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Mass Relationships of Atoms Atomic number and mass number Atomic number (Z) = the number of protons in the nucleus. Mass number (A) = the sum of the number of protons + neutrons in the nucleus. A ZX Some isotopes 1 H 1 12 C 6 16 O 8 235 U 92 2 H 1 13 C 6 17 O 8 238 U 92 3 H 1 14 C 6 18 O 8 Atomic masses - synonymous with atomic weight - is a relative scale - the isotope of carbon with 6 p+ and 6 no(carbon-12) is the reference atom and assigned an atomic mass of exactly 12 - one atomic mass unit (amu) is defined as a mass exactly equal to 1/12th the mass of one carbon-12 atom - so 1 amu is about the mass of 1 ___ p+ or 1 ___ no Average Atomic Mass and the Periodic Table Average Atomic Mass: The Story of Boron • Any chunk of Boron you dig up is a mixture of 2 isotopes: on average ~20% will be 10B, ~80% will be 11B • Say we get a small chunk of B. • 100 atoms. • 20 are B with a mass of 10 amu each • 80 are B with a mass of 11 amu each • What is the average mass of Boron? 10 11 • The Story of Boron (cont.) Average mass of a B atom • = (total mass)/(total # of B atoms) • Total mass for these 100 B atoms • = 20 x 10 amu + 80 x 11 amu • Average mass of a B atom • = (20 x 10 amu + 80 x 11 amu)/100 20 x 10 amu + 80 x 11 amu • = 100 100 • = 0.20 x 10 amu + 0.80 x 11 amu = 10.8 amu 20% 80% So to get the average atomic mass, multiply % abundance times each isotope mass and add up! The Mole - the fundamental SI measure of “amount of substance” - the amount of substance that contains as many elementary entities as there are atoms in exactly 12 g of carbon-12 - this number of atoms is 6.022045 x 1023 Avogadro’s number The Mole vs. The Dozen The Dozen - the amount of substance that contains 12 entities. The Mole - the amount of substance that contains Avogadro’s number (6.022x1023) of entities. Dozen Apples = 10 Lbs. Mole of Helium atoms = 4.0026g Dozen Apples = 12 Apples Mole of Helium atoms = 6.022x1023 atoms Converting to Dozens Example How many dozens of apples are represented by 1.3 Lbs. of apples. 1.3 Lbs x 1 dozen 10 Lbs = .13 dozen Converting to Moles Example How many moles of He are in 6.46 g of He? 6.46 gHe x 1 mol 4.003 g = 1.61 mol Mass Relationships of Atoms - Extended Example Calculate the number of grams of lead (Pb) In 12.4 moles of lead. 12.4 mol x 207.2 g 1mol = 2569 g Example What is the mass in grams of one silver atom? Solution: Convert grams per mole to grams per atom. 107.9 g 1 mol = x 1 mol 6.022 x 1023 atoms 17.91 x 10–23 g 1 atom Experimental Determination Of Atomic & Molecular Masses Atomic mass is measured by mass spectrometry e– + An atom is ionized in the instrument. Ion is deflected by magnetic field amount of deflection depends on mass to charge ratio highest m/z ratio deflected ________ least lowest m/z ratio most deflected ________ N + S Ions are detected after passage through magnetic field target N + + + + + + S Ions are detected after passage through magnetic field mixture of ions of different mass gives separate peak for each m/z target N intensity of peak proportional to percentage of each atom of different mass in mixture separation of peaks depends on relative mass S ++++ ++ The mass spectrum of the three isotopes of neon. 90.92% 8.82% 0.26% 19 20 21 22 23 Percent Composition of Compounds Percent composition is the percent by mass of each element the compound contains. Obtained by dividing the mass of each element in one mole of the compound by the molar mass of the compound and multiplying by 100% Example Calculate the percent composition by mass of H,P and O for phosphoric acid (H3PO4) Solution: Assume you have 1 mole of the compound. Molar mass = 3(1.008g) + 30.97g + 4(16.00) = 97.99 Molar mass = 3(1.008g) + 30.97g + 4(16.00) = 97.99 %H = %P = %O = 3(1.008g) 97.99g 30.97g 97.99g 4(16.00) 97.99g x 100% = 3.086% x 100% = 31.61% x 100% = 65.31% Determining Formula Determining the Formula with a CHN Analyzer Determining the Formula with a CHN Analyzer O2, and other gases CO2 , H2O, O2 , and N2 O2 0.1156 g compound containing only C,H,&N H2O absorber CO2 absorber Burning the sample completely compound + O2 H2O + CO2 + other gasses Determining the Formula Reacting the C,H,N sample with O2 , we assume all of the hydrogen and carbon present in the 0.1156g sample is converted to H2O and CO2 respectively grams collected 0.1638g CO2 0.1676g H2O (27.3% C in CO2) 12.01g C x = 0.04470 g C 44.009 g CO2 x 2.016g H = 0.01876 g H 18.015 g H2O (11.1% H in H2O) Determining the Formula Remembering the original mass of the sample the percentages of the components can be determined 0.04470 g C x 100% = x 100% = 38.67% C 0.1156 g sample 0.01876 g H 16.23% H 0.1156 g sample 38.67% C + 16.23% H + % N = 100% % N = 45.10% Levels of Structure Elemental Composition Empirical Formula Molecular Formula Constitution Configuration Conformation Elemental Composition Examples: Formaldehyde Glucose C: 40.00% C: 40.00% H: 6.73% H: 6.73% O: 53.27% O: 53.27% Levels of Structure Elemental Composition Empirical Formula Molecular Formula Constitution Configuration Conformation √ Empirical Formula The empirical formula tells us which elements are present and the simplest whole-number ratio of their atoms. Empirical Formula Examples: Formaldehyde and Glucose Elemental Composition C: 40.00% 40.00 g H: 6.73% 6.73 g O: 53.27% 53.27 g assume a 100g sample calculate atom ratios by dividing by atomic weight Calculating Empirical Formula C: H: O: 40.00 g 6.73 g 53.27 g 100 g x x x 1 mol 12.01 g 1 mol 1.00 g 1 mol 16.0 g = 3.33 mol = 6.73 mol = 3.33 mol Empirical Formula Examples: Formaldehyde and Glucose Elemental Composition C: 40.00% H: 6.73% O: 53.27% 40.00 g 3.33 mol 6.73 g 6.73 mol 53.27 g 3.33 mol assume a 100g sample calculate atom ratios by dividing by atomic weight determine the smallest whole number ratio by dividing by the smallest molar value Calculating Empirical Formula C: 40.00 g x H: 6.73 g x O: 53.27 g x 100 g 1 mol 12.01 g 1 mol 1.00 g 1 mol 16.0 g = 3.33 mol 3.33 mol = 6.73 mol 3.33 mol = 3.33 mol 3.33 mol = 1.00 = 2.02 = 1.00 Empirical Formula Examples: Formaldehyde and Glucose Elemental Composition C: 40.00% 40.00 g 6.73 g H: 6.73% O: 53.27% 53.27 g 6.73 mol 1 2 3.33 mol 1 3.33 mol Empirical Formula: CH2O Example A 1.723 g sample of aluminum oxide (which consists of aluminum and oxygen only) contains 0.912g of Al. Determine the empirical formula of the compound. 1.723 g sample - 0.912 g Al = 0.811 g O 0.912 g Al x 0.811 g O x 1 mol Al 26.98 g Al 1 mol O 16.0 g O = 0.0338 mol 0.0338 mol = 1.0 x 2 = 2 = 0.0507 mol = 0.0338 mol Al2O3 1.5 x 2 = 3 Levels of Structure Elemental Composition Empirical Formula Molecular Formula Constitution Configuration Conformation √ √ Molecular Formula (the REAL formula) determined from empirical formula and experimentally determined molecular mass Compound formaldehyde glucose True Empirical molar Formula mass CH2O CH2O 30 180 Calculation of empirical mass 1 mol C = 12.01 g 2 mol H x 2 = 2.016 g 1 mol O = 16.00g 30.026g Molecular mass glucose formaldehyde 180g =6 30g 30g =1 30g Empirical mass Molecular Formula determined from empirical formula and experimentally determined molecular mass Compound formaldehyde glucose Empirical Molar Formula mass CH2O CH2O 30 180 Molecular formula CH2O C6H12O6 Mass Relationships Stoichiometry mass relationships how much reactant is needed to yield a certain amount of product Amounts of Reactants and Products The mole method 1. Write and balance the equation. 2. Convert to moles. 3. Use the coefficients in the balanced equation to relate the number of moles of known substances to the unknown one. 4. Convert to desired units (g, L, etc.). 5. Check your answer. Stoichiometry Moles of X Mass of X (using molar ratio of Y to X) X ny nx Moles of Y Y Mass of Y Example How many grams of nitrogen dioxide can be formed by reaction of 1.44 g of nitrogen monoxide with oxygen? 2 NO + O2 2NO2 Stoichiometry Molar ratio of Y to X Moles of X 1.44 g of NO 2 NO + O2 ny nx Moles of Y Mass of NO2 2NO2 Stoichiometry Molar ratio of Y to X 0.048 Moles NO 1.44 g of NO 2 NO + O2 ny nx Moles of Y Mass of NO2 2NO2 Stoichiometry Molar ratio NO2 to NO 0.048 Moles NO 1.44 g of NO 2 NO + O2 2 2 Moles of Y Mass of NO2 2NO2 Stoichiometry Molar ratio NO2 to NO 0.048 Moles NO 1.44 g of NO 2 NO + O2 2 2 0.048 Moles NO2 Mass of NO2 2NO2 Stoichiometry Molar ratio NO2 to NO 0.048 Moles NO 1.44 g of NO 2 NO + O2 2 2 0.048 Moles NO2 2.21 g of NO2 2NO2 Example 2 NO + O2 1.44g NO x x 46g NO2 1mol NO2 2NO2 1mol NO 30g NO = x 2mol NO2 2mol NO 2.21g NO2 Limiting Reagents Limiting Reagent Reactants are not always present (or available) in “stoichiometric” quantities. One reactant may be present in quantities such that it is completely consumed while excess amounts of other reactants remain. - called “limiting reactant” or “limiting reagent” The limiting reagent will limit the amount of product produced. Example How many moles of MgCl2 will be produced? Mg + Cl2 MgCl2 Start 1 mol 1 mol Finish 0 0 0 1 mol Example How many moles of MgCl2 will be produced? Mg + Cl2 Start 1 mol 2 mol Finish 0 1 mol MgCl2 0 1 mol magnesium is the limiting reagent 1 mol of chlorine will be left unchanged Limiting Reagent Molar ratio Y to X Moles of X Moles of W Mass of X W + X Mass of W ny nx Moles of Y Y Mass of Y Compare molar ratio W to X to their coefficients in balancd equation; identify LR Molar ratio Y to LR Moles of X Moles of W Mass of X W + X Mass of W Moles of Y Y Mass of Y Example Determine the limiting reagent and the amount of PI3 produced when 6.00g P4 reacts with 25.0g of I2. P4 + 6I2 6.00g P4 x 25.0g I2 x 1mol P4 4PI3 = .0484mol P4 = .0984mol I2 124g P4 1mol I2 254g I2 Example cont... Determine how much I2 would be needed to react completely with the available amount of P4. P4 + 6I2 .0484mol P4 x 4PI3 6mol I2 = 1mol P4 but only .0984mol I2 is available I2 is the limiting reagent 0.290mol I2 amount of I2 needed Example cont... …the amount of PI3 produced from the limiting reagent... .0984mol I2 x 4mol PI3 6mol I2 27.0g PI3 x 412g PI3 1mol PI3 = another method Determine the limiting reagent and the amount of PI3 produced when 6.00g P4 reacts with 25.0g of I2. P4 + 6I2 4PI3 Compute the moles of product formed 2 ways (assuming P4 is completely consumed and then assuming I2 is completely consumed. See which reactant, if used up completely, gives the least amount of product. It’s limiting. Reaction Yield Theoretical yield the amount of product that would result if all the limiting reagent reacted Actual yield the amount of product actually obtained from the reaction Almost always less than the theoretical yield Percent Yield Actual yield x 100% %yield = Theoretical yield Determines how efficient a reaction is Example In a certain industrial operation 3.54 x 107g of TiCl4 is reacted with 1.13 x 107g of Mg. (a) Calculate the theoretical yield of Ti in grams. (b) Calculate the percent yield if 7.91 x 106g are actually obtained. TiCl4 + 2Mg Ti + 2MgCl2 Calculate theoretical yield 3.54 x 107g TiCl4 x 1.13 x 107g Mgx 1mol TiCl4 1.87 x 105mol TiCl4 = 4.65 x 105 mol Mg 187.7g TiCl4 1mol Mg 24.31g Mg 1.87 x 105mol TiCl4 x = 2mol Mg 1mol TiCl4 = 3.74 x 105 mol Mg there is more than enough Mg TiCl4 is limiting 3.54 x 107g TiCl4 x 47.88g Ti 1mol TiCl4 x 1mol Ti x 1mol TiCl4 187.7g TiCl4 = 8.93 x 106g Ti 1mol Ti %yield = Actual yield x 100% Theoretical yield 7.91 x 106g Ti 100% x 8.93 x 106g Ti = = 88.6%