9701 Chemistry Atom, Molecules & Stoichiometry Chemical Reactions The actual phenomenon that occurs when chemical interact with each other. Balanced Chemical Equations represent events that occur at the atomic level which we cannot perceive; but they explain the mass ratio of substances involved in a chemical equation (stoichiometry) CHEMICAL EQUATIONS a A + b B c C + d D a, b, c, d are stoichiometric coefficients Alonso’s Rules for BE: Easy element 1st hard elements last. One element at a time. Use fractions when necessary. The “mole” A term for a certain number of something. A “mole” of something is 6.02 x 1023 of something. 602, 000, 000, 000, 000, 000, 000, 000 Molecular Weight M.W. = the weight (in grams) of a mole of substance Round to the nearest tenth Hydrogen is 1.00797 1.0 g/mol Mole Weight is an INTENSIVE property, it doesn’t depend on amount Try these MW’s Ca 40.1 H2 2 (1.0) = 2.0 BaF2 [137.3 + 2(19.0)] = 175.3 MW of 2BaF2?? is still 175.3 1 mole of Ca weighs 40.1 grams 1 mole of H2 weighs 2.0 grams Avogadro’s Number NA = 6.02 x 1023 of anything Avogadro's law states that equal volumes of gases, at the same temperature and pressure, contain the same number of molecules. Percent Composition A. Determined from Formulas (“Accepted Value”) Is NaCl 50.0% Na by weight? No, Na is 23.0 g/mole and Cl is 35.5 g/mole To Prove, % Na = Formula Weight (FW) Sum of the atomic weights for the atoms in a chemical formula unit (ionic compound) Molecular Weight (MW) Sum of the atomic weights of the atoms in a molecule (covalent compound) For the molecule ethane, C2H6, the molecular weight would be Empirical Formula Definition: The simplest formula indicating the mole ratio of elements in a compound Examples: H2O2 HO C6H6CH N2O4? CO2 ? NO2 CO2 Empirical Formula STEPS Change grams to moles Divide by the least # moles for a RATIO Apply ratio to the formula Try This… 0.556 g Carbon and 0.0933 g Hydrogen Why is the Empirical Formula a ratio of small WHOLE numbers? Can’t have half of an atom Atoms combine as whole units Shows the simplest way that atoms can pair Try This: 70.5 % Fe and 29.5 % O ___ moles Fe, ___ moles O Assume 100g of substance: Fe2O3 1.26 moles 1.26 moles FeO1.5 Try This: 40.0% C 6.7% H 53.3% O Why doesn’t the ratio of the % give the empirical formula? Must account for differing masses of elements. Why does the ratio of the moles give the empirical formula? The ratio of the # of atoms normalizes for mass differences. Molecular Formulas Definiton: Formula of an actual compound as it exists in molecules. Benzene exists as C6H6 not CH. Hydrogen Peroxide exists as H2O2 not HO. Stoichiometry – The Big Leagues A. Define: Problem Solving involving mass-mass relationships in chemical changes Ex. How many grams of rust are formed when 12.00 g of Fe reacts with oxygen. B. Must use balanced equations for the correct mole ratios C. Coefficients yield the mole ratio!!! 2 H2 + O2 2 H2O 2 : 1 : 2 0.200mol of a hydrocarbon undergo complete combustion to give 35.2g of carbon dioxide and 14.4g of water as the only products. What is the molecular formula of the hydrocarbon? A. C2H4 B. C2H6 C. C4H4 D. C4H8 [AS Nov 2009 Paper 12 Q1] CxHy (g) + O2 (g) CO2 (g) + H2O (g) Calculating Empirical Formulas A 5.00 g sample of an unknown hydrocarbon was burned and produced 14.6 g of CO2. What is the empirical formula of the unknown compound? 5.00 g 14.6 g 2006 A ?g N = A pure hydrocarbon is used in bottled gas for cooking and heating. When 10cm3 of the hydrocarbon is burned in 70cm3 of oxygen (an excess), the final gaseous mixture contains 30cm3 of carbon dioxide and 20cm3 unreacted oxygen. All gaseous volumes were measured under identical conditions. What is the formula of the hydrocarbon? A. C2H6 B. C3H6 C. C3H8 D. C4H10 [AS June 2005 Paper I Q1] Stoichiometry (mass relationships within chemical equations) The coefficients in the balanced equation can also be interpreted as mole ratios of reactants and products Stoichiometric Calculations Limiting & Excess Reactants The limiting reactant is the reactant present in the smallest stoichiometric amount In other words, it’s the reactant you’ll run out of first Limiting H2; Excess O2 Limiting & Excess Reactants Problem If 5.0g of both Mg and O2 are used: Which is the limiting and the excess reactants? How much of the excess will be left unreacted ? How much MgO will be produced ? 2 Mg (s) + O2 (g) 2 MgO (s) Mass spectrometer A highly accurate instrument! Mass spectrometer consists of 6 parts: 1.8 Mass spectrometer (SB p.20) Mass spectrum of Cl2: 1.8 Mass spectrometer (SB p.21) m/e ratio Corresponding ion 35 35Cl+ 37 37Cl+ 70 35Cl─35Cl+ 72 35Cl ─ 37Cl+ 74 37Cl ─37Cl+ Relative atomic mass The relative atomic mass of an element is the weighted average of the relative isotopic masses of its natural isotopes on the carbon-12 scale. 1.9 Relative isotopic, atomic and molecular masses (SB p.23) 1.9 Relative isotopic, atomic and molecular masses (SB p.23) What is the relative atomic mass of Cl? The relative abundances of Cl-35 and Cl-37 are 75.77 and 24.23 respectively Mass Spectrometer – 5 Stages Once the sample of an element has been placed in the mass spectrometer, it undergoes five stages. Vaporisation – the sample has to be in gaseous form. If the sample is a solid or liquid, a heater is used to vaporise some of the sample. X (s) X (g) or X (l) X (g) Mass Spectrometer – 5 Stages Ionisation – sample is bombarded by a stream of high-energy electrons from an electron gun, which ‘knock’ an electron from an atom. This produces a positive ion: X (g) X + (g) + e- Acceleration – an electric field is used to accelerate the positive ions towards the magnetic field. The accelerated ions are focused and passed through a slit: this produces a narrow beam of ions. Mass Spectrometer – 5 Stages Deflection – The accelerated ions are deflected into the magnetic field. The amount of deflection is greater when: • the mass of the positive ion is less • the charge on the positive ion is greater • the velocity of the positive ion is less • the strength of the magnetic field is greater Mass Spectrometer If all the ions are travelling at the same velocity and carry the same charge, the amount of deflection in a given magnetic field depends upon the mass of the ion. For a given magnetic field, only ions with a particular relative mass (m) to charge (z) ration – the m/z value – are deflected sufficiently to reach the detector. Mass Spectrometer Detection – ions that reach the detector cause electrons to be released in an ion-current detector The number of electrons released, hence the current produced is proportional to the number of ions striking the detector. The detector is linked to an amplifier and then to a recorder: this converts the current into a peak which is shown in the mass spectrum. The isotopic composition of an element is indicated below. What is the relative atomic mass of the element? A. 10.2 B. 10.5 C. 10.8 D. 11.0 [AS June 2007 Paper I Q1] (c) A sample of iron has the following isotopic composition by mass: (i) Define the term relative atomic mass. (ii) By using the data above, calculate the relative atomic mass of iron to three significant figures. [5] [AS June 2005 Paper II Q1] Q: 1 FC1 is a solution containing 16.75g dm-3 of hydrated sodium carbonate, Na2CO3.xH2O. FC2 is 0.125mol dm-3 hydrochloric acid, HCl. 25.0cm3 of FC1 is pipetted into a conical flask and add a few drops of the indicator and titrate with FC2 from the burette. The titre value obtained is 23.10cm3. (a) Calculate how many moles of the acid were run from the burette into the conical flask during the titration of FC1 with FC2. 1 (b)Calculate the number of moles of anhydrous sodium carbonate, Na2CO3, in 25.0cm3 of FC1. Na2CO3 + 2HCl 2NaCl + H2O + CO2 1 (c) Calculate the concentration, in mol dm-3, of sodium carbonate, Na2CO3, in FC1. 1 (d) Calculate the mass of anhydrous sodium carbonate present in 1.00dm3 of FC1. [Ar: Na, 23.0; C, 12.0; O, 16.0.] 1 (e) Calculate the mass of water present in the hydrated sodium carbonate. 1 (f) Find the value of x. 1 [AS Nov 2001 Paper III Q1] * * * *