What can we tell from a
balanced equation?
C + O2 CO2
CH4 + 2O2 CO2 + 2H2O
CH4(g) + 2O2(g) 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.
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.
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.
How many moles and grams
are in 2.23 x 1023 atoms of Al.
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?
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.
100%whole
part Percentage
• 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%
How to determine the percent composition
for each element in a formula
Percent composition for elements in C2H5OH
• Determine the mass of each element in 1 mole of the compound C2H5OH. • 2 moles C = 2( g) = g
• 6 moles H = 6( g) = g
• 1 moles O = 1( 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, whole-number 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
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
• 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 benzopyrene, C20H12
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