1411 Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations

Stoichiometry (stoy-key-OM-uh-tree) stoichiometry - the area of study that examines the quantities of substances consumed and produced in chemical reactions Built on an understanding of atomic masses, chemical formulas, and the law of conservation of mass

“produces” or “yields” Chemical Equations “produces” or “yields” “reacts with” “and” Chemical equations are concise representations of chemical reactions.

Anatomy of a Chemical Equation CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) Reactants appear on the left side of the equation. Products appear on the right side of the equation.

Anatomy of a Chemical Equation CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) The states of the reactants and products are written in parentheses to the right of each compound. (g) = gas (l) = liquid (s) = solid (aq) = aqueuos (water) solution Sometimes, reaction conditions such as heat, symbolized Δ, are written above or below the arrow.

Anatomy of a Chemical Equation CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) Coefficients are inserted to balance the equation. balanced equation – must have equal # atoms on each side of the arrow (The coefficient 1 is left off; no coefficient = understood 1)

Subscripts and Coefficients Give Different Information Subscripts tell the number of atoms of each element in one molecule.

Subscripts and Coefficients Give Different Information Subscripts tell the number of atoms of each element in one molecule. Coefficients tell the number of molecules. To find the # of each type of atom, multiply the coefficient by each subscript.

Examples How many of each type of atom are there in the following formulas? 3 Fe2O3 4 Na2CO3 Ba(ClO3)2 3 (NH4)3PO4

Balancing Chemical Equations Steps to follow Determine the formulas of reactants and products Write an unbalanced equation Insert the coefficients that will provide equal numbers of atoms on each side of the arrow

Guidelines: Balancing Chemical Equations Guidelines for balancing equations: NEVER change or add subscripts (changes identity of chemical) Adding a coefficient only changes the AMOUNT of a substance, rather than its identity 2 H2O = 2 molecules of water 3 H2O = 3 molecules of water Balance only one element at a time. Balance elements other than H and O first Use the smallest whole-number coefficients Ex: Zn + 2 HCl  ZnCl2 + H2 good 2 Zn + 4 HCl  2 ZnCl2 + 2 H2 not good Balance as a group those polyatomic ions that appear unchanged on both sides of arrow CaO + HNO3  Ca(NO3)2 + H2O (unbalanced)

Balancing Chemical Equations: Question… Which of the following would be the correct way to balance the oxygens in a chemical equation? adding a subscript 2 to the end of CO to give CO2 or adding a coefficient in front of the formula to give 2 CO

Balancing Chemical Equations Fe3O4(s) + H2(g) Fe(s) + H2O(l) (unbalanced) Work on one element at a time. Don’t change the subscripts – only add coefficients. F: Fe3O4(s) + H2(g) 3Fe(s) + H2O(l) O: Fe3O4(s) + H2(g) 3Fe(s) + 4H2O(l) H: Fe3O4(s) + 4H2(g) 3Fe(s) + 4H2O(l) Check that all atoms are balanced, and they are!

Examples Balance the following equations: AgNO3 + CaCl2  Ca(NO3)2 + AgCl P + O2  P4O10 C9H20 + O2  CO2 + H2O

Reaction Types: Combination Reactions Flares and some fireworks: In this type of reaction two or more substances (often 2 elements) react to form one product A + B  C If reactants are metal and nonmetal, product is ionic solid. Product is determined by predicting charges of ions. Examples: 2 Al (s) + 3 Br2 (g)  2 AlBr3 (s) N2 (g) + 3 H2 (g)  2 NH3 (g) C3H6 (g) + Br2 (l)  C3H6Br2 (l)

Reaction Types: Decomposition Reactions In a decomposition one substance breaks down into two or more substances. C  A + B Often involves heat. Ex: Many metal carbonates are heated to give metal oxides and CO2 Examples: CaCO3 (s)  CaO (s) + CO2 (g) (CaO = lime) 2 KClO3 (s)  2 KCl (s) + 3 O2 (g) 2 NaN3 (s)  2 Na (s) + 3 N2 (g) (air bags)

Reaction Types: Combustion Reactions These are generally rapid reactions that produce a flame. Most often involve hydrocarbons reacting with O2(g) in the air hydrocarbon: compound with either C and H only C, H, and O If these completely combust, products are always CO2 and H2O Examples: CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g) C6H12O6 (g) + 6 O2 (g)  6 CO2 (g) + 6 H2O (g)

Examples: Reaction Types Name the type of reaction, predict the product(s), and balance the equation: CH3OH + O2 Na + S BaCO3 Al(s) + Cl2(g) When C8H8 burns in air Δ

Formula Weight (FW) Formula weight - sum of the atomic weights for the atoms in a chemical formula generally reported for ionic compounds The FW of an ionic compound is the mass of one formula unit (the empirical formula of an ionic compound) Ex: FW of one formula unit of calcium chloride, CaCl2 Ca: 1(40.078 amu) + Cl: 2(35.453 amu) 110.98 amu

Molecular Weight (MW) A molecular weight is the sum of the atomic weights of the atoms in a molecule another name for formula weight, but it is usually only called MW for molecules For the molecule ethane, C2H6, the molecular weight would be C: 2(12.0107 amu) + H: 6(1.00794 amu) 30.0690 amu

Example Determine the formula weight (FW) or molecular weight (MW) for the following: a) C5H6O b) Al2(CO3)3

Percent Composition To find the percentage of the mass of a compound that comes from each of the elements in the compound: % element = (number of atoms)(atomic weight) (FW of the compound) x 100

Percent Composition So the percentage of carbon in ethane (C2H6) is… (2)(12.0107 amu) (30.0690 amu) 24.0214 amu 30.0690 amu = x 100 = 79.9%

Example: Percent Composition What is the percent composition of hydrogen in NH4H2PO4? 5.26%

Mole Even tiny samples in the lab contain MANY molecules, atoms, or ions Ex: 1 teaspoon of water = 2 x 1023 water molecules! So chemists devised a counting unit to relate numbers of molecules, atoms, etc to lab-sized amounts…. THE MOLE Examples of other counting units (dozen = 12 objects, gross = 144 objects) mole (abbr. mol) – the amount of matter that contains as many objects as there are atoms in 12 g of 12C number of atoms = 6.02 x 1023 = “Avogadro’s number” = NA objects = atom, molecules, ions, or whatever object you’re counting… people, apples, etc

6.02 x 1023 particles = 1 mol particles Avogadro’s Number One mole of particles contains Avogadro’s number of those particles “particles” are whatever you’re looking for (or are given) the number of. usually atoms, ions, or molecules 6.02 x 1023 particles = 1 mol particles

Avogadro’s Number: A Conversion Factor! Avogadro’s number allows us to interconvert: Examples: 1 mol Ag atoms = 6.02 x 1023 Ag atoms or 1 mol H2O molecules = 6.02 x 1023 H2O molecules 1 mol NO3- ions = 6.02 x 1023 NO3- ions 6.02 x 1023 particles = 1 mol particles 6.02 x 1023 particles 1 mole of those same particles 1 mole of those same particles 6.02 x 1023 particles number of particles moles of those same particles 6.02 x 1023 Ag atoms 1 mole Ag atoms 1 mole Ag atoms 6.02 x 1023 Ag atoms (the two conversion factors) 6.02 x 1023 H2O molecules 1 mole H2O molecules 1 mole H2O molecules 6.02 x 1023 H2O molecules (the two conversion factors) 6.02 x 1023 NO3- ions 1 mole NO3- ions 1 mole NO3- ions 6.02 x 1023 NO3- ions (the two conversion factors)

Avogadro’s Number: Examples How many Cu atoms are in 0.50 mole of Cu? How many moles of CO2 are in 2.50 x 1024 molecules CO2? 3.0 x 1023 Cu atoms 4.15 moles of CO2

Subscripts: Molecular Level and Molar Level The subscripts in a formula show 2 things: the number of each type of atom in one molecule (molecular level) the moles of each element in 1 mole of compound (molar level) Glucose C6H12O6 In 1 molecule of C6H12O6 : 6 atoms C 12 atoms H 6 atoms O In 1 mole of C6H12O6 : 6 moles C 12 moles H 6 moles O

Subscripts: Another Conversion Factor! The subscripts allow us to interconvert: In 1 mole of C6H12O6 : 6 moles C 12 moles H 6 moles O We can write 3 conversion factors from the subscripts of C6H12O6 : Or we can flip them upside-down: moles of atoms or ions moles of the compound the atoms or ions are contained in 6 mol C 1 mol C6H12O6 12 mol H 1 mol C6H12O6 6 mol O 1 mol C6H12O6 1 mol C6H12O6 6 mol C 1 mol C6H12O6 12 mol H 1 mol C6H12O6 6 mol O

Subscripts: Example How many moles of O are in 0.150 mole of aspirin, C9H8O4? 0.600 mole of O

Example: Avogadro’s Number + Subscripts Calculate the number of H atoms in 2.00 mol H2O. Our given is in units of moles We want the number of particles (H atoms in this case) The only way to relate number of particles to moles of those same particles is by using Avogadro’s number. Remember: 6.02 x 1023 particles = 1 mole of those same particles So for this problem, we have: 6.02 x 1023 H atoms = 1 mol H atoms We can also relate the number of moles of H2O to number of moles of H atoms (by using the subscripts) - So, 1 mol H2O = 2 mol H atoms 2 mol H atoms 1 mol H2O 6.02 x 1023 H atoms 1 mol H atoms 2.00 mol H2O x x = 2.41 x 1024 H atoms

Examples: Avogadro’s Number + Subscripts Calculate the number of O atoms in 2.10 mol Al2(CO3)3. Calculate the number of PO43- ions in 3.0 mol Mg3(PO4)2. How many moles of Mg3(PO4)2 are in 1.450 x 1020 formula units of Mg3(PO4)2? 1.14 x 1025 O atoms 3.6 x 1024 PO43- ions 2.409 x 10-4 formula units

Molar Mass A mole is always the same number (6.02 x 1023), like a dozen is always 12, but 1-mole samples of different substances have different masses. The molar mass is the mass of 1 mole of an element or compound (unit g/mol) Equals the atomic weight (for an element) that we find on the periodic table molar mass of Li is 6.941 g/mol Or we can write, 6.941 g Li = 1 mole of Li the formula weight (for a compound) molar mass of H2O is 18.015 g/mol Or we can write, 18.015 g H2O = 1 mole of H2O The formula weight (in amu) is same number as the molar mass (in g/mol), but different units

Molar Mass: Another Conversion Factor! Molar mass allows us to interconvert: 6.941 g Li = 1 mole of Li 18.015 g H2O = 1 mole of H2O mass of a substance moles of that same substance 6.941 g Li 1 mol Li 1 mol Li 6.941 g Li 18.015 g H2O 1 mol H2O 1 mol H2O 18.015 g H2O

Molar Mass: Example How many moles of H2SO4 are in 0.2380 g of H2SO4? What is the mass in grams of 2.13 x 10-2 moles of CO2? 2.427 x 10-3 mol H2SO4 0.937 g CO2

Using Moles Moles provide a bridge from the molecular scale to the real-world scale

Avogadro’s Number + Subscripts + Molar Mass Examples: Avogadro’s Number + Subscripts + Molar Mass How many H atoms are in 3.01 x 10-2 mg of H2O? What is the mass in grams of Na+ ions in a sample of Na2O containing 1.02 x1020 Na2O formula units? 2.01 x 1018 H atoms 7.79 x 10-3 g Na+

Calculating Empirical Formulas (Review): empirical formula tells relative number of atoms of each element One can calculate the empirical formula from the percent composition. The ratio of the number of moles of each element in a compound gives the subscripts in a compound’s empirical formula. “Percent to mass, mass to mol, divide by small, multiply ‘til whole!”

Calculating Empirical Formulas The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA.

Calculating Empirical Formulas Assume 100.00 g of para-aminobenzoic acid, find mass of each element: C: 61.31 % = 61.31 g C H: 5.14 % = 5.14 g H N: 10.21 % = 10.21 g N O: 23.33 % = 23.33 g O

Calculating Empirical Formulas From mass, calculate the number of moles of each element: C: 61.31 g x = 5.105 mol C H: 5.14 g x = 5.09 mol H N: 10.21 g x = 0.7288 mol N O: 23.33 g x = 1.456 mol O 1 mol 12.01 g 14.01 g 1.01 g 16.00 g

Calculating Empirical Formulas Calculate the mole ratio by dividing by the smallest number of moles: C: = 7.005  7 H: = 6.984  7 N: = 1.000 O: = 2.001  2 5.105 mol 0.7288 mol 5.09 mol 1.458 mol Empirical formula: C7H7NO2

Molecular Formula from Empirical Formula If we are given molar mass of the compound, we can determine the molecular formula The subscripts in the molecular formula of a substance are always a whole-number multiple of the corresponding subscripts in the empirical formula: whole number multiple = molar mass of actual compound molar mass of empirical formula Example: The empirical formula of a compound is found to be NaS2O3. The molar mass of the compound is 270.4 g/mol. Determine the molecular formula. molar mass of empirical formula = 135.1 g/mol 270.4 g/mol 135.1 g/mol The molecular formula is Na2S4O6. Multiply each subscript in empirical formula by 2 to get subscripts of molecular formula = 2.001

Example: Empirical and Molecular Formulas Ibuprofen, a headache remedy, contains 75.69% C, 8.80% H, and 15.51% O by mass. Its molar mass is 206 g/mol. Determine its empirical and molecular formulas.

Example: Empirical and Molecular Formulas Ibuprofen, a headache remedy, contains 75.69% C, 8.80% H, and 15.51% O by mass. Its molar mass is 206 g/mol. Determine its empirical and molecular formulas. Empirical formula = C13H18O2 Molecular formula = C13H18O2

Combustion Analysis “Empirical” means based on observation and experiment So chemists have techniques to determine empirical formula from experiments Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this. mass of C - determined from the mass of CO2 produced mass of H - determined from the mass of H2O produced mass of O - determined by difference after the masses of C and H have been determined mass of O = mass of sample – (mass of C + mass of H) CxHyOz + O2  CO2 + H2O (not balanced)

Combustion Analysis Problem Isopropyl alcohol is composed of C, H, and O. Combustion of 0.255 g isopropyl alcohol produces 0.561 g of CO2 and 0.306 g of H2O. Determine the empirical formula of isopropyl alcohol. Plan: To determine the empirical formula of isopropyl alcohol, CxHyOz, we need to determine number of moles of C, H, and O. CxHyOz + O2  CO2 + H2O (not balanced) We will first use mole concept to find mass of C and mass of H. mass of C present in CO2 = mass of C present in isopropyl alcohol mass of H present in H2O = mass of H present in isopropyl alcohol mass of O = mass of isopropyl alcohol – (mass of C + mass of H) Finally calculate moles of C, H, and O from masses of C, H, and O to give subscripts for empirical formula

Combustion Analysis Problem Isopropyl alcohol is composed of C, H, and O. Combustion of 0.255 g isopropyl alcohol produces 0.561 g of CO2 and 0.306 g of H2O. Determine the empirical formula of isopropyl alcohol. We will first use mole concept to find mass of C and mass of H. mass of C: 0.561 g CO2 x mass of C present in CO2 = 0.153 g C = mass of C present in isopropyl alcohol mass of H: 0.306 g H2O x mass of H present in H2O = 0.0343 g H = mass of H present in isopropyl alcohol mass of O = mass of isopropyl alcohol – (mass of C + mass of H) mass of O = 0.255 g – (0.153 g + 0.0343 g) = 0.068 g O 1 mol CO2 44.0 g CO2 1 mol C 1 mol CO2 12.0 g C 1 mol C x x = 0.153 g C 1 mol H2O 18.0 g H2O 2 mol H 1 mol H2O 1.01 g H 1 mol H x x = 0.0343 g H

Combustion Analysis Problem Isopropyl alcohol is composed of C, H, and O. Combustion of 0.255 g isopropyl alcohol produces 0.561 g of CO2 and 0.306 g of H2O. Determine the empirical formula of isopropyl alcohol. Finally calculate moles of C, H, and O from masses of C, H, and O to give subscripts for empirical formula moles C = 0.153 g C x = 0.0128 mol C moles H = 0.0343 g H x = 0.03430 mol H moles O = 0.068 g O x = 0.0043 mol O CxHyOz  C3H8O 1 mol C 12.0 g C = 2.98 mol C  3 0.0043 1 mol H 1.01 g H = 7.98 mol H  8 0.0043 1 mol O 16.0 g O = 1 mol O 0.0043

Stoichiometric Calculations The coefficients in the balanced equation give the ratio of moles of reactants and products. This mole-to-mole ratio will serve as a conversion factor in calculations. Ex: For above chemical equation, some mole-to-mole ratios are: 2 mol H2 1 mol O2 2 mol H2O 2 moles H2 2 moles H2O 2 moles H2O 2 moles H2 2 moles H2 1 mole O2 1 mole O2 2 moles H2O

Stoichiometric Calculations Coefficients: Another Conversion Factor! Just like subscripts, coefficients in a chemical equation also apply to both the molecular and molar levels The coefficients in the balanced equation give the ratio of moles of reactants and products. This mole-to-mole ratio allows us to interconvert: Ex: For above chemical equation, some conversion factors are: 2 mol H2 1 mol O2 2 mol H2O moles of one substance in a chemical equation moles of another substance in the chemical equation 2 moles H2 2 moles H2O 2 moles H2O 2 moles H2 2 moles H2 1 mole O2 1 mole O2 2 moles H2O

Stoichiometric Calculations Examples: How many moles of O2 react with 3.02 moles of H2? How many moles of H2O are produced from 2.4 moles of O2? 1 mol O2 2 mol H2 3.02 mol H2 x = 1.51 mol O2 2 mol H2O 1 mol O2 2.4 mol O2 x = 4.8 mol O2

Stoichiometric Calculations: Summary This chapter’s 4 new conversion factors, always relating moles: Avogadro’s number Subscripts Molar Mass Coefficients number of particles moles of those same particles moles of atoms or ions moles of the compound the atoms or ions are contained in mass of a substance moles of that same substance moles of one substance in a chemical equation moles of another substance in the chemical equation

Stoichiometric Calculations Starting with the mass of substance A you can use the mole-to-mole ratio of the coefficients of A and B to calculate the mass of substance B formed (if it’s a product) or used (if it’s a reactant).

Stoichiometric Calculations Calculate the mass in grams of water produced when starting when 1.00 g of C6H12O6 combusts. C6H12O6 + 6 O2  6 CO2 + 6 H2O = Starting with 1.00 g of C6H12O6… we calculate the moles of C6H12O6… use the coefficients to find the moles of H2O… and then turn the moles of water to grams.

Limiting Reactants: How Many Cookies Can I Make? You can make cookies until you run out of one of the ingredients. Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat).

Limiting Reactants: How Many Cookies Can I Make? In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make.

Limiting Reactant Bread and cheese sandwich example 2 Bd + Ch  Bd2Ch - We have 10 slices of bread and 7 slices of cheese The limiting reactant is the reactant present in the smallest stoichiometric amount; the reactant that produces the least amount (fewest moles) of product!!! In other words, it’s the reactant you’ll run out of first Here, it’s the bread because 10 slices of bread makes 5 sandwiches 7 slices of cheese makes 7 sandwiches The bread limits the number of sandwiches! There will be cheese left over, so cheese would be the excess reactant - the one that produces more product (more moles) of product!!!

Limiting Reactant 2 H2 + O2  2 H2O Suppose we have 10 moles H2 and 7 moles O2. To determine which reactants are the limiting reactant and excess reactant, we calculate moles of product from the mole-to-mole ratio for both reactants and see which is less: We can determine how much of the excess reactant was actually used up and how much we have left over: 2 mol H2O 2 mol H2 10 mol H2 x = 10 mol H2O limiting reactant = H2 2 mol H2O 1 mol O2 7 mol O2 x = 14 mol H2O excess reactant = O2 1 mol O2 2 mol H2 10 mol H2 x = 5 mol O2 actually reacted We will have 2 mol O2 leftover: 7 mol O2 (started with) – 5 mol O2 (actually reacted)

Theoretical Yield The theoretical yield is the maximum amount of product that can be made. In other words it’s the amount of product possible as calculated through the stoichiometry problem. This is different from the actual yield, which is the amount one actually produces and measures.

Percent Yield Actual Yield Theoretical Yield Percent Yield = x 100 One finds the percent yield by comparing the amount actually obtained (actual yield) to the amount it was possible to make (theoretical yield). Actual Yield Theoretical Yield Percent Yield = x 100

Limiting Reactant: Example Adipic acid, H2C6H8O4, is used to produce nylon. The acid is made commercially by a controlled reaction between cyclohexane (C6H12) and O2: 2 C6H12(l) + 5 O2(g)  2 H2C6H8O4(l) + 2 H2O(g) The reaction is carried out with 25.0 g of cyclohexane and 35.0 g of O2. After the reaction, 33.5 g of adipic acid are formed. What is the limiting reactant? How much of the excess reactant remains after the reaction? What is the theoretical yield of adipic acid? What is the actual yield of adipic acid? What is the percent yield of adipic acid?

Limiting Reactant: Example Adipic acid, H2C6H8O4, is used to produce nylon. The acid is made commercially by a controlled reaction between cyclohexane (C6H12) and O2: 2 C6H12(l) + 5 O2(g)  2 H2C6H8O4(l) + 2 H2O(g) The reaction is carried out with 25.0 g of cyclohexane and 35.0 g of O2. After the reaction, 33.5 g of adipic acid are formed. What is the limiting reactant? How much of the excess reactant remains after the reaction? What is the theoretical yield of adipic acid? What is the actual yield of adipic acid? What is the percent yield of adipic acid? cyclohexane 10.5 g 43.4 g 33.5 g 77.2 %