Chapter 12 Stoichiometry

Another Analogy: (let’s get off the bike for a while and bake a cake!)  Let’s say you want to bake a cake. Here’s a recipe:  2sticks of butter + 3 cups flour + 2 cups sugar +3 eggs  1 cake  Ask yourself these questions:  What are the ingredients? The goal?  How are the ingredients and cake measured (quantified)?  How much of each ingredient do you need to make 2 cakes?  What would happen if you have 2 eggs?  What would happen if you had 6 eggs?

12.1 The Arithmetic of Equations  Chemists use balanced chemical equations as a basis to calculate how much reactant is needed or product is formed in a reaction.  Stoichiometry – the calculation of quantities in chemical reactions.

Interpreting Chemical Equations Coefficients Represent Coefficients Do Not Represent molecules moles volume mass A balanced chemical equation can be interpreted in terms of different quantities, including atoms, molecules, moles, or volume. * It is important to know what coefficients represent and what they do not represent in a chemical equation.

An example:  For the equation: N 2 (g) + 3H 2 (g)  2NH 3 (g) (nitrogen) (hydrogen) (ammonia)  Coefficients are: 1, 3, and 2  Meaning:1 mole of nitrogen gas reacts with 3 moles of hydrogen gas to produce 2 moles of ammonia.

12.2 Chemical Calculations  Remember conversion factors from chapter 10?  365 days/year  6.02 x molecules/mole  g O 2 /mole O 2  Remember that algebra allows you to use these conversion factors to multiply an expression without changing the expression.  By carefully choosing conversion factors and canceling units, you can convert any measurement into one expressed in the desired units.

12.2 Chemical Calculations  From a balanced equation, mole ratios can be determined that relate the moles of reactants to moles of product.  Mole ratios are conversion factors derived from the coefficients of a balanced chemical equation. N 2 (g) + 3H 2 (g)  2NH 3 (g) From the equation above, 3 mole ratios can be derived (and their reciprocals): 1 mol N 2 2 mol NH 3 3 mol H 2 3 mol H 2 1 mol N 2 2 mol NH 3

Mole-to-mole calculations:  If you know the number of moles of one substance, the balanced chemical equation allows you to determine the number of moles of other substances in the reaction using those mole ratios (a type of conversion factor)  Example: using the previous equation, how many moles of ammonia (NH3) are produced when.60 mol of nitrogen reacts with hydrogen. N 2 + 3H 2  2NH 3.60 mol N 2 x 2mol NH 3 = 1.2 mol NH 3 1 mol N 2

Mass-to-mass calculations  Step 1: Check that the equation is balanced.  Step 2: Start with the number given and multiply by a conversion factor with the molar mass to convert Mass to Moles.  Step 3: Multiply by the next conversion factor using coefficient (mole) ratios to convert Moles to Moles  Step 4: Multiply by a third conversion factor using the molar mass of the new compound to convert Moles to Mass. 

Mass to mass example: 1) 2AgNO 3 (aq) + Na 2 SO 4 (aq)  Ag 2 SO 4 (s) + 2NaNO 3 (aq) 2) 17.0 g AgNO 3 x 1 mol AgNO 3 Molar mass g AgNO 3 Ag = x 1 = N = 14.0 x 1 = 14.0 O = 16.0 x 3 = 48.0 TOTAL = ) 17.0 g AgNO 3 x 1 mol AgNO 3 x 1 mol Ag 2 SO g AgNO 3 2 mol AgNO 3 4) 17.0 g AgNO 3 x 1 mol AgNO 3 x 1 mol Ag 2 SO 4 x g Ag 2 SO 4 =15.6g g AgNO 3 2 mol AgNO 3 1 mol Ag 2 SO 4 Molar mass Ag = x 2 = S = 32.0 = 32.0 O = 16.0 x 4 = 64.0 TOTAL = 311.8

Other Stoichiometric Calculations  Use the diagram on p797 to solve other stoichiometric problems.  Summary: A balanced equation gives you mole ratios. With mole ratios you can calculate any measurement unit related to the mole, such as # of particles, mass, or volumes of gases (at STP)

12.3 Limiting Reactant

+=  In any chemical reaction, it is possible to “run out” of one or another reactant. This has an impact on the amount of products that can result from a reaction  To solve these problems you must identify which of the reactants is going to run out first.  This is the “limiting reactant”  The other is the “excess reactant”  Example: – If you had three yellow blocks and four red blocks, you could only make three yellow/red combinations – because there is not enough yellow blocks to make four. The yellow block is the limiting reagent and the red block is the excess reagent

Container 1 No Limiting Reactant Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 269

Before and After Reaction 1 All the hydrogen and nitrogen atoms combine. Before the reaction After the reaction Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 269

 In reaction 1 there is no limiting reactant both N 2 and H 2 are completely used in the reaction.  In reaction 2 (next) H 2 is runs out first and therefore limits the number of NH 3 molecules produced.  Excess: Notice that two N 2 molecules are left unreacted. This is called excess. If one reactant runs out the other reactant will be in excess.

Container 2 H 2 as Limiting Reactant Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 270

Before and After Reaction 2 Before the reactionAfter the reaction Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 270

19 Determining Limiting Reactant Ex: Synthesis of Ammonia (NH 3 ) N 2 + 3H 2 → 2NH 3 If 5.0 g nitrogen gas and 5.0 g hydrogen gas react to form ammonia, which is the limiting reactant?

Steps: (1) Convert units of each reactant (i.e., grams) to moles (2) Divide the moles of each reactant by the coefficient for that reactant in the equation. (3) The smaller reference number is the limiting reactant N 2 + 3H 2 → 2NH 3 The second step adjusts for the different moles that react in the equation. 20 Determining Limiting Reactant

21 Determining Limiting Reactant Calculations:

Stoichiometry  Use the limiting reactant to determine the mass of NH 3 produced.

Determining the Mass of Excess Reactant  Use the limiting reactant (N 2 ) to calculate the mass of excess reactant (H 2 ) that was consumed in the reaction  Subtract this mass from the mass available before the reaction.

Another example:  Nitrogen gas can be prepared by passing gaseous ammonia over solid copper II oxide at high temperatures. The other products of the reaction are solid copper and water vapor. If a sample containing 18.1g of NH 3 is reacted with 90.4g of CuO, which is the limiting reactant? How many grams of N 2 will be formed?

 Write the Balanced Equation:  Calculate Limiting Reactant:  Find the mass of N 2 (g) formed: LR = CuO

Percent Yield:  To calculate the percent yield, divide the actual yield by the theoretical (calculated Mass- Mass) and multiply by 100.  Example: Using the above problem and its theoretical yield, calculate the percent yield if only 10.0 g of N2 gas is formed. 10.0g/14.9g =.67  67%