Molar mass calculations: Exercises — Chemistry deep dive

Intuition How to use this page

Cover the answers. Try each problem yourself first, then open the Solution callout. The problems climb a ladder: L1 Recognition (spot the tool) → L2 Application (turn the crank) → L3 Analysis (choose the right path) → L4 Synthesis (combine several ideas) → L5 Mastery (reason backwards).

Every symbol used here is built in the parent note 1.1.13 Molar mass calculations. Quick reminders, all earned there:

$M$ = molar mass, grams that one mole weighs, units $g mol^{- 1}$ .
$n$ = number of moles (how many "scoops" you have).
$m$ = mass in grams (what a balance reads).
$N$ = number of particles (atoms, molecules, or formula units — actual count).
$N_{A} = 6.022 \times 1 0^{23}$ = Avogadro's number (see Avogadro's Number).
The three relations we lean on every single time: $n = \frac{m}{M} m = n M N = n N_{A}$

Atomic masses used throughout (from Atomic Mass and Isotopes, in $g mol^{- 1}$ ): H = 1.008, C = 12.01, N = 14.01, O = 16.00, Na = 22.99, Mg = 24.31, S = 32.07, Cl = 35.45, K = 39.10, Ca = 40.08, Fe = 55.85, Cu = 63.55.

Level 1 — Recognition

Goal: name the right tool and read a formula correctly. No long arithmetic yet.

L1.1 — Count the atoms

State how many atoms of each element are in one formula unit of $Al_{2} (SO_{4})_{3}$ .

Recall Solution

The trick is the outer bracket subscript 3: it multiplies everything inside the bracket $(SO_{4})$ , exactly like $3 (S O_{4})$ in algebra.

Al: the subscript 2 sits on Al → $2$ aluminium.
S: inside is 1 S, and the bracket 3 multiplies it → $1 \times 3 = 3$ sulfur.
O: inside is $O_{4}$ , and the bracket 3 multiplies it → $4 \times 3 = 12$ oxygen.

Answer: 2 Al, 3 S, 12 O. (Notice we did the counting before any masses — that is all L1 asks.)

L1.2 — Which operation?

You have 88.0 g of $CO_{2}$ ( $M = 44.01 g mol^{- 1}$ ). To find how many moles that is, do you multiply or divide by $M$ ? Write the relation, then state the number of moles.

Recall Solution

Grams → moles uses $n = \frac{m}{M}$ : we divide. Let the units decide — that is the surest guard: $\frac{g}{g mol ^{- 1}} = g \times \frac{mol}{g} = mol ✓$ $n = \frac{88.0}{44.01} = 2.00 mol$

Level 2 — Application

Goal: compute molar masses and run the conversion chain once.

L2.1 — Molar mass of glucose

Compute $M$ for glucose $C_{6} H_{12} O_{6}$ .

Recall Solution

Apply SAMS (Subscript, Atomic mass, Multiply, Sum) from the parent note. Multiply each atom's molar mass by its own subscript first, then add: $M = 6 (12.01) + 12 (1.008) + 6 (16.00)$ $= 72.06 + 12.096 + 96.00 = 180.16 g mol^{- 1}$

L2.2 — Grams from moles

What mass is $0.250 mol$ of glucose?

Recall Solution

Moles → grams multiplies: $m = n M$ . $m = 0.250 \times 180.16 = 45.04 g$ Units check: $mol \times g mol^{- 1} = g ✓$ .

L2.3 — Molecules from mass

How many glucose molecules are in 45.04 g?

Recall Solution

Two hops down the chain: grams $\div M$ moles $\times N_{A}$ molecules. $n = \frac{45.04}{180.16} = 0.250 mol$ $N = n N_{A} = 0.250 \times 6.022 \times 1 0^{23} = 1.51 \times 1 0^{23} molecules$

The two-hop chain is the backbone of every problem below — look at the figure:

Level 3 — Analysis

Goal: pick the correct path when the question hides which quantity is asked.

L3.1 — Atoms of one element

In 100.0 g of water $H_{2} O$ ( $M = 18.02 g mol^{- 1}$ ), how many hydrogen atoms are present?

Recall Solution

Careful: the question asks for H atoms, not water molecules. Three hops:

Grams → moles of water: $n = \frac{100.0}{18.02} = 5.549 mol$ .
Moles → water molecules: $N_{H_{2} O} = 5.549 \times 6.022 \times 1 0^{23} = 3.342 \times 1 0^{24}$ .
Each molecule has 2 H (the subscript), so multiply by 2: $N_{H} = 2 \times 3.342 \times 1 0^{24} = 6.68 \times 1 0^{24} atoms$ Why the last step? The subscript 2 is a per-molecule count of that element, so we scale by it after reaching molecules.

L3.2 — Which sample has more atoms?

You have 10.0 g of helium (He, $M = 4.003$ ) and 10.0 g of neon (Ne, $M = 20.18$ ). Without a calculator's final digit, decide which contains more atoms, then confirm numerically.

Recall Solution

Reason first: equal masses, but the lighter element packs more atoms per gram (each atom weighs less). Helium atoms are much lighter → helium should win. Confirm (both are monatomic, so atoms = molecules): $n_{He} = \frac{10.0}{4.003} = 2.498 mol, n_{Ne} = \frac{10.0}{20.18} = 0.4955 mol$ $N_{He} = 2.498 \times N_{A} = 1.505 \times 1 0^{24}, N_{Ne} = 0.4955 \times N_{A} = 2.984 \times 1 0^{23}$ Helium has about 5 times as many atoms. Prediction confirmed.

Level 4 — Synthesis

Goal: combine hydrates, brackets, and multi-step chains in one problem.

L4.1 — Water of crystallisation

Epsom salt is $MgSO_{4} \cdot 7 H_{2} O$ . (a) Find its molar mass. (b) In 61.62 g of Epsom salt, how many moles of water molecules are locked in the crystal?

Recall Solution

(a) Split into the anhydrous salt plus the attached water — the dot ( $\cdot$ ) means "plus this many waters." Anhydrous $MgSO_{4}$ : $24.31 + 32.07 + 4 (16.00) = 24.31 + 32.07 + 64.00 = 120.38$ . Water: $7 \times 18.02 = 126.14$ (with $M_{H_{2} O} = 2 (1.008) + 16.00 = 18.016 \approx 18.02$ ). $M = 120.38 + 126.14 = 246.52 g mol^{- 1}$ (b) First moles of salt: $n_{salt} = \frac{61.62}{246.52} = 0.2500 mol$ . Each formula unit carries 7 waters, so moles of water: $n_{water} = 7 \times 0.2500 = 1.750 mol$ Why multiply by 7? Exactly as a subscript scales an element's count, the " $7$ " scales the number of water molecules per formula unit.

L4.2 — Ammonium phosphate, both brackets and count

$(NH_{4})_{3} PO_{4}$ is a fertiliser. (a) Molar mass. (b) How many oxygen atoms in 4.780 g?

Recall Solution

(a) The bracket 3 multiplies the whole $NH_{4}$ group.

N: $3 \times 1 = 3$ → $3 (14.01) = 42.03$
H: $3 \times 4 = 12$ → $12 (1.008) = 12.096$
P: $1$ → $30.97$ (P = 30.97)
O: $4$ (in $PO_{4}$ ) → $4 (16.00) = 64.00$ $M = 42.03 + 12.096 + 30.97 + 64.00 = 149.10 g mol^{- 1}$ (b) Moles of compound: $n = \frac{4.780}{149.10} = 0.03206 mol$ . Formula units: $N = 0.03206 \times 6.022 \times 1 0^{23} = 1.931 \times 1 0^{22}$ . Each unit has 4 O atoms: $N_{O} = 4 \times 1.931 \times 1 0^{22} = 7.72 \times 1 0^{22} oxygen atoms$

Level 5 — Mastery

Goal: reason backwards — from a measured quantity to an unknown identity or formula.

L5.1 — Find the unknown metal

A chloride has formula $MCl_{2}$ . Exactly $0.100 mol$ of it weighs $11.10 g$ . Identify the metal M.

Recall Solution

Work backwards from moles to molar mass using $M = \frac{m}{n}$ (rearranged from $n = m / M$ ): $M_{MCl_{2}} = \frac{11.10}{0.100} = 111.0 g mol^{- 1}$ Now subtract the two chlorines to isolate M's atomic mass: $M_{M} = 111.0 - 2 (35.45) = 111.0 - 70.90 = 40.1 g mol^{- 1}$ An atomic mass of ≈ 40.1 is calcium (Ca). The compound is $CaCl_{2}$ .

L5.2 — Deduce the number of waters

A hydrate $CuSO_{4} \cdot x H_{2} O$ has molar mass $249.7 g mol^{- 1}$ . Find $x$ .

Recall Solution

Anhydrous $CuSO_{4}$ : $63.55 + 32.07 + 4 (16.00) = 159.62$ . The water contributes the rest: $249.7 - 159.62 = 90.08 g mol^{- 1}$ . Each water is $18.02$ , so: $x = \frac{90.08}{18.02} = 5.00 \approx 5$ The salt is $CuSO_{4} \cdot 5 H_{2} O$ (blue vitriol). Why round to a whole number? $x$ counts real water molecules per formula unit — it must be an integer; our 5.00 confirms clean data.

L5.3 — From percentage to identity

A pure oxide of iron is $69.94%$ iron by mass. Given its formula is $Fe_{2} O_{3}$ , verify the percentage. (Foreshadows Percentage Composition and Empirical Formula.)

Recall Solution

First the molar mass of $Fe_{2} O_{3}$ : $M = 2 (55.85) + 3 (16.00) = 111.70 + 48.00 = 159.70 g mol^{- 1}$ The mass fraction of iron is (mass of Fe in one mole) ÷ (mass of the whole mole): $% Fe = \frac{2 ( 55.85 )}{159.70} \times 100 = \frac{111.70}{159.70} \times 100 = 69.94% ✓$ Why this ratio? In one mole of compound there is one mole's worth of each element; the fraction of total mass that is iron is exactly its percentage by mass.

Recall Self-test: match each problem to its dominant relation

Which core relation did each level lean on most heavily? L1 ::: reading subscripts + $n = m / M$ (recognise the tool) L2 ::: $M = \sum (subscript \times atomic mass)$ , then $m = n M$ and $N = n N_{A}$ L3 ::: $N = n N_{A}$ then $\times$ subscript for a specific element L4 ::: hydrate split $M = M_{salt} + k \cdot M_{water}$ , then the full chain L5 ::: rearranged $M = m / n$ to solve for an unknown identity

Connections

Parent: 1.1.13 Molar mass calculations — every method used here is derived there.
The Mole Concept — the meaning of $n$ .
Avogadro's Number — needed only when counting particles.
Atomic Mass and Isotopes — source of the atomic masses in the table.
Percentage Composition and Empirical Formula — direct extension of L5.3.
Stoichiometry of Reactions — where $n = m / M$ becomes step one of every reaction calculation.
Concentration and Molarity — molar mass turns grams of solute into moles.