Chemical Equations and Reaction Stoichiometry AP Chemistry Milam

3-1 Chemical Equations To balance chemical equations you use coefficients to multiply different reactants and products until you have the same number of each type of element on each side The coefficients represent the # of molecules and the subscripts represent the # of atoms in each molecule, the product of coefficients and subscripts tells you the total # of atoms

3-1 Balancing tips –Try to keep polyatomic ions as one thing if they are present on both sides –In combustion reactions balance the oxygen gas last –If stuck just pick an element that is unbalanced and start there, you can always reduce the numbers down at the end

3-2 calculations based on chemical equations A balanced equation tells you the relative # of molecules, and since a mol is a multiple of that ratio, it also tells you the relative # of moles Ex. 2Na + 2HCl  2NaCl + H 2 This equation tells you that for every 2 moles of sodium metal and 2 moles of hydrochloric acid, you will get 2 moles of sodium chloride and 1 mole of hydrogen gas

3-2 The coefficients do not tell you the relative # of grams, or liters of a gas so in cases where these problems arise, we need to convert to moles before we can change chemicals. Ex. CH 4 + 2O 2  CO 2 + 2H 2 O If 32 grams of methane react with excess oxygen, how much water (g) is made?

grams of methane are 2 moles of methane, and for every 1 mol of methane you get 2 mol of water, so you will end up with 4 mol of water. Every mol of water is 18 grams so you will get 4x18g = 72 grams of water Excess means you have so much of that chemical you can ignore it

3-2 Stoichiometry will be used very frequently, and the problems will be made as difficult as possible, always remember that you always need to have a mol ratio in the middle of your calculation

3-3 The limiting Reactant (Reagent) Concept In these problems, rather than having an excess of one chemical, you need to determine which chemical is in excess and from there do stoichiometry problems and analyze the results. IMO, the simplest method to do this, is to perform 2 stoichiometry problems right away and from there use the results to answer all of the questions

3-3 While it is not in section 3-3 there will be problems where you must calculate the remaining leftover excess reagent. To do this, take your two product results and find the ratio of them. This ratio is how much of the excess used up Ex. If you take a ratio of.4 to 1, you use up 40% or have % = 60% leftover. So just multiply 0.6 x your original reactant and you will have your excess

3-4 Percent yields from Chemical reactions When performing a chemical, reaction you should get whatever results stoichiometry provides you with, but 99.99% of the time you get a different amount The amount you are off by is the percent error, and the amount you are on by is the percent yield

3-4 If you should get 10 grams, and instead get 9.5 grams, you have a percent error of 5% and a percent yield of 95% It is possible to have over 100% yield because sometimes you will have impurities or water will be absorbed. Human error is never an acceptable explanation of error!

3-5 Sequential reactions This is just doing a series of stoichiometry calculations where a reactant undergoes a series of chemical reactions. No tricks here, just a lot of mol ratios and molar masses to sort through.

3-6 Concentration of Solutions Solute is w/e has less mass and solvent has more Aqueous solutions are solutions with water as the solvent Concentration is how much solute in the solution or per amount of solvent There are many ways to express concentration, so be ready for some good problems to solve

3-6 Percent by mass = mass solute / (mass of solution) 6 grams of copper (II) chloride dissolved in 94 grams of water is a 6% solution Molarity = mol solute/ (L sol’n) 1 mol of NaCl in 1 liter of solution is 1 M Because you see grams and moles there will be problems switching between the two

3-6 When doing concentration problems there are so many kinds of concentrations that it is important to write down what you have to start with and what you need to end up with to simplify the process of solving. Going from percent mass to molarity could be overwhelming, but changing grams into moles and going from grams to Liters is much easier.

3-7 Dilution of solutions Since molarity is moles solute over Liters of solution, if you multiply molarity * liters you will get moles of solute This is handy when diluting because when you add more solvent your number of moles of solute does not change. Therefore we can take the original molarity and volume and it will equal the product of the new molarity and volume after dilution

3-7 Molarity * Volume = Molarity * Volume M 1 V 1 = M 2 V 2 The volume does not need to be in Liters for the equation to work, but the volume units must be the same for both sides

3-8 Using solutions in Chemical Reactions In this section we are doing more stoichiometry only now we are using molarity as a conversion factor to get from volume to moles or from moles to volume