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The Mathematics of Chemistry

The Mathematics of Chemistry. Stoichiometry. The Mole. 1 mole of an element or compound is equal to its atomic mass in grams. Calculations. P A R T I C L E S. G R A M S. Avogadro’s Number. Molar Mass. MOLE. Molar Mass. The mass in grams of 1 mole of the compound.

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The Mathematics of Chemistry

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  1. The Mathematics of Chemistry Stoichiometry

  2. The Mole 1 mole of an element or compound is equal to its atomic mass in grams.

  3. Calculations P A R T I C L E S G R A M S Avogadro’s Number Molar Mass MOLE

  4. Molar Mass The mass in grams of 1 mole of the compound C10 H6 O3 10 C = 10 X 12.01g = 120.10 6 H = 6 X 1.00g = 6.00 3 O = 3 X 16.00g = 48.00 TOTAL = 174.1 grams

  5. Molar Mass The mass in grams of 1 mole of the compound CaCO3 Ca 1 X 40.0 grams = 40.0 grams C 1 X 12.0 grams = 12.0 grams O 3 X 16.0 grams = 48.0 grams TOTAL = 100.0 grams

  6. Chemical Equations • Recipe for a Chemical Reaction • Relative number of Reactants and Products • Coefficients – relative numbers CH4 + O2 CO2 + H2O

  7. BalancingChemical Equations • Atoms are Conserved in a Chemical Reaction • When an Equation is Balanced • Never change the IDENTITIES of the Reactants or Products C2H5OH (l) + O2 (g)  CO2 (g) + H2O (g)

  8. BalancingChemical Equations C2H5OH (l) + O2 (g)  CO2 (g) + H2O (g) Count Atoms: Carbon 2 Carbon 1 Hydrogen 6 Hydrogen 2 Oxygen 3 Oxygen 3

  9. BalancingChemical Equations C2H5OH (l) + O2 (g)  CO2 (g) + H2O (g) C2H5OH (l) + 3O2 (g)  2CO2 (g) + 3H2O (g)

  10. Stoichiometric CalculationsThe Plan • Write the Chemical Equation. • Balance the Chemical Equation. • Grams known Moles known • Moles known Moles unknown • 5. Moles unknown Grams unknown

  11. Stoichiometric CalculationsThe Problem Lithium hydroxide is used in an outer space environment to remove excess exhaled carbon dioxide from the living environment. The products of the reaction are lithium carbonate and water. If 48.0 grams of lithium hydroxide are used in a small scale experimental device, how much carbon dioxide will the device process? The Plan 1. Write the Chemical Equation. 2. Balance the Chemical Equation. 3. Grams known  Moles known 4. Moles known  Moles unknown 5. Moles unknown Grams unknown

  12. Stoichiometric CalculationsThe Plan 1. Write the Chemical Equation. LiOH + CO2 (g)  Li2CO3 + H2O (g) 2. Balance the Chemical Equation. 2 LiOH + CO2 (g)  Li2CO3 + H2O (g)

  13. Stoichiometric CalculationsThe Plan 3. Grams known Moles known 48 grams X 1 mole = 2.00 moles of LiOH 24 grams 4. Moles known Moles unknown 2 LiOH + CO2 (g)  Li2CO3 (g) + H2O (g) 2.00 mole LiOH X 1 mole CO2 = 1.00 moles of CO2 2 mole LiOH

  14. Stoichiometric CalculationsThe Plan 5. Moles unknown  Grams unknown 1.00 mole CO2 X 44.0 grams = 44.0 grams of CO2 1 mole

  15. Stoichiometric CalculationsThe Problem Lithium hydroxide is used in an outer space environment to remove excess exhaled carbon dioxide from the living environment. The products of the reaction are lithium carbonate and water. If 48.0 grams of lithium hydroxide are used in a small scale experimental device, how much carbon dioxide will the device process? When 48.0 grams of lithium hydroxide are available for use in a reaction, 44 grams of carbon dioxide can be processed by the reaction.

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