1. Chapter 3: Chemical Proportions in Compounds

2. Terms: u The law of definite proportions describes that elements in a given compound are always present in the same proportions by mass. u The Percentage composition of a compound refers to the relative mass of each element in the compound.

3. Finding Percentage composition from a chemical formula: 1. Find the mass of the compound, then find the mass of the individual parts (atoms). 2. Divide the atoms by the compound and multiply by 100% For example: H 2 O H 2 = 2 x 1.01 = 2.02 +O = +16 Total molecular mass = 18.02  Percentage by mass of H in water = 2.02/18.02 =.11 x 100% = 11%  Percentage by mass of O in water = 16/18.02 =.888 x 100% = 89%

4. Percentage Composition from mass 1.Take the mass of the atom and divide by the total mass of the compound, then multiply by 100% Example: Mass of compound is 48.72g, it contains 32.69g of zinc and 16.03g of sulfur %Zn = (32.69g/ 48.72g) x 100% = 67.10% %S = (16.03g/ 48.72g) x 100% = 32.90%

5. Practice: Calculate the mass percentage of nitrogen in each compound: a)N 2 O b)Sr(NO 3 ) 2 c)NH 4 NO 3 d)HNO 3 e)NH 4 f)KNO 3

6. Empirical Formulas  The empirical formula is simply the simplest whole number ratio of the elements in a compound.  For example, the empirical formula of ethene (C 2 H 4 ) is CH 2 and the empirical formula of butene (C 4 H 8 ) is also CH 2

7. Comparing NameMolecularEmpiricalLow ratio Hydrogen peroxide H2O2H2O2 HO1:1 GlucoseC 6 H 12 O 6 CH 2 O1:2:1 Benzene C6H6C6H6 CH1:1 Ethene C2H4C2H4 CH 2 1:2 Butene C4H8C4H8 CH 2 1:2 Aniline C6H7NC6H7NC6H7NC6H7N 6:7:1 WaterH2OH2OH2OH2O2:1

8. Finding Empirical formula: Calculate empirical formula of a compound that is 85.6% carbon and 14.4% hydrogen. First, assume that you have 100g of the substance, so you really have 85.6g of C and 14.4g of H. Next, convert it to moles using molar mass. Finally, find the lowest whole ration between the two.

9. 1.85.6g C and 14.4g H 2.mol C = 85.6g x mol = 7.13mol C 12.01g mol H = 14.4g x mol = 14.3 mol H 1.01g 3.Lowest mole ratio: C = 7.13 /7.13 = 1 H = 14.3/ 7.13 = 2.01, so 2 ANSWER: CH 2

10. Practice: Find the empirical formula for the following: 1.17.5% hydrogen and 82.4% nitrogen 2.46.3% lithium and 53.7% oxygen 3.15.9% boron and 84.1% fluorine 4.52.51% chlorine and 47.48% sulfur

11. Finding Molecular Formula The empirical formula is CH 2 O, and its molar mass is 150g/mol. What is molecular formula?  Find empirical molar mass of CH 2 O: 12.01 + 2.02 + 15.99 g/mol =30.02 g/mol  Divide real molar mass by empirical molar mass: 150/30.02 = 5  Multiply subscripts of empirical formula through by ratio (5): C 5 H 10 O 5

12. Hydrated Ionic compounds  Many ionic compounds crystallize from a water solution with water molecules incorporated into their crystal structure  This is called a HYDRATE  Ex: CuCl 2 2H 2 O The dot does NOT stand for multiplying by water but shows a weak bond between the waters and the compound.

13. MgSO 4. 7H 2 O  Every formula unit of MgSO 4, there are 7 water molecules attached to it.  Naming: magnesium sulfate-7-water OR magnesium sulfate heptahydrate  2=di, 3=tri, 4=tetra, 5=penta, 6=hexa, 7=hepta, 8=octa, 9=nona, 10=deca

14. Determining the formula of a hydrate  A 50.0 g sample of a hydrated Ba(OH)2 contains 27.2 g of Ba(OH)2. (a) calculate the % by mass of water (b) find the number of waters in the hydrate

15. Ba(OH) 2. xH 2 O % mass = (total mass) – (mass of Ba(OH) 2 ) x 100% of water(total mass of sample) (a)% mass = (50.0 g – 27.2 g) x 100% =45.6% 50.0 g

16. Ba(OH) 2. xH 2 O (b)Moles Ba(OH) 2 =27.2 g Ba(OH) 2 x mol Ba(OH) 2 171.3 g Ba(OH) 2 = 0.159 mol Ba(OH) 2 Moles H 2 O =(50-27.2 g H 2 O ) x mol H 2 O 18.02 g H 2 O =1.27 mol H 2 O Do mole ratio between the two: Ba(OH) 2 : 0.159/0.159 =1 H 2 O: 1.27/0.159 = 8 The formula is Ba(OH) 2. 8H 2 O

17. Advanced Level Question: 1.Organic compound, X, with a molar mass of approximately 88 g/mol contains 54.5 % carbon, 36.3 % oxygen and 9.2% hydrogen by mass. a)Determine the empirical formula of X. b)Determine the molecular formula of X. c)X is a straight chained carboxylic acid. Draw its structural formula. d)Draw the structural formula of an isomer of X which is an ester. c) and d) require ORGANIC UNIT to be answered.

18. Practice  Pg. 82 #1-3 (Percent composition)  Pg. 89 # 9-12 (empirical formula)  Pg. 97 #17-19 (molecular formula)  Pg 103 # 23-25 (hydrates)