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 1: the route on the map

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

Worked example 2: the route on the map

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, and its real units, g²/mol, are not moles; writing mol on it hides the flip. D skipped the molar mass: 65.0 g relabeled as 65.0 mol; grams become moles only through the g/mol factor.

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

Worked example 3: the route on the map

Practice 2: answer C

B stopped at moles: 8.00 / 32.00 = 0.250 counts moles; writing molecules on it does not finish the conversion. 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 4: solution

Three conversion factors are needed.

Step 2 · Write the route

g H₂O → mol H₂O → molecules H₂O → atoms of H. Molar mass, then Avogadro's number, then the subscript.

Worked example 4: solution

The subscript is an equality too: 1 molecule H₂O = 2 atoms H. Only one orientation cancels molecules:

Worked example 4: solution

Worked example 4: solution

Worked example 4: the route on the map

Practice 3: answer D

A counted every C atom as a molecule: 1.20 × 10²⁴ ÷ 6.022 × 10²³ = 1.99 mol, × 180.16 = 359. B stopped at moles: 0.332 mol of glucose, never converted to grams. C flipped the subscript factor: × 6 instead of ÷ 6 gives 11.96 mol and 2.15 × 10³ g.

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 1: the route on the map

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

Worked example 2: the route on the map

Practice 1: answer C

A counted one N of three: 2.00 × 10³ × 14.01 ÷ 149.10 = 188. B stopped halfway: 2.00 × 10³ g is the ammonium phosphate in the sack, not its nitrogen. D flipped part and whole: 2.00 × 10³ × 149.10 ÷ 42.03 = 7.09 × 10³, more nitrogen than compound. E skipped the 40.0%: 5.00 × 10³ × 42.03 ÷ 149.10 = 1.41 × 10³ counts the filler as fertilizer.

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 1: the route on the map

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

Worked example 2: the route on the map

Practice 1: answer A

B stopped at moles: 0.0375 mol is the amount in the sample, not an amount per liter. C multiplied by the liters: 0.0375 × 0.1250 = 0.00468. D divided by milliliters: 0.0375 ÷ 125.0 = 0.000300, in mol/mL.

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

Worked example 3: the route on the map

Practice 2: answer D

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. C stopped at moles: 10.1 ÷ 101.11 = 0.0999 mol is the KNO₃ needed, not the volume that carries it.

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 1: the route on the map

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

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

Worked example 2: the route on the map

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: scale to the molar mass

Steps 1 to 4 · Reach the empirical formula

Percents to grams to moles: 3.33 mol C, 6.65 mol H, 3.33 mol O. Divided by the smallest: 1 : 2 : 1, already whole.

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

Worked example 3: the route on the map

Practice 2: answer C

A stopped at CH: 13.02 g/mol, not 78.11. B used H₂'s 2.016 for hydrogen: 7.7 ÷ 2.016 = 3.82 mol → C₂H, n = 78.11 ÷ 25.03 = 3.12 → C₆H₃. D doubled the H moles: 2 × 7.64 = 15.3 → CH₂, n = 78.11 ÷ 14.03 = 5.57 → C₆H₁₂. Neither wrong n landed whole.

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 1: the route on the map

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

Worked example 2: the route on the map

Practice 1: answer A

B divided by the water alone: 20.0 / 230.0 × 100 = 8.70. C flipped part and whole: 250.0 / 20.0 × 100 = 1.25 × 10³, 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

Worked example 3: the route on the map

Practice 2: answer B

A skipped the density: 250. × 28.0 / 100 = 70.0 treats mL of solution as grams. C stopped halfway: 250. × 0.900 = 225 g is the whole solution, not the NH₃ in it. D flipped the density: 250. ÷ 0.900 = 278 g, then × 28.0 / 100 = 77.8.

Worked example: solution

Step 1 · Mass fraction, same units

Worked example: solution

Step 1 · Mass fraction, same units

Worked example: solution

Step 1 · Mass fraction, same units

Worked example: the route on the map

Practice: answer B

A stopped at the mass fraction: 0.00416 ÷ 208.0 = 2.00 × 10⁻⁵, never scaled by 10⁶. C used the water shortcut: 4.16 mg ÷ 0.1600 L = 26.0 mg/L, read as ppm; that needs 1 L ≈ 1 kg. D flipped the density: 160.0 ÷ 1.30 = 123 g, and 0.00416 ÷ 123 × 10⁶ = 33.8.

Extra practice 1: answer D

Extra practice 1: answer D

Extra practice 2: answer E

Three conversion factors are needed.

Extra practice 2: answer E

Extra practice 3: answer A

Two conversion factors carry the dose to milligrams of fluoride.

Extra practice 3: answer A