Lecture #7 - (a) The Mole Concept, (b) Formula of an Unknown Chemistry 142 B James B. Callis, Instructor Autumn Quarter, 2004 Figure 3.1: Mass spectrometer Figure 3.2 (a): Peaks of neon injected Figure 3.2 (b): Bar graph of neon injected Problem 7-1: Isotopic Composition The two isotopes of potassium with significant abundance in nature are 39K (isotopic mass 38.9637 amu, 93.258%) and 41K (isotopic mass 40.9618 amu, 6.730%). Fluorine has only one naturally occurring isotope, 19F (isotopic mass 18.9984 amu). Use this information to calculate the formula mass of potassium fluoride. Solution: MOLE Definition: The amount of substance that contains as many elementary particles (atoms, molecules, ions, or other ?) as there are atoms in exactly 12 grams of carbon 12. 1 Mole = 6.022145 x 1023 particles (atoms, molecules, ions, electrons, or…) = NA particles NA is Avogado’s Number. (~100 million x 100 million x 100 million) The Mole is a Chemical Concept It represents a fixed number of chemical entities A mole of a chemical entity has a fixed, unique mass. (Molar Mass) Thus, the mole allows the mass balance to count chemical entities. Counting objects of fixed relative mass 12 red marbles @ 7g each = 84g 12 yellow marbles @4e each=48g 55.85g Fe = 6.022 x 1023 atoms Fe 32.07g S = 6.022 x 1023 atoms S Mole - Mass Relationships of Elements Element Atomic Mass Molar Mass Number of Atoms 1 atom of H = 1.008 amu 1 mole of H = 1.008 g = 6.022 x 1023 atoms 1 atom of Fe = 55.85 amu 1 mole of Fe = 55.85 g = 6.022 x 1023atoms 1 atom of S = amu 1 mole of S = g = atoms 1 atom of O = amu 1 mole of O = g = atoms Molecular mass: 1 molecule of O2 = amu 1 mole of O2 = g = molecules 1 molecule of S8 = amu 1 mole of S8 = g = molecules Mole - Mass Relationships of Elements Element Atomic Mass Molar Mass Number of Atoms 1 atom of H = 1.008 amu 1 mole of H = 1.008 g = 6.022 x 1023 atoms 1 atom of Fe = 55.85 amu 1 mole of Fe = 55.85 g = 6.022 x 1023 atoms 1 atom of S = 32.07 amu 1 mole of S = 32.07 g = 6.022 x 1023 atoms 1 atom of O = 16.00 amu 1 mole of O = 16.00 g = 6.022 x 1023 atoms Molecular mass: 1 molecule of O2 = 16.00 x 2 = 32.00 amu 1 mole of O2 = 32.00 g = 6.022 x 1023 molecule 1 molecule of S8 = 32.07 x 8 = 256.56 amu 1 mole of S8 = 256.56 g = 6.022 x 1023 molecules Molecular Mass - Molar Mass ( M ) The Molecular mass of a compound expressed in amu is numerically the same as the mass of one mole of the compound expressed in grams, called its molar mass. For water: H2O Molecular mass = (2 x atomic mass of H ) + atomic mass of O = 2 ( amu) + amu = amu Mass of one molecule of water = amu Molar mass = ( 2 x molar mass of H ) + (1 x molar mass of O) = 2 ( g ) + g = g g H2O = 6.022 x 1023 molecules of water = 1 mole H2O The Molecular mass of a compound expressed in amu is numerically the same as the mass of one mole of the compound expressed in grams , called its molar mass. For water: H2O Molecular mass = (2 x atomic mass of H ) + atomic mass of O = 2 ( 1.008 amu) + 16.00 amu = 18.02 amu Mass of one molecules of water = 18.02 amu Molar mass = ( 2 x molar mass of H ) + (1 x molar mass of O) = 2 ( 1.008 g ) + 16.00 g = 18.02 g 18.02 g H2O = 6.022 x 1023 molecules of water = 1 mole H2O Molecular Mass - Molar Mass ( M ) One mole of common substances CaCO3 100.09 g Oxygen, O2 32.00 g Copper 63.55 g Water 18.02 g Calculating the Number of Moles and Atoms in a Given Mass of Element Problem 7-2: Tungsten (W) is the element used as the filament in light bulbs, and has the highest melting point of any element 3680oC. How many moles of tungsten, and atoms of the element are contained in a 35.0 mg sample of the metal? Plan: Convert mass into moles by dividing the mass by the atomic mass of the metal, then calculate the number of atoms by multiplying by Avogadro’s number. Solution: Converting from mass of W to moles: Moles of W = No. of W atoms = Calculating the Moles and Number of Formula Units in a given Mass of Compound Problem 7-3: Trisodium phosphate is a component of some detergents. How many moles and formula units are in a 38.6 g sample? Plan: We need to determine the formula, and the molecular mass from the atomic masses of each element multiplied by the coefficients. Solution: The formula is Na3PO4. Calculating the molar mass: MM = Converting mass to moles: # Formula units = Flow Chart of Mass Percentage Calculation Moles of X in one mole of Compound Mass % of X Mass fraction of X Mass (g) of X in one mole of compound Multiply by M (g / mol of X) Divide by mass (g) of one mole of compound Multiply by 100 % Calculating Mass Percentage and Masses of Elements in a Sample of a Compound - I Problem 7-4: Sucrose (C12H22O11) is common table sugar. ( a) What is the mass percent of each element in sucrose? ( b) How many grams of carbon are in 24.35 g of sucrose? (a) Determining the mass percent of each element: mass of C per mole sucrose = mass of H / mol = mass of O / mol = total mass per mole = Finding the mass fraction of C in Sucrose & % C : mass of C per mole mass of 1 mole sucrose = To find mass % of C = Mass Fraction of C = = Calculating Mass Percents and Masses of Elements in a Sample of Compound - II 7-4 (a) continued Mass % of H = x 100% = Mass % of O = x 100% = 7-4 (b) Determining the mass of carbon: Mass (g) of C = mass of sucrose x ( mass fraction of C in sucrose) Mass (g) of C = mol H x M of H mass of 1 mol sucrose mol O x M of O mass of 1 mol sucrose Empirical and Molecular Formulas Empirical Formula - The simplest formula for a compound that agrees with the elemental analysis. The smallest set of whole numbers of atoms. Molecular Formula - The formula of the compound as it exists. It may be a multiple of the Empirical formula. Some Examples of Compounds with the same Elemental Ratios Empirical Formula Molecular Formula CH2(unsaturated Hydrocarbons) C2H4 , C3H6 , C4H8 OH or HO H2O2 S S8 P P4 Cl Cl2 CH2O (carbohydrates) C6H12O6 Steps to Determine Empirical Formulas Mass (g) of Element Moles of Element Preliminary Formula Empirical Formula ÷ M (g/mol ) Use no. of moles as subscripts. Change to integer subscripts: ÷ smallest, conv. to whole #. Determining Empirical Formulas from Masses of Elements - I Problem 7-5: The elemental analysis of a sample compound gave the following results: 5.677g Na, 6.420 g Cr, and 7.902 g O. What is the empirical formula and name of the compound? Plan: First we have to convert mass of the elements to moles of the elements using the molar masses. Then we construct a preliminary formula and name of the compound. Solution: Finding the moles of the elements: Moles of Na = Moles of Cr = Moles of O = Determining Empirical Formulas from Masses of Elements - II Constructing the preliminary formula: Converting to integer subscripts (dividing all by smallest subscript): Rounding off to whole numbers: CnHm + (n+ )O2(g) n CO2(g) + H2O(g) m 2 m 2 Determining a Chemical Formula from Combustion Analysis - I Problem 7-6: Erythrose (M = 120 g/mol) is an important chemical compound used often as a starting material in chemical synthesis, and contains Carbon, Hydrogen, and Oxygen. Combustion analysis of a 700.0 mg sample yielded: 1.027 g CO2 and 0.4194 g H2O. From this data calculate the molecular formula. Plan: We find the masses of Hydrogen and Carbon using the mass fractions of H in H2O, and C in CO2. The mass of Carbon and Hydrogen are subtracted from the sample mass to get the mass of Oxygen. We then calculate moles, and construct the empirical formula, and from the given molar mass we can calculate the molecular formula. Determining a Chemical Formula from Combustion Analysis - II Calculating the mass fractions of the elements: Mass fraction of C in CO2 = Mass fraction of H in H2O = Calculating masses of C and H: Mass of Element = mass of compound x mass fraction of element Determining a Chemical Formula from Combustion Analysis - III Mass (g) of C = Mass (g) of H = Calculating the mass of O: Calculating moles of each element: C = H = O = Answers to Problems in Lecture #7 58.09 amu 1.90 x 10 - 4 mol, 1.15 x 1020 atoms 0.23545 mol, 1.46 x 1023 formula units (a) 42.10% C, 6.479% H, 51.417% O; (b) 10.25 g C Na2CrO4 C4H8O4