Home / Learn / Chemistry / Mole bridgeTool #L40 (formerly #1085)

Mole concept: grams to moles to particles to gas volume, empirical formula and limiting reactant

Type an amount of any substance in grams, moles, particles or litres of gas and see the other three, with the path across the mole bridge drawn and every step written out. Work out an empirical and molecular formula from percentages, and see which reactant runs out first. Ends with fresh practice questions.

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    One mole is 6.022 × 10²³ particles (Avogadro's number). Counting one particle a second, a mole would take about 1.9 × 10¹⁶ years, over a million times the age of the universe. The gas volume only makes sense for a gas, treated as an ideal gas at 0 °C.

    Empirical formula from percentages

    Which runs out first?

    Try this

    • How many molecules are in 18 g of water? In 36 g? Watch the particles box double.
    • Type 1 mol for O₂ and for CO₂. The mass differs, but the number of molecules and the gas volume are the same.
    • Switch to Class 11: why does 1 mol of gas fill 22.7 L instead of 22.4 L?
    • In the empirical formula card, change the molar mass from 180 to 60. What substance could that be?
    • Make water from 4 mol of H₂ and 3 mol of O₂. Which gas is left over, and how much?

    Check yourself

      What it does

      Chemists count particles by weighing them. The mole is the bridge: one mole of anything is 6.022 × 10²³ particles, and it has a mass in grams equal to the formula mass. This page makes the bridge visible.

      • The bridge. Type a formula (or tap one: H₂O, CO₂, NaCl …), an amount and its unit: grams, moles, particles or litres of gas. The picture shows four boxes (mass, moles, particles, gas volume) with the right operation on each link: divide by the molar mass M to go from grams to moles, multiply by Avogadro’s number Nₐ to go from moles to particles, multiply by the molar volume to get litres of gas. The link you start from is highlighted, and the steps below show every sum with your numbers.
      • Empirical formula. Enter the mass percentage of each element. The table works out the moles in 100 g, divides by the smallest and scales to whole numbers. Give a molar mass and it also finds the molecular formula.
      • Which runs out first? Pick a reaction and the moles of each reactant. The circles show which reactant is used up (the limiting reactant), how much product forms and what is left over.
      • Level. Class 9 uses rounded atomic masses (H 1, C 12, O 16 …) and 22.4 L per mole of gas; Class 11 uses IUPAC atomic weights and 22.7 L per mole at STP (0 °C and 1 bar).

      It follows Class 9 Science (atoms and molecules: the mole concept, molar mass, number of particles) and Class 11 Chemistry (some basic concepts of chemistry: mole concept and molar masses, percentage composition, empirical and molecular formula, stoichiometry and limiting reagent). Find formulas first with the formula writer, get exact molar masses for any formula in the molar mass calculator, and balance full equations in the equation balancer.

      How to use it

      1. Choose the level.
      2. Type a formula and an amount, and choose the unit.
      3. Read the four boxes and the steps.
      4. Fill in percentages in the empirical formula card. Add element adds a row; × removes one.
      5. Pick a reaction and amounts in the last card.

      Worked example: 36 g of water

      Formula H₂O, amount 36, unit grams, Class 9.

      • Molar mass: 2 × 1 + 1 × 16 = 18 g/mol.
      • Moles = 36 ÷ 18 = 2 mol.
      • Particles = 2 × 6.022 × 10²³ = 1.204 × 10²⁴ molecules.
      • As a gas (steam treated as an ideal gas at 0 °C, just for practice): 2 × 22.4 = 44.8 L.

      Empirical formula: carbon 40.0 %, hydrogen 6.7 %, oxygen 53.3 %. In 100 g there are 40 ÷ 12 = 3.33 mol C, 6.7 ÷ 1 = 6.7 mol H and 53.3 ÷ 16 = 3.33 mol O. Dividing by 3.33 gives 1 : 2 : 1, so the empirical formula is CH₂O (formula mass 30). With a molar mass of 180 g/mol, n = 180 ÷ 30 = 6 and the molecular formula is C₆H₁₂O₆, glucose.

      Limiting reactant: 5 mol H₂ and 2 mol O₂ in 2H₂ + O₂ → 2H₂O. H₂ could react 5 ÷ 2 = 2.5 times, O₂ only 2 ÷ 1 = 2 times, so oxygen runs out first. 4 mol of water form and 1 mol of hydrogen is left.

      Tips

      • If ratios come out like 1 : 1.5 or 1 : 1.33, multiply all of them by 2 or 3. The table does this and says so.
      • Percentages that do not add up to about 100 % usually mean an element (often oxygen) is missing.
      • The gas volume only applies to gases. 22.4 L is at 1 atm (the older standard); 22.7 L is at 1 bar, the standard used in newer books.

      FAQ

      Why are my answers slightly different from my book? Books round atomic masses and Avogadro’s number differently. Switch the level to match yours.

      Is anything saved? No. Everything runs in your browser and nothing is stored or sent.

      Is this an official textbook tool? No. It is an independent practice page that follows the same school topics.

      Where the numbers come from

      • Avogadro constant 6.022 140 76 × 10²³ mol⁻¹ (exact, SI 2019) and the gas constant R (CODATA, NIST); molar volumes 22.414 L (0 °C, 1 atm) and 22.711 L (0 °C, 1 bar, IUPAC STP) worked out from them.
      • Rounded school atomic masses and IUPAC/CIAAW standard atomic weights from fibub's shared element data.

      Standard scientific values, such as constants, atomic weights and melting points, are facts found in every textbook; we name where ours were checked. If you are an author or publisher and want a value, a credit or a table changed or removed, tell us and we will act quickly.

      Last updated 2026-11-09.

      Change log for Mole bridge

      1. Fix pictures in #1082, #1084, #1085, #1087, #1089, #1090 Details on GitHub
      2. Add #1085 Mole bridge Details on GitHub

      Every change to fibub

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