Mole – chemical unit of amount

900 views

Published on

Published in: Technology
0 Comments
0 Likes
Statistics
Notes
  • Be the first to comment

  • Be the first to like this

No Downloads
Views
Total views
900
On SlideShare
0
From Embeds
0
Number of Embeds
7
Actions
Shares
0
Downloads
19
Comments
0
Likes
0
Embeds 0
No embeds

No notes for slide

Mole – chemical unit of amount

  1. 1. MOLE – Chemical unit of amount • Refers to the amount of a substance which contains the same number of pieces as there are the number of atoms of Carbon in 12 grams of C12 • 12 grams C12 = 6.02 x 1023 atoms of Carbon = 1 mole
  2. 2. • 1 mole any substance = 6.02 x 10 23 pieces • • 1 mole of O atoms = 6.02 x 10 23 O atoms • 1 mole Na atoms = 6.02 x 10 23 Na atoms • 1 mole of H2O = 6.02 x 10 23 H2O molecules • 1 mole of apples = 6.02x 1023 apples
  3. 3. Molar Mass – mass in grams of 1 mole of a substance • G atomic mass- mass in grams of 1 mole of atoms that is numerically equal to its atomic mass • 22.9Na11
  4. 4. • g-molecular mass – mass in grams of 1 mole of molecules that is numerically equal to its molecular mass (sum of all atomic masses of atoms comprising the molecule if the formula of the molecule is known) H2O 1 mole x15.9 =15.9 2 molesx1 = 2 17.9 g H2O 1mole x16 =16 2 molesx1 = 2 18 g or NaCl 1 x35.45 =35.45 1 x23 = 23 58.45 g G formula mass G molecular mass
  5. 5. Molar Mass – mass in grams of 1 mole of a substance • g – formula mass – mass in grams of 1 mole of formula units that is numerically equal to its formula mass ( sum of all atomic masses of atoms comprising the formula unit if formula of the compound is known) H2O 1 mole x15.9 =15.9 2 molesx1 = 2 17.9 g H2O 1mole x16 =16 2 molesx1 = 2 18 g or NaCl 1 x35.45 =35.45 1 x23 = 23 58.45 g G formula mass
  6. 6. Molar Mass G atomic mass Gram molecular mass Gram formula mass Can be said to be substance Element molecule Formula unit when
  7. 7. Is a MOLE a number or a mass ? • It is both. • MOLE = 6.02 X 10 23 = Molar mass
  8. 8. Exercise VII: Chemical Quantities Exercise Vii Problem Categories Of the Mole Concept I. Conversion between G,moles, pieces II. Calculation Involving Composition III. Stoichiometry in a balanced chemical equation
  9. 9. Category I : Conversions between mole , grams, atoms, molecules Calculation where information about the quantity of the substance is sought from a given quantity of the same substance mole pieces Gram 1 mole = molar mass 1 mole = Avogadro’s Number Equivalent factor
  10. 10. Calculation where information about the quantity of the substance is sought from a given quantity of the same substance Category I : Conversions between mole , grams, atoms, molecules of a given substance • How many moles of sulfur atoms are there in 8.23 grams of sulfur, S ? mole Gram 1 mole = molar mass Equivalent factor ? S 8.23 g S 1 mole S = 32 g S mole S = 8.23 g S 1 mole S ______ 32 g S X
  11. 11. Calculation about Composition • Calculations where informations about the quantities of a substance are sought from the given amounts of the compound containing the substance. • conversion factors • ?gram pieces mole mole gram pieces compound Element / component Formula subscripts ? ? ?
  12. 12. • E.g. How many grams of Cl are contained in 20 grams of CCl4 ?
  13. 13. 7.76 g C8H10N4O2 (ca ffeine) X 1 mole caffeine _______ 194 g caffeine X 8 moles C _____ 1 mole caffeine X 12 g C ____ 1 mole C = 3.84 g C How many grams of C are present in 7.76 g caffeine ?
  14. 14. Other calculations • Percent Composition • Emperical Formula Determination • Molecular Formula Determination
  15. 15. • Emperical Formula = formula which shows the smallest whole number ratio of atoms of the elements comprising the compound. • Molecular Formula = formula which shows the actual number of atoms of the elements comprising the compound.
  16. 16. Stoichiometry- Quantitative Calculations of substances in a balanced chemical equation 1. Mole- molecalculation 2. Mole-mass calculation 3. Mass-mass calculation 4. Limiting Reactant Calculation
  17. 17. Stoichiometry- Quantitative Calculations of substances in a balanced chemical equation • Determination of the Quantity of a substance From a known quantity Of another substance In a balanced chemical • equation reactant reactant product reactant product product Can be Can be Individual amounts Of reactants needed To react Individual amounts of products produced tells
  18. 18. • 1 mole CH4= 2 moles O2 • Individual moles of reactants needed to react • 1 mole CO2 = 2 moles H2O • Individual moles of products produced CH4 2 O2+ CO2 2 H2O+ To be used as conversion factors
  19. 19. Conversion Map : category III gram pieces mole gram mole pieces Balanced equation Reactant versus reactant Reactant versus product Product versus product ?
  20. 20. C2H5OH O2 CO2 H2O+ +3 2 3 Mole O2 = 10. O mole C2H 5 O H X 3 moles O2 _____ 1 mole C2H5OH CO2mole = C2H5OH50.0 g X C2H5OH1 mole C2H5OH46 g CO22 moles C2H5OH1 mole X = 30 moles = 2.17 moles CO 2 Mole ? 50.0 g
  21. 21. C2H5OH O2 CO2 H2O+ +3 2 3 G ?10 g G O2 = 10 g C2H5OH C2H5OH X 1 mole 46 g C2H5OH 3 moleO2 1 mole C2H5OH x 32 g O2 1 mole O2 X 20.86 grams of O2

×