Transcript File

Unit 9:
Covalent Bonding
Chapters 8 & 9
Chemistry 1K
Cypress Creek High School
Percent Composition
• Percent Composition is the percentage by mass of
each element in a compound
– The percent composition is found by using the
following formula:
Total mass of element
% Mass 
100
molar mass of compound
Percent Composition
• What is the percent of C & H in C2H6?
– 79.89% carbon & 20.11% hydrogen
HYDROGEN
CARBON
Percent Composition
• What is the percent of each element in sodium sulfite, Na2SO3?
– 2 mole Na x 22.990 g/mol Na = 45.980 g Na
– 1 mole S x 32.066 g/mol S =
32.066 g S
– 3 moles O x 15.999 g/mol O = + 47.997 g O
126.043 g
Molar Mass of Na2SO3 =
• Sodium
– % = (45.980 g / 126.043 g) x 100 = 36.486 % Na
• Sulfur
– % = (32.066 g / 126.043 g) x 100 = 25.441 % S
• Oxygen
– % = (47.997 g / 126.043 g) x 100 = 38.080 % O
Empirical Formula
• An Empirical Formula is the LOWEST whole
number ratio of the elements in a compound. The
subscript cannot be a fraction.
• Ex: The empirical formula for caffeine (C8N4O2H10)
is C4N2OH5
– C8N4O2H10 is a completely different
chemical than C4N2OH5!
• What is the empirical formula
= Carbon
of glucose (C6H12O6)? Back
Blue = Nitrogen
CH2O
Red = Oxygen
Gray = Hydrogen
Empirical Formula Calculations
• There are 3 types of empirical calculations:
–Mole Ratio
–Grams
–Percent
Empirical Formula from Mole Ratios
• To calculate the empirical formula from mole ratios, follow
these steps:
1) Divide each mole quantity by the smallest mole quantity
2) Assign ratios as subscripts
3) If any of the ratios are not even (ex: 0.5, 1.25, 1.3),
multiply them all by the lowest common multiple to
achieve whole numbers then place these numbers as the
subscripts.
Empirical Formula from Mole Ratios
• Find the empirical formula for a compound
containing 2 mole carbon and 6 mole hydrogen.
•C = 2/2 = 1
•H = 6/2 = 3
LOWEST
empirical formula = CH3
Empirical Formula from Grams
• To calculate the empirical formula from grams, follow these steps:
1) Convert the grams to moles (divide grams by molar mass)
2) Divide each mole quantity by the smallest mole quantity
3) Assign ratios as subscripts
4) If any of the ratios are not even (ex: 0.5, 1.25, 1.3), multiply
them all by the lowest common multiple to achieve whole
numbers then place these numbers as the subscripts
Empirical Formula from Grams
•
Find the empirical formula of a compound that
contains 13.5 grams of calcium, 4.05 grams of carbon,
and 16.2 grams of oxygen.
calcium
carbon
oxygen
13.5 g
= 0.337
40.08 g/mol mol
4.05 g
16.2 g
= 0.337
= 1.01
12.011 g/mol mol 15.999 g/mol mol
LOWEST
0.337 = 1
0.337
0.337 = 1
0.337
1.01
=3
0.337
empirical formula = CaCO3
Empirical Formula from Percents
• To calculate the empirical formula from percent composition, follow
these steps:
1) Change % sign to grams. You assume 100 g of the compound
2) Convert the grams to moles (divide grams by molar mass)
3) Divide each mole quantity by the smallest mole quantity
4) Assign ratios as subscripts
5) If any of the ratios are not even (ex: 0.5, 1.25, 1.3), multiply them
all by the lowest common multiple to achieve whole numbers then
place these numbers as the subscripts
Empirical Formula from Percents
• Calculate the empirical formula of a compound
containing 18.8% nickel and 81.2% iodine.
nickel
18.8 g
58.69 g/mol
iodine
81.2 g
126.905 g/mol
= 0.320
mol
= 0.640
mol
LOWEST
0.320
0.320
=1
0.640
0.320
=2
empirical formula = NiI2
Molecular Formulas
• For many compounds, the empirical formula is not
the true formula.
– A molecular formula tells the exact number of atoms
of each element in a molecule or formula unit of a
compound.
• The actual formula
– The molecular formula for a compound is always a
whole-number multiple of the empirical formula.
Determining Molecular Formulas
1) Determine the empirical formula
2) Find the empirical formula mass (you are given
the molecular formula mass)
3) Divide the molecular formula mass by the
empirical formula mass. This is the scalar
(multiplier).
4) Multiply the subscripts in the empirical formula
by the multiplier.
Calculating Molecular Formula
• A compound has an empirical formula of Sc2O3. The
molecular mass is 414 g/mol. What is the molecular
formula?
– Empirical Formula = Sc2O3
– Empirical Formula Mass =
• 2 mol Sc x 44.956 g/mol Sc =
89.912 g Sc
• 3 mol O x 15.999 g/mol O = + 47.997 g O
molar mass = 137.909 g Sc2O3
– Find the multiplier…
• 414 g / 137.909 g = 3 (the multiplier is 3)
– Molecular Formula = Sc6O9
Practice Molecular Formula
• A molecule has 85.6% carbon and 14.5%
hydrogen. If the molecule has a mass of 42.1
grams, what is the molecular formula?
THE END.
Be Prepared for Unit 9 Test.