Moles, formulas, and yield: conserve the reacting material

MCAT content library / Chemistry & physics / Moles, formulas, and yield: conserve the reacting material

Convert measurements to amounts, use the balanced reaction, and distinguish the material-accounting ceiling from recovered product.

About 15 minutes, including the worked examples and checks.

From solutions to buffers · Lesson 1 of 6 · Continue in this path

On this page

Concept: count the chemical entities before comparing masses
Pattern: mass ratios must become mole ratios
Rule: balance first, then compare each reactant’s possible yield
Application: account for what remains and what was recovered
Check your reasoning
Continue learning
Sources

Concept: count the chemical entities before comparing masses

A reaction connects numbers of reacting particles. A balance measures mass, a pipette measures volume, and an instrument may report composition. Stoichiometry connects those observations through amounts in moles. AAMC category 4E includes the mole, formulas, mass composition, balanced equations, limiting reactants, and theoretical yield.¹

One mole contains exactly 6.02214076 × 10²³ specified entities. Specify which entities: molecules of CO₂, atoms of carbon, and formula units of an ionic compound are different counts. Molar mass converts between mass and amount: n = m/M_m, where n is in mol, m in g, and M_m in g/mol. Particle number is N = nN_A.²

For example, 0.0200 mol of CO₂ contains about 1.20 × 10²² molecules. Each molecule has two oxygen atoms, so that sample contains 0.0400 mol of oxygen atoms. The molecular count and the oxygen-atom count differ even though they describe the same sample. Multiplying by a formula subscript changes the entity being counted; it does not create more material.

Pattern: mass ratios must become mole ratios

An empirical formula gives the smallest whole-number ratio of elements. A molecular formula gives the actual atom counts in one molecule. Elemental mass percentages alone do not distinguish molecular formulas that are whole-number multiples of the same empirical formula; a measured molar mass supplies that extra constraint.³

The datasets below are illustrative. Use rounded atomic masses C = 12.0, H = 1.00, and O = 16.0 g/mol. A 4.40 g sample containing only these elements has the composition shown.

Element	Mass	Amount of atoms	Ratio to 0.100 mol
C	2.40 g	0.200 mol	2
H	0.400 g	0.400 mol	4
O	1.60 g	0.100 mol	1

Original composition example: convert each element separately

The empirical formula is C₂H₄O. Reading the gram amounts directly as subscripts would overrepresent heavier atoms. Carbon contributes 2.40/4.40 × 100% = 54.5% of the sample mass, but it represents only two of the seven atoms in the empirical formula.

The empirical-formula molar mass is 2(12.0) + 4(1.00) + 16.0 = 44.0 g/mol. If the measured molecular molar mass is 88.0 g/mol, the multiplier is 88.0/44.0 = 2. The molecular formula is therefore C₄H₈O₂. This establishes composition, not how the atoms connect: different structures can share that formula.

When data are percentages, a hypothetical 100 g sample makes each percentage numerically equal to grams. After conversion to moles, divide all amounts by the smallest. If the resulting ratio is close to 1:1.5, multiply every entry by two; rounding 1.5 to either one or two would change the composition. Precision matters at the ratio step, not just in the final formula.

Rule: balance first, then compare each reactant’s possible yield

Changing a coefficient changes the amount of a substance; changing a subscript changes the substance. Balance an equation by adjusting coefficients until every element has the same count on both sides. In an ionic equation, total charge must also match. The balanced coefficients supply mole ratios, not gram ratios.⁴

Consider a separate synthesis of methyl propanoate, one compound with the molecular formula above:

C₃H₆O₂ + CH₄O ⇌ C₄H₈O₂ + H₂O

All coefficients are one. Both sides contain four carbons, ten hydrogens, and three oxygens. For the theoretical calculation, assume pure reactants and count only this target reaction. The reversible arrow does not promise complete conversion: theoretical yield is the material-accounting ceiling, before equilibrium, side reactions, and recovery losses are considered.⁵

Reactant	Supplied measurement	Mass	Amount
Propanoic acid	7.40 g; M_m = 74.0 g/mol	7.40 g	0.100 mol
Methanol	3.00 mL; density = 0.800 g/mL; M_m = 32.0 g/mol	2.40 g	0.0750 mol

Original synthesis inventory; use the supplied rounded density

Methanol’s volume must first become mass: 3.00 mL × 0.800 g/mL = 2.40 g. Dividing by 32.0 g/mol gives 0.0750 mol. A volume in milliliters cannot be compared directly with the acid’s mass or amount.

The acid could supply 0.100 mol of ester if methanol were unlimited. The methanol could supply only 0.0750 mol. Methanol therefore limits the theoretical production. At an ester molar mass of 88.0 g/mol, the theoretical yield is 0.0750 × 88.0 = 6.60 g. For a reaction with different coefficients, convert each reactant to the same product amount before comparing; the smallest raw mass or raw mole amount need not be limiting.

Application: account for what remains and what was recovered

At the calculated ceiling, 0.0750 mol of acid reacts and 0.0250 mol remains. The reaction also produces 0.0750 mol of water, or 1.35 g. The mass account closes: 5.55 g of consumed acid plus 2.40 g of methanol becomes 6.60 g of ester plus 1.35 g of water. The unreacted 1.85 g of acid is still part of the initial inventory.

If the isolated, pure ester mass is 5.28 g, its percent yield is (5.28/6.60) × 100% = 80.0%. Actual yield describes the recovered target product; it does not revise the starting amount of limiting reactant. A crude sample can contain solvent or other substances, so its total mass is not automatically the product mass.⁵

Now change one condition: double the methanol to 6.00 mL while keeping the acid unchanged. Methanol becomes 0.150 mol, so the acid limits the target to 0.100 mol, or 8.80 g. Doubling one input raises the ceiling by only one-third in this case. Once another reactant becomes limiting, adding still more methanol cannot raise that material-accounting ceiling.

A useful final check is to name both the conserved quantity and the requested quantity. Atoms and total mass are conserved in this account; molecule counts can change. The question may ask for a formula, a remaining reactant, a maximum product amount, or a recovered percentage. Keep the conversion chain tied to that request.

Check your reasoning

Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.

Check 1

An illustrative sample contains 1.80 g C, 0.400 g H, and 0.800 g O, with no other elements. Use C = 12.0, H = 1.00, and O = 16.0 g/mol. Which empirical formula fits these measurements?

A. CH₂O

Review this choice. This gives a 1:2:1 atom ratio. The measured amounts are 0.150 mol C, 0.400 mol H, and 0.0500 mol O, which reduce to 3:8:1.

B. C₂H₆O

Review this choice. Two carbons and six hydrogens per oxygen would require different elemental amounts. Dividing every measured mole amount by 0.0500 gives 3:8:1.

C. C₃H₈O

Correct. Convert mass to moles before taking ratios: 1.80/12.0 = 0.150; 0.400/1.00 = 0.400; 0.800/16.0 = 0.0500. Dividing by 0.0500 gives 3:8:1.

D. C₆H₁₆O₂

Review this choice. This has the correct element ratio but is not reduced to the smallest whole numbers. It is a multiple of the empirical formula; molecular-mass information would be needed to support it as a molecular formula.

The principle to carry forward

An empirical formula expresses the simplest mole ratio, not a ratio of masses or an untested molecular size.

Check 2

A target reaction is 2 Al + 3 Cl₂ → 2 AlCl₃. A sample contains 0.300 mol Al and 0.360 mol Cl₂. Assuming only this reaction, which pair correctly identifies the limiting reactant and the theoretical amount of AlCl₃?

A. Al; 0.240 mol AlCl₃

Review this choice. The product amount is the ceiling imposed by chlorine, not aluminum. Aluminum could form 0.300 mol AlCl₃ if sufficient chlorine were present.

B. Al; 0.300 mol AlCl₃

Review this choice. Comparing raw mole amounts misses the coefficients. Consuming 0.300 mol Al requires 0.450 mol Cl₂, but only 0.360 mol is available.

C. Cl₂; 0.240 mol AlCl₃

Correct. Chlorine permits 0.360 × (2/3) = 0.240 mol AlCl₃; aluminum permits 0.300 × (2/2) = 0.300 mol. The smaller possible product amount identifies chlorine as limiting.

D. Cl₂; 0.360 mol AlCl₃

Review this choice. Chlorine is limiting, but its conversion to AlCl₃ is not 1:1. Three moles of Cl₂ support two moles of product.

The principle to carry forward

Compare reactants after expressing each as a possible amount of the same product.

Check 3

A separate synthesis has a theoretical target-product yield of 6.60 g. The recovered mixture weighs 7.20 g, and an analysis shows that 80.0% of its mass is the target product. What is the target-product percent yield?

A. 80.0%

Review this choice. The 80.0% value describes purity within the recovered mixture. Yield compares the mass of target product with the theoretical target-product mass.

B. 87.3%

Correct. The recovered target mass is 7.20 × 0.800 = 5.76 g. Dividing by 6.60 g and multiplying by 100% gives 87.3%.

C. 100%

Review this choice. A recovered mixture heavier than the theoretical product does not establish complete conversion. Correct for its measured composition before calculating yield.

D. 109%

Review this choice. This is approximately 7.20/6.60 × 100%, which counts nonproduct material as product. The composition measurement supplies the necessary correction.

The principle to carry forward

Purity and yield have different denominators: recovered-mixture mass for purity, theoretical target-product mass for yield.