Mass Relationships and Avogadro’s Number 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 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 table contains data obtained from equal volumes of several gases 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. Gas Mass of 1L of Gas Mass of Gas Relative to Hydrogen hydrogen 0.08g 1 nitrogen 1.12g 14 oxygen 1.31g 16 fluorine 1.52g 19 chlorine 2.80g 35

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 10.87g Fe x 1mol Fe = 0.195 mol Fe 55.8g Fe 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.

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 mol HCl 1L soln