The Mole

Posted on February 25, 2009

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>Atomic mass provides a means to count atoms by measuring the mass of a sample
The periodic table on the inside cover of the text gives atomic masses of the elements
The mass of an atom is called its atomic mass
When using atomic masses, retain a sufficient number of significant figures so the atomic mass data contributes only slightly to the uncertainty of the result
The molecular mass allows counting of molecules by mass
The molecular mass is the sum of atomic masses of the atoms in the compounds formula
For example the molar mass of water, H2O, is twice the mass of hydrogen (1.008) plus the mass of oxygen (15.999) = 18.015
Strictly speaking, ionic compounds do not have a “molecular mass” because they don’t contain molecules
The mass of the formula unit is called the formula mass
Formula masses are calculated the same way as molecular masses
For example the formula mass of calcium oxide, CaO, is the mass of calcium (40.08) plus the mass of oxygen (15.999) = 56.08
One mole of a substance contains the same number of formula units as the number of atoms in exactly 12 g of carbon-12
One mole of a substance has a mass in grams numerically equal to its formula mass
The mass of one mole of a substance is also called its molar mass
One mole of any substance contains the same number of formula units
This number is called Avogadro’s number or constant
Counting formula units by moles is no different than counting eggs by the dozen (12 eggs) or pens by the gross (144 pens)
Avogadro’s number is huge because atoms and molecules are so small: a huge number of them are needed to make a lab-sized sample
Avogadro’s number links moles and atoms, or moles and molecules and provides an easy way to link mass and atoms or molecules
Using water (molar mass 18.015) as an example:
1 mole H2O  6.022 x 1023 molecules H2O
1 mole H2O  18.015 g H2O
18.015 g H2O  6.022 x 1023 molecules H2O
Within chemical compounds, moles of atoms always combine in the same ratio as the individual atoms themselves so:
1 mole H2O  2 mole H
1 mole H2O  1 mole O
Stoichiometry is the study of the mass relationships in chemical compounds and reactions
A common use for stoichiometry is to relate the masses of reactants needed to make a compound
These calculations can be solved using the factor-label method and equivalence relations relating molecular masses and/or formula masses
Example: How many grams of iron are in a 15.0 g sample of iron(III) oxide?
ANALYSIS: 15.0 g Fe2O3  ? g Fe
LINKS: 1 mol Fe2O3  2 mol Fe
1 mol Fe2O3  159.7 g Fe2O3
1 mol Fe  55.85 g Fe
SOLUTION:
The usual form for describing the relative masses of the elements in a compound is a list of percentages by mass
This is called the percentage composition or percentage composition by mass
The percentage by mass is the number of grams of the element in 100 g of the compound and can be calculated using:
Example: A sample was analyzed and found to contain 0.1417 g nitrogen and 0.4045 g oxygen. What is the percentage composition of this compound?
ANALYSIS: Find sample mass and calculate %
LINKS: whole sample = 0.5462 g
SOLUTION:
Hydrogen peroxide consists of molecules with the formula H2O2
This is called the molecular formula
The simplest formula for hydrogen peroxide is HO and is called the empirical formula
It is possible to calculate the empirical formula for a compound from mass data
The goal is to produce the simplest whole number mole ratio between atoms
Example: A 2.012 g sample of a compound contains 0.522 g of nitrogen and 1.490 g of oxygen. Calculate its empirical formula
ANALYSIS: We need the simplest whole number mole ratio between nitrogen and oxygen
SOLUTION:
Empirical formulas may also be calculated indirectly
When a compound made only from carbon, hydrogen, and oxygen burns completely in pure oxygen only carbon dioxide and water are produced
This is called combustion
Empirical formulas may be calculated from the analysis of combustion information
Example: The combustion of a 5.217 g sample of a compound of C, H, and O gave 7.406 g CO2 and 4.512 g of H2O. Calculate the empirical formula of the compound.
ANALYSIS: This is a multi-step problem. The mass of oxygen is obtained by difference:
g O = 5.217 g sample – ( g C + g H )
The masses of the elements may then by used to calculate the empirical formula of the compound
SOLUTION:
The formula for ionic compounds is the same as the empirical formula
For molecules, the molecular formula and empirical are usually different
If the experimental molecular mass is available, the empirical formula can be converted into the molecular
The molecular formula will be a common multiplier times all the coefficients in the empirical formula
Example: The empirical formula of hydrazine is NH2, and its molecular mass is 32.0. What is its molecular formula?
ANALYSIS: The molecular mass (32.0) is some simple multiple of the mass calculated from the empirical formula (16.03)
SOLUTION:
The coefficients of a balanced chemical equation provide the mole-to-mole ratios for the substances involved in the reaction
Whenever a problem asks you to convert between different substances, the calculation usually involves a mole-to-mole relationship
How to detect unbalanced equations will be covered shortly
Example: If 0.575 mole of CO2 is produced by the combustion of propane, C3H8, how many moles of oxygen are consumed? The balanced equation is:
C3H8 + 5 O2  3 CO2 + 4 H2O
ANALYSIS: Relating two compounds usually requires a mole-to-mole ratio
SOLUTION:
In many problems you will also need to perform one or more mole-to-mass conversions
These types of stoichiometry problems are summarized with the flowchart:
Example: How many grams of Al2O3 are produced when 41.5 g Al react?
2Al(s) + Fe2O3(s)  Al2O3(s) + 2 Fe(l)
ANALYSIS:
41.5 g Al  ? g Al2O3
SOLUTION:
Chemical equations provide quantitative descriptions of chemical reactions
Conservation of mass is the basis for balancing equations
To balance an equation:
Write the unbalanced equation
Adjust the coefficients to get equal numbers of each kind of atom on both sides of the arrow
Guidelines for Balancing Equations:
Balance elements other than H and O first
Balance as a group any polyatomic ions that appears unchanged on both sides of the arrow
Balance separately those elements that appear somewhere by themselves
As a general rule you should use the smallest whole-number coefficients when writing balanced chemical equations
All reactions eventually use up a reactant and stop
The reactant that is consumed first is called the limiting reactant because it limits the amount of product that can form
Any reagent that is not completely consumed during the reactions is said to be in excess and is called an excess reactant
The computed amount of product is always based on the limiting reagent
Example: How many grams of NO can form when 30.0 g NH3 and 40.0 g O2 react according to:
4 NH3 + 5 O2  4 NO + 6 H2O
ANALYSIS: This is a limiting reactant problem
SOLUTION:
The amount of product isolated from a chemical reactions is almost always less than the calculated, or maximum, amount
The actual yield is the amount of the desired product isolated
The theoretical yield is the amount that would be recovered if no loss occurred (the calculated, maximum amount)
The percentage yield is the actual yield as a percentage of the theoretical yield
When working with percentage yield:
Remember they involve a measured (actual yield) and calculated (theoretical yield) quantity
The calculation may be done in either grams or moles
The result can never be a number larger than 100%

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