AP Chemistry - Notes - Chapter 3

AP/IB Chemistry Mr. Root - Chandler High School Name: AP/IB Chemistry - Notes - Chapter 3 : Stoichiometry

A. Stoichiometry - the study of quantities of substances consumed and produced in chemical reactions. 1. Atomic mass - the average mass of an atom of an element in atomic mass units (amu) a. Atomic masses are based on 12C ("carbon twelve"), which is assigned a value of exactly 12 atomic mass units. b. The atomic masses of other elements are determined by comparison to 12C. c. Mass spectrometer - currently best method for determination of atomic masses of atoms - Procedure : - atoms or molecules are placed in abeam of high-speed electrons (knocks electrons off the atoms or molecules giving them a positive charge - cations) - an electric field is then applied accelerating these cations which causes each cation to create a magnetic field - cations pass through an applied magnetic field - cations with the least mass are deflected more than heavier cations and masses are determined by comparing amounts of deflection e.g. If , in a mass spectrometer, 13C is found to have a mass 1.0836 times that of 12C, then the atomic mass of 13C = 1.0836 (12 amu) = 13.003 amu d. The mass spectrometer can also be used to determine the isotopic composition of an element. When a sample of an element is placed in a mass spectrometer, a mass spectrum can be obtained which indicates the relative amounts of the various isotopes present in the sample. e. The average mass of an atom (or mole) of an element is the sum of the fractions of each isotope times their mass. e.g. Naturally occurring carbon is composed of two isotopes, 12C (98.89%) and 13C (1.11%). The average atomic mass for an atom of carbon = ( .9889 x 12 amu) + ( .0111 x 13.003) = 12.01 amu. 2. The Mole (abbrev. mol) - SI unit for the amount of a substance and is equal to the number of atoms in exactly 12 grams of pure 12C. - Mass spectrometry has determined this number to be 6.02214 x 1023 (Avogadro's number) (We will use 4 sig.fig : 6.022 x 1023). a. Some things to keep in mind about the mole 1. 1 mole = 6.022 x 1023 particles, atoms, molecules (Avogadro's number) 2. 1 mole of an element = atomic mass of that element 3. 1 mole of a compound = molar mass of that compound 4. 1 mole of a gas = 22.4 liters 3. Molar mass - the mass in grams of one mole of a substance (equals the atomic mass in grams) a. Determining the molar mass 1. Add together the atomic mass of each element in the compound. 2. Remember to multiply the atomic mass by the number of atoms in the compound. 3. Determine the molar mass of Barium sulfate

b. When converting between grams of a substance and moles of a substance, use the molar mass for the conversions. 1. Determine the number of moles in 48.3 g of Barium sulfate

c. When converting between moles of a substance and number of particles or molecules, use Avogadro's number for the conversions. 1. How many molecules are there in the number of moles of Barium sulfate previously determined? AP/IB Chemistry Mr. Root - Chandler High School Name:

d. When converting between the grams of a substance and the number of molecules or particles, remember to convert to moles first! 1. How many grams of Copper (II) chloride are represented by 4.82 x 1025 molecules?

4. Mass percent composition - the percent composition of a substance by mass A. To determine the mass percent of a particular element in a compound 1. Find the mass of that element in the compound, 2. Find the molar mass, 3. Divide the mass contributed by that element by the molar mass.

Sample Problem : What is the percent composition of glucose (C6H12O6) mass of carbon 72.06g mass percent of carbon  x 100%  x 100%  39.99% mass of compound 180.18g mass of hydrogen 12.12g mass percent of hydrogen  x 100%  x 100%  6.727 % mass of compound 180.18g mass of oxygen 96.00g mass percent of oxygen  x 100%  x 100%  53.28 % mass of compound 180.18g

5. Determining the Formula of a Compound a. Empirical formula - lowest whole number ratio of the atoms in a compound - Determination from percent composition data : Step 1 : Determine mass composition (if not given) from percent composition (assume 100g and then percents convert to grams) Step 2 : Determine moles of each element in the compound by dividing the number of grams of each element by THAT element's atomic mass. Step 3 : Find the ratio of the number of moles of each element to the number of moles of the other elements. Compare each element to the element with the smallest number of moles. Step 4: If the mole ratios are not whole numbers, multiply each number by an integer so that they are whole numbers. Step 5: Write the formula using the ratio of moles for each element as the subscript on that element AP/IB Chemistry Mr. Root - Chandler High School Name:

b. Determining the Molecular Formula from Mass Percent and Molar Mass - Molecular formula = Exact formula of a compound; gives actual number of each element in the compound. Molecular formulas represent the actual ratio of the atoms of elements in a compound and are whole number multiples of their empirical formulas. - All of the steps are the same as determining the empirical formula to the point that the empirical formula has been determined: Step 1: Determine the molar mass of the empirical formula. Step 2: Find the ratio of the molar mass of the empirical formula to the actual molar mass of the compound. Step 3: If the ratio is one to one, the empirical formula is the molecular formula. If the ratio is a number greater than one, multiply the number of each element in the compound by that number.

6. Chemical Equations - represent what happens in a chemical reaction a. reactant  products (read "Reactants yield products") b. conservation of atoms (mass) - atoms are neither created nor destroyed in chemical reactions, they are recombined to form different substances - mass is neither created nor destroyed chemical reactions (as opposed to nuclear reactions) - chemical reactions must therefore be balanced - have same kinds and numbers of atoms on both sides of the yields sign () c. The physical states of the substances involved in a chemical reaction are represented by symbols : (aq) - aqueous (dissolved in water) (g) - gas (l) - liquid (s) - solid 7. Balancing chemical equations - usually by inspection - it helps to do the most complex substance first AP/IB Chemistry Mr. Root - Chandler High School Name: - never change the correct formulas of reactants or products. i.e. never change the subscript in a formula because this changes the identity of the substance. (Instead, change the amounts of the substances represented by changing the coefficient) - so balance by placing coefficients in front of the reactant or product species - always save oxygen and hydrogen until the very end.

8. Stoichiometric Calculations : Amounts of Reactants and Products a. The following diagram represents the basic conversions used in stoichiometry :

- Calculations between moles and mass and moles and numbers of particles have already been covered (moles and volume conversion will be covered later). The only new thing here is the use of the mole ratio in a balanced equation to convert moles of one substance (known) to moles of another (unknown) .

Sample : From the balanced equation for the Haber process (N2 + 3H2  2 NH3) we see that one mole of nitrogen is needed for every three moles of hydrogen present. How many moles of nitrogen are needed if 1.2 moles of hydrogen are present?

1 mol N 2 moles nitrogen needed  1.2 mol H 2    .40 mol N 2 3 mol H 2

b. Mole Relationships – BCA Method [NOT in your Textbook] - In order to determine how much of a substance is needed or is produced when a reaction is carried out with specified amounts of reactant(s) or product(s), it is important to keep in mind the particle-to-particle (and therefore mole-to-mole) ratio that is characteristic of chemical change. - The goal is to analyze the reaction during three stages of the reaction: how much of each reactant and product is present in the reaction vessel just before the reaction, how much change occurs in each substance during the reaction, and what is present in the reaction vessel immediately after the reaction is complete. - E xample: 1. Lead will react with hydrochloric acid to produce lead chloride and hydrogen gas . How many moles of hydrochloric acid are needed to completely react with 4.0 mole of lead? AP/IB Chemistry Mr. Root - Chandler High School Name:

Equation: 1Pb + 2HCl  1PbCl2 + 1H2

Before: 4.0mol xs mol 0mol 0mol

Change -4.0mol -8.0mol +4.0mol +4.0mol

After 0mol xs mol 4.0mol 4.0mol

8.0 moles of HCl is required to react with the 4.0 moles of Pb. - Since you are asked in this question to determine the number of moles of hydrochloric acid will be consumed (changed) during this reaction, the answer will be found in the Change line under the HCl (-8.0mol). - The sign on the reactant amounts in the Change line is negative to indicate that this amount was consumed (rearranged) during the reaction and is no longer in that form. - Other Stoichiometry Hints: - When starting amounts are given in quantities other than moles (i.e. grams, molarity, volume of gas, etc.) convert to moles before using BCA table. This is because the table focuses on mole relationships. - When solving for a quantity other than moles, convert your answer (in moles) after using the BCA table to the desired quantity.

9. Calculations Involving a Limiting Reactant. a. Stoichiometric quantities - the exact amounts of reactants needed so that no amount of any reactant will be left unused b. Limiting Reactant - a reactant that there is proportionally less of so that it is used up first and limits the amount of product that can be produced - you must first determine which reactant is the limiting reactant when doing stoichiometry problems involving limiting reactants - to determine which reactant is limiting arbitrarily choose one reactant and use the mole ratio of reactants to see if you have an excess or not of the other reactant - e.g. Lets say you have 100.0 g of hydrogen and 200.0 grams of nitrogen and you are going to combine them to form ammonia according to the Haber Process :

N2 + 3H2  2NH3 You can start with either the 100.0 g of H2, or the 200.0 g of N2 and convert to the other reactant. Starting with grams of N2 and converting to moles of H2 :

1 mol N 2 3 mol H 2 200.0 g N 2     21.41 mol H 2 28.02 g 1 mol N 2

Converting grams of H2 to moles H2 : 1 mol H 100.0 g H  2   49.5 mol H 2 2.02 g 2

Since the 200.0 g of N2 equates to a smaller quantity of H2 than what is available, N2 is the limiting reactant. Or, to put it another way, since the 100.0g of H2 present is much more than needed for the 200.0 g of N2, H2 is the excess reactant. c. Excess reactant -opposite of limiting reactant - some will remain at the completion of a reaction d. Theoretical yield - the amount of product a reaction should yield if 100% of the limiting reactant is converted into the product e. Actual yield - the amount of product that is actually produced when the reaction takes place f. Percent yield - a measure of the efficiency of a chemical reaction or process actual yield Percent yield  x 100% (theoretical yield)