1. Chapter 7 “Chemical Formulas and Chemical Compounds” Yes, you will need a calculator for this chapter!

2. Using Formulas How many oxygen atoms in the following? CaCO 3 Al 2 (SO 4 ) 3 How many ions in the following? CaCl 2 NaOH Al 2 (SO 4 ) 3 3 atoms of oxygen 12 (3 x 4) atoms of oxygen 3 total ions (1 Ca 2+ ion and 2 Cl 1- ions) 2 total ions (1 Na 1+ ion and 1 OH 1- ion) 5 total ions (2 Al 3+ + 3 SO 4 2- ions)

3. Review Problems   How many molecules of CO 2 are in 4.56 moles of CO 2 ?   How many moles of water is 5.87 x 10 22 molecules?   How many atoms of carbon are in 1.23 moles of C 6 H 12 O 6 ?   How many moles is 7.78 x 10 24 formula units of MgCl 2 ? 2.75 x 10 24 molecules 0.0975 mol (or 9.75 x 10 -2 ) 4.44 x 10 24 atoms C 12.9 moles

4. More Practice   How many grams is 2.34 moles of carbon?   How many moles of magnesium is 24.31 g of Mg?   How many atoms of lithium is 1.00 g of Li?   How many grams is 3.45 x 10 22 atoms of U? 28.1 grams C 1.000 mol Mg 8.68 x 10 22 atoms Li 13.6 grams U

5. What about compounds?   in 1 mole of H 2 O molecules there are two moles of H atoms and 1 mole of O atoms (think of a compound as a molar ratio)   To find the mass of one mole of a compound – –determine the number of moles of the elements present – –Multiply the number times their mass (from the periodic table) – –add them up for the total mass

6. Calculating Formula Mass Calculate the formula mass of magnesium carbonate, MgCO 3. 24.31 g + 12.01 g + 3 x (16.00 g) = 84.32 g Thus, 84.32 grams is the formula mass for MgCO 3.

7. Molar Mass   Molar mass is the generic term for the mass of one mole of any substance (expressed in grams/mol)   The same as: 1) Gram Molecular Mass (for molecules) 2) Gram Formula Mass (ionic compounds) 3) Gram Atomic Mass (for elements) – – molar mass is just a much broader term than these other specific masses

8. Examples   Calculate the molar mass of the following: Na 2 S N 2 O 4 C Ca(NO 3 ) 2 C 6 H 12 O 6 (NH 4 ) 3 PO 4 = 78.05 g/mol = 92.02 g/mol = 12.01 g/mol = 164.1 g/mol = 180.156 g/mol = 149.096 g/mol

9. Since Molar Mass is…   The number of grams in 1 mole of atoms, ions, or molecules,   We can make conversion factors from these. - To change between grams of a compound and moles of a compound.

10. For example   How many moles are in 5.69 g of NaOH?

11. For example How many moles are in 5.69 g of NaOH?

12. For example How many moles are in 5.69 g of NaOH? l We need to change 5.69 grams NaOH to moles

13. For example How many moles are in 5.69 g of NaOH? l We need to change 5.69 grams NaOH to moles l 1mole Na = 22.99 g 1 mol O = 16.00 g 1 mole of H = 1.008 g

14. For example How many moles are in 5.69 g of NaOH? l We need to change 5.69 grams NaOH to moles l 1mole Na = 22.99 g 1 mol O = 16.00 g 1 mole of H = 1.008 g l 1 mole NaOH = 39.998 g

15. For example How many moles are in 5.69 g of NaOH? l We need to change 5.69 grams NaOH to moles l 1mole Na = 22.99 g 1 mol O = 16.00 g 1 mole of H = 1.008 g l 1 mole NaOH = 39.998 g

16. For example How many moles are in 5.69 g of NaOH? l We need to change 5.69 grams NaOH to moles l 1mole Na = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.008 g l 1 mole NaOH = 39.998 g

17. Calculating Percent Composition of a Compound   Like all percent problems: part whole 1) 1)Next, divide by the total mass of the compound; then x 100 2) 2)Find the mass of each of the components (the elements), x 100 % = percent

18. Example   Calculate the percent composition of a compound that is made of 29.0 grams of Ag with 4.30 grams of S. 29.0 g Ag 33.3 g total X 100 = 87.1 % Ag 4.30 g S 33.3 g total X 100 = 12.9 % S Total = 100 %

19. Getting it from the formula   If we know the formula, assume you have 1 mole,   then you know the mass of the elements and the whole compound (these values come from the periodic table!).

20. Examples   Calculate the percent composition of C 2 H 4.   How about aluminum carbonate? 85.63% C, 14.37 % H 23.1%Al, 15.4% C, and 61.5 % O

21. Formulas Example: molecular formula for benzene is C 6 H 6 (note that everything is divisible by 6) Example: molecular formula for benzene is C 6 H 6 (note that everything is divisible by 6) Therefore, the empirical formula = (the lowest whole number ratio) Therefore, the empirical formula = CH (the lowest whole number ratio) Empirical formula: the lowest whole number ratio of atoms in a compound. Molecular formula: the true number of atoms of each element in the formula of a compound.

22. Formulas (continued) Formulas for ionic compounds are ALWAYS empirical (the lowest whole number ratio = cannot be reduced). Examples: NaClMgCl 2 Al 2 (SO 4 ) 3 K 2 CO 3

23. Formulas (continued) Formulas for molecular compounds might be empirical (lowest whole number ratio). Molecular: H2OH2O C 6 H 12 O 6 C 12 H 22 O 11 Empirical: H2OH2O CH 2 OC 12 H 22 O 11 (Correct formula) (Lowest whole number ratio)

24. Calculating Empirical   Just find the lowest whole number ratio C 6 H 12 O 6 CH 4 N   A formula is not just the ratio of atoms, it is also the ratio of moles.   In 1 mole of CO 2 there is 1 mole of carbon and 2 moles of oxygen.   In one molecule of CO 2 there is 1 atom of C and 2 atoms of O. = CH 2 O = this is already the lowest ratio.

25. Calculating Empirical   We can get a ratio from the percent composition. 1) 1)Assume you have a 100 g sample - the percentage becomes grams (75.1% = 75.1 grams) 2) 2)Convert grams to moles. 3) 3)Find lowest whole number ratio by dividing each number of moles by the smallest value.

26. Example   Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N.   Assume 100 g sample, so   38.67 g C x 1mol C = 3.220 mol C 12.01 g C   16.22 g H x 1mol H = 16.09 mol H 1.008 g H   45.11 g N x 1mol N = 3.220 mol N 14.01 g N Now divide each value by the smallest value

27. Example   The ratio is 3.220 mol N = 1 mol N 3.220 mol N   The ratio is 3.220 mol C = 1 mol C 3.220 mol N   The ratio is 16.09 mol H = 5 mol H 3.220 mol N = C 1 H 5 N 1 which is = CH 5 N

28. Practice   A compound is 43.64 % P and 56.36 % O. What is the empirical formula? P2O5P2O5P2O5P2O5   Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula? C4H5N2OC4H5N2OC4H5N2OC4H5N2O

29. Empirical to molecular   Since the empirical formula is the lowest ratio, the actual molecule would weigh more.   By a whole number multiple.   Divide the actual molar mass by the empirical formula mass – you get a whole number to increase each coefficient in the empirical formula   Caffeine has a molar mass of 194 g. what is its molecular formula? = C 8 H 10 N 4 O 2

30. Hydrates Hydrate: a compound that has a specific number of water molecules bound to its atoms Hydrate: a compound that has a specific number of water molecules bound to its atoms You can determine the formula of a hydrate by heating it until the water is gone, and analyzing the data You can determine the formula of a hydrate by heating it until the water is gone, and analyzing the data

31. BaCl 2  ?H 2 O Mass of the compound before heating (hydrate): 5.00 g Mass of the compound before heating (hydrate): 5.00 g Mass after heating (anhydrous): 4.26 g Mass after heating (anhydrous): 4.26 g Find the mass of the water lost by subtracting 5.00g – 4.26 g = 0.74 g H 2 O Find the mass of the water lost by subtracting 5.00g – 4.26 g = 0.74 g H 2 O The mass of just the BaCl 2 is 4.26 g The mass of just the BaCl 2 is 4.26 g

32. Continued Convert the mass of BaCl 2 and the mass of H 2 O to mol Convert the mass of BaCl 2 and the mass of H 2 O to mol 4.26 g BaCl 2  1 mol BaCl 2 = 0.02046 mol 4.26 g BaCl 2  1 mol BaCl 2 = 0.02046 mol 208.2 g BaCl 2 208.2 g BaCl 2 0.74 g H 2 O  1 mol H 2 O = 0.04107 mol 0.74 g H 2 O  1 mol H 2 O = 0.04107 mol 18.016 g H 2 O 18.016 g H 2 O Divide by the smallest molar value Divide by the smallest molar value

33. Continued 0.02046 mol BaCl 2 = 1 0.02046 mol BaCl 2 = 1 0.02046 mol BaCl 2 0.02046 mol BaCl 2 0.04107 mol H 2 O = 2 0.04107 mol H 2 O = 2 0.02046 mol BaCl 2 0.02046 mol BaCl 2 Formula of the hydrate: BaCl 2  2H 2 O Formula of the hydrate: BaCl 2  2H 2 O

34. Another Practice FePO 4  ? H 2 O FePO 4  ? H 2 O Mass before heating: 7.61 g Mass before heating: 7.61 g Mass after heating: 5.15 g Mass after heating: 5.15 g Mass of water: 2.46 g Mass of water: 2.46 g Formula of the hydrate: FePO 4  4H 2 O Formula of the hydrate: FePO 4  4H 2 O