Worked example 1 — solution

Step 1 · Find the molar mass

The subscript outside the parentheses multiplies everything inside: 2 N and 6 O.

Worked example 1 — solution

Worked example 1 — solution

Worked example 2 — solution

One conversion factor is needed.

A common first attempt

No unit cancels, and the result is not a count of anything.

Worked example 2 — solution

g NaCl → mol NaCl. One arrow, one factor: the equality 1 mol NaCl = 58.44 g NaCl.

Step 3 · Pick the orientation that cancels

Worked example 2 — solution

Worked example 2 — solution

Practice 1 — answer: A

B put the molar mass on top: 95.21 / 65.0 = 1.46. C used the factor written 95.21 g over 1 mol: 65.0 × 95.21 = 6.19 × 10³ g²/mol. D kept converting past moles: 0.683 × 6.022 × 10²³ = 4.11 × 10²³, a count of formula units, not moles.

Worked example 3 — solution

Two conversion factors are needed.

Step 2 · Write the route

g H₂O → mol H₂O → molecules. Molar mass covers the first arrow; Avogadro's number covers the second.

Worked example 3 — solution

Avogadro's number is an equality too: 1 mol = 6.022 × 10²³ molecules, so it gives two factors. Only one cancels moles:

Worked example 3 — solution

One continuous chain; each factor cancels the unit before it:

Worked example 3 — solution

Practice 2 — answer: C

B stopped at moles: 8.00 / 32.00 = 0.250, a count of moles, not molecules. D used the factor written 32.00 g over 1 mol: 8.00 × 32.00 × 6.022 × 10²³ = 1.54 × 10²⁶. A used Avogadro's number upside down: 0.250 ÷ (6.022 × 10²³) = 4.15 × 10⁻²⁵, a fraction of one molecule.

Worked example 1 — solution

Step 1 · Molar mass

Worked example 1 — solution

Step 1 · Molar mass

Each subscript is 1, so each element contributes one atom's mass: 39.10 g of K and 35.45 g of Cl in one mole.

Worked example 1 — solution

Step 1 · Molar mass

Worked example 1 — solution

Step 1 · Molar mass

Worked example 2 — solution

Step 1 · Molar mass

Worked example 2 — solution

Step 1 · Molar mass

The first attempt took 14.01 g — the mass of one N. The subscript outside the parentheses doubles the group, so the formula holds two.

Worked example 2 — solution

Step 1 · Molar mass

Worked example 2 — solution

Step 1 · Molar mass

Practice 1 — answer: C

A counted one N of the two: 14.01 ÷ 132.15 × 100 = 10.60. B counted atoms, not mass: 2 of its 15 atoms are N, and 2 ÷ 15 × 100 = 13.33. D computed everything except nitrogen: 104.13 ÷ 132.15 × 100 = 78.80.

Worked example 3 — molar mass and element mass

Wanted: the mass percent of N in each compound. Two percents are needed, one per compound.

Step 1 · Molar mass

Worked example 3 — molar mass and element mass

Step 1 · Molar mass

Worked example 3 — molar mass and element mass

Step 1 · Molar mass

Worked example 3 — divide and scale

Step 3 · Divide and scale

Worked example 3 — divide and scale

Step 3 · Divide and scale

Practice 2 — answer: D

A counted atoms, not mass: 1 of 5 = 20.00%, 1 of 6 = 16.67%, reversing the ranking. B ignored the whole: the same 14.01 g of N is a share of 53.49 g in one compound, then of 101.11 g in the other. C used the mole's size, not the element's share.

Practice 2 — answer: D

Worked example 1 — solution

Step 1 · Grams → moles

The mole count is given directly: 0.350 mol NaCl, no mass conversion needed.

Worked example 1 — solution

Worked example 1 — solution

Worked example 1 — solution

Worked example 2 — solution

Step 1 · Grams → moles

The molar mass converts the mass to moles: 9.35 g ÷ 74.55 g/mol = 0.1254 mol KCl.

Worked example 2 — solution

The unit came out mol/mL. Molarity is mol/L — the volume must enter in liters.

Worked example 2 — solution

The mark reads 250.0 mL of solution, which is 0.2500 L. Divide:

Worked example 2 — solution

Practice 1 — answer: B

A divided by milliliters: 0.500 ÷ 500 = 0.00100. C divided grams by liters: 20.0 ÷ 0.5000 = 40.0, in g/L. D multiplied by the molar mass: 20.0 × 40.00 ÷ 0.5000 = 1600.

Worked example 3 — solution

Three conversion factors are needed.

Step 1 · Grams → moles

Worked example 3 — solution

Two orientations exist. Only one cancels mol KOH:

Worked example 3 — solution

Worked example 3 — solution

Practice 2 — answer: C

A stopped at liters and relabeled: 10.1 ÷ 101.11 ÷ 0.400 = 0.250 L, which is 250. mL, not 0.250 mL. B flipped the molarity fraction: 10.1 ÷ 101.11 × 0.400 × 1000 = 40.0, and the units, mol²/L, are not a volume. D skipped the molar mass: 10.1 ÷ 0.400 = 25.3 L, or 25300 mL. Grams must become moles first.

Worked example 1 — grams, then moles

Three molar-mass conversions are needed — one for each element.

Step 1 · Percents → grams

In a 100 g sample, each percent reads directly as grams: 40.0 g C, 6.7 g H, 53.3 g O.

Worked example 1 — grams, then moles

Each element converts with its own molar mass:

Worked example 1 — grams, then moles

Worked example 1 — the smallest ratio

Step 3 · Divide by the smallest

The smallest mole count, 3.33, divides into every count:

Worked example 1 — the smallest ratio

1.00 : 2.00 : 1.00 is already whole — nothing to multiply.

Worked example 1 — the smallest ratio

Worked example 2 — grams, then moles

Two molar-mass conversions are needed.

Step 1 · Percents → grams

In a 100 g sample: 72.4 g Fe and 27.6 g O.

Worked example 2 — grams, then moles

Worked example 2 — grams, then moles

Worked example 2 — clearing the fraction

Step 3 · Divide by the smallest

Worked example 2 — clearing the fraction

Tempting: 1.33 is nearly 1 — call the ratio 1 : 1 and write FeO. The rounded formula can be tested against the data:

Worked example 2 — clearing the fraction

1.33 is the fraction 4⁄3. Multiplying every count by 3 clears it:

Worked example 2 — clearing the fraction

Practice 1 — answer: B

A compared the grams directly: 52.9 ÷ 47.1 = 1.12 reads as 1 : 1, but grams are not counts. C rounded 1.50 up to 2; 1.50 is 3⁄2, and fractions multiply away. D crossed the mole amounts: O's 2.94 mol is the larger count, so O takes the larger subscript.

Worked example 3 — the empirical formula

All five steps are needed: four reach the empirical formula, the fifth scales it.

Step 1 · Percents → grams

A 100 g sample holds 40.0 g C, 6.7 g H, 53.3 g O.

Worked example 3 — the empirical formula

Step 1 · Percents → grams
Step 2 · Grams → moles

Worked example 3 — the empirical formula

Step 1 · Percents → grams
Step 2 · Grams → moles

Worked example 3 — the empirical formula

Step 1 · Percents → grams
Step 2 · Grams → moles

Worked example 3 — scale to the molar mass

Step 4 · Multiply to whole numbers

1 : 2 : 1 is already whole — nothing to multiply.

Worked example 3 — scale to the molar mass

n counts the empirical units in one molecule; every subscript multiplies by n:

Worked example 3 — scale to the molar mass

Practice 2 — answer: C

A stopped at the empirical formula — CH weighs 13.02 g/mol, not 78.11. B took the mass ratio as an atom ratio: 92.3 ÷ 7.7 = 12.0. D scaled twice: 12 × 13.02 = 156.2 g/mol, double the given.

Worked example 1 — solution

A common first attempt

That ratio compares the sucrose to the water alone. A mass percent compares it to the whole solution.

Worked example 1 — solution

The whole is everything in the beaker, solute plus solvent: 25.0 g + 100.0 g = 125.0 g of solution.

Worked example 1 — solution

Both measurements are masses: m/m, grams over grams. Part over whole, × 100.

Worked example 1 — solution

Worked example 1 — solution

Worked example 2 — solution

One conversion factor is needed.

Step 2 · Match the units

m/m pairs grams of solute with grams of solution, so the label declares 3.0 g H₂O₂ = 100 g solution.

Worked example 2 — solution

The equality gives two orientations. Only one cancels the given unit, g solution:

Worked example 2 — solution

Worked example 2 — solution

Practice 1 — answer: B

A divided by the water alone: 20.0 / 230.0 × 100 = 8.70. C flipped part and whole: 250.0 / 20.0 × 100 = 1250, and no percent concentration tops 100. D stopped at the decimal fraction: 20.0 / 250.0 = 0.0800; the × 100 makes it 8.00 per hundred.

Worked example 3 — solution

One conversion factor is needed.

Step 2 · Match the units

m/v pairs grams of solute with milliliters of solution — the one percent that crosses between mass and volume, the way a density does.

Worked example 3 — solution

The given is a mass of NaCl, so g NaCl belongs in the denominator: 100 mL solution over 0.90 g NaCl.

Worked example 3 — solution

Worked example 3 — solution

Practice 2 — answer: B

A flipped the factor: 14.0 × 70.0 / 100 = 9.80, the alcohol inside 14.0 mL of solution, not the solution that carries 14.0 mL. C divided by the raw percent: 14.0 / 70.0 = 0.200. D multiplied by the raw percent: 14.0 × 70.0 = 980; the label means 70.0 per 100.