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Honors Chemistry

Honors Chemistry. Chapter 3: Mass relationships in Chemical Reactions. 3.1 Atomic Mass. Too tiny to mass individual atoms Unit needed to compare masses of atoms Atomic Mass Unit Early amu had several standards Now defined as 1/12 the mass of carbon-12 Labeled u to signify unified amu

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Honors Chemistry

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  1. Honors Chemistry Chapter 3: Mass relationships in Chemical Reactions

  2. 3.1 Atomic Mass • Too tiny to mass individual atoms • Unit needed to compare masses of atoms • Atomic Mass Unit • Early amu had several standards • Now defined as 1/12 the mass of carbon-12 • Labeled u to signify unified amu • Average Atomic Mass • Weighted average of all isotopes of element • Used for most calculations

  3. 3.2 Avogadro’s Number • Amedeo Avogadro • Looking for a convenient relationship between grams and number of particles • Varies by atomic mass • 1 atomic mass in grams = 6.022 x 1023 atoms • Avogadro’s number: NA = 6.022 x 1023 • The Mole • 1 mol = NA particles • Unit of counting particles, similar to the unit “dozen”

  4. 3.3 Molecular Mass • Sum of all atoms in a molecule • E.g., H2O = 2 (1.008) + 1 (16.00) = 18.02 u • Try this: Find M of NaHCO3 • Mole Conversions • 1 mol = NA particles • 1 mol = (molecular mass) g • mass  moles  particles M NA • Solve using dimensional analysis

  5. 3.3 Molecular Mass • How many grams are in 0.75 mol NH3? • M = 1 (14.01) + 3 (1.008) = 17.03 g/mol • Therefore, 1 mol = 17.03 g • 0.75 mol 17.03 g ------------ x ----------- = 13 g 1 1 mol • Try this… • How many moles are in 15.0 g N2O?

  6. 3.3 Molecular Mass • How many molecules are in 125 g CH3COOH? • M = 2 (12.01) + 4 (1.008) + 2(16.00) • M = 60.05 u • 125 g 1 mol 6.022 x 1023 molecule-------- x ---------- x ------------------------- 1 60.05 g 1 mol • = 1.25 x 1024 molecules

  7. 3.4 Mass Spectrometer • Device that separates particles by mass • helped verify the existence of isotopes • Used to identify unknown substances • Not terribly important to us!

  8. 3.5 Percent Composition • Percent by mass of each element in the compound • Find % composition of CaCl2 • 1 Ca = 40.082 Cl = 70.90 • 110.98 u • % Ca = 40.08 / 110.98 = 36.11 % • % Cl = 70.90 / 110.98 = 63.89 %

  9. 3.6 Empirical Formula • Based on laboratory data • Simplest whole number ratio of elements • E.g., glucose is C6H12O6 • Empirical formula is CH2O • Subscripts represent ratio by mole, not mass

  10. 3.6 Empirical Formula • A compound contains 0.900 g Ca and 1.60 g Cl. Find the empirical formula. • Ca: 0.900 g 1 mol ---------- x ---------- = 0.0225 mol 1 40.08 g • Cl: 1.60 g 1 mol -------- x ---------- = 0.0451 mol 1 35.45 g • Divide by lowest value to get whole numbers • 0.0451 / 0.0225 = 2, so formula is CaCl2

  11. 3.6 Empirical Formula • Try this… • Find the empirical formula for a compound that is composed of 53.73% Fe and 46.27% S. • Remember that it is percent by mass, so those percents can be treated as grams! • There is a trick at the end of this one.

  12. 3.6 Molecular Formula • Need to know the molecular mass • Divide molecular mass by empirical mass to find how much “bigger” the molecular formula is • Multiply empirical formula by that number • Try this… • A compound is 40.0% C, 6.67% H, 53.3% O. • Molecular mass = 180.1 u. • Find molecular formula.

  13. 3.7 Chemical Equations • Chemical reaction • Process in which a substance is changed into one or more new substances • Chemical equation • Arrangement of chemical symbols to represent what happens during a chemical reaction • Example reaction • Carbon reacts with oxygen to yield carbon dioxide • C + O2 CO2

  14. 3.7 Chemical Equations • States of substances • (s) = solid • (l) = liquid • (g) = gas • (aq) = aqueous solution • HgO (s)  Hg (l) + O2 (g) • Note that the number of oxygens is not conserved. We have to fix that!

  15. 3.7 Balancing Equations • Consider: H2 + O2 H2O • We’ve lost an oxygen! • Can we change it to H2 + O2  H2O2 ? • Can we add an O: H2 + O2 H2O + O ? • A better solution: H2 + O2  2 H2O • Now fix the hydrogens: 2 H2 + O2  2 H2O • Two H2 molecules react with one O2 molecule to form two water molecules

  16. 3.7 Balancing Equations • Try these… • NaClO3 NaCl + O2 • 2 NaClO3 2 NaCl + 3 O2 • CO2 + H2O  C6H12O6 + O2 • 6 CO2 + 6 H2O  C6H12O6 + 6 O2 • CaCl2 + Na3PO4 Ca3(PO4)2 + NaCl • 3 CaCl2 + 2 Na3PO4Ca3(PO4)2 + 6 NaCl

  17. 3.8 Stoichiometry • Quantitative study of chemical reactions • Realize that reaction coefficients are really moles! • 15.0 g oxygen react with excess hydrogen. How much water is produced? • 2 H2 + O2 2 H2O • 15.0g O2 1 mol O2 2 mol H2O 18.0 g H2O------------ x ------------- x -------------- x ------------- 1 32.0 g O2 1 mol O2 1 mol H2O • = 16.9 g H2O produced

  18. 3.9 Limiting Reagents • When reactants are not present in the exact stoichiometric ratio, one reactant will be used up before the others • Limiting reagent – reactant used up first • Excess reagent – reactant that is present in larger quantity than needed • Convert both to moles to find out which is the limiting reagent

  19. 3.9 Limiting Reagents • Try this… • 5.0 g hydrogen react with 20.0 g nitrogen. How much ammonia is produced? • N2 + 3 H2 2 NH3 • Find how many mol NH3 can be made from the H2 • Find how many mol NH3 can be made from the N2 • Use the lesser amount (limiting reagent) • The other substance is present in excess

  20. 3.10 Reaction Yield • Theoretical yield • Amount you should have gotten from reaction • Result of mol calculation • Actual Yield • What you actually got from the reaction • Percent Yield • Actual / theoretical x 100%

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