.
Verify: 5675 L ≈ 5.7 m 3
Worked example Example 10 — which balloon rises, and by how much lighter?At STP, compare helium (M = 4 M ˉ = 29
Forecast: helium is famously light — expect its density far below air's, and it should be roughly 85–90% lighter.
Density of each via ρ = M / V m ρ H e = 22.7 4 = 0.1762 g L − 1 ρ ai r = 22.7 29 = 1.2775 g L − 1 Why this step? Same V m M
Percentage lighter: ρ ai r ρ ai r − ρ H e × 100% = 1.2775 1.2775 − 0.1762 × 100% Why this step? "X % relative to air .
= 1.2775 1.1013 × 100% = 86.2%
Verify: shortcut — since V m 1 − 29 4 = 1 − 0.1379 = 86.2%
Recall Which bridge(s) does each case need?
Grams → moles ::: molar-mass bridge n = m / M
Solid mass → gas volume at STP ::: two bridges, n = m / M V = n V m
Gas volume off STP ::: P V = n R T V m
Density of a liquid's one mole ::: V = m / ρ m = M
Gas density → identity ::: M = ρ V m
Mnemonic Solve any of them in 3 questions"Where am I? Where do I want to be? Which bridge crosses the gap?" Island = mass / volume / moles. That's the entire method.
