7.1 The Mole
The mole is a counting unit of particles
such as a mole of a helium atom or a mole
of water molecules or a mole nitrogen.
molecules. In each case the number of
particles is the same, namely 6.022 x 1023
which is known as Avogadros number.
Named after the famous Italian scientist,
Amedeo Avogadro.
*There is one mole of a substance in one
molar mass of that substance.
Example: How many moles of calcium are in
22.0g of calcium?
This is your conversion factor for going from grams
to moles. Be sure to always place the desired units
in the numerator and the ones to cancel out in the
denominator. Now lets finish this problem.
Molar Mass is the formula mass in grams
of the substance. There are 6.02 x 1023
molecules or formula units of that substance
in the molar mass of the material.
The molar mass of water is 18.01g/mol.
therefore in 18.01 grams of water there is one
mole of water and there are 6.022 x 1023
molecules of water in that 18.01 grams of
water.
The percent composition of a compound is
also the mass percent of each element in that
compound.
Example: CO2 Formula weight(molar mass)=
(C x 1) + (O x 2)= 12.01 g/mol + 16.00 g/mol=
44.01 g/mol
7.4 Empirical Formula vs. Molecular
Empirical formula is the simplest formula.
Molecular formula is the formula with the
actual numbers of atoms in the molecule.
Example: NH2 is the empirical formula for
hydrazine which has the molecular formula
of N2H4.
7.5 Calculation of Empirical Formulas
Example: A compound is made up of
11.91% hydrogen, and 88.79% oxygen..
Determine the molecular formula.
H=11.91% O=88.79%
Step 1 : Convert percentage to grams in
a 100-gram sample of the substance.
In 100-gram sample there must be
11.19 grams of H atoms and 88.79 grams
of O atoms.
Step 2 : Convert the grams to moles.
Step 3 : Change the numbers to whole
intergers by dividing each by the smallest
number of the two.
The Emperical Formula is H2O
7.6 Molecular Calculation From Emperical
Follow the same steps as in 7.5
plus two more are added.
Example (with all steps): A compound is
composed of 92.26% carbon and
7.743% hydrogen.
*The molecular mass was found to be
78.11g/mol.
Determine the molecular Formula.
Step 1 : Convert percentage to grams
in a 100-gram sample of the substance.
99.26 g C and 7.743 g H
Step 2 : Convert grams to moles.
(92.26 g C) / (12.01 g/mol)= 7.682 mol C
This is no different then the last example only the
position is. If you prefer to set it up the other way
the idea of grams to grams to get to moles is the
same.
(7.743 g H) / (1.008 g/mol)= 7.681 mol H
Step 3 : Change the numbers to whole
intergers by dividing each by the smallest
number.
7.682 mol C / 7.681 mol H= 1.000
C/H= 1/1 C1H1
Step 4 : Divide the molecular mass by the
emperical mass derived from the empirical
formula.
Emperical mass= (1xC) + (1xH)=
12.01 + 1.008= 13.02 g/mol