Guided example: solution

One conversion factor is needed.

Step 2 · Write the route

mol Al → g Al. One arrow, one factor: the equality 1 mol Al = 26.98 g Al.

Guided example: solution

The given unit is mol, so mol goes in the denominator:

Guided example: solution

Guided example: solution

Guided example: the route on the map

Practice 1: answer D

A divided by the molar mass: 2.50 ÷ 110.98 = 0.0225, the grams-to-moles direction. B put the molar mass over the moles: 110.98 ÷ 2.50 = 44.4. C dropped the subscript 2: 40.08 + 35.45 = 75.53 g/mol, and 2.50 × 75.53 = 189.

Practice 1: the route on the map

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

Your turn: the route on the map

Practice 2: 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.

Practice 2: the route on the map

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 3: 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.

Practice 3: the route on the map

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 4: 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.

Practice 4: the route on the map

Practice 5: answer B

D stopped at formula units: 2.64 × 10²². A used aluminum's subscript: × 2 gives 5.28 × 10²². E counted every ion, 2 Al³⁺ + 3 SO₄²⁻ = 5 per formula unit: 1.32 × 10²³. C missed the parentheses throughout: Al₂SO₄ at 150.03 g/mol and one sulfate per formula unit: 15.0 ÷ 150.03 × 6.022 × 10²³ = 6.02 × 10²².

Practice 5: the route on the map

Extra practice: answer A

B read PO₄ once: 3(40.08) + 30.97 + 4(16.00) = 215.21 g/mol, and 0.0150 × 215.21 = 3.23. D applied the 2 to P but not to the O inside: 246.18 g/mol gives 3.69. C divided by the molar mass: 0.0150 ÷ 310.18 = 4.84 × 10⁻⁵.

Extra practice: the route on the map

Guided example: solution

Step 1 · Molar mass

Add the mass of every atom, all three oxygens included:

Guided example: solution

Calcium's subscript is 1: 1 × 40.08 = 40.08 g Ca in one mole. The 3 belongs to oxygen.

Guided example: solution

Guided example: solution

Guided example: the route on the map

Practice 1: answer B

A divided by the lead instead of the whole sample: 2.01 ÷ 12.99 × 100 = 15.5. C is lead's share: 12.99 ÷ 15.00 × 100 = 86.6. D flipped part and whole: 15.00 ÷ 2.01 × 100 = 746, a share above 100.

Practice 1: the route on the map

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 2: answer D

A is magnesium's share: 72.93 ÷ 100.95 × 100 = 72.24. B counted atoms: 2 of 5 atoms is 40.0%, but an N atom weighs less than an Mg atom. C divided by magnesium's mass instead of the whole: 28.02 ÷ 72.93 × 100 = 38.42.

Practice 2: the route on the map

Worked example 3: solution

Step 1 · Molar mass Step 2 · Element mass Step 3 · Divide and scale

Worked example 3: solution

One conversion factor is needed. The percent, written as grams over 100 grams, cancels g CaCO₃:

Worked example 3: solution

Worked example 3: the route on the map

Practice 3: answer A

B used one K: 250. × 39.10 ÷ 138.21 = 70.7. C is the carbon and oxygen: 250. × 60.01 ÷ 138.21 = 109. D stopped at the percent, 56.6, before using it on the bag. E counted atoms: 2 of 6, 250. × 2 ÷ 6 = 83.3.

Practice 3: the route on the map

Practice 4: 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.

Practice 4: the route on the map

Practice 5: answer E

A put 4 O on top but kept the right whole: 64.00 ÷ 342.17 × 100 = 18.70. B read the formula as Al₂SO₄ everywhere: 64.00 ÷ 150.03 × 100 = 42.66. C counted atoms: 12 of 17 = 70.6. D is the aluminum and sulfur: 150.17 ÷ 342.17 × 100 = 43.89.

Practice 5: the route on the map

Guided example: grams, then moles

Step 1 · Percents → grams

Take a 100 g sample. Each percent reads as grams: 40.05 g S and 59.95 g O.

Guided example: grams, then moles

Two conversion factors are needed: one molar mass per element.

Guided example: grams, then moles

Guided example: the smallest ratio

Step 3 · Divide by the smallest

The smallest count, 1.249 mol S, divides into both:

Guided example: the smallest ratio

1.00 : 3.00 is already whole: nothing to multiply.

Guided example: the smallest ratio

Guided example: the route on the map

Practice 1: answer A

B divided by the smallest subscript, 6: 14 ÷ 6 = 2.33, then rounded the fraction away to CH₂. C kept the molecular formula: 6 and 14 still share a factor of 2. D set every subscript to 1: the empirical formula keeps the ratio, not one of each.

Practice 1: the route on the map

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 2: 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.

Practice 2: the route on the map

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 3: 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.

Practice 3: the route on the map

Practice 4: answer D

Practice 4: answer D

Practice 4: the route on the map

Extra practice: answer B

A rounded 1.25 down to 1: CH. C multiplied by 2: 1.25 × 2 = 2.50, then rounded up to C₂H₃. D read grams as counts: 1.810 ÷ 0.190 = 9.53, about 10 C per H. E crossed the subscripts: C took hydrogen's 5.

Extra practice: the route on the map

Guided example: solution

Step 2 · Match the units

m/v pairs grams of solute with milliliters of solution. The whole converts to milliliters:

Guided example: solution

Part over whole, × 100, with the whole in milliliters. Left in liters, the fraction runs far above 100:

Guided example: solution

Guided example: solution

Guided example: the route on the map

Practice 1: answer C

A compared the ethanol to the other 140. mL alone: 260. / 140. × 100 = 186. B stopped at the decimal fraction: 260. / 400. = 0.650. D flipped part and whole: 400. / 260. × 100 = 154. E added the ethanol to a volume that already held it: 260. / 660. × 100 = 39.4.

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 2: 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 3: answer D

Two conversion factors are needed.

A read the label as 15.0 g per 100 mL of water, as if 115 parts made the solution: 500. × 15.0 / 115 = 65.2. B skipped liters to milliliters: 0.500 × 15.0 / 100 = 0.0750. C flipped the label factor: 500. × 100 / 15.0 = 3.33 × 10³. E used the raw percent: 500. × 15.0 = 7.50 × 10³.

Practice 3: the route on the map

Practice 4: 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.

Practice 4: the route on the map

Guided example: solution

Step 1 · Grams → moles

The amount is already in moles: 0.300 mol glucose. No molar mass is needed.

Guided example: solution

The volume is already in liters: 1.50 L of solution.

Guided example: solution

Guided example: solution

Guided example: the route on the map

Practice 1: answer D

A divided by the water added: 0.270 ÷ 1.15 = 0.235, but the dissolved sugar takes up room too. B flipped the fraction: 1.20 ÷ 0.270 = 4.44, in L/mol. C multiplied: 0.270 × 1.20 = 0.324, in mol·L.

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 2: answer B

A stopped at moles: 34.8 ÷ 101.11 = 0.344 mol is the amount in the sample. C divided by milliliters: 0.344 ÷ 225 = 0.00153, in mol/mL. D found the percent: 34.8 ÷ 225 × 100 = 15.5% (m/v). E divided grams by liters: 34.8 ÷ 0.225 = 155 g/L.

Practice 3: answer A

B divided by the whole batch: 0.0200 ÷ 2.000 = 0.0100, but the 2.38 g came from only 25.00 mL. C stopped at moles: 0.0200 mol is the amount in the portion. D divided by milliliters: 0.0200 ÷ 25.00 = 0.000800, in mol/mL. E multiplied by the volume instead of dividing: 0.0200 × 0.02500 = 0.000500.

Worked example 3: solution

Two conversion factors are needed.

Step 1 · Grams → moles

Worked example 3: solution

Two orientations exist. Only one cancels mol NaCl:

Worked example 3: solution

Worked example 3: solution

Worked example 3: the route on the map

Worked example 4: solution

Three conversion factors are needed.

Step 1 · Grams → moles

Worked example 4: solution

Worked example 4: solution

Worked example 4: the route on the map

Practice 4: 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 5: solution

Two conversion factors are needed.

Liters → moles: the molarity

Worked example 5: solution

Worked example 5: solution

Worked example 5: the route on the map

Worked example 5: making the solution

Weigh 4.0 g of NaOH. Dissolve it in part of the water in a 1.0 L volumetric flask. Add water to the mark, then mix.

Worked example 5: making the solution

Weigh 4.0 g of NaOH. Dissolve it in part of the water in a 1.0 L volumetric flask. Add water to the mark, then mix.

Practice 5: answer B

A left out the volume: 0.150 × 142.04 = 21.3 g is the mass in a full liter. C flipped the molarity: 0.5000 ÷ 0.150 × 142.04 = 473, and the units do not cancel. D skipped mL → L: 500.0 × 0.150 × 142.04 = 1.07 × 10⁴.

Worked example 6: solution

Two conversion factors are needed for the mass.

Liters → moles: the molarity

Worked example 6: solution

Worked example 6: solution

Worked example 6: the percent

Grams per 100 mL: the % (m/v)

Keep the unrounded 20.6 g in the calculator. The blood is the solution: 5.6 L = 5.6 × 10³ mL.

Worked example 6: the percent

Grams per 100 mL: the % (m/v)

Keep the unrounded 20.6 g in the calculator. The blood is the solution: 5.6 L = 5.6 × 10³ mL.

Worked example 6: the percent

Grams per 100 mL: the % (m/v)

Keep the unrounded 20.6 g in the calculator. The blood is the solution: 5.6 L = 5.6 × 10³ mL.

Worked example 6: the route on the map

Practice 6: answer C

A filled one flask: 0.2500 × 0.200 × 159.61 = 7.98. B flipped the molarity: 1.500 ÷ 0.200 × 159.61 = 1.20 × 10³. D stopped at moles: 1.500 × 0.200 = 0.300 mol. E skipped mL → L: 1500. × 0.200 × 159.61 = 4.79 × 10⁴.

Practice 6: the route on the map

Guided example: solution

Step 1 · List the knowns

Before, the stock: M₁ = 10.0 M, and V₁ is the unknown. After, the flask: M₂ = 0.200 M and V₂ = 250. mL.

Guided example: solution

Guided example: solution

Both concentrations are in M, so they cancel. V₂ is in mL, so V₁ comes out in mL.

Guided example: solution

At the bench: measure 5.00 mL of the stock, then add water to the 250. mL mark.

Guided example: solution

Guided example: the route on the map

Practice 1 answer: D

A flipped the ratio: 300. × (12.0 ÷ 0.600) = 6.00 × 10³ mL, twenty times the batch. B stopped at the solute: 300. × 0.600 = 180. mmol of HCl, not a volume. C reported the water: 300. − 15.0 = 285 mL goes in after the stock.

Practice 1: the route on the map

Worked example 1: solution

Step 1 · List the knowns

M₁ = 6.00 M. V₁ = 25.0 mL. V₂ = 150. mL. The unknown is M₂.

Worked example 1: solution

Worked example 1: solution

Both volumes are in mL, so the volume ratio cancels to a pure number.

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

A common first attempt

A dilution that ends with less liquid than it started. The concentration ratio is upside down.

Worked example 2: solution

M₁ = 6.00 M. V₁ = 30.0 mL. M₂ = 0.500 M. The unknown is V₂.

Worked example 2: solution

Solve M₁V₁ = M₂V₂ for V₂: it equals V₁ scaled by the concentration ratio M₁ ÷ M₂.

Worked example 2: solution

V₁ is in mL, so V₂ comes out in mL. The starting concentration sits on top:

Worked example 2: solution

Worked example 2: the route on the map

Practice 2 answer: B

A stopped at the total volume: 500. mL still counts the 40.0 mL of acid already in the flask. C flipped the ratio: 40.0 × (0.400 ÷ 5.00) = 3.20 mL, less than the start. D stopped at the solute: 40.0 × 5.00 = 200. mmol of HNO₃, not a volume.

Practice 2: the route on the map

Worked example 3: solution

A common first attempt

50.0 mL is only the water added. A percent counts per 100 mL of solution, and the solution is all 60.0 mL: stock plus water.

Worked example 3: solution

C₁ = 9.00% (m/v). V₁ = 10.0 mL. V₂ = 60.0 mL, the whole solution. The unknown is C₂.

Worked example 3: solution

C₂ = C₁ × V₁ ÷ V₂. Both volumes are in mL, so C₂ comes out in % (m/v), the unit of C₁.

Worked example 3: solution

Worked example 3: solution

Worked example 3: the route on the map

Practice 3 answer: A

B divided by the water alone, per solvent: 8.00 × (15.0 ÷ 25.0) = 4.80. C flipped the volume ratio: 8.00 × (40.0 ÷ 15.0) = 21.3, stronger than the start. D stopped at the solute: 15.0 mL × 8.00 g/100 mL = 1.20 g of KCl, not a percent.

Practice 3: the route on the map

Worked example 4: solution

A common first attempt

A droplet cannot dilute to fill 2.00 L. The result came out in liters, because the volume entered in liters.

Worked example 4: solution

M₂ = 0.150 M. V₂ = 2.00 L. M₁ = 6.00 M. The unknown is V₁, wanted in mL.

Worked example 4: solution

Solve M₁V₁ = M₂V₂ for V₁: it equals V₂ scaled by the concentration ratio M₂ ÷ M₁.

Worked example 4: solution

V₂ entered in liters, so V₁ lands in liters. Convert to milliliters at the end:

Worked example 4: solution

Worked example 4: the route on the map

Practice 4 answer: C

A stopped at the stock: 80.0 mL is what gets measured, and water fills the rest. B stopped at the solute: 0.900 × 0.400 = 0.360 mol of KOH, not a volume. D subtracted in liters: 0.900 − 0.0800 = 0.820 L, not mL; convert both volumes first.

Practice 4: the route on the map

Practice 5 answer: E

A stopped at the stock: 150. mL is what gets measured. B read 0.900% per solvent, 27.0 g NaCl per 3.00 × 10³ mL of water; the stock fills part of the 3.00 L. C subtracted in liters: 3.00 − 0.150 = 2.85 L, not mL. D stopped at the solute: 3.00 × 10³ mL × 0.900 g/100 mL = 27.0 g of NaCl.

Practice 5: the route on the map

Extra practice 1 answer: A

B divided by the water alone: 2.40 × (35.0 ÷ 85.0) = 0.988; the 35.0 mL of stock is part of the solution too. C stopped at the solute: 35.0 × 2.40 = 84.0 mmol of NaCl, not a molarity. D flipped the volume ratio: 2.40 × (120.0 ÷ 35.0) = 8.23, stronger than the stock.

Extra practice 2 answer: D

A mixed the units: 4.00 × (28.0 ÷ 0.350) = 320., eighty times the stock; mL over L does not cancel. B flipped the volume ratio: 4.00 × (350. ÷ 28.0) = 50.0, stronger than the stock. C stopped at the solute: 28.0 × 4.00 = 112 mmol of NaOH, not a molarity. E read 0.350 L as the water added: 4.00 × (28.0 ÷ 378) = 0.296; 0.350 L is the final volume, stock included.

Extra practice 3 answer: E

Two moves: the stock's molarity, then the dilution.

Extra practice 3 answer: E