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CH 3 Mass Relations in Chemistry; Stoichiometry

CH 3 Mass Relations in Chemistry; Stoichiometry. Atomic Mass. Indicates how heavy an element is compared to another element. Units AMU---Atomic Mass Unit Defined as 1/12 of the mass of a C-12 atom. Atomic Mass from isotope composition. Isotopic Abundance: the natural of an isotope.

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CH 3 Mass Relations in Chemistry; Stoichiometry

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  1. CH 3 Mass Relations in Chemistry;Stoichiometry

  2. Atomic Mass • Indicates how heavy an element is compared to another element. • Units AMU---Atomic Mass Unit • Defined as 1/12 of the mass of a C-12 atom

  3. Atomic Mass from isotope composition • Isotopic Abundance: the natural of an isotope.

  4. Mass of individual atoms • Mass of 1 atom = molar mass/ NA (Avogadro's #) Reacquaint yourself with the mole wheel.

  5. The Mole • 1 mol= 6.022 x 1023 items • 1mol H = 6.022 x 1023 H atoms = 1.008g • 1mol Cl= 6.022 x 1023 Cl atoms = 35.45g • 1mol Cl2= 6.022 x 1023 Cl2 molecules = 70.90g

  6. Molar mass (MM) • Molar mass is numerically equal to the sum of the atomic masses.

  7. Mass % from formula • % composition of K2CrO4 Use part / whole, assume you have 1 mole of compound. (the math is easier) 1mol = K2CrO4194.20 g %K = 78.90/194.20 = %Cr = %O =

  8. Simplest / empirical formula • Simplest whole number ratio of atoms present in a compound.

  9. Simplest (empirical) formulafrom % composition • Steps: • Find the mass of each element in the sample compound, assume 100g total. • Find the numbers of moles of each compound. • Divide each by the smallest # of moles and look for obvious ratios. Use the following K= 26.6%, Cr= 35.4%, O = 38.0%

  10. Simplest formula (empirical) from analytical data An organic sample containing only C, H, O atoms weighs 1.000g Burning the sample gives 1.466g CO2, 0.6001g H2O Find the simplest formula… • All the carbon from the sample is “locked up” in CO2 • The portion of CO2 that is carbon can be determined by the mass ratio in the formula (12.01/44.01) • This multiply this by the mass of CO2 and you find grams of carbon in the sample.

  11. Yield of product in a reaction • Ordinarily, reactants are not present in the exact ratio required for reaction. Usually 1 is in excess; some left when reaction is over. 1 is limiting; completely consumed to give the theoretical yield of product.

  12. Calculating theoretical yield • Calculate the yield expected if the first reactant is limiting. • Repeat the calculation for the second reactant • The theoretical yield is the SMALLER of these two quantities. The reactant that gave the smaller theoretical yield is the limiting reactant.

  13. 2 Ag (s) + I2 (s)  2 AgI (s) • Calculate the theoretical yield of AgI and determine the limiting reactant. • There is 1.00g Ag, and 1.00g I2.

  14. % Yield • Suppose the actual yield is 1.50g AgI, what was the % yield? Actual/Theoretical (x100) = % Yield

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