Determining Isotopic Composition and Molar Mass: Empirical & Molecular Formulas, Study notes of Chemistry

Download Determining Isotopic Composition and Molar Mass: Empirical & Molecular Formulas and more Study notes Chemistry in PDF only on Docsity! 1 Chapter 3 - Stoichiometry 3.1 Atomic Masses 3.2 The Mole 3.3 Molar Mass 3.4 Percent Composition of Compounds 3.5 Determining the Formula of a Compound 3.6 Chemical Equations 3.7 Balancing Chemical equations 3.8 Stoichiometric Calculations: Amounts of Reactants and Products 3.9 Calculations Involving a Limiting Reactant Atomic Definitions: amu and 12C Standard Atomic Mass Unit (amu) = 1/12 the mass of a 12C atom (isotope) one amu = 1.66054 x 10 -24 g. One 12C atom has a mass of 12 amu (exact number). Isotopic mass of 13C = 13.0034 amu Atomic Mass = “Atomic Weight” of an element is the average of the masses of its naturally occurring isotopes weighted according to their abundances. 12 mass of isotope atom= 12 mass I o sotopic Mass (am f C atom u) × 1 6 8 O 1 7 8 O 1 8 8 O 8 Protons 8 Neutrons 99.759% 15.99491462 amu 8 Protons 9 Neutrons 0.037% 16.9997341 amu 8 Protons 10 Neutrons 0.204 % 17.999160 amu Element #8 : Oxygen, isotopes Periodic table lists average molecular mass (weighted average): 15.99491462 x 0.99759 +16.9997341 x 0.00037 +17.999160 x 0.00204 = 15.9974 amu → 16.00 amu Determine isotopes and their abundance using a mass spectrometer Figure 3.2 (a): Isotopes of Neon detected. Each isotope yields one peak. Three Isotopes for Neon, mass number = 20 the most abundant. Figure 3.2 (b): Bar graph of neon Mass Spec data indicated relative percentage of each isotope 2 Isotopic Composition Problem EP 68: 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 to 4 significant figures . MOLE • Definition: The mass of an element or compound that contains as many elementary particles (atoms, molecules, ions, or other?) as there are atoms in exactly 12 grams of 12C. • 1 Mole = 6.022145 x 1023 particles (atoms, molecules, ions, electrons, or…) = NA particles NA is Avogado’s number. The Mole Is a Chemical Concept •A mole represents a fixed number of chemical entities (atoms, molecules or ions) •A mole of a chemical entity has a fixed, unique mass. (molar mass or molecular weight) •Thus, the mole allows masses to be converted to the number of 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 1023 atoms 1 atom of S = 1 mole of S = = 6.022 x 1023 atoms 1 atom of O = 1 mole of O = = 6.022 x 1023 atoms Molecular mass: 1 molecule of O2 = 16.00 x 2 = 1 mole of O2 = = 6.022 x 1023 molecule 1 molecule of S8 = 32.07 x 8 = 1 mole of S8 = = 6.022 x 1023 molecules 32.07 amu 32.07 g 16.00 amu 16.00 g 32.00 amu 32.00 g 256.6 amu 256.6 g S8 O2 S 6.022x102355.85 g55.85 amuFe 6.022x10231.008 g1.008 amuH Number of Atoms Molar mass (g) Molecular mass (amu) Element 5 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 # of moles as subscripts Change to integer subscripts: ÷ smallest, conv. to whole #. EP 72: 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? Assume that the empirical and molecular formulas are the same and name the compound. Strategy: Determining Empirical Formulas from Masses of Elements - I = moles of element(mass of element)/(molar mass element) =(5.677 g)/(22.99 g mol-1)Na =(6.420 g)/(52.00 g mol-1)Cr =(7.902 g)/(16.00 g mol-1)O Preliminary formula: Divide all numbers by smallest number Result: Round off if very close to integer If we assume that empirical formula is also the molecular formula, CnHm + (n+ )O2(g) n CO2(g) + H2O(g) m 2 m 2 6 Determining a Chemical Formula from Combustion Analysis - I EP 73: Erythrose (M = 120.1 g/mol) is an important chemical compound used often as a starting material in chemical synthesis, and contains only 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. Strategy: (1) Find the masses of hydrogen and carbon using the mass fractions of H in H2O, and C in CO2. (2)Subtract the mass of carbon and hydrogen from the sample mass to get the mass of oxygen. (3) Calculate moles of H, C, and O, and construct empirical formula (4) Use the given molar mass and empirical formula to calculate the molecular formula. Calculate the mass fraction of C in CO2 and H in H2O 2 C 2 CO mol C×Mmass fraction of C in CO = M 2 H 2 H O mol H×Mmass fraction of H in H O = M Calculate the masses of H and C in the CO2 and H2O formed from erythrose Calculate the mass of O in erythrose Calculate the number of moles of each element Preliminary formula is Divide all numbers by smallest number Result for empirical formula is Round off Molar mass of unit is g, but molecular mass is . Therefore molecular formula is Chemical Equations Reactants Products Qualitative Information: States of Matter: (s) solid (l) liquid (g) gaseous (aq) aqueous 2 H2 (g) + O2 (g) 2 H2O (g) Phases Used in Chemical Equations Solid (s) Liquid (l) Gas (g) Dissolved in Water (in aqueous solution) (aq) 7 Balancing Chemical Equations Mass balance (atom balance)- Need to have same number of each element on each side of the equation: a C8H18 (l) + b O2 (g) c CO2 (g) + d H2O (l) (1) start with one element 1 C8H18 (l) + O2 (g) 8 CO2 (g) + H2O (l) (2) progress to next element: H 1 C8H18 (l) + O2 (g) 8 CO2 (g) + 9 H2O (l) (4) progress to next element: O 1 C8H18 (l) + 25/2 O2 (g) 8 CO2 (g) + 9 H2O (l) (5) Make all coefficients integers, omit “ones” 2 C8H18 (l) + 25 O2 (g) 16 CO2 (g) + 18 H2O (l) (6) Recheck numbers of atoms on each side 16C, 36H, 50O 16C, 36H, 50O Calculating Reactants and Products in a Chemical Reaction - I EP 74: 65.80 g of Al2S3 undergoes the following reaction with water. Al2S3 (s) + 6 H2O(l) 2 Al(OH)3 (s) + 3 H2S(g) a) How many moles of water are required for the reaction? b) What mass of H2S & Al(OH)3 would be formed? Strategy (1) Calculate moles of aluminum sulfide using its molar mass. (2) From the balanced equation, calculate the moles of water consumed. (3) Calculate the moles of hydrogen sulfide from the balanced equation. (4) Calculate the mass of hydrogen sulfide using its molecular weight. (5) Calculate the mass of Al(OH)3 as in (3) and (4). (1) Calculate moles of aluminum sulfide using its molar mass. molar mass of aluminum sulfide = 150.17 g / mol moles Al2S3 = (2) From the balanced equation, calculate the moles of water consumed. (3) Calculate the moles of hydrogen sulfide from the balanced equation (4) Calculate the mass of hydrogen sulfide using its molecular weight. (5) Calculate the moles of Al(OH)3 from the balanced equation (6) Calculate the mass of Al(OH)3 using its molecular weight.