Molar mass calculations: Worked examples — Chemistry deep dive

Once we can build $M$ , the parent's three conversion tools take over — all in terms of $n$ = moles, $m$ = mass in grams, $N$ = number of particles: $n = \frac{m}{M}, m = n M, N = n N_{A}$ Nothing new — just every angle of attack.

The scenario matrix

Chemistry problems don't have "quadrants" like angles do — but they have case classes: each one is a place where a beginner slips. Here is the full map. Every worked example below is tagged with the cell it covers.

#	Case class	What makes it tricky	Example
A	Plain compound	just SAMS (subscript, atomic mass, multiply, sum)	Ex 1
B	Brackets / nested subscript	outer number distributes over the group	Ex 2
C	Hydrate (the dot `·`)	add attached water	Ex 3
D	Reverse: particles → grams	run the chain backwards	Ex 4
E	Zero / degenerate input	$0$ g, $1$ atom, monatomic gas — does the machine still work?	Ex 5
F	Limiting / very large & very small	tiny masses, huge counts, keeping sig figs	Ex 6
G	Real-world word problem	strip a story down to $m$ , $M$ , $N_{A}$	Ex 7
H	Exam twist: unknown subscript	molar mass is given, solve for the count	Ex 8

Figure: the two-way conversion chain. Ex 6 and Ex 7 walk it left-to-right (top arrows); Ex 4 walks it right-to-left (bottom arrows).

Case A — plain compound

Worked example Ex 1 (cell A) — molar mass of glucose

$C_{6} H_{12} O_{6}$ Forecast: guess before computing — is it above or below 100 g/mol? (It has 24 atoms, and most weigh 12–16 u each, so... predict big.)

Count atoms: 6 C, 12 H, 6 O. Why this step? The formula is literally a headcount of atoms — read it left to right.
Subscript × atomic mass, for each element: $6 (12.01) = 72.06, 12 (1.008) = 12.096, 6 (16.00) = 96.00$ Why this step? The subscript multiplies only its own atom, done before adding — the #1 mistake is multiplying the total instead.
Sum: $M = 72.06 + 12.096 + 96.00 = 180.16 g/mol$ Why this step? Mass is additive: a molecule's mass is its atoms stacked together.

Verify: rounded whole-number check $6 (12) + 12 (1) + 6 (16) = 72 + 12 + 96 = 180$ . Our precise answer $180.16$ hugs it. ✓ And $180 > 100$ — forecast confirmed.

Case B — brackets

Worked example Ex 2 (cell B) — ammonium sulfate

$(NH_{4})_{2} SO_{4}$ Forecast: how many hydrogen atoms — 4 or 8? (The outer 2 doubles the whole $NH_{4}$ group, so predict 8.)

Expand the bracket first. The outer subscript $2$ distributes over everything inside $(NH_{4})$ : $2 \times N = 2 N, 2 \times H_{4} = 8 H$ Why this step? Just like $2 (x + y) = 2 x + 2 y$ in algebra — the bracket subscript hits every atom inside, not just the last one.
Add the $SO_{4}$ part: 1 S, 4 O. Final atom tally: 2 N, 8 H, 1 S, 4 O. Why this step? The $SO_{4}$ group sits outside the bracket, so the outer 2 does not touch it — we count its atoms once and merge them with the expanded bracket to get the full formula-unit tally.
SAMS — subscript × atomic mass, then sum: $2 (14.01) + 8 (1.008) + 1 (32.07) + 4 (16.00)$ $= 28.02 + 8.064 + 32.07 + 64.00 = 132.15 g/mol$ Why this step? Same additive rule; the only new skill was the bracket expansion in step 1.

Verify: whole-number sanity $2 (14) + 8 (1) + 32 + 4 (16) = 28 + 8 + 32 + 64 = 132$ . Matches $132.15$ . ✓ H count = 8 — forecast confirmed.

Case C — hydrate (the dot)

Worked example Ex 3 (cell C) — washing soda

$Na_{2} CO_{3} \cdot 10 H_{2} O$ Forecast: the water alone is $10 \times 18.02 = 180.2$ . Will the water outweigh the salt itself? (Predict yes — 10 waters is a lot.)

Anhydrous salt $Na_{2} CO_{3}$ : $2 (22.99) + 12.01 + 3 (16.00) = 45.98 + 12.01 + 48.00 = 105.99$ Why this step? Handle the salt as an ordinary compound (case A) before touching the water.
Water of crystallisation: the dot · means "plus this many attached water molecules." $10 \times 18.02 = 180.20$ Why this step? The crystal physically contains those waters — their mass is really on the balance, so we add it.
Total: $M = 105.99 + 180.20 = 286.19 g/mol$ Why this step? The dot means "and also attached to," so mass is still just additive — we sum the salt part and the water part to get the mass of one full mole of crystal.

Verify: water fraction $= 180.20/286.19 = 0.630 = 63.0%$ — the water genuinely outweighs the salt. ✓ Forecast confirmed.

Case D — run the chain backwards

Worked example Ex 4 (cell D) — you have

$3.01 \times 1 0^{23}$ molecules of $CO_{2}$ . What mass is that? Forecast: $3.01 \times 1 0^{23}$ is half of $N_{A}$ , so half a mole. Half a mole of a $\sim 44$ g/mol gas → predict about 22 g.

Molar mass of $CO_{2}$ (case A): $12.01 + 2 (16.00) = 44.01 g/mol$ . Why this step? Before we can turn a count into a mass we need the "one-scoop weight" $M$ — it's the bridge between the moles world and the grams world, so we build it first.
Particles → moles. The forward chain was $n \times N_{A} N$ ; reverse it by dividing: $n = \frac{N}{N _{A}} = \frac{3.01 \times 1 0 ^{23}}{6.022 \times 1 0 ^{23}} = 0.500 mol$ Why this step? Every step of the chain is reversible — go the opposite direction with the opposite operation ( $\times$ becomes $\div$ ).
Moles → grams. Now use $m = n M$ (multiply this way — grams are what we want): $m = 0.500 \times 44.01 = 22.0 g$ Why this step? $m = n M$ has units $mol \times g/mol = g$ — cancellation confirms the direction.

Verify: units $mol \times \frac{g}{mol} = g$ ✓. And $22.0 \approx 22$ — forecast confirmed.

Case E — zero and degenerate inputs

Worked example Ex 5 (cell E) — three "does the machine break?" checks

Forecast: will $n = m / M$ and $N = n N_{A}$ still behave for $0$ grams, a single atom, and a monatomic gas?

(i) Zero mass. You have $0$ g of gold. $n = \frac{0}{197.0} = 0 mol, N = 0 \times N_{A} = 0 atoms$ Why this step? Nothing weighed = nothing there. The formula gives $0$ cleanly — no division-by-zero, because $M$ (never zero) is in the denominator, not the numerator.

(ii) A single atom. One neon atom — what's its mass? $m = n M = \frac{1}{N _{A}} \times M = \frac{20.18}{6.022 \times 1 0 ^{23}} = 3.35 \times 1 0^{- 23} g$ Why this step? "One atom" is $n = 1/ N_{A}$ moles. The machine handles absurdly small counts too — it's just $M$ divided by Avogadro's number.

(iii) Monatomic gas. Molar mass of argon (Ar) — no subscript at all. $M = 39.95 g/mol$ Why this step? A "formula" with one atom and no subscript means $x = 1$ : $M = 1 \times M_{Ar}$ . The compound rule $M = x M_{A}$ degenerates gracefully to "just the element."

Verify: (i) $0$ g $\to 0$ atoms ✓. (ii) $20.18/6.022 \times 1 0^{23} = 3.35 \times 1 0^{- 23}$ g ✓. (iii) $M_{Ar} = 39.95$ , single atom, no multiply ✓.

Case F — limiting: very tiny and very huge

Worked example Ex 6 (cell F) — how many atoms in

$1.00 ng$ ( $1.00 \times 1 0^{- 9}$ g) of carbon? Forecast: a nanogram sounds like "almost nothing." But atoms are unimaginably tiny — predict the count is still enormous (billions of billions).

Molar mass of carbon: $M = 12.01 g/mol$ . Why this step? Carbon is monatomic here, so its molar mass is just its atomic mass carried into g/mol — and we need $M$ as the bridge to divide grams into moles in the next step.
Grams → moles (divide): $n = \frac{1.00 \times 1 0 ^{- 9}}{12.01} = 8.33 \times 1 0^{- 11} mol$ Why this step? Even a tiny mass gives a tiny mole count — but "tiny in moles" is still gigantic in atoms, because one mole is $6 \times 1 0^{23}$ .
Moles → atoms (multiply by $N_{A}$ ): $N = 8.33 \times 1 0^{- 11} \times 6.022 \times 1 0^{23} = 5.02 \times 1 0^{13} atoms$ Why this step? $N = n N_{A}$ scales the tiny mole number back up. $5 \times 1 0^{13}$ = fifty trillion atoms in one nanogram.

Verify: $8.33 \times 1 0^{- 11} \times 6.022 \times 1 0^{23} = 5.02 \times 1 0^{13}$ ✓. Forecast confirmed — "almost nothing" holds fifty trillion atoms.

Case G — real-world word problem

Worked example Ex 7 (cell G) — a story to strip down

A patient is given a $0.900 g$ dose of table salt $NaCl$ dissolved in water. How many sodium ( $Na^{+}$ ) ions enter their bloodstream?

Forecast: less than a mole of salt (under 60 g), so predict fewer than $N_{A}$ ions — somewhere in the $1 0^{22}$ range.

Strip the story to symbols. Given $m = 0.900$ g. Wanted: $N (Na)$ . Bridge: molar mass. Why this step? Every word problem is the same three quantities ( $m$ , $M$ , $N_{A}$ ) wearing a costume. Undress it first.
Molar mass of $NaCl$ : $22.99 + 35.45 = 58.44 g/mol$ . Why this step? NaCl is a plain two-atom formula unit (case A), so SAMS applies; we need $M$ before we can turn the weighed grams into moles.
Grams → moles of NaCl (divide): $n = \frac{0.900}{58.44} = 0.01540 mol$ Why this step? Grams → moles always divides by $M$ ; the units $g \div (g/mol) = mol$ confirm we're dividing, not multiplying.
Moles → formula units → Na ions. Each NaCl unit releases exactly one $Na^{+}$ , so moles of Na = moles of NaCl: $N = 0.01540 \times 6.022 \times 1 0^{23} = 9.27 \times 1 0^{21} ions$ Why this step? The subscript on Na is 1, so multiply by $N_{A}$ once. (If it were $Na_{2} SO_{4}$ , we'd multiply by 2 here.)

Verify: $9.27 \times 1 0^{21} < 6.022 \times 1 0^{23} = N_{A}$ ✓ (fewer than one mole of ions), matching the forecast.

Case H — exam twist: solve for an unknown subscript

Worked example Ex 8 (cell H) — molar mass given, find the count

A metal chloride $MgCl_{x}$ has molar mass $95.21 g/mol$ . Find $x$ .

Forecast: Mg is $\approx 24$ , Cl is $\approx 35.5$ . If $x = 2$ that's $24 + 71 = 95$ — predict $x = 2$ .

Write molar mass with $x$ unknown (case A, but leave $x$ as a letter): $M = M_{Mg} + x M_{Cl} = 24.31 + 35.45 x$ Why this step? Molar mass is a linear equation in the subscript. Given $M$ , we can solve for $x$ — the machine runs in either direction.
Set equal to the given value and solve: $24.31 + 35.45 x = 95.21 \Rightarrow 35.45 x = 70.90 \Rightarrow x = 2.00$ Why this step? Isolate $x$ like any algebra equation. A near-integer answer is your sanity signal — atom counts must be whole numbers.
Formula is $MgCl_{2}$ . Why this step? An atom count must be a whole number (you can't have half an atom in a formula), so $x = 2.00$ rounds confidently to $x = 2$ — the compound is magnesium chloride, $MgCl_{2}$ .

Verify: rebuild it: $24.31 + 2 (35.45) = 24.31 + 70.90 = 95.21 g/mol$ ✓. Integer $x = 2$ — forecast confirmed.

Recall

Recall Which case classes divide by

$M$ and which multiply? Grams → moles divides; moles → grams multiplies; moles ↔ particles uses $\times N_{A}$ / $\div N_{A}$ . ::: Cases A–C build $M$ ; D and G run grams→particles; H solves for a subscript. In a hydrate, what does the · add? ::: The full mass of the attached water molecules ( $k \times 18.02$ for $\cdot k H_{2} O$ ). Why doesn't $0$ g break $n = m / M$ ? ::: Because $M$ (never zero) is the denominator; $0/ M = 0$ cleanly. How do you find an unknown subscript from a given molar mass? ::: Write $M$ as a linear equation in the subscript and solve; the answer must round to a whole number.

Connections

Molar mass calculations — the parent: derives $M = x M_{A} + y M_{B}$ and the conversion chain used everywhere here.
The Mole Concept — defines $n$ and the "one scoop" idea behind cases D–G.
Avogadro's Number — the $\times N_{A}$ step in every particle-count example.
Atomic Mass and Isotopes — source of the element masses plugged into SAMS.
Percentage Composition and Empirical Formula — Ex 3's water-fraction check is a mini version of this.
Stoichiometry of Reactions — Ex 7's dose problem is stoichiometry's first step.
Concentration and Molarity — Ex 7 continues here once the salt is dissolved.

Worked examples — Molar mass calculations

The scenario matrix

Case A — plain compound

Case B — brackets

Case C — hydrate (the dot)

Case D — run the chain backwards

Case E — zero and degenerate inputs

Case F — limiting: very tiny and very huge

Case G — real-world word problem

Case H — exam twist: solve for an unknown subscript

Recall

Connections

Case Map