Introduction to stoichiometry and moles

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  1. What is stoichiometry?
  2. Ratios in chemical reactions.
  3. Avogadro's number, moles and molar mass.
  4. Percentage composition by mass.
  5. Example: Find percentage composition and empirical formula.
  1. Find the reaction stoichiometry and use it to predict expected mass of products.
    Consider the reaction: CH4+_4 + 2O2 _2 \,   \, CO2+_2 + 2H2_2O
    1. How many moles of H2_2O were produced if 2.5 moles of CO2_2 were produced?
    2. What is the mass of 2.5 moles of CO2_2?
    3. In a repeat experiment, 132 grams of CO2_2 were produced. What was the mass of water also produced?
  2. Find the reaction stoichiometry and use it to predict the mass of chemicals used in a reaction.
    Consider the reaction: HCI+ + NaOH  \,   \, NaCl+ + H2_2O
    1. How many moles of HCI are required to react with 4.75 moles of NaOH?
    2. What would be the mass of 4.5 moles of NaOH?
    3. In a repeat experiment, 54.75g HCI (dissolved in solution) reacted completely with some NaOH added. What was the mass of NaOH used in this experiment?
  3. Find the reaction stoichiometry and use it to predict the mass of chemicals used and produced in a reaction.
    Consider the reaction: 2C2_2H6+_6 + 7O2 _2 \,   \, 4CO2+_2 +6H2_2O
    1. How many moles of O2_2 would react with 1 mole of C2_2H6_6?
    2. If 15 moles of H2_2O were produced in this experiment, how many moles of C2_2H6_6 were used?
    3. What is the mass of this amount of C2_2H6_6 used?
    4. A total of 0.19 mol of reactants and products combined were involved in this reaction. What is the number of moles of H2_2O produced?
  4. Find the percentage composition by mass of chemical compounds and use percentage composition to find empirical formula.
    1. Find the percentage composition by mass of the following atoms in these compounds:
      1. Ca in CaCO3
      2. C in CO2
      3. H in C6H6
    2. A compound has the following percentage composition by mass: C 52.2%; H 13.0%; O 34.8%. What is the empirical formula of this compound?
Topic Notes
In this lesson, we will learn:
  • The mole concept and its importance to the amount of chemical substance.
  • To predict amounts of product made in a reaction with a given amount of reactant.
  • To apply and convert moles to find mass of reactants and products in reactions.
  • To calculate percentage composition by mass of atoms in a chemical substance.


  • The stoichiometry of a reaction is the ratio of different amounts of each reactant that is needed to run a chemical reaction.
    As you should have seen previously, recall that:
    • Matter is particulate in nature; we call these matter particles atoms.

    • The 100 or so known unique atoms (the elements) can combine in unique arrangements of fixed ratios to form new chemical substances. This is what a chemical reaction is, and these combined arrangements of atoms are called molecules. The ratios of the different atoms combining does not depend on how the substance is prepared. This principle is known as the law of definite proportions or Proust’s law.

    • Reactions take place between fixed integer ratios of the reactants. For example, the reactants are two molecules of A and one atom of B which combine to make one molecule of product C. To make C, you need A and B in that ratio.
      This is like making a coffee: if the liquid in a coffee takes four parts hot water and one part milk, to make two servings you increase the amounts to make more but keep the ratio of them the same.

    We have two problems when finding amounts of reactants in chemistry:
    • Real atoms and molecules are far too small to count in an ordinary laboratory without special equipment.
    • Different atoms and molecules have different masses. 50 grams of one substance will not contain the same number of molecules as 50 grams of another.

    Stoichiometry is about knowing the ratios of different reactants in chemical processes and using measures like mass, volume and concentration to find the correct amounts we need to use to get the products we want.

  • Recall from Introduction to chemical equations that a chemical equation tells you the fixed ratio of reactants the reaction uses, and products that the reaction produces.
    • In the example below, we are being told that to produce 1 molecule of C, the reaction needs 1 molecule of A and 2 molecules of B in that fixed ratio. Again, like coffee with one part milk and four parts hot water: if you want to make coffee for two, you must add twice as much milk and water in the same ratio. This is what the β€œstoichiometry of the reaction” is.

    A + 2B   \,   \, C

    The coefficients (the number before the chemical formulae) tell you how many molecules of that chemical are used in the reaction as a ratio with the others. If no number is there, it’s one molecule.

  • As said above, atoms are so small they cannot be counted with regular laboratory equipment, so chemists use mass which is easily measurable to get the amounts they need. The mass of a substance is related to the number of moles of the substance.

  • A mole is a fixed number of atoms known as Avogadro’s number – 6.02x1023 (six hundred and two billion trillion!). It is a number in exactly the same way that a dozen is the number 12 or a pair is the number 2. It’s just a very big number because atoms are tiny!

    In chemistry the molar mass of a substance is measured in grams per mole (gmol-1), for example carbon has the molar mass 12 g mol-1, meaning one mole of carbon atoms has a mass of 12 grams.
    Molar mass is known as atomic mass (Ar) for atoms and the periodic table contains this information for every element (e.g. 12 for C, 16 for O). Molar mass is known as molecular mass (Mr) for molecules, and it can be found by just combining the atomic masses of the atoms in the molecule (e.g. for O2, 16+16=32 g mol-1 for an O2 molecule).

    Using mass and molar mass we can find the number of moles in a sample:
    • For example, carbon dioxide or CO2 has a molecular mass of 44 grams per mole (g mol-1).
      That means that 1 mole (six hundred billion trillion molecules) of carbon dioxide has a mass of 44 grams. A sample of 22 grams of CO2 would be 0.5 moles of CO2.
    • The mole is important because we can now find out the correct amounts of substance needed in a reaction by using molar mass (from the periodic table) and mass which is very easy to measure.

    The equation to calculate number of moles is:

    moles (mol) = moles \, (mol) \, = \, mass (g)Mr (g molβˆ’1)\large \frac{mass \, (g) } {M_{r} \, (g \, mol^{-1}) }

    Use the chemical equation to find the ratio of reactant moles to product moles (the reaction stoichiometry) and then find their relative masses. Now, you can predict the mass of any reactants required or products expected for any reaction scale.

  • The idea of fixed ratios of amounts of substance applies to atoms within compounds too. Compounds are formed by fixed ratios of elemental atoms combining and the method of preparation does not affect this (Proust’s law), so a given compound will always have the same composition by mass of its elements.
    • For example, no matter how you make CO2, its molar mass is 44 g mol-1 and O2 provides 32 g mol-1 of this mass. This is 73% of the mass of CO2 from oxygen; all CO2 molecules will be 73% oxygen as a percentage composition by mass.

    The equation to find percentage composition by mass is:

    % by mass = \, by \, mass \, = \, total mass of elements in compoundtotal molecular mass of compoundβ€‰βˆ—β€‰100 \frac{total\, mass \, of \, elements \, in \, compound} {total\, molecular \, mass \, of \, compound} \, * \, 100

    If given the percentage composition by mass, the equation can be re-arranged to help identify compounds by their empirical formula.

    For example, knowing the percentage composition of mass of a compound containing C and O would tell you if you have CO (a highly toxic gas!) or CO2 (a relatively harmless gas) judging by the mass ratios.

  • WORKED EXAMPLE: Using percentage composition to find empirical formula.

  • A compound containing has the following percentage compositions by mass:
    • Na, 43.5%
    • C, 11.3%
    • O, 45.2%

    Find the empirical formula of this compound.

    To find the empirical formula is found by dividing the percentages by mass by the atomic mass of the element. This is going to find the relative proportions of each atom, accounting for their different masses:




    % by mass




    Atomic mass




    Relative proportion




    Empirical formula




    Accounting for mass, the ratios of Na : C : O are 2 : 1 : 3, so the empirical formula of this compound is Na2CO3. This is sodium carbonate.