Molar mass calculations: Question bank — Chemistry deep dive

The picture below anchors these four symbols to a physical scene — a balance, a scoop, and a swarm of particles — so you can see which arrow multiplies and which divides.

This bank drills the ideas built in the parent note. Where a term needs its deeper source, that's flagged inline — but every trap below is self-contained on this page.

Spot the error

Each statement hides one flaw. Name it.

Ca(NO_{3})_{2}

has 1 Ca, 2 N, 3 O."

The outer subscript

2

distributes over the whole bracket, so oxygens are

3 \times 2 = 6

, not

3

. Correct count: 1 Ca, 2 N, 6 O.

"To find moles

n

from 20 g of a gas, multiply 20 by its molar mass

M

Grams → moles divides by

M

. Multiplying gives units

g \times g mol^{- 1}

, which is nonsense; only

g \div (g mol^{- 1}) = mol

works.

H_{2} SO_{4}

molar mass =

2 + 32 + 4 = 38

Each subscript must multiply that atom's molar mass, not just the count of atoms:

2 (1.008) + 32.07 + 4 (16.00) \approx 98.08

. You cannot add raw subscript numbers.

"Number of O atoms in one mole of

CO_{2}

equals

N_{A}

Each molecule has 2 O, so it is

2 N_{A}

oxygen atoms. Multiply moles

n

of compound by the element's subscript before applying

N_{A}

"Since molar mass has units

g mol^{- 1}

, dividing grams by it gives grams squared."

Units divide too:

g \div (g mol^{- 1}) = g \times mol / g = mol

. The grams cancel, leaving moles.

CuSO_{4} \cdot 5 H_{2} O

: I'll add

5 \times 2 = 10

H and

5

O for the water, then move on."

You must also include the anhydrous salt's own atoms; the water is added to

CuSO_{4}

, not used instead of it.

"Molar mass of

O_{2}

16.00 g mol^{- 1}

because oxygen's atomic mass is 16."

O_{2}

is two oxygen atoms bonded:

M = 2 \times 16.00 = 32.00 g mol^{- 1}

. Elemental oxygen gas is diatomic, not atomic.

Why questions

Prompt yourself, then reveal the reasoning.

Why do we never memorise two separate numbers "12 u for one atom, 12 g for a mole"?

Because

N_{A}

was chosen so 1 u of particle mass maps to exactly 1

g mol^{- 1}

. The atomic-mass number and molar-mass number are the same by definition (carbon-12 anchors both).

Why does the subscript act before summing, not after?

A subscript is a count of that specific atom in the unit; you tally each element's contribution (

x M_{A}

) first, then stack them (

+ y M_{B}

). Multiplying the finished sum would double-count everything.

Why must we check that units cancel in

n = m / M

Unit cancellation is a free error-detector: if the answer doesn't come out in mol, you divided the wrong way. It replaces memorising "multiply or divide."

Why does an outer-bracket subscript distribute like algebra's

2 (a + b)

The bracket groups atoms that repeat as a unit; the outer number says "this whole group appears that many times," so every atom inside is multiplied — exactly like the distributive law.

Why does a compound's molar mass

M

equal the sum of its atoms' molar masses?

Mass is additive: one formula unit's mass is its atoms stacked, and multiplying by

N_{A}

(the same for all) distributes over the sum, giving

M = x M_{A} + y M_{B}

Why is

M

for the mole-to-particles step irrelevant — we use

N_{A}

instead?

M

converts mass↔moles (a weighing question);

N_{A}

converts moles↔count (a counting question). Different questions need different bridges.

Edge cases

Boundary and degenerate inputs — the topic still must work here.

What is the molar mass of a single element that exists as isolated atoms, like helium?

Just its atomic mass with

g mol^{- 1}

units:

\approx 4.00 g mol^{- 1}

. No subscript, no bond — one atom is the whole formula unit.

Does

n = m / M

still make sense at

m = 0

Yes:

n = 0/ M = 0

mol. Zero mass means zero moles means zero particles — the chain stays consistent all the way to a count of

0

If a "compound" is really just one atom (e.g. the noble gas Ar), does the compound formula

M = x M_{A} + y M_{B}

break?

No — it reduces to a single term with

x = 1

and no second element, giving

M = M_{Ar}

. The general rule contains the single-element case.

For an isotopically pure sample (say pure carbon-12), is molar mass still the tabulated average?

No — pure carbon-12 has

M = 12.000

exactly. Tabulated

12.011

is the natural-abundance average; a pure isotope uses that isotope's own mass.

What is the molar mass "of an electron gas" or free electrons — does molar mass apply?

Yes in principle:

N_{A}

electrons have a mass of about

5.5 \times 1 0^{- 4}

g, so

M \approx 0.00055 g mol^{- 1}

. Molar mass is defined for any representative particle, not only atoms.

Can two different subscripts give the same molar mass contribution?

Yes — e.g. 12 H atoms (

12 \times 1.008 \approx 12.10

) and 1 C atom (

12.01

) contribute nearly equal mass. Equal mass contribution does not imply equal atom count.

Recall Self-check: which single habit kills most of these traps?

Writing the units next to every number and letting them cancel. It catches multiply-vs-divide errors, exposes bracket-distribution mistakes, and confirms whether you built the right per-mole quantity.

Connections

Molar mass calculations — the parent note these traps drill.
The Mole Concept — where $n$ and the mole are first defined.
Avogadro's Number — the source of $N_{A}$ , the moles↔particles bridge.
Atomic Mass and Isotopes — source of the weighted-average element masses used above.
Percentage Composition and Empirical Formula — where these conceptual habits get applied.
Stoichiometry of Reactions — every mass problem opens with $n = m / M$ .

Question bank — Molar mass calculations

True or false — justify