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## Stoichiometry

• Definition
• The word  Stoichiometry comes from the Greek stoicheion, which means to measure the elements
• A good definition of the term’s meaning in the study of chemistry is the “quantitative study of reactants and products in a chemical reaction” (from Chemistry by Raymond Chang)
• Stoichiometry allows one to calculate how much of a given product a reaction is expected to produce based on how much of the reactants are available
• Given the mass, volume and density, or the number of moles of reactants, one can calculate the mass, volume (if the density is known) or moles of product
• Molar Ratios
•  Calculations using stoichiometry depend on the molar relationships in chemical equations; this is why a properly balanced chemical equation is so important
• A properly balanced chemical equation shows the molar ratios of each of the species present, whether they are reactants or products
• Take the combustion of propane as an example: C3H8 + 5O2 --> 3CO2 + 4H2O
• The ratios found in this equation are as follows:
1 mol propane:5 mol oxygen
(each mole of C3H8 requires five moles of O2 to burn completely)
1 mol propane:3 mol carbon dioxide
(each mole of completely burned C3H8 produces three moles of CO2)
1 mol propane:4 mol water
(each mole of completely burned C3H8 produces four moles of H2O)
5 mol oxygen:3 mol carbon dioxide
(for every five moles of O2 consumed, three moles of CO2 are produced)
5 mol oxygen:4 mol water
(for every five moles of O2 consumed, four moles of H2O are produced)
3 mol carbon dioxide:4 mol water
(for every three moles of CO2 produced, 4 moles of H2O are produced)
• Each of these ratios can be written in reverse order and can be thought of as a fraction or a conversion factor
• The key to using stoichiometry is to calculate using moles and to convert whatever information you are given into moles, whether it be mass or volume and density
• It is easy to calculate the number of moles needed or moles produced if the starting information is given in moles: just find the appropriate molar ratio and multiply by the moles in the starting information (be sure to write the ratio with the quantity you start with on the bottom of the ratio)
• It only requires one additional step to find the mass of a substance that is needed or produced given the number of moles of one of the substances: just convert the answer you find in moles into grams using the molar mass of the substance
• Given the mass of one of the substances you can find the mass of reactants needed or products produced by first converting the given mass into a number of moles, finding the number of moles of the other substance using the proper molar ratio, then converting that number of moles into grams

### Molar Mass

The molar mass of a compound is the total of the molar masses of its constituent atoms. The molar mass of an element is the mass given in the periodic table expressed in grams. This is the average atomic mass and takes into account the mass of each elements’ isotopes and their abundance.

The molar mass of NaCl is 58.44 g/mol because 22.99 g/mol (Na) + 35.45 g/mol (Ca) = 58.44 g/mol.
CaCl2: Ca, 40.08 g/mol; Cl, 35.45 g/mol
(1 × 40.08 g/mol) + (2 × 35.45 g/mol) = 110.98 g/mol

When a compound contains a polyatomic ion or a repeated unit, multiply the subscripts inside the parentheses by the subscript outside the parentheses:
Ca3(PO4)2: Ca, 40.08 g/mol; P, 30.97 g/mol; O, 15.99 g/mol
(3 × 40.08 g/mol) + (2 × 1 × 30.97 g/mol) + (2 × 4 × 15.99 g/mol) =

CH3(CH2)3Cl: C, 12.01 g/mol; H, 1.01 g/mol; Cl, 35.45 g/mol
(1 × 12.01 g/mol) + (3 × 1.01 g/mol) + (3 × 12.01 g/mol) + (3 × 2 × 12.01 g/mol) + (1 × 35.45 g/mol) = 92.58 g/mol

Others to try: XeF2 (169.29 g/mol), SnCl4 (260.51 g/mol), SnCl2 (189.61 g/mol), CO2 (44.01 g/mol), NH3 (17.04 g/mol)

### Stoichiometry

The key to stoichiometry is to always convert everything to moles before you even begin. When you have found the answer you can always convert to mass or volume.
Here are some sequences with which to work in order to solve different types of problems:
1. Starting with the number of moles of one of the substances in the balanced chemical equation you can easily find the number of moles of one of the other substances using the relevant molar ratio. Make sure to put the substance you know on the bottom of the ratio and the substance you need to find on the top of the ratio before you multiply.
2. Starting with a mass requires a further step: convert that mass into a number of moles using the molar mass of the substance. Then simply follow the same steps used above.
3. If the answer required is a mass, simply convert the number of moles you find to the mass using the molar mass of the substance in question.
4. If you are given a volume and a density, simply find the mass using D = m/V and then calculate moles and complete the problem. Or simply use this five-step method:
1. Write correct formulas for all reactants and products, and balance the resulting equation.
2. Convert the quantities of some or all given or known substances (usually reactants) into moles.
3. Use the coefficients in the balanced equation to calculate the number of moles of the sought or unknown quantities (usually products) in the problem.
4. Using the calculated numbers of moles and the molar masses, convert the unknown quantities are required (usually grams).
C3H8 + 5O2 → 3CO2 + 4H2O
Moles Example: You are told that a bottle contains 40 mol C3H8 (propane) and asked how many moles of O2 you would need to burn it all.
The molar ratio is 5 mol O2 to 1 mol C3H8.
40 mol C3H8 × 5 mol O2/1 mol C3H8 = 200 mol O2
Mass Example: Given 32 g of O2 and an excess of C3H8 how much water would the reaction produce, in grams?
Convert mass to moles: 32 g O2 × 1 mol/16 g = 2 mol O2
Use molar ratio (4 mol H2O to 5 mol O2) to find moles of water: 2 mol O2 × 4 mol H2O/5 mol O2 = 1.6 mol H2O
Convert moles back to mass: 1.6 mol H2O × 18.02 g/1 mol = 28.83 g
Others to try: S + 6HNO3 → H2SO4 + 6NO2 2H2O
starting w/14 g S; how many grams of HNO3 are needed? (165.07 g)
3 mol H2O produced; how many grams of S reacted? (48.098 g)
10.0 L NO2 produced (D = 2.05 g/L); how many moles of S reacted? (3.74 ×10-1 mol)

Last updated: Apr 29, 2008       Home