Formulas, Equations and Moles

Chapter 3 Formulas, Equations and Moles Interpreting Chemical Equations You can interpret a balanced chemical equation in many ways. On a microscopic level, two molecules of H 2 react with one molecule of O 2 to produce two molecules of H 2 O On a macroscopic level, these formulas and equations represent large-scale behaviors of atoms and molecules that result in observable properties! A chemical equation can mean whatever you want it to mean depending on the context. Chapter 3 2 1

Moving from Molecules to Mass A balanced chemical equation is the scientist s guide to the preparation or formation of product compounds. What does this equation tell us about the preparation of Ethyl Chloride? Chapter 3 3 Atomic Mass The atomic mass of any element on the periodic table is expressed in atomic mass units (amu) This value is listed on the periodic table and represents the mass of a single atom of an element. This relationship can be used to write conversion factors For example, the atomic mass of iron is 55.85 amu, so: 55.85 amu 1 Fe atom OR 1 Fe atom 55.85 amu Chapter 3 4 2

Formula Mass The formula mass of a compound is the sum of the atomic masses of all the atoms in its formula. To determine the formula mass, you multiply each element s atomic mass by its formula subscript and then add them all up. Chapter 3 5 Avogadro s Number How do we keep track of atoms or molecules? They are very small, so we group them in a large bunch We use Avogadro s Number (N A ) to represent this bunch of atoms or molecules. Avogadro s Number was experimentally determined to be the number of atoms in 12.01 grams of carbon. Its numerical value is 6.022 10 23. Therefore, a 12.01 g sample of carbon contains 6.022 10 23 carbon atoms. Chapter 3 6 3

The Mole The mole (mol and abbreviated n) is a unit of measure that allows us to make comparisons between substances that have different masses A mole is Avogadro s number of atoms, that is 6.022 10 23 atoms. 1 mol = 6.022 10 23 atoms Notice how this mole relationship resembles one of our unit equations. Therefore, we can use to write conversion factors to convert between the number of atoms and the mass of a substance. Chapter 3 7 The atomic mass of any substance expressed in grams per mole (g/mol) is the molar mass (MM) of that substance. Molar Mass Molar Mass = Mass mole = m n If the atomic mass of iron is 55.85 amu, then the molar mass of iron is 55.85 g/mol. Chapter 3 8 4

Calculating Molar Mass The molar mass of a substance is the sum of the molar masses of each element present in the substance. What is the molar mass of potassium phosphate? Chapter 3 9 Mole Calculations Grams to Moles What is the mass of 2.87 mol of sulfuric acid? Step 1: Determine what you have: 2.87 mol sulfuric acid Step 2: Determine what you want:??? g sulfuric acid Step 3: Write a plan to convert from what you have to what you want. Step 4: Select conversion factor(s) that allows you to perform your plan: 1 mol SA = 98.09 g SA Grams Molar Mass Moles 2.87 mole SA 98.09 g SA 1 mole SA = 282 g SA Chapter 3 10 5

Mole Calculations Grams to Atoms Now we will use the molar mass of a compound to convert between grams of a substance and particles of a substance. 6.022 10 23 particles = 1 mol = molar mass If we want to convert particles to mass, we must first convert particles to moles and then we can convert moles to mass. Grams Molar Mass Moles Avogadro s Number Atoms Chapter 3 11 Stoichiometry: Mole - Mole Relationships We can use a balanced chemical equation to write mole ratios which can be used as unit factors: N 2 (g) + O 2 (g) 2 NO(g) Since 1 mol of N 2 reacts with 1 mol of O 2 to produce 2 mol of NO, we can write the following mole relationships: 1 mol N 2 1 mol O 2 1 mol N 2 2 mol NO 1 mol O 2 2 mol NO 1 mol O 2 2 mol NO 2 mol NO 1 mol N 2 1 mol N 2 1 mol O 2 Chapter 3 12 6

Stoichiometry Problems In a mass-mass stoichiometry problem, we will convert a given mass of a reactant or product to an unknown mass of reactant or product. There are four steps: Step 1: Check that the equation is balanced!! Step 2: Convert the given mass to moles using the molar mass as a conversion factor. Step 3: Convert the moles of given to moles of the unknown using the coefficients in the balanced equation as a conversion factor. Step 4: Convert the moles of unknown to grams using the molar mass as a conversion factor. A + B C + D Molar Mole Molar Mass of A Ratio Mass of B Grams of A Moles of A Moles of B Grams of B Chapter 3 13 Stoichiometry Problem Aqueous solutions of NaOCl (household bleach) are prepared by the reaction of NaOH with Cl 2 : NaOH (aq) + Cl 2 (g) NaOCl (aq) + NaCl (aq) + H 2 O (l) How many grams of NaOH are needed to react with 25.0 g of Cl 2? Step 1: Balance the equation!!! Step 2: Convert grams Cl 2 to moles Cl 2 using the molar mass of Cl 2. Step 3: Convert moles Cl 2 to moles NaOH using the balanced equation. Step 4: Convert moles NaOH to grams NaOH using the molar mass. Molar Mole Molar Mass of Cl 2 Ratio Mass of NaOH Grams of Cl 2 Moles of Cl 2 Mol of NaOH Grams of NaOH 25.0 g Cl 2 1 mol Cl 2 70.90 g Cl 2 2 mol NaOH 1 mol Cl 2 28.2 g NaOH 40.00 g NaOH 1 mol NaOH Chapter 3 14 7

Percent Yield: Efficiency of Reaction When you perform a laboratory experiment, the amount of product collected is the actual yield. The amount of product calculated from the reaction is the theoretical yield. The percent yield is a measure of the efficiency of the lab reaction by comparing the amount of the actual yield to the theoretical yield. actual yield theoretical yield 100 % = percent yield Chapter 3 15 Calculations with Percent Yield Dichloromethane (CH 2 Cl 2 ) is prepared by reaction of methane (CH 4 ) with chlorine (Cl 2 ) giving hydrogen chloride as a by-product. How many grams of dichloromethane result from the reaction of 1.85 kg of methane if the yield is 43.1%? Chapter 3 16 8

Limiting Reactant Concept Say you re making grilled cheese sandwiches. You need 1 slice of cheese and 2 slices of bread to make one sandwich. 1 Cheese + 2 Bread 1 Sandwich If you have 5 slices of cheese and 8 slices of bread, how many sandwiches can you make? With 5 slices of cheese you could make 5 sandwiches (a 1:1 ratio) With 8 slices of bread you could make 4 sandwiches (a 2:1 ratio) You would run out of bread after 4 sandwiches so that s all you can make! Chapter 3 17 Limiting Reactant Concept Since you run out of bread first, bread is the ingredient that limits how many sandwiches you can make. In a chemical reaction, the limiting reactant is the reactant that controls the amount of product you can make. A limiting reactant is used up before the other reactants. The other reactants are present in excess. Chapter 3 18 9

Limiting Reactant Concept Which Reactant below is limiting? How do you know? Chapter 3 19 Limiting Reactant Problems There are three steps to a limiting reactant problem: Step 1: Calculate the moles of product that can be produced from the first reactant. mass reactant #1 mol reactant #1 mol product Step 2: Calculate the moles of product that can be produced from the second reactant. mass reactant #2 mol reactant #2 mol product Step 3: The limiting reactant is the reactant that produces the least amount of product. Chapter 3 20 10

Limiting Reactants: Table Method One way to keep your limiting reactant problems organized is to use a table. You do the exact same calculations as shown before but you organize your answers in the table as shown below. This helps to see the limiting reactant and how much product is made. g initial MM n initial g final Li 2 O(s) + H 2 O(l) 2.50 g 29.88 g/mol 0.0837 mol 0 g 2.50 g 18.02 g/mol 0.139 mol 0.997 g 2 LiOH (s) 23.95 g/mol 0 mol n- / n+ - 0.0837-0.0837 + 0.0837 n final 0 mol 0.0553 mol 0.0837 mol Limiting Reactant 2.00 g Chapter 3 21 0 g Determining the Limiting Reactant Lithium oxide is a drying agent used on the space shuttle. If 80.0 g of water is to be removed and 65.0 g of lithium oxide is available, which reactant is limiting? Li 2 O(s) + H 2 O(l) LiOH (s) Chapter 3 22 11

Limiting Reactant Problem Cisplatin is an anti-cancer agent prepared as follows: K 2 PtCl 4 + NH 3 Pt(NH 3 ) 2 Cl 2 + KCl If 10.0 g of K 2 PtCl 4 and 10.0 g of NH 3 are allowed to react: (a) Which is the limiting reagent? (b) How many grams of the excess reagent are consumed? (c) How many grams of cisplatin are formed? Chapter 3 23 Solution Concentration The molar concentration, or molarity (M), is the number of moles of solute per liter of solution, is expressed as moles/liter. moles of solute liters of solution = M Chapter 3 24 12

Calculating Molarity What is the molarity of a solution containing 27.5 g of KOH in 0.100 L of solution? Step 1: We have 27.5 g of KOH so we need to convert to moles of KOH. Step 2: We want molarity of the solution (mol/l), so check your units. Step 3: Divide the moles of KOH by the volume of solution (0.100 L). moles of solute liters of solution = M Chapter 3 25 Molar Concentration Problem How many grams of Ca(OH) 2 would you use to prepare 500.0 ml of a 1.25 M calcium hydroxide solution? Step 1: We have ml of solution and molarity Ca(OH) 2 Step 2: We want grams of Ca(OH) 2 Step 3: Use the molarity formula to solve for moles of Ca(OH) 2. Step 4: Convert the moles of Ca(OH) 2 to grams of Ca(OH) 2 using its molar mass. Chapter 3 26 13

Dilution of a Solution Dilution is the process of reducing a solution s concentration by adding more solvent. Rather than prepare a dilute solution by dissolving a solid in water, we can prepare a solution by diluting a more concentrated solution. When performing a dilution, the amount of solute does not change, only the amount of solvent. M concentrated x V concentrated = M dilute x V dilute Chapter 3 27 Dilution Problem What volume of 18.0 M H 2 SO 4 is required to prepare 250.0 ml of 0.500 M aqueous H 2 SO 4? Step 1: Determine your unknown (V C ) Step 2: Identify M C, M D and V D Step 3: Plug these values into the dilution equation and solve for V C M C V C = M D V D (18.0 M) V C = (0.500 M) (250.0 ml) V C = (0.500 M) (250.0 ml) 18.0 M = 6.94 ml Chapter 3 28 14

Solution Stoichiometry A solution stoichiometry problem uses molarity as a conversion factor between volume and moles of a substance in solution. There are three steps: Step 1: Convert the given volume of solution to moles using the molarity (mol/l) as a unit factor. Step 2: Convert the moles of given to moles of the unknown using the coefficients in the balanced equation. Step 3: Convert the moles of unknown to molarity by dividing by the volume of the solution. Chapter 3 29 Solution Stoichiometry Problem Stomach acid, a dilute solution of HCl in water, can be neutralized by reaction with sodium hydrogen carbonate, NaHCO 3. How many milliliters of 0.125 M NaHCO 3 solution are needed to neutralize 18.0 ml of 0.100 M HCl? NaHCO 3 (aq) + HCl (aq) CO 2 (g) + H 2 O (l) + NaCl (aq) Chapter 3 30 15

Summary of Stoichiometry Problems Chapter 3 31 Acid-Base Titrations A titration is used to analyze an acid solution using a solution of a base. A measured volume of base is added to the acid solution. When all of the acid has been neutralized, the ph is 7. One extra drop of base solution after the endpoint increases the ph dramatically. Chapter 3 32 16

Acid Base Titration Problem What is the molarity of a sulfuric acid solution if a 25.0 ml sample is titrated to equivalence with 50.0 ml of 0.150 M potassium hydroxide solution? Step 1: Write the balanced neutralization reaction. H 2 SO 4 (aq) + KOH (aq) K 2 SO 4 (aq) + H 2 O (l) Step 2: We want concentration sulfuric acid (SA), we have concentration and volume of potassium hydroxide (KOH) so we need to convert from potassium hydroxide to sulfuric acid. Volume Mole Volume of KOH Ratio of SA Conc of KOH Moles of KOH Moles of SA Conc of SA Chapter 3 33 Redox Titrations As with acids and bases, a titration can be used to analyze the concentration of oxidizing and reducing agents. Chapter 3 34 17

Percent Composition A percent, %, expresses the amount of a single portion compared to an entire sample. portion of interest ("Part") % = 100% total sample ("Whole") The percent composition of a compound lists the mass percent of each element.0 % = Mass of element Mass of compound X 100% Chapter 3 35 Calculating Percent Composition Saccharin has the molecular formula C 7 H 5 NO 3 S. What is the percentage composition of saccharin? Step 1: Determine the molar mass of the compound (this will be your whole ) Step 3: Determine the mass of each element in the compound (you should have this from your molar mass calculation). These will be your parts. Step 3: Find the percent composition of the compound by dividing the mass of the part (the element) by the mass of the whole (saccharin) then multiplying by 100 Chapter 3 36 18

Empirical Formulas The empirical formula of a compound is the simplest whole number ratio of elements in a formula unit The molecular formula of a compound is some multiple of the empirical formula However, the molecular formula and empirical formula can be the same for a compound too! Compound Formula Empirical Formula Hydrogen peroxide H 2 O 2 Benzene C 6 H 6 Ethylene C 2 H 4 Propane C 3 H 8 Chapter 3 37 Determining Empirical Formula: Mass Percents A compound s empirical formula can be determined from its percent composition. Rhyme to Remember: Percent to Mass Mass to mole Divide by smallest Multiply til whole Chapter 3 38 19

Determining Empirical Formula: Component Mass Combustion analysis is one of the most common methods for determining empirical formulas. A weighed compound is burned in oxygen and its products analyzed by a gas chromatogram. It is particularly useful for analysis of hydrocarbons (Combustion Reactions). Chapter 3 39 Converting Decimals to Whole Numbers When calculating empirical formulas, you don t always get a nice whole number. Sometimes the result of dividing by the smallest number of moles gives a decimal instead. Decimal values that are close to a whole number can be rounded to that number: 2.04 becomes 2.00 and 6.98 becomes 7.00 However, a decimal that is greater than 0.1 or less than 0.9 has to be multiplied by a small integer: Chapter 3 40 20

Determining Empirical Formulas A compound was analyzed to be 82.67% carbon and 17.33% hydrogen by mass. What is the empirical formula of the compound? Chapter 3 41 Determining Empirical Formulas Combustion analysis of 45.62 mg of toluene gives 35.67 mg of H 2 O and 152.5 mg of CO 2. What is the empirical formula of toluene? Chapter 3 42 21

Determining Empirical Formulas Menthol, a flavoring agent obtained from peppermint oil, contains carbon, hydrogen and oxygen. Combustion analysis of 1.00 g of menthol yields 1.161 g of H 2 O and 2.818 g of CO 2. What is the empirical formula of menthol? Chapter 3 43 Molecular Formulas The empirical formula for benzene is CH. This represents the ratio of C to H atoms in a benzene molecule. The actual molecular formula is some multiple (f) of the empirical formula, (CH) f. To determine f, we divide the mass of the molecular formula by the mass of the empirical formula: Mass of Molecular Formula Mass of Empirical Formula = f Chapter 3 44 22

Finding Molecular Formulas There are 4 steps to determining the molecular formula of a compound: Step 1: Determine the empirical formula of the compound. May need to calculate yourself or may be given in the problem Step 2: Calculate the mass of the empirical formula. Step 3: Divide the mass of the molecular formula (usually given in the problem!) by the mass of the empirical formula to determine the multiplier factor (f). Step 4: Multiply all the subscripts in the empirical formula by the factor f to get the molecular formula. Chapter 3 45 Molecular Formulas A compound was analyzed to be 82.67% carbon and 17.33% hydrogen by mass. An osmotic pressure experiment determined that its molar mass is 58.11 g/mol. What is the molecular formula of the compound? Step 1: Determine the empirical formula. We did so in earlier problem: C 2 H 5 Step 2: Determine the mass of the empirical formula C 2 H 5 : 29.06 g/mol Step 3: Divide the mass of the molecular formula (given above: 58.11 g/mol) by the mass of the empirical formula to determine f Step 4: Multiply all the subscripts in the empirical formula by f (C H ) = 58.11 g/mol 2 5 f C 2 H 5 29.06 g/mol f = 2 and the molecular formula is C 4 H 10 Chapter 3 46 23