How do you find what fraction of a compound is each element, and its simplest formula?
Percent composition and empirical formulas: calculate the percent composition by mass of a compound and determine its empirical and molecular formulas from composition data.
A focused Virginia SOL Chemistry answer on composition under CH.3: calculating percent composition by mass from a formula, finding the empirical formula from percent data, and scaling the empirical formula to the molecular formula using the molar mass.
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What this topic is asking
Within standard CH.3, Virginia expects you to calculate the percent composition by mass of a compound from its formula, and to work backward from composition data to the empirical formula and then the molecular formula. These skills connect a formula to measurable masses and are a common SOL constructed-response task.
Percent composition
For example, in (molar mass ), carbon is and oxygen is . The percentages of all elements add to , which is a quick check. Percent composition is by mass, not by the number of atoms, so a light element present in many atoms can still be a small mass fraction.
Empirical formula
To find it from percent data, assume a g sample so each percentage becomes a mass in grams, convert each mass to moles (divide by the atomic mass), and divide every mole value by the smallest. The results give the subscripts. If a value comes out close to a simple fraction (like ), multiply all of them by the small whole number that clears the fraction (here ).
Molecular formula
The molecular formula gives the actual number of each atom in a molecule and is always a whole-number multiple of the empirical formula. To find it, calculate the empirical-formula mass, divide the compound's actual molar mass by that mass to get the multiplier, and multiply all the empirical subscripts by it. If the molar mass equals the empirical-formula mass, the molecular and empirical formulas are the same.
Try this
Q1. Calculate the percent by mass of nitrogen in ammonia, (molar mass ). [2 points]
- Cue. .
Q2. A compound has the empirical formula and a molar mass of g/mol. State its molecular formula. [1 point]
- Cue. Empirical mass ; , so the molecular formula is .
Exam-style practice questions
Practice questions written in the style of VDOE exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
SOL (multiple choice)1 marksWhat is the percent by mass of oxygen in water, (molar mass g/mol)? (A) (B) (C) (D) Show worked answer →
The answer is (C) .
Percent composition is the mass of the element divided by the molar mass, times . Oxygen contributes g per mole of water, so . The two hydrogen atoms make up the remaining .
The trap is using the number of atoms ( of atoms is oxygen, suggesting ); percent composition is by mass, not by atom count, and oxygen is much heavier than hydrogen.
SOL (tech-enhanced, fill in the blank)3 marksA compound is carbon, hydrogen and oxygen by mass. (a) Determine the empirical formula. (b) If the molar mass is g/mol, determine the molecular formula.Show worked answer →
A 3-point empirical-to-molecular item.
(a) Empirical formula (2 points): assume g, so g C, g H, g O. Convert to moles: C , H , O . Divide by the smallest (): C , H , O , so the empirical formula is .
(b) Molecular formula (1 point): the empirical formula mass is g/mol; , so the molecular formula is .
Markers reward converting mass to moles, dividing by the smallest to get the whole-number ratio, and scaling by the molar-mass ratio.
Related dot points
- The mole and molar mass: use the mole, molar mass and Avogadro's number to convert between mass, moles and number of particles.
A focused Virginia SOL Chemistry answer on the mole under CH.3: Avogadro's number, finding the molar mass from the periodic table, and converting between mass, moles and number of particles, the master skill behind all chemical calculations.
- Balancing equations and conservation of mass: balance chemical equations by adjusting coefficients to satisfy the law of conservation of mass.
A focused Virginia SOL Chemistry answer on chemical equations under CH.3: the law of conservation of mass, why only coefficients (not subscripts) may change, and a reliable method for balancing equations including combustion.
- Stoichiometry and the mole ratio: use the mole ratio from a balanced equation to convert between moles and masses of reactants and products, including gas volumes at STP.
A focused Virginia SOL Chemistry answer on stoichiometry under CH.3: reading the mole ratio from a balanced equation, mole-to-mole and mass-to-mass calculations, and using the molar volume of a gas at STP, with the full three-step chain.
- Naming compounds and writing formulas: name and write formulas for ionic compounds (including polyatomic ions), binary molecular compounds and simple acids.
A focused Virginia SOL Chemistry answer on nomenclature under CH.3: writing formulas for ionic compounds by balancing charges (the crossover method), using polyatomic ions and roman numerals, and naming binary molecular compounds with prefixes and simple acids.
- Limiting reactants and percent yield: identify the limiting and excess reactants, calculate the theoretical yield, and calculate the percent yield.
A focused Virginia SOL Chemistry answer on yield under CH.3: identifying the limiting and excess reactants, calculating the theoretical yield of product from the limiting reactant, and finding the percent yield from the actual yield.
Sources & how we know this
- 2018 Science Standards of Learning - Chemistry — Virginia Department of Education (2018)
- Chemistry Curriculum Framework — Virginia Department of Education (2018)