Stoichiometry

The Mole and Stoichiometry

Counting atoms by weighing them, and predicting how much a reaction makes

Stoichiometry (stoy-kee-OM-uh-tree) is the math of how much. How many grams of oxygen does a candle need? How much ammonia can a factory make from a tank of hydrogen? A balanced equation tells you the recipe in particles; the mole turns particles into grams you can weigh.[9]

This unit builds on Chemical Reactions, so you should be comfortable balancing equations. Try the limiting-reactant simulator and the mole-map calculator, then take the quiz.

Why We Count in Moles

Bakers count eggs by the dozen and paper comes in reams of 500. Chemists need a much bigger batch, because atoms are so small. That batch is the mole. Since 2019 it has been defined by a fixed number: one mole contains exactly 6.022 140 76 × 1023 particles, the Avogadro number.[1][3] Written out, that is 602 214 076 000 000 000 000 000.

The particles can be atoms, molecules, ions or anything else you name, just as a dozen can be eggs or donuts. For example, NIST notes that 18 grams of water, about one mole, contains roughly 600 billion trillion molecules.[4]

Molar Mass from the Periodic Table

The atomic mass on the periodic table does double duty. One carbon atom averages 12.011 atomic mass units, and one mole of carbon atoms has a mass of 12.011 grams. The mass of one mole is the molar mass, in g/mol.[8] For a compound, add up the atoms in its formula:[5]

SubstanceWorkingMolar mass
H2O2(1.008) + 15.99918.015 g/mol
CO212.011 + 2(15.999)44.009 g/mol
NH314.007 + 3(1.008)17.031 g/mol
NaCl22.990 + 35.4558.44 g/mol
C6H12O6 (glucose)6(12.011) + 12(1.008) + 6(15.999)180.156 g/mol

The Mole Map

Every conversion goes through moles. Grams to particles? Grams → moles → particles. Use the arrows:

MOLESmol MASSgrams PARTICLESatoms, molecules ÷ molar mass ↓× molar mass ↑ × 6.022 × 10²³ ↓÷ 6.022 × 10²³ ↑ GASL at STP ÷ 22.4 L/mol →← × 22.4 L/mol

Which STP? This page uses STP = 0 °C and 1 atm, where one mole of an ideal gas takes up 22.4 L (NIST: 22.414 L).[6] Some textbooks use 0 °C and 1 bar instead, which gives 22.7 L (22.711 L).[7] Check which one your class uses. The shortcut works only for gases, and real gases are close to ideal but not exactly.

Worked example: how many molecules in 9.0 g of water?
  1. Grams to moles: 9.0 g ÷ 18.015 g/mol = 0.50 mol
  2. Moles to molecules: 0.50 mol × 6.022 × 1023 /mol = 3.0 × 1023 molecules
Worked example: what volume do 2.00 mol of O2 fill at STP?

2.00 mol × 22.4 L/mol = 44.8 L (at 0 °C and 1 atm).

Mole-Map Calculator

Pick a substance, type an amount and choose its unit. Every step is shown. Very large or small numbers can be typed like 3.01e23.

9.0 g of water is 0.4996 mol.

Results are rounded to 4 significant figures; your own number is shown as you typed it. Liter results, and anything worked out from liters, use 22.4 L/mol, which has only 3 significant figures, so read those to 3 significant figures.

  1. Moles = 9.0 g ÷ 18.015 g/mol = 0.4996 mol
  2. Particles = 0.4996 mol × 6.022 140 76 × 10²³ /mol = 3.009 × 10²³ molecules

Mole Ratios and Gram-to-Gram Problems

The coefficients of a balanced equation are a recipe in moles. For making ammonia,

N2(g) + 3H2(g) → 2NH3(g)

1 mol of N2 reacts with 3 mol of H2 to give 2 mol of NH3. Any pair gives a mole ratio, such as 2 mol NH3 / 3 mol H2.[9] A gram-to-gram problem uses three steps: grams of A → moles of A → moles of B → grams of B.

Worked example 1: how much NH3 from 10.0 g of H2?
  1. 10.0 g H2 ÷ 2.016 g/mol = 4.960 mol H2
  2. 4.960 mol H2 × (2 mol NH3 / 3 mol H2) = 3.307 mol NH3
  3. 3.307 mol × 17.031 g/mol = 56.3 g NH3

Assumes there is enough N2.

Worked example 2: how much CO2 from burning 8.0 g of methane?

CH4 + 2O2 → CO2 + 2H2O

  1. 8.0 g ÷ 16.043 g/mol = 0.499 mol CH4
  2. Ratio 1 mol CO2 / 1 mol CH4, so 0.499 mol CO2
  3. 0.499 mol × 44.009 g/mol = 22 g CO2

The CO2 outweighs the methane because it also contains oxygen taken from the air.

Limiting and Excess Reactants

Start with sandwiches. A cheese sandwich needs 2 slices of bread and 1 slice of cheese. You have 10 slices of bread and 3 slices of cheese. The bread could make 5 sandwiches, but the cheese can make only 3. You make 3 sandwiches, and 4 slices of bread are left over. The cheese is the limiting reactant: it runs out first and sets how much product forms. The bread is in excess.[10]

Reactions work the same way. To find the limiting reactant, work out how much product each reactant could make on its own. The one that makes less is limiting.

Worked example: 4.0 g of H2 with 16.0 g of O2

2H2 + O2 → 2H2O

  1. Moles: H2 4.0 ÷ 2.016 = 1.98 mol; O2 16.0 ÷ 31.998 = 0.500 mol
  2. From H2: 1.98 mol H2O. From O2: 0.500 × 2 = 1.00 mol H2O. O2 makes less, so O2 is limiting.
  3. Water formed: 1.00 mol × 18.015 = 18.0 g
  4. H2 used: 1.00 mol; left over 0.98 mol, about 2.0 g

Check: 4.0 g + 16.0 g = 20.0 g before; 18.0 g + 2.0 g = 20.0 g after.

Limiting-Reactant Simulator

Choose a reaction or the sandwich recipe and set how many particles of each reactant you start with.

BeforeAfter

H₂ is the limiting reactant: 5 H₂ ÷ 2 = 2.5, less than 4 O₂ ÷ 1 = 4. 4 H₂O molecules form. Excess: 2 O₂ left over. 1 H₂ is also left, too few for another reaction.

Worked answer for 5 H₂ + 4 O₂

Each reaction uses 2 H₂ and 1 O₂. 5 H₂ is enough for 2 reactions (5 ÷ 2 = 2, remainder 1); 4 O₂ is enough for 4. So H₂ limits: 2 reactions make 4 H₂O, using 4 H₂ and 2 O₂. Left over: 1 H₂ and 2 O₂.

Theoretical, Actual and Percent Yield

The amount stoichiometry predicts is the theoretical yield. What you really collect is the actual yield, and it is usually less: some product sticks to glassware, side reactions happen, or the reaction doesn't finish.[10]

percent yield = (actual yield ÷ theoretical yield) × 100%

Worked example

The limiting-reactant example predicts 18.0 g of water. If you collect 15.3 g, the percent yield is 15.3 ÷ 18.0 × 100% = 85.0%.

Percent Composition and Empirical Formulas

Percent composition is the share of a compound's mass from each element. Water: hydrogen is 2.016 ÷ 18.015 = 11.19% and oxygen 88.81%.[11]

An empirical formula gives the simplest whole-number ratio of atoms. Glucose, C6H12O6, has the empirical formula CH2O.[11]

Worked example: a compound is 40.0% C, 6.7% H and 53.3% O
  1. Assume 100 g: 40.0 g C, 6.7 g H, 53.3 g O
  2. Moles: C 40.0 ÷ 12.011 = 3.33; H 6.7 ÷ 1.008 = 6.65; O 53.3 ÷ 15.999 = 3.33
  3. Divide by the smallest (3.33): C 1, H 2.0, O 1
  4. Empirical formula: CH2O. Knowing the molar mass (for example 180 g/mol for glucose) gives the molecular formula, 6 × CH2O = C6H12O6.

How Do We Know?

Avogadro: equal volumes, equal numbers (1811)

Amedeo Avogadro proposed that equal volumes of gases, at the same temperature and pressure, contain equal numbers of molecules. Few chemists paid attention: in the Science History Institute's words, “In 1811 Avogadro put forward a hypothesis that was neglected by his contemporaries for years.”[12]

Cannizzaro sorts out the atomic weights (1858–1860)

Stanislao Cannizzaro used Avogadro's idea to build a consistent set of atomic weights for his course at the University of Genoa in 1858. At the first international chemical congress, in Karlsruhe in 1860, many chemists went home with a printed copy of his course outline. Within a decade his atomic weights had spread, and “Mendeleev acknowledged that his periodic table was inspired by Cannizzaro’s work.”[13]

Perrin shows molecules are real (1908)

In the 1880s, whether atoms and molecules physically existed was still a matter of dispute. Jean Perrin reasoned that if molecules were real, tiny particles in a liquid should not all sink but spread out in a predictable way, and in 1908 he confirmed it by experiment. He also confirmed Einstein's explanation of Brownian motion, the jiggling of small particles struck by molecules. These experiments also let him estimate Avogadro's number. In his Nobel lecture he reported that his most careful sedimentation measurements “gave a value for N of 68 X 1022”, that is, about 6.8 × 10²³, close to today's 6.022 × 10²³; other methods in the same lecture gave similar values, such as 62 × 10²² and 65 × 10²², all roughly 6 to 7 × 10²³.[15] His 1926 Nobel Prize in Physics was “for his work on the discontinuous structure of matter, and especially for his discovery of sedimentation equilibrium.”[14]

Fixing the number (2019)

For decades the Avogadro number had to be measured. Scientists counted the atoms in near-perfect silicon spheres, and those measurements let the number be fixed exactly.[4] On May 20, 2019, the mole was redefined in terms of that fixed constant, along with the kilogram, kelvin and ampere.[3][1]

Test Yourself

Ten questions. The calculator and simulator are allowed!

The Recipe Challenge

Think About It

Questions to puzzle over in class or at home.

  1. Why do you think chemists picked such an enormous number for the mole instead of, say, a million?
  2. A factory makes ammonia from nitrogen and hydrogen. Which reactant would you put in excess, and why might the cost of each matter?
  3. A student gets a percent yield of 110%. What might have gone wrong?
  4. Avogadro's idea was ignored for decades. What does it take for a correct idea in science to be accepted?

Next unit: Acids and Bases →

Back to all chemistry lessons

Sources

  1. BIPM, “SI base unit: mole.” Quote: “One mole contains exactly 6.022 140 76 x 1023 elementary entities.” https://www.bipm.org/en/si-base-units/mole
  2. NIST, “CODATA Value: Avogadro constant.” https://physics.nist.gov/cgi-bin/cuu/Value?na
  3. NIST, “SI Redefinition.” Quote: “On May 20, 2019, four of them — the kilogram, kelvin, ampere and mole — were redefined in terms of constants of nature.” https://www.nist.gov/si-redefinition
  4. NIST, “Redefining the Mole.” Quote: “18 grams of water … contains approximately 600 billion trillion individual water molecules.” https://www.nist.gov/si-redefinition/mole
  5. PubChem, “Periodic Table of Elements” (atomic masses). https://pubchem.ncbi.nlm.nih.gov/periodic-table/
  6. NIST, “CODATA Value: molar volume of ideal gas (273.15 K, 101.325 kPa)”: 22.413 969 54… × 10−3 m3 mol−1. https://physics.nist.gov/cgi-bin/cuu/Value?mvolstd
  7. NIST, “CODATA Value: molar volume of ideal gas (273.15 K, 100 kPa)”: 22.710 954 64… × 10−3 m3 mol−1. https://physics.nist.gov/cgi-bin/cuu/Value?mvol
  8. OpenStax, Chemistry 2e, “3.1 Formula Mass and the Mole Concept.” https://openstax.org/books/chemistry-2e/pages/3-1-formula-mass-and-the-mole-concept
  9. OpenStax, Chemistry 2e, “4.3 Reaction Stoichiometry.” https://openstax.org/books/chemistry-2e/pages/4-3-reaction-stoichiometry
  10. OpenStax, Chemistry 2e, “4.4 Reaction Yields” (limiting reactant, percent yield). https://openstax.org/books/chemistry-2e/pages/4-4-reaction-yields
  11. OpenStax, Chemistry 2e, “3.2 Determining Empirical and Molecular Formulas.” https://openstax.org/books/chemistry-2e/pages/3-2-determining-empirical-and-molecular-formulas
  12. Science History Institute, “Amedeo Avogadro.” Quote: “In 1811 Avogadro put forward a hypothesis that was neglected by his contemporaries for years.” https://www.sciencehistory.org/education/scientific-biographies/amedeo-avogadro/
  13. Science History Institute, “Stanislao Cannizzaro.” Quote: “Mendeleev acknowledged that his periodic table was inspired by Cannizzaro’s work.” https://www.sciencehistory.org/education/scientific-biographies/stanislao-cannizzaro/
  14. NobelPrize.org, “Jean Baptiste Perrin – Facts.” Prize motivation: “for his work on the discontinuous structure of matter, and especially for his discovery of sedimentation equilibrium.” https://www.nobelprize.org/prizes/physics/1926/perrin/facts/
  15. Jean Baptiste Perrin, “Discontinuous Structure of Matter,” Nobel Lecture, NobelPrize.org. Quote: “My most careful measurements made with an emulsion the rarefaction of which doubled with each rise of 6 microns, gave a value for N of 68 X 1022.” https://www.nobelprize.org/prizes/physics/1926/perrin/lecture/