Name:
Period:
Date:______/_____/______
Chapter 10: The Mole
How did Avogadro determine that particles have different masses? (1811)
What is the actual mass of a carbon atom?
Mass of proton : 1,6726 x 10-27 kg
Mass of neutron: 1,6749 x 10-27 kg
Average mass of a single
carbon atom (in amu)
What are the relative masses of each of the elements?
To make calculating easier, scientists assigned 1 au ≈ 1.660538782(83) × 10−27 kg
Chapter 9
Stoichiometry
How much is a mole?
The value of a mole was determined as the number of atoms in exactly 12.0 g of
carbon-12. (A mole allows chemists to work in grams rather than amu’s)
The molar mass of a substance is the mass of one mole of that
substance.
What is the mass of 6.02 x 1023 atoms of Na?
Al?
Rounding Values…mass of Cl?
Converting number of particles to moles…
A balloon contains 8.74 x 1023 atoms of
He. Determine the number of moles of
He.
From moles to number of particles…
Determine the number of atoms in 0.36
moles of Al.
Calculating Molar Mass of Compounds.
H2O = 2 atoms of H @ 1 g + 1 atom of O @ 16 g = 18 g
KCl (Potassium Chloride) = K (40 g) + Cl (35.5 g) = 75.5 g
What about the mass of 1 mole of NaOH?
HCl?
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How to Convert grams to moles…
Find the mass of 0.650 moles of P2O5.
How to Convert moles to grams….
To carry out a chemical reaction you
need 3.20 moles of zinc nitrate. What
is the mass of 3.20 moles of zinc
nitrate?
A bottle of NaNO3 contains 100.0 g of the compound. How many moles of
NaNO3 does it contain?
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Multistep Conversions: Moles and Gases
At Standard Temperature (O0 C) and Pressure (1 atm) conditions, 1 mole of gas
occupies 22.4 L.
Empirical vs. Molecular Formulas
Molecular Formula Empirical Formula
H2O
H2O
CH3COOH
CH2O
CH2O
CH2O
C6H12O6
CH2O
What is the difference between an Empirical Formula and a Molecular Formula?
Notice two things:
1. The molecular formula and the empirical formula can be identical.
2. You scale up from the empirical formula to the molecular formula by a whole
number factor.
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A Simple Rhyme for a Simple Formula
by Joel S. Thompson
Percent to mass
Mass to mole
Divide by small
Multiply 'til whole
Example: A compound was determined to consist of 72.2% magnesium and 27.8%
nitrogen by mass. What is the empirical formula?
(1) Percent to mass: Assume 100 g of the substance, then 72.2 g magnesium and 27.8 g
nitrogen.
(2) Mass to moles:
Mg: 72.2 g Mg x (1 mol Mg/24.3 g Mg) = 2.97 mol Mg
N: 27.8 g N x (1 mol N/14.0 g N) = 1.99 mol N
(3) Divide by small:
Mg: 2.97 mol / l.99 mol = 1.49
N: 1.99 mol / l.99 mol = 1.00
(4) Multiply 'til whole:
Mg: 2 x 1.49 = 2.98 (i.e., 3)
N: 2 x 1.00 = 2.00
The formula of the compound is Mg3N2.
© 1988 by the Division of Chemical Education of the American Chemical Society, Inc.
Journal of Chemical Education
Vol. 65, No. 8; August 1988, p. 704
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Molecular Formula
Percent to mass
Mass to mole
Divide by small
Multiply 'til whole
A compound is analyzed and found to contain 68.54% carbon, 8.63% hydrogen, and
22.83% oxygen. The molecular weight of this compound is known to be approximately
140 g/mol. What is the molecular formula?
To determine the molecular formula, you need to first determine the empirical
formula
1) Percent to mass. Assume 100 grams of the substance is present, therefore its
composition is:
carbon: 68.54 grams
hydrogen: 8.63 grams
oxygen: 22.83 grams
(2) Mass to moles. Divide each mass by the proper atomic weight.
carbon: 68.54 / 12.011 = 5.71 mol
hydrogen: 8.63 / 1.008 = 8.56 mol
oxygen: 22.83 / 16.00 = 1.43 mol
(3) Divide by small:
carbon: 5.71 ÷ 1.43 = 3.99
hydrogen: 8.56 ÷ 1.43 = 5.99
oxygen: 1.43 ÷ 1.43 = 1.00
(4) Multiply 'til whole. Not needed since all values came out whole.
The empirical formula of the compound is C4H6O.
To determine the molecular formula, simply divide the molecular weight by the
empirial weight to determine the ratio.
1) Calculate the "empirical formula weight." C4H6O = 70.092.
2) Divide the molecular weight by the empirical formula weight: 140 ÷ 70 = 2
3) Multiply the subscripts of the empirical formula by the factor just computed.
C4H6O times 2 gives a formula of C8H12O2.
This is the molecular formula.
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Empirical Formula Practice Problems
1) A compound is found to have (by mass) 48.38% carbon, 8.12% hydrogen and the rest oxygen.
What is its empirical formula?
2) A compound is found to have 46.67% nitrogen, 6.70% hydrogen, 19.98% carbon and 26.65%
oxygen. What is its empirical formula?
3) A compound is known to have an empirical formula of CH and a molar mass of 78.11 g/mol.
What is its molecular formula?
4) Another compound, also with an empirical formula if CH is found to have a molar mass of
26.04 g/mol. What is its molecular formula?
5) A compound is found to have 1.121 g nitrogen, 0.161 g hydrogen, 0.480 g carbon and 0.640 g
oxygen. What is its empirical formula? (Note that masses are given, NOT percentages.)
6) A compound containing only carbon, hydrogen and oxygen is found to be 48.38% carbon and
8.12% hydrogen by mass. What is its empirical formula?
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