3. According to the Law of Definite Proportions, a chemical compound always contains the same proportion of elements by mass. Consequently, if we know the formula of a compound, we can then calculate the percentage by mass of each element. We can compare the mass of a single element, to the mass of the cmpd.

4. Example: Let’s calculate the percentage by mass of each element in CO 2. We have to compare the mass of each element, to the total mass.

5. - First, calculate the mass of the whole compound (the gram formula mass). 1 x C = 1 x 12 = 12 2 x O = 2 x 16 = 32 GFM = 44 g/mol Of this 44 grams, how many were due to the presence of carbon? So the percentage of carbon in the compound is equal to : 12 x 100% = 27% carbon 44

6. To calculate the percentage of oxygen, we can do it one of two ways: Since there are only two elements, we can subtract the 27% calculated above from 100% to get 73%. Or We can calculate it the same way we did the carbon. Show Work: 32 x 100% = 73% 44

7. Example: Calculate the percentage of each element in CaClO 3. Calculate the GFM. 1 x Ca = 1 x 40.1 = 40.1 1 x Cl = 1 x 35.5 = 35.5 3 x O = 3 x 16.0 = 48.0 GFM = 123.6 g/mol 40.1 x 100% = 32.4% calcium 123.6 35.5 x 100% = 28.7% chlorine 123.6 48.0 x 100% = 38.8 % oxygen 123.6

8. Note that the total percentages might not add up to exactly 100% due to rounding.

9. Example: What is the percent by mass of nitrogen in ammonium phosphate? FORMULA: Answer: (NH 4 ) 3 PO 4 GFM: 149 g Mass of N: 42 g = 28%

10. Ex: A compound containing nitrogen and oxygen has a a mass of 80.0 grams. Experiments show that the 80.0 grams is made up of 56.0 grams of oxygen and 24.0 grams of nitrogen. What is the % composition, by mass, of each element? This is an example of “experimental” determination. You do not need the formula, the PT, or the GFM. You just need brains. Like the scarecrow. SOLUTION:

12. A hydrate is an ionic compound with a fixed ratio of water molecules loosely bonded into the crystal structure. (Trapped inside the crystal lattice) When the water is removed (by heating the substance)… the salt that remains is called the anhydrous salt.

13. When calculating the gram formula mass of a hydrate, think of the dot as a plus sign. We add in the mass of the water molecules to the mass of the salt. TIP: When calculating the percent composition of a hydrate, treat the water molecule as a single unit.

14. Examples: CuSO 4. 5H 2 Ocopper (II) sulfate pentahydrate MgSO 4. 7H 2 Omagnesium sulfate heptahydrate Na 2 CO 3. 10H 2 O sodium carbonate decahydrate

15. Example 1: What is the GFM of Na 2 CO 3. 10H 2 O ?

16. Example 2: What is the gram formula mass of MgSO 4. 7H 2 O? Answer: 1 x Mg = 1 x 24.3 = 24.3 1 x S = 1 x 32.1 = 32.1 4 x O = 4 x 16.0 = 64.0 7 x H 2 O = 7 x 18.0 = 126.0 246.4 g/mol

17. Example 3: What is the percentage of water in MgSO 4 7H 2 O? Answer: % water = mass of water x 100% gram formula mass 126.0 x 100% = 51.1% water 246.4

18. Sometimes, you are asked to analyze hydrates given data collected in a laboratory setting… For example: A 10.40 gram sample of hydrated crystal is heated to a constant mass of 8.72 grams. What is the percent water? Why has the mass decreased? –The water evaporating has caused the mass of the salt to decrease. –This missing mass is ALWAYS the mass of the water.

19. Solution: 1. First, find the missing mass to determine the mass of the water. 2. Then plug into the “equation” below: 10.40 g – 8.72 g = 1.68 g Mass of Water Mass of Hydrate x 100 1.68 g 10.40 g x 100 = 16.1%

20. Example 3: 5.00 grams of a hydrated salt is heated to constant mass. Only 1.75 grams of the anhydrous (without water) salt remains. What is the percentage of water in the hydrate? Mass of water = mass of hydrate – mass of anhydrous salt 5.00 grams – 1.75 grams = 3.25 grams of water % water = grams of water x 100% grams of hydrate 3.25 grams x 100% = 65.0% water 5.00 grams

21. Ex. An empty evaporating dish is found to have a mass of 29.993 grams. A sample of hydrated crystal is placed into the evaporating dish, the combined mass is 39.486 grams. The dish is removed from the flame, and weighed, then reheated, then weighed again, until the mass remains a constant 38.378 grams. What is the mass of the hydrate? What is the mass of the water driven off by heating? What is the percent by mass of water?

22. Challenge Question: A hydrated salt of CuCO 3 is found to contain 22.6% water. What is the value of X in the formula CuCO 3 XH 2 O? 1.Assume you have 100 grams of the hydrate: Therefore: The mass of the water in the hydrate is 22.6 grams The mass of the anhydrous salt is 77.4 grams 2. Recall that the coefficients give us the MOLE ratios… find the moles of the water and anhydrous salt to get the formula:

24. Formulas can be interpreted differently depending on what information we need to gather. The ratio of subscripts can be interpreted differently, depending on the situation. Ex. S 2 O 3 This can interpreted as 2 atoms of sulfur combine with 3 atoms of oxygen. This can interpreted as 2 moles of sulfur combine with 3 moles of oxygen.

25. For us to be able to calculate the empirical and molecular formulas… we most of the time find it most useful to treat the subscripts as MOLES.

26. Empirical and Molecular Formulas: What’s the Difference? EMPIRICAL: the lowest, whole number ratio of elements in a formula. The subscripts can mean either atoms or moles of each respective element. MOLECULAR: whole number multiples of the empirical formula. They represent the ACTUAL number of atoms in a given formula, if we could count atoms ourselves

27. Example: Glucose Molecular formula = C 6 H 12 O 6 –This means a molecule of glucose contains 6 carbons, 12 hydrogens and 6 oxygens Empirical formula = CH 2 O –This is the simplest ratio, found using the greatest common factor of all subscripts.

28. An empirical formula is the lowest whole number ratio of atoms. Remember that this ratio of atoms is the same as the ratio of moles in the formula.

29. “True Formulas”

30. The relationship between empirical and molecular formulas: Molecular formulas are always a multiple of the empirical formula.

31. Although almost all ionic compounds are expressed as empirical formulas, the formula of a covalent compound is the actual formula of its molecules. This may not be lowest whole number ratio.

32. To determine the molecular formula, you have to know what integer to multiply the empirical formula by: This is accomplished by: 1. Determine the GFM of the empirical formula 2. Divide the molecular mass by the empirical mass to get a WHOLE NUMBER INTEGER. - This tells you how much larger the molecular (true) formula is than the empirical formula. 3. Multiply the subscripts of the empirical formula by the integer.

33. Example 1: The formula of a compound is C 6 H 14. What is its empirical formula? Answer: Reduce to lowest whole number ratio: C 3 H 7

34. The empirical formula of a compound is P 2 0 5. The GFM mass of the compound is 283.88 grams. What is the molecular formula? Example 2: The empirical formula of a compound is P 2 0 5. The GFM mass of the compound is 283.88 grams. What is the molecular formula? Step 1: Determine the GFM of empirical formula GFM = 142 grams. Step 2: Divide molecular mass by empirical GFM 283.88 / 142 grams = 2 Step 3: Multiply the empirical by the integer and round to the nearest whole number. P 2 O 5  P 4 O 10

35. Example 3: The empirical formula of a compound is C 2 H 3 and the molecular mass is 54.0 g/mol. What is the molecular formula? Step 1: Determine the GFM of empirical formula GFM =27.0 g/mol Step 2: Divide molecular mass by empirical GFM 54.0 / 27.0 = 2 Step 3: Multiply the empirical by the integer and round to the nearest whole number. C 2 H 3  C 4 H 6

36. Example 4: What is the molecular formula of a compound with an empirical formula of CH and a molecular mass of 52 g/mol? Sol’n: C 4 H 4