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The Mole

The Mole. Chapter 11. 11.1 Measuring Matter. Substances react according to definite ratios of numbers of particles (atoms, ions, formula units or molecules). Balanced chemical equations tell us what the ratios of these particles is.

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The Mole

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  1. The Mole Chapter 11

  2. 11.1 Measuring Matter • Substances react according to definite ratios of numbers of particles (atoms, ions, formula units or molecules). • Balanced chemical equations tell us what the ratios of these particles is. • If we want to know how many of these particles we have, we can find it by measuring the mass .

  3. Avogadro’s Number • Scientists have found that the atomic mass of an element contains 6.02 X 1023 atoms of that element. • This number is known as Avogadro’s number. • This number is also equivalent to 1 mole of particles.

  4. The Mole • The mole is the SI base unit for the amount of substance. • It is a quantity, just like a dozen, a pair, or a gross!! • It is defined as the number of particles in exactly 12 g of pure carbon-12. • When dealing with moles, the simplest unit of any substance is called a representative particle. • The particle may be an atom, a molecule, a formula unit, or an ion.

  5. Example • How many molecules are in 2.25 moles of Br2? • Bromine is an element that consists of diatomic molecules. One mole of bromine contains 6.02 X 1023 bromine molecules. • Using dimensional analysis, 2.25 molBromine x 6.02 X 1023 molecules Bromine 1 molBromine = 1.35 X 1024 molecules Br2

  6. Practice • Calculate the number of molecules in 15.7 moles of carbon dioxide. • Calculate the number of molecules in 0.0544 moles of H2O. • Calculate the number of moles in 9.22 X 1023 atoms of iron. • Calculate the number of moles of sucrose in a sample that contains 2.05 X 1022 sucrose molecules.

  7. 11.2 Mass and the Mole • One mole of a monatomic element consists 6.02 X 1023 atoms of that element. • The mass of a mole of any substance is called the molar mass of the substance. • The molar mass of a monoatomic element is numerically equal to the atomic mass of element, but expressed in grams.

  8. Practice • Calculate the mass of 6.89 mol of antimony. • A chemist needs 0.0700 mol selenium for a reaction. What mass of selenium should the chemist use? • A sample of sulfur has a mass of 223 g. How many moles of sulfur are in the sample? • Calculate the number of atoms in 2.00g of platinum. • How many sulfur atoms are in a metric ton (1.00 106) of sulfur?

  9. 11.3 Moles of Compounds • Remember that a mole of a compound contains as many moles of each element as are indicated by the subscripts in the formula of the compound. • For example, a mole of ammonia (NH3) contains one mole of nitrogen and three moles of hydrogen.

  10. Molar Mass of a Compound • The molar mass of a compound is the mass of a mole of the representative particles of the compound. • To find the molar mass, add up the molar masses of all the atoms in the representative particle.

  11. Compound’s Molar Mass • For example, in NH3 you would add up the molar mass of nitrogen, and three times the molar mass of hydrogen. • Mmammonia = 14.01 + 3 (1.01) = 14.01 + 3.03 = 17.04 g

  12. Practice • Calculate the number of moles in 17.2 g of benzene (C6H6). • Calculate the number of moles in 350.0 g of potassium chlorate. • Determine the mass of 0.187 mol of tin (II) sulfate (SnSO4).

  13. 11.4 Empirical and Molecular Formulas • Remember that every chemical compound has a definite composition—a composition that is always the same wherever that compound is found. • The composition of a compound is usually stated as the percent by mass of each element in the compound.

  14. Percent Composition • The percent composition of an element can be found in the following way. % mass = mass of element in compound X 100% Molar mass of compound

  15. Example • Determine the percent composition of calcium chloride (CaCl2). • %Ca = 40.078 g CaX 100% = 36.112% Ca 110.98 g CaCl2 • %Cl = 70.906 gCl X 100% = 63.888% Cl 110.984 g CaCl2

  16. Empirical Formulas • The empirical formula is the simplest whole-number ratio of atoms of elements in the compound.

  17. Example • The percent composition of an unknown compound is found to be 38.43% Mn, 16.80% C and 44.77% O. Determine the compound the compound’s empirical formula.

  18. Practice • The composition of acetic acid is 40.00 % carbon, 6.71% hydrogen, 53.29 % oxygen. Calculate the empirical formula for acetic acid. • Calculate the empirical formula for a compound whose analysis is 74.97 % aluminum and 25.03% carbon.

  19. Molecular Formula • For many compounds, the empirical formula is not the true formula. • A molecular formula tells the exact number of atoms of each element in a molecule or formula unit. • The molecular formula is always a whole-number multiple of the empirical formula. • To find, you must know the molar mass of the compound and its empirical formula.

  20. Example • Maleic acid is a compound that is widely used in plastics and textiles. The composition of maleic acid is 41.39% carbon, 3.47% hydrogen and 55.14% oxygen. Its molar mass is 116.1 g/mol. Calculate the molecular formula.

  21. Practice • Ricinine is one of the poisonous compounds found in the castor plant. The composition of ricinine is 58.54% carbon.91% hydrogen, 17,06% nitrogen and 19.49% oxygen. Ricinine’s molar mass is 164.16 g/mol. Find its molecular formula. • A compound of phosphorous and sulfur contains 27.87% phosphorous and 72.13% sulfur. The molar mass of the compound is 222.3 g/mol. Calculate its molecular formula.

  22. 11.5 The Formula for a Hydrate • Many compounds, particularly ionic compounds, include a specific number of water molecules into their crystals when they crystallize from aqueous solution. • Compounds that include water in their crystal structures are called hydrates. • The formula is written by adding the formula for water to the formula for the compound, and the number of waters is shown with a coefficient.

  23. Example • A hydrate of aluminum bromide is composed of 71.16% AlBr3 and 28.84% H2O. What is the formula of the hydrate?

  24. Practice • Cerium (III) iodide (CeI3) occurs as a hydrate with the composition 76.3% CeI3 and 23.7% H2O. Calculate the formula of the hydrate. • Lead (II) acetate (Pb(C2H3O2)2) exists as a crystalline hydrate that is 14.25% water by mass. What is the molecular formula for this hydrate?

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