The Mole

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# The Mole - PowerPoint PPT Presentation

The Mole. Mass Relationships and Avogadro’s Number. Relative Mass . It is possible to determine the mass of an atom without knowing the mass of a single atom. This procedure involves comparing the masses of equal numbers of atoms. Ex: Oranges: 2160g = 3 = 0.600 Grapefruit: 3600g 5

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### The Mole

Relative Mass
• It is possible to determine the mass of an atom without knowing the mass of a single atom.
• This procedure involves comparing the masses of equal numbers of atoms.

Ex: Oranges: 2160g = 3 = 0.600

Grapefruit: 3600g 5

• Since there are an equal number of oranges and grapefruit, the mass of one orange is 3/5 or 0.600 that of the mass of one grapefruit.
• 0.600 is the Relative mass of the orange.
• Relative mass of any object is expressed by comparing it mathematically to the mass of another object.

The mass of an atom is called its atomic mass.

• At first the masses of atoms were compared to hydrogen, as it was the lightest atom.
• The nitrogen was 14 times as heavy, and so on.
• Later, Oxygen with a mass of 16 was used.
• Eventually Carbon with a mass of 12.0000 was chosen as the standard.
• The masses of individual atoms are assigned a unit of relative measurement known as the atomic mass unit (amu).
• The atomic mass unit is defined as 1/12 of the mass of a carbon-12 atom.
Combining Volumes of Gases and Avogadro’s Hypothesis
• Joseph Louis Gay-Lussac (1778-1850) was a French chemist who performed experiments to investigate how gases combined to form compounds that were also gases.
• He did much of his work with nitrogen and oxygen.
• For all compounds formed, the ratios of the volumes of gases used were simple, whole-number ratios. (Table 4-1, p. 96)
• The volumes of gases that react to form each compound can be expressed as a simple ratio: 2 to 1, 1 to 1, or 1 to 2.

The importance of Gay-Lussac’s results was recognized by Amadeo Avogadro, an Italian scientist.

• In 1811 Avogadro wrote what became known as Avogadro’s Hypothesis: Equal volumes of gases (at the same temperature and pressure) contain equal numbers of particles.
• The table contains data obtained from equal volumes of several gases
• The relative mass of one atom of nitrogen (14) compared to one atom of hydrogen is the same as the relative mass found when comparing all of the molecules in a liter of each gas.
How Many is a Mole?
• A mole (mol) is simply the amount of a substance that contains 6.02 x 1023 particles.
• 6.02 x 1023 is known as Avogadro’s number.
• The particle can be anything: atoms, molecules, or baseballs.
• Relative masses of atoms do not change when you consider individual atoms or moles of atoms.
• One mole of carbon-12 atoms has a mass in grams that equals the atomic mass, in amu’s, of a single atom of carbon-12: 12.00 grams.
Formula Calculations
• Actual formulas of compounds are determined by laboratory analysis.
• Experiments are done to measure the amount of each element in the compound.
• The interpretation makes use of molar masses
• The analysis provides the simplest ratio of atoms in the compound
Finding an Empirical Formula
• Empirical means based on experiment.
• An empirical formula is one that is obtained from experimental data and represents the smallest whole number ratio of atoms in a compound.
• CO2 represents one carbon atom for every two oxygen atoms
• A mole of CO2 has 6.02 X 1023 molecules. There are 6.02 X 1023 carbon atoms in a mole of CO2, and 2(6.02 X 1023) oxygen atoms.
• 44g of CO2 contains 12g of carbon and 32g of oxygen
• One molecule CO2

mass= 44amu

1 atom C 2 atoms O

mass=12amu mass=32amu

One mole of CO2

Mass=44g

1mole C 2 moles O

Mass=12g mass=32g

• To determine the formula for a compound, it is not necessary to count the atoms in a single molecule.
• The information is obtained by finding the number of moles of each element in a mole of the compound

Example 1:

• A charcoal briquette of carbon has a mass of 43.2g. It is burned and combines with oxygen and the resulting compound has a mass of 159.0g. What is the empirical formula for the compound?
• 159.0g – 43.2g carbon = 115.8g oxygen
• Now find the number of moles of C and O in the compound.
• 43.2g C x 1mol C = 3.60 mol C

12.0g C

115.8g O x 1mol O = 7.24 mol O

16.0 g O

• There are 2.01 moles of oxygen for every 1.0 mole of carbon (7.24mol O/3.60mol C=2.01mol O/1mol C)
• We can assume that the formula is CO2
Summary of Steps
• The mass of each element in a sample of the compound is determined.
• The mass of each element is divided by its molar mass to determine the number of moles of each element in the sample of the compound.
• The number of moles of each element is divided by the smallest number of moles to give the ratio of atoms in the compound.
Example 2
• Charcoal is mixed with 15.53g of rust and heated in a covered crucible to keep air out until all of the oxygen atoms in the rust combine with carbon. When this process is complete, a pellet of pure iron with a mass of 10.87g remains. Empirical formula for rust?
• Mass of rust: 15.53g
• Mass of pure iron: 10.87g
• Mass of oxygen in rust: 15.53g-10.87g=4.66g
• 10.87g Fe x 1mol Fe = 0.195 mol Fe

55.8g Fe

4.66g O x 1mol O = 0.291 mol O

16.0g O

• 0.195mol Fe = 1.00 0.291mol O = 1.49 mol O/mol Fe

0.195mol Fe 0.195mol Fe

• Multiply both numbers to get a whole number ratio:
• Fe2O3
Molecular Formula/Percent Composition
• Molecular Formula: Always some multiple of the empirical formula
• Divide the molar mass of the compound by the molar mass of the empirical formula. (see Ex. 4-14)
• Percent Composition: comparison of the elements in a compound by percentage, rather than by masses (Ex 4-17)
Molarity
• Many compounds are stored, measured, and used as solutions.
• Concentration describes how much solute is in a given amount of solution.
• % by mass or volume is a convenient way of expressing concentration.
• Chemists commonly describe concentration by indicating the number of moles of solute dissolved in each liter of solution.
• The symbol for molarity is M.
• 4.90M solution of NaCl means that there are 4.90 moles of NaCl in one litre of the solution.

Example:

 How many moles of HCl are contained in 1.45L of a 2.25M solution?

2.25M = 2.25 mol/ 1L of solution

1.45L X 2.25mol = 3.26 molHCl

1L soln