The Mole

1. The Mole Mass Relationships and Avogadro’s Number

2. 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.

3. 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.

4. 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.

5. 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.

6. 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.

7. 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

8. 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

9. 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

10. 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

11. 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.

12. 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

13. 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)

14. 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.

15. Molarity is the concentration of a solution in moles per litre. • 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