Empirical Formulas and Molecular Formulas Ch 3.5
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
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
Empirical Formula of Eugenol a Component of Clove Oil? is 73.14% C, 7.37% H and 19.49 g O Remember, % is the same as grams (g)
Empirical Formula of Eugenol, continued Next, convert to the central unit, the mole 73.14 g C x 1 mole = 6.09 mol C 12.01 g 7.37 g H x 1 mole H = 7.31 moles H 1.0079 g 19.49 g O x 1 mole O = 1.22 moles O 15.9994 g
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
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
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
True Formulas The molar mass of Eugenol is 164.2 g/mol, what s the molecular formula of Eugenol? Use: True amu empirical amu 164.2 g/mol = 164.2 g/mol = 2 C 5 H 6 O 82 g/mol Therefore 2(C 5 H 6 O) = C 10 H 12 O 2
Complete Practice Problems #1-#18 Quiz Next Class Period, THEN begin Lab!... Whew! Finally!!
Hydrated Compounds
Hydrates A compound that is hydrated is called a hydrate since they form solids that include water in their crystal structure.
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 249.7 g/mol. The dot does NOT mean to multiple the amu masses.
Notice the color difference of the anhydrous crystals & hydrated crystals Cobalt (II) Chloride Copper (II) Sulfate
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
Number Prefixes Mono- = 1 Di- = 2 Tri- = 3 Tetra- = 4 Penta- = 5 Hexa- = 6 Hepta- = 7 Octa- = 8 Nona- = 9 Deca- = 10
Play Movie By simply heating the solid, water can be driven from a hydrate to leave an anhydrous compound.
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
Calculate The Empirical Formula Of Ca(NO 3 ) 2 H 2 O
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: 1.023 g of CuSO 4 x H 2 O Step 2: Subtract the dehydrated crystal mass from the initial crystal mass = mass of water 1.023 g of CuSO 4 x H 2 O 0.654 g of CuSO 4 = 0.369 g water
Units of Hydration Continued Step 3: Determine the number of moles 0.369 g H 2 O x 1 mol 18.02 g = 0.0205 mol H 2 O 0.654 g CuSO 4 x 1 mol/159.6 g = 0.00410 mol CuSO 4 Step 4: Determine the molar ratio (see above) 0.0205 mol H 2 O 0.00410 mol CuSO 4 0.00410 mol CuSO 4 0.00410 mol CuSO 4 1 CuSO 4 5 H 2 O
Combustion Formula of a Compound Ch 3.5
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
Combustion Analysis C x H y (any hydrocarbon) + O 2 H 2 O + CO 2
0.1156 g of a compound is reacted with O 2 & 0.1638 g of CO 2 & 0.1676 g of H 2 O is collected. The unknown compound has C, H and N, what s the empirical & molecular formula molar mass = 31.06 g/mol? Go in the order of C, H, O, (N) Remember %C in CO 2 (part/whole x 100% = % comp.): %C: 12.01 g C x 0.1638 g CO 2 = 0.04470 g C 44.01 g CO 2 % H: 2.016 g H(note 2 hydrogen)x0.1676 g H 2 O = 0.01875 g of H 18.02 g H 2 O 0.1156 g compound = 0.04470 g C + 0.01875 g H + g N = 0.05211 g N
0.04470gC x 1mol 0.01875 g H x 1mol 0.05211g N x 1mol 12.01 g 1.01 g 14.01 g = 0.003722 mol C = 0.01860 mol H = 0.003719 mol N Divide by Smallest number of moles = 0.003719 mol 1 C : 5 H : 1 N CH 5 N empirical amu = 31.07 True (given) amu = 31.06 31.06/31.07 = 1 (CH 5 N ) = CH 5 N is true formula
Percent Yield
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.
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.
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