Concentration units — mass %, volume %, ppm, ppb, molarity (M), molality (m), mole fraction: Worked examples

Parent: Concentration units

Before we start, one word we lean on constantly is density. Density $ρ$ (read "rho") is just mass packed into each millilitre: if $ρ = 1.14$ g/mL, then every 1 mL of liquid weighs 1.14 g. It is the bridge that lets us swap between "how much volume" and "how much mass". See Density. We also convert grams to moles using Molar Mass and Formula Mass via $n = \frac{mass}{M _{w}}$ , where $n$ is the count of moles (The Mole and Avogadro's Number).

The scenario matrix

Every concentration problem you will ever meet falls into one of these cells. The goal below is to hit all of them at least once.

Cell	What makes it that case	Example that hits it
A. Simple ratio	Straight numerator ÷ denominator × base	Ex 1 (mass %)
B. Grams → moles first	Must convert mass to moles before dividing	Ex 2 (molarity)
C. Solvent-mass unit	Denominator is kg of solvent, not solution	Ex 3 (molality)
D. Sum-to-one unit	Mole fraction, must total 1	Ex 4 (mole fraction)
E. Density-bridge conversion	Need $ρ$ to link M ↔ m ↔ x	Ex 5 (M → m)
F. Trace / dilute limit	ppm, ppb, the $M \approx m$ limit	Ex 6 (ppm)
G. Degenerate / zero input	Pure solvent, or "add nothing"	Ex 7 (zero solute)
H. Real-world word problem	Hidden numbers, everyday framing	Ex 8 (seawater)
I. Exam twist	Reverse direction, or trap built in	Ex 9 (find density from m and M)

Cell A — the plain ratio

Cell B — grams to moles before you divide

Cell C — the kg-of-solvent unit

Cell D — the unit that must sum to 1

Cell E — the density bridge (the hard one)

The master formula from the parent note is $m = \frac{M \times 1000}{1000 ρ - M M _{w}} .$ Let us see where every piece comes from geometrically before using it.

Figure — Concentration units — mass %, volume %, ppm, ppb, molarity (M), molality (m), mole fraction

The red bar is 1 litre of solution — our chosen basis. Its total mass is $1000 ρ$ g. A slice of it (grey) is the solute, weighing $M \cdot M_{w}$ g. What is left over is the solvent, weighing $1000 ρ - M M_{w}$ g. Molality then divides the moles $M$ by that leftover mass (in kg).

Worked example Ex 5 — molarity → molality (Cell E)

A 2.0 M sulfuric acid solution has density $ρ = 1.14$ g/mL and $M_{w} = 98$ g/mol. Find its molality.

Forecast: will $m$ come out bigger or smaller than $M = 2$ ? (Hint: is the solvent less than 1 kg?)

Take 1 L of solution. Then moles of solute $= M = 2.0$ mol. Why this step? Choosing exactly 1 L makes "moles per litre" become just "moles".
Mass of whole solution: $1000 mL \times 1.14 g/mL = 1140$ g. Why? Density turns volume into mass.
Mass of solute: $2.0 \times 98 = 196$ g. Why? moles × molar mass = grams.
Mass of solvent $= 1140 - 196 = 944$ g $= 0.944$ kg. Why subtract? solvent = solution − solute (see the red bar figure).
Molality: $m = \frac{2.0}{0.944} = 2.12$ m.

Verify: plug into the master formula: $\frac{2 \times 1000}{1000 ( 1.14 ) - 2 ( 98 )} = \frac{2000}{944} = 2.12$ ✓. And $m > M$ because we divided by 0.944 kg, which is less than 1 kg — exactly as forecast.

Cell F — trace amounts and the dilute limit

Cell G — degenerate and zero inputs

Worked example Ex 7 — the zero-solute and pure-solvent cases (Cell G)

(a) You add 0 g of salt to 500 g of water. (b) You have pure ethanol (no water at all). Find every concentration you can.

Forecast: which quantities are 0, which are 1, and which are simply undefined?

(a) Zero solute: numerator of every unit is the solute amount $= 0$ .
- mass % $= \frac{0}{500} \times 100 = 0%$ ; molarity $= 0$ M; molality $= 0$ m. Why? Zero over any nonzero denominator is zero — the water is just water.
(b) Pure ethanol, mole fraction: here ethanol is the only component, so $n_{total} = n_{eth}$ . $x_{eth} = \frac{n _{eth}}{n _{eth}} = 1, x_{water} = \frac{0}{n _{eth}} = 0.$ Why? A mole fraction of a pure substance is always 1 — it is 100% of itself.

Verify: (a) $0 + 500 = 500$ g solution, 0% of it is solute ✓. (b) $x_{eth} + x_{water} = 1 + 0 = 1$ ✓ — still sums to one, the defining check. Trap: molarity of a pure substance divides by its own volume — it is a valid number, but molality of a pure substance is undefined (no solvent, so denominator $= 0$ ).

Cell H — a real-world word problem

Worked example Ex 8 — seawater salinity (Cell H)

Seawater is about 3.5% salt by mass and has density $ρ = 1.025$ g/mL. Treating all the salt as NaCl ( $M_{w} = 58.5$ g/mol), estimate its molarity.

Forecast: ocean salt "feels" strong — do you expect roughly 0.6 M, 6 M, or 0.06 M?

Pick 1 L of seawater. Its mass: $1000 mL \times 1.025 = 1025$ g. Why this basis? Molarity is per litre, so start from exactly 1 L.
Mass of salt in it: $3.5%$ of $1025$ g $= 0.035 \times 1025 = 35.875$ g. Why? Mass % tells us how many grams of solute sit inside that mass of solution.
Grams → moles: $n = \frac{35.875}{58.5} = 0.613$ mol. Why? Molarity counts moles, so convert with molar mass.
Molarity: $M = \frac{0.613}{1 L} = 0.613$ M.

Verify: dimensional path grams → moles → mol/L is clean; answer ≈ 0.6 M, matching the real ocean value of ~0.6 M NaCl ✓.

Cell I — the exam twist (reverse direction)

Worked example Ex 9 — back out the density (Cell I)

A solution is 1.0 M and 1.09 m in a solute with $M_{w} = 60$ g/mol. Find the density $ρ$ .

Forecast: this reverses Ex 5 — instead of finding $m$ from $ρ$ , we solve the same equation for $ρ$ .

Start from the master formula: $m = \frac{M \times 1000}{1000 ρ - M M _{w}} .$ Why this step? The same relation links all four quantities; we simply treat $ρ$ as the unknown.
Isolate the denominator. Multiply both sides by $(1000 ρ - M M_{w})$ then divide by $m$ : $1000 ρ - M M_{w} = \frac{M \times 1000}{m} .$ Why? We move the unknown out of the fraction so we can solve linearly.
Plug numbers: $\frac{1.0 \times 1000}{1.09} = 917.4$ , and $M M_{w} = 1.0 \times 60 = 60$ . So $1000 ρ = 917.4 + 60 = 977.4 ⟹ ρ = 0.977 g/mL .$

Verify: feed $ρ = 0.977$ back into the forward formula: $\frac{1 \times 1000}{1000 ( 0.977 ) - 60} = \frac{1000}{917} = 1.09$ m ✓. The trap: students forget the $- M M_{w}$ term and get $ρ = 0.917$ , which is wrong — the solute mass is real and must be subtracted.

Recall Self-test: match the cell before you compute

For each, name the matrix cell first, then solve. Given moles + mL of solution, find concentration ::: Cell B → molarity. Given M, ρ, and molar mass, find m ::: Cell E → density-bridge master formula. Given a pure substance's fraction of itself ::: Cell G → mole fraction = 1. Given salt-by-mass and density, find molarity ::: Cell H → real-world, use 1 L basis.