- The law of definite proportions states that a chemical compound always contains the same elements in the same proportions by mass, regardless of sample size. This allows determination of a compound's mass and elemental percentages. - Molar mass is the mass of one mole of a substance and can be calculated by adding the masses of each element in a compound based on its chemical formula. Common units are grams per mole (g/mol). - Percent composition by mass can be determined by calculating the mass of each element in a compound and expressing it as a percentage of the total mass of the compound.

1. Molar Mass and Percent Composition

2. Law of Definite Proportions John Dalton proposed that regardless of sample size a compound is always made up of elements combined in the same proportion by mass

3. This law allows us to determine the mass of a compound from the masses of its component elements. It also allows us to determine the percentage of every element in the compound.

4. This law allows us to have generic/store brand medications. Although the name seems different, the essential compounds in the medication are chemically the same. For example: the brand name of Tylenol ™ is the same as the generic brand of acetaminophen. There are numerous other brand names! Acetaminophen is C 8 H 9 NO 2

5. Atomic Mass - Element The mass of an element is found on the periodic table. We will round all masses off the table to two decimal places before using in any calculations!

6. Compound Mass – Counting Atoms For the mass of a compound, you have to consider both the number and types of atoms! A subscript (number appearing below) indicates the number of atoms (if molecule) or ions (if formula unit)

7. Example CH 4 (methane) contains 1 atom of carbon and 4 atoms of hydrogen AlPO 4 (aluminum phosphate)contains 1 ion of aluminum and 1 ion of phosphate but it also contains 1 atom of aluminum, 1 atom of phosphorus and 4 atoms of oxygen. We need to use the numbers of atoms in our calculation of molar mass.

8. Compound Mass – Counting Atoms If subscripts appear outside of parentheses, you need to multiply! Example: Mg(NO 3 ) 2 contains: 1 atom Magnesium 2 atoms of Nitrogen(2 x 1) 6 atoms of Oxygen (2 x 3)

9. Compound Mass – Counting Atoms If you have a hydrate (crystal with water enclosed) the formula will look something like this: CuSO 4 . 5H 2 O You need to multiply the number in front of the water molecule to get the correct # of atoms: Copper – 1 atom, Sulfur – 1atom, Oxygen – 4 atoms, water – 5 molecules or Cu -1, S – 1, O - (4 + 5) = 9atoms, H – 10atoms (The dot indicates addition or contains, not multiplication!)

10. Mass Calculation - Compound Multiply the # of atoms of each element in the compound by its corresponding mass from the periodic table (round to two decimal places) Round the product to the same # of significant digits as the mass Add the products and round based on rule – least # of decimal places Unit is amu

11. Examples! BaCl 2  ( 1 atom Ba x 137.33 ) + (2 atoms Cl x 35.45 ) = 137.33 + 70.90 = 208.23 amu Mg(NO 3 ) 2  (1 atom Mg x 24.30) + ((2x1) atoms nitrogen x 14.01) + ((2x3) atoms oxygen x 16.00) = 24.30 + 28.02 + 96.00 = 148.32 amu NOTE : I am typing these across the page, but on your paper it should be written down the page! See example on the board!

12. Molar Mass - Element (Text Reference Chapter 11) The mass of one mole of an element is equivalent to its atomic mass. Example: one atom of sodium has a mass of 22.99 amu; 22.99 grams Na = 1 mole Na Molar mass is the mass of one mole of any substance. Molar Mass Sodium = 22.99 g/mol

13. Molar Mass - Compound The mass of one mole of a compound is also based on the number and type of atoms. The calculation is the same as the mass of a compound! The unit becomes g/mol

14. Examples! BaCl 2  ( 1 atom Ba x 137.33 ) + (2 atoms Cl x 35.45 ) = 137.33 + 70.90 = 208.23 g/mol Mg(NO 3 ) 2  (1 atom Mg x 24.31) + ((2x1) atoms nitrogen x 14.01) + ((2x3) atoms oxygen x 16.00) = 24.31 + 28.02 + 96.00 = 148.33 g/mol

15. Hydrates Learn the molar mass of water! H2O  (2 atoms Hydrogen x 1.01) + (1atom oxygen x 16.00) = 2.02 + 16.00 = 18.02 g/mol CuSO 4 . 5H 2 O  (1 atom Cu x 63.55) + (1 atom Sulfur x 32.06) + (4 atoms Oxygen x 16.00) + (5 molecules of water x 18.02) = 63.55 + 32.06 + 64.00 + 90.10 = 249.71 g/mol

16. Molar Mass Practice! Calculate the molar mass of: 1. Sodium chloride, NaCl

17. Continued Practice 2. Lithium Phosphate, Li 3 PO 4

18. Continued Practice 3. Manganese (VII) carbonate, Mn 2 (CO 3 ) 2

19. Continued practice 4. Ferric chloride hexahydrate, FeCl 3 . 6H 2 O

20. Check it! 1.) NaCl  (1 x 22.99) + ( 1 x 35.45) = 58.44 g/mol 2. )Li 3 PO 4  (3 x 6.94) + (1 x 30.97) + (4 x 16.00) = 20.8 +30.97 + 64.00 = 115.77  115.8g/mol 3. )Mn 2 (CO 3 ) 2  (2 x 54.94) + (2 x 12.01) + (6 x 16.00) = 109.9 + 24.02 + 96.00 = 229.92  229.9 g/mol 4.) FeCl 3 . 6H 2 O  (1 x 55.85) + (3 x 35.45) + (6 x 18.02) = 55.85 + 106.4 + 108.1 = 270.35  270.4 g/mol Li3PO4

21. Mole Conversions 1 mole = molar mass (g) This equivalent relationship can be used to convert between moles and grams and reverse. Must calculate the molar mass first if a compound .

22. Example: How many moles are in 65.0 grams of sodium chloride? Given: 65.0g NaCl Want: moles NaCl Relationship – molar mass NaCl (calculated before): 58.44 g/mol  Means 58.44 g NaCl = 1 mol NaCl

23. Possible conversion factors: 58.44 g NaCl/1mol NaCl or 1mol NaCl/58.44g NaCl Set up: 65.0g NaCl x 1mol NaCl = 1.11mol NaCl 58.44 g NaCl

24. Example: How many grams are in 1.25 moles of lithium phosphate? Given: 1.25 moles Li 3 PO 4 Want: grams Li 3 PO 4 Relationship: molar mass Li 3 PO 4 (calculated earlier): 115.8g/mol  Means 115.8 g Li 3 PO 4 = 1mole Li 3 PO 4

25. Possible conversion factors: 115.8 g Li 3 PO 4 /1mole Li 3 PO 4 or 1mole Li 3 PO 4 / 115.8 g Li 3 PO 4 Set up: 1.25 moles Li 3 PO 4 x 115.8 g Li 3 PO 4 = 145g Li 3 PO 4 1mole Li 3 PO 4

26. Practice! 5. How many moles are in 3.4g Mg(OH) 2 ?

27. Continued Practice 6. How many grams are in 0.00500 moles of Ca 3 N 2 ?

28. Answers: 5. Need molar mass of Mg(OH) 2 : (1 x 24.30) + ((2x1)x16.00) + ((2 x 1) x 1.01) = 24.30 + 32.00 + 2.02 = 58.32 g/mol Means: 58.32 g Mg(OH) 2 = 1mol Mg(OH) 2 3.4 g Mg(OH) 2 x 1mole = 0.058mol 58.32 g

29. Answers (cont.) 6. Need molar mass of Ca 3 N 2 : (3 x 40.08) + (2 x 14.01) = 120.2 + 28.02 = 148.2 g/mol 0.00500 moles Ca 3 N 2 x 148.2 g = 0.741g 1 mole

30. Multiple Conversions Note: use previously learned conversion factors: 1 mole = 6.02 x 10 23 particles , 1 mole = molar mass(g) 7. How many grams are in 5.75 x 10 19 formula units of NaCl?

31. Continued Practice 8. How many molecules of nitrogen, N 2 , are in 165 grams of the gas?

32. Answers: 7. Need molar mass of NaCl: 22.99 + 35.45 = 58.44 g/mol 5.75 x 10 19 f.unit NaCl x 1 mole x 58.44 g 6.02 x 10 23 1 mole formula units = 5.58 x 10 -3 g

33. Answers (cont.) 8. Need the molar mass of N 2 : 2 x 14.01 = 28.02 g/mol 165 g N 2 x 1 mole x 6.02 x 10 23 molecules 28.02 g 1 mole = 3.54 x 10 24 molecules

34. Percent by Mass/Percent Composition of Compound You can find the mass of any element in a compound by first calculating the mass and using the general relationship: part/whole x 100 = %. The part is the element mass, the whole is the compound mass. If you find the percent of every element you are finding the percent composition.

35. Example BaCl 2  ( 1 atom Ba x 137.33 ) + (2 atoms Cl x 35.45 ) = 137.33 + 70.90 = 208.23 amu Percent Composition: 137.33/208.23 x 100 = 65.951% Ba 70.90/208.23 x 100 = 34.05% Cl

36. Example Mn 2 (CO 3 ) 2  (2 x 54.94) + (2 x 12.01) + (6 x 16.00) = 109.9 + 24.02 + 96.00 = 229.92  229.9 g/mol What is the percent by mass of manganese in the above compound? 109.0gMn/229.9g Mn 2 (CO 3 ) 2 x 100 = 47.80% Mn

37. Practice! 9. Find the percent composition of Acetaminophen: C 8 H 9 NO 2

38. Continued practice 10. Find the percent by mass of iron in Iron(III)sulfate: Fe 2 (SO 4 ) 3 .

39. Answers: 9. ) C 8 H 9 NO 2 (8 x 12.01) + (9 x 1.01) + (1 x 14.01) + (2 x 16.00) = 96.08gC + 9.09gH + 14.01gN + 32.00gO = 151.18g C 8 H 9 NO 2 96.08/ 151.18 x 100 = 63.29%C 9.09/ 151.18 x 100 = 6.01%H 14.01/ 151.18 x 100 = 9.267%N 32.00/ 151.18 x 100 = 21.17%O

40. 10. )Fe 2 (SO 4 ) 3 (2 x 55.85) + ((3x1) x 32.06) + ((3 x 4) x 16.00) = 111.7gFe + 96.18gS+ 192.0gO = 399.88  399.9g Fe 2 (SO 4 ) 3 111.7/399.9 x 100 = 27.93%Fe

41. Hydrates - % Water You can calculate the percent of water in the hydrate: CuSO 4 . 5H 2 O  (1 atom Cu x 63.55) + (1 atom Sulfur x 32.06) + (4 atoms Oxygen x 16.00) + (5 molecules of water x 18.02) = 63.55 + 32.06 + 64.00 + 90.10 gH 2 O = 249.71 g/mol CuSO 4 . 5H 2 O 90.10/249.71 x 100 = 36.08% H 2 O

42. Practice 11. Determine the percent of water in sodium sulfate decahydrate (Na 2 SO 4 . 10H 2 O)

43. Check Answer: 11.) Na 2 SO 4 . 10H 2 O (2 x 22.99) + (1 x 32.06) + (4 x 16.00) + (10 x 18.02) = 95.98 + 32.06 + 64.00 + 180.2 = 372.24 = 372.2 g/mol(1 decimal place) 180.2/372.2 x 100 = 48.41%H 2 O