Percent Composition, Empirical & Molecular Formulas — Print · Grade 11 Science
Chapter: Reactions, Moles & CompositionPercent Composition, Empirical & Molecular Formulas
Learning goals
- I can convert mass composition to a simplest mole ratio.
- I can use molar mass to distinguish empirical and molecular formulas.
Readings for this unit
Before this unit
Molar mass, percent calculations, and proportional reasoning.
Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.
Start with something to explain
Can two molecules share the same elemental percentages but have different molecular formulas?
Illustrative evidence case. CH₂O and C₂H₄O₂ have the same simplest atom ratio.
Notice and wonder
Record two observations using the stated quantities or relationships. Keep observations separate from possible causes. Write one question that a new measurement could help answer.
Can two molecules share the same elemental percentages but have different molecular formulas?
Your first model
Separate the ratio inferred from composition from the number of atoms in one molecule.
Build the language from the idea
- empirical formula
- Simplest whole-number atom ratio.
- molecular formula
- Actual atom counts in a molecule.
- percent composition
- Elemental mass divided by total mass, expressed as a percent.
Percent composition describes mass contribution: percent of an element = mass of that element in one mole of compound divided by the compound molar mass ×100%. Multiply each atomic mass by its subscript first. For CO₂, carbon contributes 12.01 of about 44.01 g/mol, or 27.29% by mass.
An empirical formula gives the simplest whole-number atom ratio. A molecular formula gives the actual atom count in a molecule. Glucose C₆H₁₂O₆ has empirical formula CH₂O; reducing the subscripts does not make glucose chemically identical to formaldehyde, which has that molecular formula.
To infer an empirical formula from measured composition, convert each elemental mass to moles and divide all by the smallest mole amount. If values are near whole numbers within experimental uncertainty, use that ratio. A 1:1.5 ratio needs multiplication by two; do not round 1.5 to 2 without checking.
If percentages are given, assume 100 g as a convenient mathematical basis. The percentage numbers then become grams, but they are still mass data. Converting to moles is necessary because atomic masses differ. A 1:1 mass ratio rarely gives a 1:1 atom ratio.
A known molecular molar mass allows recovery of a molecular formula. Calculate empirical-formula mass, divide molecular molar mass by it, and multiply every subscript by the resulting whole number. If the multiplier is not near an integer, examine measurement uncertainty or an incorrect starting model.
Chemical formulas inferred from bulk composition alone do not specify how atoms are connected. Distinct isomers can have the same formula. Keep composition, molecular count, and structure as different questions, with different evidence needed to answer each.
Build your understanding
Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.
Part 1 · Distinguish percent mass from atom ratio
Mass percent is component mass divided by total sample mass times 100%. It is not an atom-number percentage. A formula can contain many light hydrogen atoms while hydrogen contributes a relatively small mass fraction. To compare a composition analysis with a formula, account for each element’s atomic mass and the measured sample boundary. Percentages should sum to approximately 100% only if all components are included and rounding/measurement uncertainty are understood. Missing components or water can change the inference.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
Why is a 40% carbon mass fraction not proof that 40% of the atoms are carbon?
Visual reasoning — after your attempt
Carbon and other atoms have different masses; a mass ratio must be converted through molar masses to compare atom numbers.
Mass percent is component mass divided by total sample mass times 100%. Atom ratios require mole conversion because equal masses of different elements contain different atom counts. An empirical formula is the simplest whole-number ratio, not necessarily the molecular formula.
A 1: 1 mass ratio of carbon and hydrogen is not a 1: 1 atom ratio because their molar masses differ. Use the supplied atomic masses before simplifying.
Pause and explain
Why not use percentage numbers directly as subscripts?
Check your reasoning after answering
They represent masses, not atom amounts. Divide by atomic molar masses before finding a particle ratio.
Part 2 · Use a convenient 100 g basis
An empirical formula is the simplest whole-number atom ratio consistent with a composition model. Taking a 100 g basis turns each mass percent into grams without claiming the actual specimen weighs 100 g. Divide by the appropriate molar mass to get proportional mole amounts, then divide by the smallest. Here the approximate ratio is 1:2:1, giving CH₂O. The method assumes the listed elements account for the analyzed sample and that measurement/rounding permit a simple ratio.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
Why divide the masses by atomic molar masses before reducing the ratio?
Visual reasoning — after your attempt
The empirical formula counts atoms; converting to proportional mole amounts removes the unequal per-atom masses.
For percentage composition, a hypothetical 100 g sample turns each percentage into grams. Divide each element mass by its molar mass, then divide all amounts by the smallest. This basis is a calculation convenience and does not assert that an actual 100 g sample was measured.
40% C, about 6.7% H, and about 53.3% O can give an approximately 1: 2: 1 atom ratio using suitable molar masses. Retain intermediate precision.
Pause and explain
Why does choosing 100 g not change the final ratio?
Check your reasoning after answering
Scaling all component masses equally scales all mole amounts equally, so their ratios remain unchanged.
Part 3 · Resolve near-fractional ratios carefully
An empirical ratio near 1:1.5 can support a 2:3 atom ratio after multiplying all terms by two. Premature rounding to 1:2 changes the composition substantially. Look for simple fractional relationships and compare the reconstructed percent composition with the original data and uncertainty. A complicated multiplier chosen only to force an answer can overfit noisy measurements. If several formulas fit within error, report that ambiguity and request more evidence rather than treating the nearest decimal as an exact atomic count.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
Why is multiplying the entire 1:1.5 ratio by two preferable to rounding only 1.5?
Visual reasoning — after your attempt
It preserves the measured proportional relationship, yielding 2:3; rounding one term changes the ratio.
Ratios near simple fractions may require multiplying all terms by a common integer. Do not round 1.5 to 2 without considering a 2: 3 relationship. Measurement uncertainty and approximate percentages can shift values slightly, so explain the simplest ratio supported by the data.
A ratio 1: 1.5 suggests 2: 3 after multiplying both terms by two. A ratio far from a simple rational pattern may need checking for conversion or sample errors.
Pause and explain
What is wrong with independently rounding every ratio to the nearest integer?
Check your reasoning after answering
It can change a supported fractional relationship. Multiplying all ratios by a common factor preserves their relative values.
Part 4 · Find a molecular multiplier from molar mass
A molecular formula gives the actual atom counts in a discrete molecule. Composition alone gives only an empirical ratio; an independent molar mass can determine its integer multiplier. The selected CH₂O ratio has empirical formula mass about 30.0 g/mol, so a molecule near 180 g/mol has a multiplier of six and formula C₆H₁₂O₆. Several structures can still share that molecular formula. Ionic and network substances do not automatically have discrete molecules whose sizes this calculation can determine.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
Does obtaining C₆H₁₂O₆ identify one unique molecular structure?
Visual reasoning — after your attempt
No. It gives composition per molecule under the model; multiple structural arrangements can share the same molecular formula.
Calculate the empirical-formula mass, then divide measured molecular molar mass by it. A near-integer multiplier scales every subscript. This applies when a discrete molecular formula and suitable molar mass are supplied; ionic or network materials need a different interpretation.
An empirical formula CH₂O has approximate mass 30 g/mol. A supplied molecular molar mass 180 g/mol gives multiplier six and C₆H₁₂O₆.
Pause and explain
Does an empirical formula uniquely establish a molecular formula?
Check your reasoning after answering
No. A molecular molar mass or equivalent evidence is needed to determine the whole-number multiplier.
Apply the whole unit · Infer a formula from a composition record
Illustrative composition: C40.0%, H6.67%, O53.33%; use C12.0,H1.00,O16.0 g/mol and molecular molar mass 180 g/mol.
Your task
Calculate an empirical formula, derive the molecular formula, and check the mass composition. Explain why percentages alone cannot uniquely distinguish every molecule with that empirical ratio.
What your response should show
- Choose the appropriate chemical model or relationship and state its conditions.
- Show substitutions, units, particle or charge counts, and intermediate reasoning as relevant.
- Use supplied evidence to justify the conclusion; distinguish observation from inference.
- Identify a meaningful limitation, uncertainty, or next measurement.
Response guidance — after your attempt
Use a 100 g basis: amounts are C3.333,H6.67,O3.333 mol.
Divide by 3.333 to obtain approximately 1: 2: 1, giving CH₂O.
Empirical mass=30.0 g/mol; multiplier=180/30.0=6.
Molecular formula=C₆H₁₂O₆.
The formula is supported by the supplied composition and molecular mass, with small deviations reflecting rounded inputs. Rechecking C gives 72/180=40.0%, H gives 12/180≈6.67%, and O gives 96/180≈53.33%. The formula does not specify molecular connectivity or identify a unique isomer without structural evidence.
Scientific references and source-model foundations
Parent / teacher teaching notes
Require a percent-to-mass-to-mole table. Keep guard digits and ask what additional information distinguishes composition from structure. Ask the learner to explain each particle symbol, axis, unit and arrow before calculating. Distinguish hypothetical models and calculated curves from measured records. Public OpenSciEd excerpts here are selected foundation prompts, not complete high-school teacher editions or an endorsement. All required advanced explanations and practice are provided locally.
The key idea
Convert elemental mass to mole ratios; use independent molar mass to find molecular subscripts.
Worked examples
Example 1
- C:12/12 = 1 mol; O:32/16 = 2 mol.
- Divide both by 1.
Example 2 · Worked application
- Assume 100 g: moles are 3.33 C, 6.7 H and 3.33 O.
- Divide by the smallest to obtain approximately 1:2:1.
- Write the simplest ratio, not an automatically known molecular formula.
Percentages describe mass, so convert to moles before atom ratios.
Example 3
- 180/30 = 6.
- Multiply every subscript by 6.
Example 4 · Find and fix the mistake
- Test the claim against the stated conditions: Percentages describe mass, so convert to moles before atom ratios.
- Different elements have different molar masses; equal masses do not contain equal atom counts.
Percentages describe mass, so convert to moles before atom ratios.
Example 5 · Reason from the visual model
- An empirical ratio near 1:1.5 can support a 2:3 atom ratio after multiplying all terms by two. Premature rounding to 1:2 changes the composition substantially. Look for simple fractional relationships and compare the reconstructed percent composition with the original data and uncertainty. A complicated multiplier chosen only to force an answer can overfit noisy measurements. If several formulas fit within error, report that ambiguity and request more evidence rather than treating the nearest decimal as an exact atomic count.
- It preserves the measured proportional relationship, yielding 2:3; rounding one term changes the ratio.
Original fractional-ratio demonstration without an assigned chemical species or measurement uncertainty.
Avoid these mistakes
- Mass percentage is not atom percentage.
Check your understanding
What extra information distinguishes empirical from molecular formula?
Discuss the answer
A molecular molar mass determines the integer multiple of the empirical formula.
Return to the phenomenon
Return to the initial case. Can two molecules share the same elemental percentages but have different molecular formulas? Revise your drawing or explanation using two named ideas and one stated limitation.
Compare explanations after your attempt
Composition can establish an empirical ratio; an independent molar mass provides the multiplier. Structural identity needs still more evidence.
What should we investigate next?
Which comparison would test a remaining uncertainty? State what you would change, keep comparable and measure.
Investigate, calculate and revise
Keep a first model and an evidence-supported revision. Attempt each task before opening guidance; all required explanations and supplied-data routes are available locally.
Activity 1 · Return to the opening case
Use the opening phenomenon, unit definitions and explained Parts 1–4.
Your task
Return to the initial case. Can two molecules share the same elemental percentages but have different molecular formulas? Revise your drawing or explanation using two named ideas and one stated limitation.
Success criteria
- Use the stated observations and appropriate units.
- Connect a labeled model with the chemical mechanism.
- State assumptions and an evidence limitation.
- Explain a substantive revision or next test.
Hint
Keep species identity, material transfer and energy account distinct.
Response guidance — discuss after your attempt
Composition can establish an empirical ratio; an independent molar mass provides the multiplier. Structural identity needs still more evidence.
Activity 2 · Transfer to a changed case
Cover the worked answer and use the explained models, relationships and notebook.
Your task
A selected oxide sample has 70.0 g Fe and 30.0 g O. Use Fe=56.0 and O=16.0 g/mol to find the simplest formula. State one assumption.
Success criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the chemical mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Hint
Name known quantities and conditions, show intermediate steps and check conservation.
Response guidance — discuss after your attempt
Amounts are 1.25 mol Fe and 1.875 mol O, ratio 1:1.5, hence Fe₂O₃. Assume the analyzed sample is one oxide accounted for by those elements and the input precision supports that ratio.
Words to know
- empirical formula
- Simplest whole-number atom ratio.
- molecular formula
- Actual atom counts in a molecule.
- percent composition
- Elemental mass divided by total mass, expressed as a percent.
In the real world
Use the chapter investigation to connect this unit to measured evidence and a practical decision.
Exit check
What extra information distinguishes empirical from molecular formula?
Check your explanation after answering
A molecular molar mass determines the integer multiple of the empirical formula.
Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.
Problem 1
Hints
Write the relationship first; substitute with consistent units.
Answer and explanation
27.27 %
12/44 ×100 = 27.27%.
Problem 2
Answer and explanation
Subscripts count atoms; convert mass to moles because elements have different atomic masses.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 3
Hints
List the given quantities, select a relationship, and check units before substituting.
Answer and explanation
4
60/15=4.
Problem 4
Hints
Write the relationship first; substitute with consistent units.
Answer and explanation
2
92/46 = 2.
Problem 5
Hints
Percentages describe mass, so convert to moles before atom ratios.
Answer and explanation
90/30=3; multiply subscripts to get C₃H₆O₃.
Problem 6
Hints
State what the supplied observations or conditions establish, and separate that from any assumption.
Answer and explanation
Moles C=2, H=4; ratio 1:2, giving CH₂.
Problem 7
Answer and explanation
It preserves the measured proportional relationship, yielding 2:3; rounding one term changes the ratio.
An empirical ratio near 1:1.5 can support a 2:3 atom ratio after multiplying all terms by two. Premature rounding to 1:2 changes the composition substantially. Look for simple fractional relationships and compare the reconstructed percent composition with the original data and uncertainty. A complicated multiplier chosen only to force an answer can overfit noisy measurements. If several formulas fit within error, report that ambiguity and request more evidence rather than treating the nearest decimal as an exact atomic count.
It preserves the measured proportional relationship, yielding 2:3; rounding one term changes the ratio.
Response criteria
- Interpret labels, structures, axes, units and directions correctly.
- Explain the chemistry relationship or mechanism.
- Respect stated assumptions and distinguish observation from inference.
Problem 8
Answer and explanation
Amounts are 1.25 mol Fe and 1.875 mol O, ratio 1:1.5, hence Fe₂O₃. Assume the analyzed sample is one oxide accounted for by those elements and the input precision supports that ratio.
Amounts are 1.25 mol Fe and 1.875 mol O, ratio 1:1.5, hence Fe₂O₃. Assume the analyzed sample is one oxide accounted for by those elements and the input precision supports that ratio.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the chemical mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.