Honors1011 molar mass and percent composition
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  • 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