Empirical Formulas and Molecular Formulas - Ms. Lara La...
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Transcript of Empirical Formulas and Molecular Formulas - Ms. Lara La...
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:
• C6H6 = CH
• H2O2 = HO
• C6H12O6 = CH2O
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 C5H6O
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
C5H6O 82 g/mol
Therefore 2(C5H6O) = C10H12O2
• 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.
CuSO4 • 5 H2O
When figuring the molar mass should you add the amu of water?
Yes, therefore
CuSO4 • 5 H2O 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
CuSO4 • 5 H2O
Number Prefixes
• Mono- = 1
• Di- = 2
• Tri- = 3
• Tetra- = 4
• Penta- = 5
• Hexa- = 6
• Hepta- = 7
• Octa- = 8
• Nona- = 9
• Deca- = 10
• By simply heating the solid, water can be driven from a hydrate to leave an anhydrous compound.
Play Movie
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 H2O
CuSO4 • 5 H2O
Calculate The Empirical Formula Of Ca(NO3)2 _____ H2O
Units of Hydration
• A student heats hydrated crystals of CuSO4, how many moles of water are associated with the crystals?
• Step 1: Find the mass of the crystals:
1.023 g of CuSO4 • x H2O
• Step 2: Subtract the dehydrated crystal mass from the initial crystal mass = mass of water
1.023 g of CuSO4 • x H2O – 0.654 g of CuSO4
= 0.369 g water
Units of Hydration Continued…
• Step 3: Determine the number of moles 0.369 g H2O x 1 mol
18.02 g = 0.0205 mol H2O
0.654 g CuSO4 x 1 mol/159.6 g = 0.00410 mol CuSO4
• Step 4: Determine the molar ratio (see above) 0.0205 mol H2O 0.00410 mol CuSO4
0.00410 mol CuSO4 0.00410 mol CuSO4
1 CuSO4 • 5 H2O
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 O2 producing CO2 and H2O, these products are then collected and from here one can determine the %C in CO2 and %H in H2O…
Combustion Analysis CxHy(any hydrocarbon) + O2 H2O + CO2
• 0.1156 g of a compound is reacted with O2 & 0.1638 g of CO2 & 0.1676 g of H2O 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 CO2 (part/whole x 100% = % comp.):
%C: 12.01 g C x 0.1638 g CO2 = 0.04470 g C
44.01 g CO2
% H: 2.016 g H(note 2 hydrogen)x0.1676 g H2O = 0.01875 g of H
18.02 g H2O
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
CH5N empirical amu = 31.07
True (given) amu = 31.06
31.06/31.07 = 1 (CH5N ) = CH5N 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