Chapter 12 Stoichiometry

12.1 The Arithmetic of Equations Objectives Explain how balanced equations apply to both chemistry and everyday life Interpret balanced chemical equations in terms of moles, representative particles, mass and gas volume at STP Identify the quantities that are always conserved in chemical reactions

COOKIES! Who makes the best chocolate chip cookies? WHY? When baking cookies, a recipe is usually used, telling the exact amount of each ingredient If you need more, you can double or triple the amount of each ingredient Thus, a recipe is much like a balanced equation!

Using Balanced Chemical Equations We use balanced chemical equations as a basis to calculate how much reactant is needed or product is formed in a reaction Stoichiometry Greek for “measuring elements” Defined as the calculation of quantities in chemical reactions based on a balanced equation

Using Balanced Chemical Equations Stoichiometry allows chemists to tally the amounts of reactants and products using ratios of moles or representative particles There are four ways to interpret a balanced chemical equation. In terms of: Particles Moles Mass Volume

In Terms of Particles At the atomic level, a balanced equation indicates that the number and type of each atom that makes up each reactant also makes up each product Reminders Element = atoms Molecular compound = molecules Ionic compound – formula units

In Terms of Particles 2H2 + O2  2H2O Two molecules of hydrogen and one molecule of oxygen form two molecules of water 2Al2O3  4Al + 3O2 2 formula units of Al2O3 form 4 atoms of Al and 3 molecules of O2 Try this 2Na + 2H2O  2NaOH + H2

** Most important information ** In Terms of Moles A balanced chemical equation tells us the number of moles of reactants and products The coefficients tell us how many moles of each substance 2Al2O3  4Al + 3O2 2Na + 2H2O  2NaOH + H2 A balanced equation is a Molar Ratio ** Most important information **

In Terms of Mass 2.02 g H2 2 moles H2 = 4.04 g H2 1 mole H2 32.00 g O2 A balanced chemical equation obeys the law of conservation of mass Using mole relationship, we can relate mass to the number of atoms in a chemical equation 2.02 g H2 2 moles H2 = 4.04 g H2 1 mole H2 + 32.00 g O2 32.00 g O2 1 mole O2 = 1 mole O2 36.04 g H2 + O2 36.04 g H2 + O2

In Terms of Mass 2H2 + O2 ® 2H2O 2H2 + O2 ® 2H2O 18.02 g H2O = 2 moles H2O 1 mole H2O 2H2 + O2 ® 2H2O 36.04 g H2 + O2 = 36.04 g H2O Although the number of moles of reactants does not equal the number of moles of product, the total number of grams of reactant does equal total number of grams of product

In Terms of Volume At STP, one mole of any gas = 22.4 L 2H2 + O2  2H2O (2 x 22.4 L H2) + (1 x 22.4 L O2)  (2 x 22.4 L H2O) NOTE – Mass and atoms are always conserved – however, molecules, formula units, moles and volume will not necessarily be conserved

Practice Show that the following equation follows the Law of Conservation of Mass (show the atoms balance and the mass on both sides is equal) 2Al2O3  4Al + 3O2

Section 12.2 Chemical Calculations Objectives: Construct mole ratios from balanced chemical equations, and apply these ratios in mole-mole stoichiometric calculations Calculate stoichiometric quantities from balanced chemical equations using units of moles, mass, representative particles, and volumes of gases at STP

Writing and Using Mole Ratios Conversion factor derived from the coefficients of a balanced chemical equation interpreted in terms of moles In chemical calculations, mole ratios are used to convert- between moles of reactant and moles of product between moles of reactants between moles of products

Writing and Using Mole Ratios N2(g) + 3H2(g)  2NH3(g) 3 mole ratios: 1 mole N2 2 mole NH3 3 mole H2 3 mole H2 1 mole N2 2 mole NH3

Mole to Mole Conversions 2 Al2O3  4Al + 3O2 each time we use 2 moles of Al2O3 we will also make 3 moles of O2 2 moles Al2O3 3 mole O2 or 3 mole O2 2 moles Al2O3 These are the two possible conversion factors

Mole to Mole Conversions How many moles of O2 are produced when 3.34 moles of Al2O3 decompose? 2 Al2O3  4Al + 3O2 3 mol O2 3.34 mol Al2O3 = 5.01 mol O2 2 mol Al2O3

Practice: 2C2H2 + 5 O2  4CO2 + 2 H2O If 3.84 moles of C2H2 are burned, how many moles of O2 are needed? (9.6 mol) How many moles of C2H2 are needed to produce 8.95 mole of H2O? (8.95 mol) If 2.47 moles of C2H2 are burned, how many moles of CO2 are formed? (4.94 mol)

How do you get good at this?

Calculating Stoichiometric Problems No laboratory balance can measure substances directly in moles Instead the amount of a substance is usually determined by measuring its mass in grams Using the molar mass of the substances we can convert grams to moles and then back to grams

Calculating Stoichiometric Problems Balance the equation Convert mass in grams to moles Set up mole ratios Use mole ratios to calculate moles of desired chemicals Convert moles back into grams, if necessary

Mass-Mass Problem: 4 Al + 3 O2  2Al2O3 12.3 g Al2O3 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed? 4 Al + 3 O2  2Al2O3 6.50 g Al 1 mol Al 2 mol Al2O3 101.96 g Al2O3 ? g Al2O3 = 1 mol Al2O3 26.98 g Al 4 mol Al 12.3 g Al2O3 (6.50 x 2 x 101.96) ÷ (26.98 x 4) =

Another example: Answer = 17.2 g Cu 2Fe + 3CuSO4 ® Fe2(SO4)3 + 3Cu If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how much solid copper would form? 2Fe + 3CuSO4 ® Fe2(SO4)3 + 3Cu Answer = 17.2 g Cu

Volume-Volume Calculations: How many liters of CH4 at STP are required to completely react with 17.5 L of O2 ? CH4 + 2O2  CO2 + 2H2O 22.4 L O2 1 mol O2 1 mol CH4 1 mol CH4 22.4 L CH4 1 mol O2 22.4 L CH4 17.5 L O2 22.4 L O2 2 mol O2 1 mol CH4 = 8.75 L CH4 Notice anything concerning these two steps?

Avogadro told us… Equal volumes of gas, at the same temperature and pressure contain the same number of particles Moles are numbers of particles You can treat reactions as if they happen liters as a time, as long as you keep the temperature and pressure the same 1 mole = 22.4 L @ STP

Shortcut for Volume-Volume: How many liters of CH4 at STP are required to completely react with 17.5 L of O2? CH4 + 2O2  CO2 + 2H2O 1 L CH4 17.5 L O2 = 8.75 L CH4 2 L O2 Note: This only works for Volume-Volume problems.

Other Stoichiometric Calculations In a typical stoichiometric problem, the given quantity is - first converted to moles Then the mole ratio from the balanced equation is used to calculate the number of moles of the wanted substance Finally, the moles are converted to any other unit of measurement(grams, number of particles, liters) related to the unit mole, as the problem requires Remember your conversion factors from chap. 10 1 mole = molar mass (masses of elements from Periodic Table) 1 mole = 6.02 x 1023 representative particles 1 mole = 6.02 x 1023 molecules 1 mole = 6.02 x 1023 atoms 1 mole = 22.4 L