Week 3: Stoichiometry
Week 3: Stoichiometry CHEM 1315 - 002
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Popular in Chemistry
This 3 page Class Notes was uploaded by Julia Tang on Sunday September 13, 2015. The Class Notes belongs to CHEM 1315 - 002 at University of Oklahoma taught by Fares Z Najar in Fall 2015. Since its upload, it has received 33 views. For similar materials see General Chemistry in Chemistry at University of Oklahoma.
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Date Created: 09/13/15
Stoichiometry 1 Atomic Weight Atoms are composed of three subatomic particles protons neutrons and electrons For convenience charges of subatomic particles are expressed as multiples of the electronic charge Atomic weights Atomic mass unit used instead of describing mass in grams such small masses Relatives masses of elements 1amu 1665054 x 103924 grams very close to the mass of the neutrons or proton 1 g 602214 x 1023amu 0 Water appeared to contain 8 times as much oxygen by mass than hydrogen Average atomic mass atomic mass or atomic weight 0 Since most element have more than one isotope a weighted average must be calculated to be representative of isotopic distributions found naturally O The average atomic mass of an element is determined by using the masses of neutrons and protons Isotopes of an element have identical chemical properties 0 To calculate atomic mass 0 Multiply the decimal fraction of the natural abundance of each isotope by its mass roughly equal to mass 0 35Cl 7577 0 37Cl 2423 39 35 7577 I 37 2423 I 3548 Atomic weight of the element Charges of subatomic particles are expressed as multiples of the electronic charge 0 Mass of proton almost equals the mass of the neutron 0 Since the mass of either the protons or the neutron is very small scientist decided to use the atomic mass unit 0 1 amu 112 of the mass of carbon12 isotope Mole The average atomic mass of an element is shown on the periodic table 0 What about molecules 0 Ionic substances Formula weight FW the sum of the atomic masses of each atom in its chemical formula By using the atomic weights from the periodic table and the chemical formula the formula weight may be calculated 0 Formula weight of sulfuric acid H2SO4 o 2 atoms of H 210079 o 1 atom S 13206 o 4 atoms of 0 4 x 16 I Add them up 980758g 21 Avogadro39s number 0 amu is too tiny so we measure things to grams 0 Ex C is 12 grams per mole 0 There is 6022x1023 molecules in one mole 0 Atomic weight gmol o 18g of H20 1 mol of H20 0 Molar mass 0 The lighter the atom the less mass in one mole 0 Conversion factors may be developed to convert between the number of atoms molecules mass grams and moles 3 Percent Composition of atoms of element x AtomicWeight of element x 100 formula weight of the compound 0 element O eg CaN032 0 40078 2140067 6159994 164088gmol Whatever youre looking for total mass 2x140067mi01 171 164088 Determine molecular formulas from experimental analysis data 0 Empirical formula simplest ratio whole number of the types of atoms present in a compound 0 Cr203 2 atoms of Cr 3 atoms of oxygen 0 H20 2 atoms of hydrogen one atom of oxygen 0 To Find the Empirical Formula 0 g of each element gt mol of each element gt Molecular ratio of elements gt Simplest molar ratio gt g of each element mol of each element gt Molecular ratio of elements If a sample of photophor is analyzed and found to be composed of 37 104 g of Ca and 19074 9 of P what is the empirical formula Step l COhvert eaclh mass to its correspOhdihg quantity in moles 1 mol Ca Ca 37104 9 of Ca x 092579 lTIOl of Ca 40078 9 1 mol P P 19074 9 of P x 06158l lmol of P 309738 9 Step 2 Divide each mole value by the SMALLEST mole value and round to the nearest positive value or mixed fraction ie 25 15 1333 From step 1 092579 mol Ca 061581 mol P Since 061581 is the smallest molar quantity this is the quantity that we divide by Ca 092579 mol 061581 15 mol Ca P 061581 mol 061581 2 1 molP Step 3 If you obtain a fraction multiply each mole value by Ehe denominator of the fractim in order to get rid of the raction Examples 125 511 715 31 1133 41 167 512 From step 2 Ca 15 mol Ca P 1 mol l3 Ill Ca 312 x 2 3 P l x 2 2 use these values as the subscripts m the empirical formula Ca3P2 0 Molecular formula describes actual numbers of the atoms of each element in a molecule 0 Glucose 0 Molecular formula C5H1205 0 Empirical formula CHZO 0 Hydrogen peroxide 0 Molecular formula H202 0 Empirical formula HO 0 Molecular Formula is calculated by molecular weight given formula weight Suppose we know that the molecular weight of the compound glucose is 180 gmol The formula weight of the empirical formula is 30 gmol Divide the molecular weight by the empirical formula weight to find a multiple 130 331131 30 gma The molecular formula is a multiple of 6 times the empirical formula Cl x 6 H2 x 6 01 x 6 which becomes C5H1205
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