Empirical Formulas and Molecular Formulas. Ch 3.5

Transcription

1 Empirical Formulas and Molecular Formulas Ch 3.5

2 Empirical Formulas are the simplest (lowest) whole number ratio of atoms in a molecule or ionic compound Molecular Formulas are true formulas. For example: C 6 H 6 = CH H 2 O 2 = HO C 6 H 12 O 6 = CH 2 O

3 Empirical Formulas can be determined from % composition, here is the process: 1. % is the same as grams 2. Convert from grams to moles 3. Next divide by the smallest # of moles 4. this gives the empirical formula

4 Empirical Formula of Eugenol a Component of Clove Oil? is 73.14% C, 7.37% H and g O Remember, % is the same as grams (g)

5 Empirical Formula of Eugenol, continued Next, convert to the central unit, the mole g C x 1 mole = 6.09 mol C g 7.37 g H x 1 mole H = 7.31 moles H g g O x 1 mole O = 1.22 moles O g

6 Empirical Formula of Eugenol, continued Finally divide by the smallest # of moles 6.09 mol C 7.31 moles H 1.22 moles O 1.22 mol 1.22 mol 1.22 moles C: 4.99 H: 5.99 O: 1.00 Or C: 5 atoms H: 6 atoms O: 1.00 atoms Therefore C 5 H 6 O is Eugenol s empirical formula

7 Eugenol Count the number of carbon atoms, hydrogen atoms and oxygen atoms, does this fit the empirical formula that we just derived? No, because we did not find the molecular formula

8 Molecular Formulas Molecular formulas are also known as the true formula of a molecule. To derive this use amu: Molecular Formula = True amu empirical amu

9 True Formulas The molar mass of Eugenol is g/mol, what s the molecular formula of Eugenol? Use: True amu empirical amu g/mol = g/mol = 2 C 5 H 6 O 82 g/mol Therefore 2(C 5 H 6 O) = C 10 H 12 O 2

10 Complete Practice Problems #1-#18 Quiz Next Class Period, THEN begin Lab!... Whew! Finally!!

11 Hydrated Compounds

12 Hydrates A compound that is hydrated is called a hydrate since they form solids that include water in their crystal structure.

13 CuSO 4 5 H 2 O When figuring the molar mass should you add the amu of water? Yes, therefore CuSO 4 5 H 2 O has an amu of g/mol. The dot does NOT mean to multiple the amu masses.

14 Notice the color difference of the anhydrous crystals & hydrated crystals Cobalt (II) Chloride Copper (II) Sulfate

15 Naming Hydrates 1. Name the criss-cross compound 2. Use number prefixes to indicate the number of waters 3. Example: Copper (II) Sulfate Pentahydrate CuSO 4 5 H 2 O

16 Number Prefixes Mono- = 1 Di- = 2 Tri- = 3 Tetra- = 4 Penta- = 5 Hexa- = 6 Hepta- = 7 Octa- = 8 Nona- = 9 Deca- = 10

17 Play Movie By simply heating the solid, water can be driven from a hydrate to leave an anhydrous compound.

18 Naming Hydrates To name hydrates: 1. Name the compound 2. Plus the word hydrate use prefixes to indicate how many waters are associated with the compound 3. Example: Copper (II) Sulfate pentahydrate 4. To write their formulas Write: the name of the compound number of H 2 O CuSO 4 5 H 2 O

19 Calculate The Empirical Formula Of Ca(NO 3 ) 2 H 2 O

20 Units of Hydration A student heats hydrated crystals of CuSO 4, how many moles of water are associated with the crystals? Step 1: Find the mass of the crystals: g of CuSO 4 x H 2 O Step 2: Subtract the dehydrated crystal mass from the initial crystal mass = mass of water g of CuSO 4 x H 2 O g of CuSO 4 = g water

21 Units of Hydration Continued Step 3: Determine the number of moles g H 2 O x 1 mol g = mol H 2 O g CuSO 4 x 1 mol/159.6 g = mol CuSO 4 Step 4: Determine the molar ratio (see above) mol H 2 O mol CuSO mol CuSO mol CuSO 4 1 CuSO 4 5 H 2 O

22 Combustion Formula of a Compound Ch 3.5

23 The AP Exam will most likely give you a combination of work to complete by using a combustion device which analyzes substances containing C and H. It is burned in excess O 2 producing CO 2 and H 2 O, these products are then collected and from here one can determine the %C in CO 2 and %H in H 2 O

24 Combustion Analysis C x H y (any hydrocarbon) + O 2 H 2 O + CO 2

25 g of a compound is reacted with O 2 & g of CO 2 & g of H 2 O is collected. The unknown compound has C, H and N, what s the empirical & molecular formula molar mass = g/mol? Go in the order of C, H, O, (N) Remember %C in CO 2 (part/whole x 100% = % comp.): %C: g C x g CO 2 = g C g CO 2 % H: g H(note 2 hydrogen)x g H 2 O = g of H g H 2 O g compound = g C g H + g N = g N

26 gC x 1mol g H x 1mol g N x 1mol g 1.01 g g = mol C = mol H = mol N Divide by Smallest number of moles = mol 1 C : 5 H : 1 N CH 5 N empirical amu = True (given) amu = /31.07 = 1 (CH 5 N ) = CH 5 N is true formula

27 Percent Yield

28 Theoretical Yield The amount of product formed is controlled by the limiting reactant products stop forming when on reactant runs out. The amount of product calculated in this way is called the theoretical yield. This is the amount of product predicted from the amount of reactants used.

29 Actual Yield However, the amount of product predicted (the theoretical yield) is seldom obtained. One reason for this is the presence of side reactions (other reactions that consume one or more of the reactants or products). The actual yield of product, is the amount of product actually obtained.

30 Percent Yield The comparison of the product actually obtained and theoretically obtained is called the percent yield: Percent Yield = Actual Yield x 100% Theoretical Yield