Chem 103, Section F0FUnit IV - Stoichiometry of Formulas and EquationsLecture 11•The concept of a mole, which is a very large group of atoms or molecules•Determining the formulas for a compoundLecture 11 - StoichiometryReading in Silberberg•Chapter 3, Section 1 The Mole•Chapter 3, Section 2 Determining the Formula of an Unknown Compound2Lecture 11 - IntroductionStoichiometry is the study of the quantitative aspects of chemical formulas and chemical reactions.•Using the tools of stoichiometry, you can predict the quantities of reactants and products that can be consumed or produced in a chemical reaction.•These calculations will require working with chemical formulas and balanced chemical reactions.3chemicalformulafor methaneCH4+2 O2CO2+2 H2OCH4+2 O2CO2+2 H2Obalanced chemicalequation for thecombustion of methaneLecture 11 - The MoleWe usually quantify objects either by counting them or weighing them.•For liquids we also measure volumes.4Lecture 11 - The MoleIn chemistry we often need to deal with numbers of molecules•For example, when working with balanced chemical equations.-For the combustion of methane!1 molecule of methane reacts with!2 molecules of oxygen to produce!1 molecule of carbon dioxide and!2 molecules of water.5CH4+2 O2CO2+2 H2Obalanced chemicalequation for thecombustion of methaneLecture 11 - The MoleThe problem is, molecules are too small to count in the lab.The concept of the mole allows us to count molecules by weighing them.6Lecture 11 - The MoleBecause any pure substance contains either identical atoms, molecules (covalent compounds) or formula units (ionic compounds) (see Dalton’s postulates)•a given mass of a pure substance will always contain the same number of either identical atoms, molecules (covalent compounds) or formula units (ionic compounds).•For example, 10 g of methane contains 3.754 x 1023 methane molecules.-How did I know this?7Lecture 11 - The MoleA mole (mol) is defined as•The amount of a substance that contains the same number of entities as there are atoms in exactly 12 g of carbon-12.•This number is called Avagodro’s Number and is equal to 6.022 x 1023For example:•1 mol of carbon-12 contains 6.022 x 1023 carbon-12 atoms•1 mol of H2O contains 6.022 x 1023 water molecules.•1 mol of NaCl 6.022 x 1023 NaCl formula units.8Alert!!! Alert!!!Tomorrow is Mole DayAlert!!! Alert!!!Tomorrow is Mole DayLecture 11 - The MoleA mole is more than just a number; in chemistry it has a special meaning:•1 mol of atoms for an element has a mass in grams that is numerically equivalent to the average mass of an atom of the element in amu’s (atomic mass units).-1 atom of C has an average mass of 12.01 amu!1 mol of C atoms has a mass of 12.01 g-1 atom of H has an average mass of 1.008 amu!1 mol of H atoms has a mass of 1.008 g9Lecture 11 - The MoleThe mass of 1 mol of a substance made up of molecules or formula units can be calculating using the chemical formula to determine the numbers for each type of atom in a molecule of a substance:•For example, 1 mol of methane (CH4) has a mass of101 ! 12.01 g of carbon (C)+ 4 ! 1.008 g of hydrogen (H)= 16.042 g of CH4Lecture 11 - The MoleMolar mass is defined as•the mass of mol of atoms, molecules or formula units of a substance.-The units are g/mol•For example, the molar mass of methane 16.042 g/mol11Lecture 11 - The Mole12Lecture 11 - Clicker Question 1Calculate the molar mass of BF3The molar mass of BF3 isA) 57.81 g/molB) 48.21 g/molC) 67.81 g/molD) 29.81 g/mol13Lecture 11 - The MoleCalculating molar masses•Elements-Determine whether the element exists as either individual atoms (metals and noble gases), or as molecules (nonmetals).-For individual atoms, the molar mass is numerically equal to the molecular mass of an atom in amu-For molecules multiply the molecular mass of one atom by the number of atoms in a molecule).14Lecture 11 - The MoleCalculating molar masses•Compounds-The molar mass is the sum of themolar mases of the atoms of the elements in the chemical formula. -For example, glucose (C6H12O6):15Lecture 11 - The MoleThe molar mass, along with Avogadro’s number and the chemical formula, allows us to interconvert between•the mass (g)•number of mols of a substance•number of atoms, molecules or formula units.16Lecture 11 - The Mole•For elements:17Lecture 11 - The Mole•For compounds:18Lecture 11 - The MoleMass Percent from the Chemical Formula•The molecular or formula unit for a compound can be used to calculate the mass percent of an element in a quantity of a substance.19Lecture 11 - Question 2Calculate the Mass % of sodium in sodium carbonate.20Lecture 11 -Formula of an Unknown CompoundEmpirical Formulas•There ways that a chemist can decompose a compound and determine experimentally (empirically) the mass percent of its constituent elements•From these data, an empirical formula for a substance can be determined.•The empirical formula gives the whole number ratio of the elements in a substance.-For an ionic compound, the formula unit and the empirical formula are the same thing.-For a covalent compound, the molecular formula is either equal to the empirical formula, or is an integer multiple of the empirical formula.21Lecture 11 -Formula of an Unknown CompoundThe empirical formula is found by•First dividing the the moles for each by the element with the smallest number of moles,•multiplying these ratios by increasing integers: 1, 2, 3, ... until a near integer is obtained for all of the elements.These integers are the subscripts for each element in the empirical formula.•This is probably best demonstrated by example.22Lecture 11 - Question 3Find the empirical formula of the following compound:9.903 g of phosphorus (P) combines with 6.99 g of bromine (Br).23Lecture 11 -Formula of an Unknown CompoundMolecular Formulas•The molar mass of a substance can be combined with the empirical formula to determine the molecular formula or formula unit for a substance.-Divide the molar mass by empirical molar mass!This should result in an integer-Multiply the subscripts in the empirical formula by this integer to get the molecular formula.As an example, the empirical and molecular formulas can be determined for hydrocarbons by doing a combustion analysis.24Lecture 11 -Formula of an Unknown CompoundFor example•The empirical and molecular formulas can be determined for hydrocarbons by doing a combustion analysis.25Lecture 11 - Question 4A