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Transcript
```Dr. Saidane
Chem. 152
The mole Concept
Skills you should have mastered
Conceptual
1. Give the definition of a mole.
2. Explain what is meant by a mole ratio and state how it is used.
Problem-solving
1. Use the Avogadro constant to convert between number of moles and the
number of atoms, molecules, or ions in a sample.
2. Calculate the average molar mass of an element, an ion, or a compound.
3. Convert between mass and number of moles by using the molar mass.
4. Convert between volume and number of moles by using the molar volume.
5. Calculate the empirical formula of a compound from its mass percentage
composition.
6. Determine the molecular formula of a compound from its empirical formula
and its molar mass.
7. Carry out mole-to-mole, mass-to-mole, and mass-to-particles, mass-to-volume
calculations for any two species.
Composition Stoichiometry is the quantitative analysis of substances. It deals with
the mass relationships of elements in compounds.
1. A mole (mol) is the amount of a substance that contains as many particles as there
are atoms in exactly 12 g of carbon-12. The mole is the SI unit for amount of
substance. The number of particles in a mole has been experimentally determined
as 6.022 x 1023, which is also known as Avogadro’s number. The mole is a
counting unit, just as a dozen is. We all know that a dozen of pencils is the same
as 12 pencils. Similarly, a chemist uses 1 mole of iron, which means 6.022 x 10 23
atoms of iron.
2. The molar mass is the mass of one mole of a pure substance. Molar mass is
usually written in units of g/mol. The molar mass of an element is equal to the
numerical atomic mass of the element in atomic mass unit, amu (which can be
found in the periodic table). The molar mass of a compound is equal to the
numerical sum of the average atomic masses of all the atoms present in the
compound.
3. The molar volume is the volume of one mole of a gas substance. Molar volume
is usually written in units of L/mol. The molar volume of a gas is equal to 22.4 L.
4. Converting moles to number of particles, to mass or to volume. We will be
using the Unit Analysis Method again. Recall:
 First we write down the unit asked for
 Second we write down the given value
 Third we apply unit factor(s) to convert the given units to the desired units
The unit factors are based on the three interpretations for the mole:
o 1 mol = 6.02 × 1023 particles
o 1 mol = molar mass in grams (from the periodic table)
o 1 mol = molar volume in liters (22.4 L at STP for a gas)
5. The law of definite composition states that “Compounds always contain the same
elements in a constant proportion by mass”.
 Sodium chloride is always 39.3% sodium and 60.7% chlorine by mass, no
matter what its source.
 Water is always 11.2% hydrogen and 88.8% oxygen by mass.
6. Chemical Formulas
 A chemical formula expresses the number and types of atoms in a molecule.
 The number of each type of atom in a molecule is indicated with a subscript in
a chemical formula. If there is only one atom of a certain type, no ‘1’ us used.
 A molecule of the vitamin niacin has 6 carbon atoms, 6 hydrogen atoms, 2
nitrogen atoms, and
1 oxygen atom. What is the chemical formula?
C6H6N2O.
7. Percent Composition

The percent composition of a compound lists the mass percent of each
element. For example, the percent composition of water, H2O is:
(2g/18 g) x 100 = 11% hydrogen and (16g /18 g) x100 = 89% oxygen
 All water contains 11% hydrogen and 89% oxygen by mass.
8. Determination of chemical formulas
a) Mass percentage composition. The percentage by mass of each element in a
compound is known as the percentage composition of the compound.
Mass percentage of element = (mass of element in sample/total mass of sample) x 100%
b) Empirical formulas. An empirical formula consists of the symbols for the
elements combined in a compound, with subscripts showing the smallest
whole-number mole ratio of the different atoms in the compound. The
following steps are used to determine the empirical formula:
 First convert the percentage composition to a mass composition. If a
compound is 78% B and 22% H, we can assume that 100 g of the
compound is made of 78 g of B and 22 g of H.
 Next, convert the mass composition of each element to a mole
composition by dividing by the appropriate molar mass.
78 g B x 1 mol B/10.81 g B = 7.22 mol B
22 g H x 1 mol H/1.01 g H = 21.7 mol H.

Find a mole ratio by dividing each number of moles by the smallest
number in the existing ration.
7.22 mol B / 7.22 = 1 mol B
21.7 mol H / 7.22 = 3.01 ~ 3 mol H.
The empirical formula is BH3
The mole ratio must be close to a whole number. If one of the mole ratios is not a whole
number, then multiply them by the same integer until they all are whole numbers.
c) The molecular formula of a molecular compound is the real formula of the
compound and is a multiple of the empirical formula. The relationship
between a compound’s empirical formula and its molecular formula can be
written as follows: n x empirical formula = molecular formula, with n an
integer.
```
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