Download Molar Mass and Formulas

Chemical Composition:
Molar Mass and Formulas
Chem 101
Lectures 12 and 13
What can we tell from a
balanced equation?
C + O2
CH4 + 2O2
CH4(g) + 2O2(g)
CO2
CO2 + 2H2O
CO2(g) + 2H2O(l)
What else can we figure out
from a balanced equation?
• If you know how many C atoms at start
– how many O2 molecules you will need
– how many CO2 molecules you can make
• Count the atoms
• Weigh the carbon and calculate the
number of atoms.
• Atomic masses allow us to convert
weights into numbers of atoms.
Moles
• The mass of 1.0 mole of an element is
equal to the atomic mass in grams.
• One mole element = 6.022 x 1023 atoms.
This number is called Avogadro’s
number.
• Example: 1 mole of C atoms weighs 12.0
g and has 6.02 x 1023 atoms.
A 1-mol sample of graphite
(a form of carbon) weighs 12.01 g.
Hmco Photo File
Equalities
• For any element:
• 1 mole = 6.02 x 1023 atoms = Atomic mass
• 1 mole of calcium =
– 6.02 x 1023 atoms of calcium
– 40.078 grams of calcium
• 1 mole of oxygen (O) atoms =
– 6.02 x 1023 atoms of oxygen
– 15.999 grams of oxygen
• 1 mole of oxygen gas (O2) =
– 12.04 x 1023 atoms of oxygen
– 31.998 grams of oxygen
All of these examples of
pure elements contain the same number
(a mole) of atoms: 6.02 x 1023 atoms.
Hmco Photo File
One-mole samples of iron (nails),
iodine crystals, liquid mercury, and powdered
sulfur.
Hmco Photo File
How many moles and atoms are
in 10.0 g of Al?
1. What are the equalities?
2. Use the atomic mass as a conversion
factor for grams-to-moles.
3. Use Avogadro’s number as a conversion
factor for moles-to-atoms.
How many moles and grams
are in 2.23 x 1023 atoms of Al.
1. What are the equalities?
2. Use Avogadro’s number as a conversion
factor for atoms-to-moles.
3. Use atomic mass as a conversion factor
for moles-to-grams.
Various numbers of methane molecules
and their constituent atoms.
One mole of SO2 contains
1 mole of S and 2 moles of O.
Molar Mass
• Molar mass: the mass in grams of one mole of a
compound
• The relative weights of molecules can be calculated
from atomic masses
Water = H2O = 2(1.008 g) + 16.00 g = 18.02 g
• 1 mole of H2O will weigh 18.02 g, therefore the
molar mass of H2O is 18.02 g
• 1 mole of H2O will contain 16.00 g of oxygen and
2.02 g of hydrogen
Molar Mass
• For any molecule:
• 1 mole = 6.02 x 1023 molecules = molar mass
• 1 mole of hydrogen gas = 1 mole of H2
– 6.02 x 1023 molecules of H2
– 2.016 grams of H2
– 2 x 6.02 x 1023 atoms of H for a total of
12.04 x 1023 atoms or 1.204 x 1024 atoms
Molar Mass
• For any molecule:
• 1 mole = 6.02 x 1023 molecules = molar mass
• 1 mole of water = 1 mole of H2O
– 6.02 x 1023 molecules of H2O
– 18.02 grams of H2O
– How many atoms?
• 1 mole of hydrogen gas = 1 mole of H2
– 6.02 x 1023 molecules of H2
– 2.016 grams of H2
– How many atoms?
Practice Mole Problems
1.
How many moles are in 5.5 x 1025 atoms of S?
2.
What is the mass in g of 1.505 x 1024 atoms of Cl?
3.
How many moles of Ne are in 25.0 g of Ne?
4.
How many atoms of Na are in 75.0 g of Na?
5. Which contains more atoms: 50.0 g of Al or
50.0 g of Fe?
6. Which contains more Ni: 20.0g
2.85 x 1023 atoms
0.45 mol?
7. How many molecules are in 0.5 moles of CaCl2?
Chemical Composition: Part 2
• Percent composition
• Empirical formulas
• Molecular formulas
What is Percent Composition?
• Percentage by mass of each element in a
compound
• Can be determined from the formula of the
compound or by experimental mass analysis of
the compound
• The percentages may not always total to 100%
due to rounding.
Percentage
part
100%
whole
How to determine the percent composition
for each element in a formula
•
Determine the mass of each element in 1
mole of the compound.
•
Determine the molar mass of the compound
by adding the masses of the elements.
•
Divide the mass of each element by the molar
mass of the compound and multiply by 100%
What is the percent composition for
each element in Ethanol? C2H5OH
Percent composition for elements in C2H5OH
• Determine the mass of each element in 1 mole
of the compound C2H5OH.
• 2 moles C = 2(
• 6 moles H = 6(
• 1 moles O = 1(
g) =
g) =
g) =
g
g
g
• Determine the molar mass of the compound by
adding the masses of the elements.
• 1 mole C2H5OH =
g
• Divide the mass of each element by the molar mass of
the compound and multiply by 100%
• C
• H
• O
Empirical Formulas
• Empirical formula: the simplest, wholenumber ratio of atoms in a molecule
– Can be determined from percent composition
or by combining masses
– Different substances can have the same
empirical formula
• Molecular formula: a multiple of the
empirical formula
Same Empirical Formula
CH2O
Formaldehyde
Acetic acid
Dextrose
Benzopyrene, C20H12
• Benzopyrene is found in nature from the eruption of
volcanoes and forest fires. It is also produced by burning
plants, wood, coal, and operating cars, trucks and other
vehicles.
• The major indoor sources of benzopyrene in the air are
wood-burning fireplaces and stoves, and tobacco
smoking.
• Benzopyrene can be found in surface water, tap water,
rainwater, groundwater, wastewater and sewage sludge.
• No known industry production or use.
• Skin and eye irritant, Carcinogenic
Determine the empirical formula
of benzopyrene, C20H12
•
Find the greatest common factor (GCF) of the
subscripts.
factors of 20 = (10 x 2), (5 x 4)
factors of 12 = (6 x 2), (4 x 3) GCF = 4
•
Divide each subscript by the GCF to get the
empirical formula.
C20H12 = (C5H3)4
Empirical Formula = C5H3
Determine the empirical formula of acetic
anhydride by percent composition:
47% carbon, 47% oxygen, and 6.0% hydrogen.
Convert the percentages to grams by assuming you have
100 g of the compound.
Convert the grams to moles
Divide by the smallest number of moles
If any of the ratios is not a whole number, multiply all the
ratios by a factor to make it a whole number.
Use the ratios as the subscripts in the empirical formula.
Molecular Formulas
• The molecular formula is a multiple of
the empirical formula.
• To determine the molecular formula you
need to know the empirical formula and
the molar mass of the compound.
Determine the molecular formula of
benzopyrene if it has a molar mass of 252 g
and an empirical formula of C5H3
• What is the empirical formula: C5H3
• Determine the molar mass of C5H3
• 5 C =_____g, 3 H =_____g, so C5H3 =______g
• Divide the given molar mass of the compound by
the molar mass of the empirical formula
• Round to the nearest whole number
• Multiply the empirical formula by the calculated
factor to give the molecular formula
(C5H3)__ = C H
Empirical and Molecular
Formulas
• Different compounds may have the same
empirical formula
• The molecular formula and structure
identifies the different compounds