Understanding Atomic Masses and Empirical Formulas in Chemistry, Slides of Chemistry

Download Understanding Atomic Masses and Empirical Formulas in Chemistry and more Slides Chemistry in PDF only on Docsity! Chemical Composition docsity.com Atomic Masses • Balanced equation tells us the relative numbers of molecules of reactants and products C + O2  CO2 1 atom of C reacts with 1 molecule of O2 to make 1 molecule of CO2 • If I want to know how many O2 molecules I will need or how many CO2 molecules I can make, I will need to know how many C atoms are in the sample of carbon I am starting with docsity.com Example #1 • Determine the mass of 1 Al atom 1 atom of Al = 26.98 amu • Use the relationship as a conversion factor amu 2024 atomAl1 amu26.98x atoms Al 75  Calculate the Mass (in amu) of 75 atoms of Al docsity.com Chemical Packages - Moles • We use a package for atoms and molecules called a mole • A mole is the number of particles equal to the number of Carbon atoms in 12 g of C-12 • One mole = 6.022 x 1023 units • The number of particles in 1 mole is called Avogadro’s Number • 1 mole of C atoms weighs 12.01 g and has 6.02 x 1023 atoms docsity.com Figure 8.2: One- mole samples of iron (nails), iodine crystals, liquid mercury, and powdered sulfur docsity.com  Use Avogadro’s Number to determine the number of atoms in 1 mole 1 mole Al = 6.02 x 1023 atoms  Use this as a conversion factor for atoms-to-moles Al mol 0.370 atoms10x 6.02 Almol1x atoms Al 10x 2.23 23 23  Compute the number of moles and mass of 2.23 x 1023 atoms of Al Example #3 docsity.com  Use the Periodic Table to determine the mass of 1 mole of Al 1 mole Al = 26.98 g  Use this as a conversion factor for moles-to-grams Al g 9.99 Almol1 g26.98x Al mol 0.370  Compute the number of moles and mass of 2.23 x 1023 atoms of Al Example #3 docsity.com Molar Mass • The molar mass is the mass in grams of one mole of a compound • The relative weights of molecules can be calculated from atomic masses water = H2O = 2(1.008 amu) + 16.00 amu = 18.02 amu • 1 mole of H2O will weigh 18.02 g, therefore the molar mass of H2O is 18.02 g • 1 mole of H2O will contain 16.00 g of oxygen and 2.02 g of hydrogen docsity.com  Determine the mass of each element in 1 mole of the compound 2 moles C = 2(12.01 g) = 24.02 g 6 moles H = 6(1.008 g) = 6.048 g 1 mol O = 1(16.00 g) = 16.00 g  Determine the molar mass of the compound by adding the masses of the elements 1 mole C2H5OH = 46.07 g Determine the Percent Composition from the Formula C2H5OH Example #4 docsity.com  Divide the mass of each element by the molar mass of the compound and multiply by 100% 52.14%C100% 46.07g 24.02g  13.13%H100% 46.07g 6.048g  34.73%O100% 46.07g 16.00g  Determine the Percent Composition from the Formula C2H5OH Example #4 docsity.com Empirical Formulas • The simplest, whole-number ratio of atoms in a molecule is called the Empirical Formula – can be determined from percent composition or combining masses • The Molecular Formula is a multiple of the Empirical Formula % A mass A (g) moles A 100g MMA % B mass B (g) moles B 100g MMB moles A moles B docsity.com 1. Convert the percentages to grams by assuming you have 100 g of the compound – Step can be skipped if given masses 47gC 100g 47gC100g  47gO 100g 47gO100g  6.0gH 100g 6.0gH100g  Determine the Empirical Formula of Acetic Anhydride if its Percent Composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen Example #6 docsity.com 2. Convert the grams to moles C mol 3.9 12.01g Cmol1C 47g  O mol 2.9 16.00g Omol1O g 47  Hmol 6.0 1.008g Hmol1 Hg 6.0  Determine the Empirical Formula of Acetic Anhydride if its Percent Composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen Example #6 docsity.com 3. Divide each by the smallest number of moles 1.3 2.9Cmol3.9  1 2.9Omol2.9  2 2.9H mol6.0  Determine the Empirical Formula of Acetic Anhydride if its Percent Composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen Example #6 docsity.com Molecular Formulas • The molecular formula is a multiple of the empirical formula • To determine the molecular formula you need to know the empirical formula and the molar mass of the compound docsity.com  Determine the empirical formula • May need to calculate it as previous C5H3  Determine the molar mass of the empirical formula 5 C = 60.05 g, 3 H = 3.024 g C5H3 = 63.07 g Determine the Molecular Formula of Benzopyrene if it has a molar mass of 252 g and an empirical formula of C5H3 Example #7 docsity.com  Divide the given molar mass of the compound by the molar mass of the empirical formula – Round to the nearest whole number 4 07.63 252  g g Determine the Molecular Formula of Benzopyrene if it has a molar mass of 252 g and an empirical formula of C5H3 Example #7 docsity.com