1. Chapter 7: Chemical Formula Relationships + + 4 Wood Boards6 Nails 1 Fence Panel + 4 Dozen Wood Boards 6 Dozen Nails 1 Dozen Fence Panels For a 12 panels fence…

2. Chapter 7: Chemical Formula Relationships + + 4C3H 2 C4H6C4H6 + 4 moles of C 3 moles of H 2 1 mole of C 4 H 6 For a mole of 1,3-Butadiene …

3. Chapter 7: Chemical Formula Relationships 1 Dozen = 12 items The mole is defined (since 1960) as the amount of substance of a system that contains as many entities as there are atoms in 12 g of carbon-12. Symbol: mol. Coined by Wilhelm Ostwald in 1893 1 mol = 12 g of carbon-12 1 mole = 6.0221415×10 23 items # of Molecules = # of moles X 6.022×10 23 A mole of carbon contains 6.0221415×10 23 atoms of carbon, but the same is true for any other element or molecule; in general: 6.0221415×10 23 is the Avogadro’s Number

4. Chapter 7: Chemical Formula Relationships m = number of nails X mass of 1 nail Example: 500g = 100 nails X 5g m = number of moles X mass of 1 mole of the substance m = n X M.M. Where: m = mass (g) n = number of moles (mol) M.M. = Molecular Mass (g/mol) = mass of 1 mole of the substance Sometime the Molecular Mass is measured in Daltons (1Da = 1g/mol)

5. Chapter 7: Chemical Formula Relationships The molar mass of an element is numerically the same as the atomic weight of the element. They are not however the same; they have different units: The atomic weight is defined as one twelfth of the mass of an isolated atom of carbon-12 and is therefore dimensionless The molar mass is measured in g/mol. The molar mass of a compound is given by the sum of the atomic weights of the atoms which form the compound. Example: molar mass of Ca(NO 3 ) 2

6. Write a Solution Map for converting the units : Information Given:1.1 x 10 22 Ag atoms Find:? moles Conv. Fact.: 1 mole = 6.022 x 10 23 Example: A silver ring contains 1.1 x 10 22 silver atoms. How many moles of silver are in the ring? atoms Ag moles Ag

7. Check the Solution: 1.1 x 10 22 Ag atoms = 1.8 x 10 -2 moles Ag The units of the answer, moles, are correct. The magnitude of the answer makes sense since 1.1 x 10 22 is less than 1 mole. Information Given:1.1 x 10 22 Ag atoms Find:? moles Conv. Fact.: 1 mole = 6.022 x 10 23 Sol’n Map: atoms  mole Example: A silver ring contains 1.1 x 10 22 silver atoms. How many moles of silver are in the ring?

8. Write a Solution Map for converting the units : Information Given:1.75 mol H 2 O Find:? g H 2 O C F: 1 mole H 2 O = 18.02 g H 2 O mol H 2 O g H 2 O Example: Calculate the mass (in grams) of 1.75 mol of water

9. Check the Solution: 1.75 mol H 2 O = 31.5 g H 2 O The units of the answer, g, are correct. The magnitude of the answer makes sense since 31.5 g is more than 1 mole. Information Given:1.75 mol H 2 O Find:? g H 2 O C F: 1 mole H 2 O = 18.02 g H 2 O Sol’n Map:mol  g Example: Calculate the mass (in grams) of 1.75 mol of water

10. Chemical Formulas as Conversion Factors 1 spider  8 legs 1 chair  4 legs 1 H 2 O molecule  2 H atoms  1 O atom

11. Mole Relationships in Chemical Formulas since we count atoms and molecules in mole units, we can find the number of moles of a constituent element if we know the number of moles of the compound Moles of CompoundMoles of Constituents 1 mol NaCl1 mole Na, 1 mole Cl 1 mol H 2 O2 mol H, 1 mole O 1 mol CaCO 3 1 mol Ca, 1 mol C, 3 mol O 1 mol C 6 H 12 O 6 6 mol C, 12 mol H, 6 mol O

12. Example: Carvone, (C 10 H 14 O), is the main component in spearmint oil. It has a pleasant odor and mint flavor. It is often added to chewing gum, liquers, soaps and perfumes. Find the mass of carbon in 55.4 g of carvone.

13. Write a Solution Map for converting the units : g C 10 H 14 O Information Given:55.4 g C 10 H 14 O Find: g C CF: 1 mol C 10 H 14 O = 150.2 g 1 mol C 10 H 14 O  10 mol C 1 mol C = 12.01 g Example: Find the mass of carbon in 55.4 g of carvone, (C 10 H 14 O). mol C 10 H 14 O mol C gCgC

14. Check the Solution: 55.4 g C 10 H 14 O = 44.3 g C The units of the answer, g C, are correct. The magnitude of the answer makes sense since the amount of C is less than the amount of C 10 H 14 O. Information Given:55.4 g C 10 H 14 O Find: g C CF: 1 mol C 10 H 14 O = 150.2 g 1 mol C 10 H 14 O  10 mol C 1 mol C = 12.01 g SM: g C 10 H 14 O  mol C 10 H 14 O  mol C  g C Example: Find the mass of carbon in 55.4 g of carvone, (C 10 H 14 O).

15. Percent Composition Percentage of each element in a compound – By mass Can be determined from 1.the formula of the compound 2.the experimental mass analysis of the compound The percentages may not always total to 100% due to rounding

16. Mass Percent as a Conversion Factor the mass percent tells you the mass of a constituent element in 100 g of the compound – the fact that NaCl is 39% Na by mass means that 100 g of NaCl contains 39 g Na this can be used as a conversion factor – 100 g NaCl  39 g Na

17. Example - Percent Composition from the Formula C 2 H 5 OH 1.Determine the mass of each element in 1 mole of the compound 2 moles C = 2(12.01 g) = 24.02 g 6 moles H = 6(1.008 g) = 6.048 g 1 mol O = 1(16.00 g) = 16.00 g 2.Determine the molar mass of the compound by adding the masses of the elements 1 mole C 2 H 5 OH = 46.07 g

18. Sample - Percent Composition from the Formula C 2 H 5 OH 3.Divide the mass of each element by the molar mass of the compound and multiply by 100%

19. Empirical Formulas Hydrogen Peroxide Molecular Formula = H 2 O 2 Empirical Formula = HO Benzene Molecular Formula = C 6 H 6 Empirical Formula = CH Glucose Molecular Formula = C 6 H 12 O 6 Empirical Formula = CH 2 O

20. Finding an Empirical Formula 1)convert the percentages to grams a)skip if already grams 2)convert grams to moles a)use molar mass of each element 3)write a pseudoformula using moles as subscripts 4)divide all by smallest number of moles 5)multiply all mole ratios by number to make all whole numbers a)if ratio ?.5, multiply all by 2; if ratio ?.33 or ?.67, multiply all by 3, etc. b)skip if already whole numbers

21. All these molecules have the same Empirical Formula. How are the molecules different? NameMolecular Formula Empirical Formula glyceraldehydeC3H6O3C3H6O3 CH 2 O erythroseC4H8O4C4H8O4 CH 2 O arabinoseC 5 H 10 O 5 CH 2 O glucoseC 6 H 12 O 6 CH 2 O

22. All these molecules have the same Empirical Formula. How are the molecules different? NameMolecular Formula Empirical Formula Molar Mass, g glyceraldehydeC3H6O3C3H6O3 CH 2 O90 erythroseC4H8O3C4H8O3 CH 2 O120 arabinoseC 5 H 10 O 5 CH 2 O150 glucoseC 6 H 12 O 6 CH 2 O180

23. Molecular Formulas The molecular formula is a multiple of the empirical formula To determine the molecular formula you need to know the empirical formula and the molar mass of the compound