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The mole (abbreviation: mol) is the amount of substance equal to 6.02 x 10 23 particles These particles can be atoms, ions, formula units,molecules, electrons,

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Presentation on theme: "The mole (abbreviation: mol) is the amount of substance equal to 6.02 x 10 23 particles These particles can be atoms, ions, formula units,molecules, electrons,"— Presentation transcript:

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2 The mole (abbreviation: mol) is the amount of substance equal to 6.02 x 10 23 particles These particles can be atoms, ions, formula units,molecules, electrons, etc. The Mole (mol) Avogadro’s number 1 mole of Carbon= 6.02 x 10 23 carbon atoms 1 mole water= 6.02 x 10 23 water molecules 1 mole of H + = 6.02 x 10 23 H + ions 1 mole NaCl= 6.02 x 10 23 NaCl formula units IN GENERAL: 1 mole of a substance = 6.02 x 10 23 particles

3 Examples (# of particles) How many atoms of Lithium are in 0.43 moles of Li atoms? 2.00 x 10 32 ions of Na + mol of Na + 1 6.02 x 10 23 0.43 mol of Li 1 6.02 x 10 23 mol of Li x Atoms of Li = 2.6 x 10 23 atoms of Li How many moles of sodium ions are if we have 2.00 X 10 32 sodium ions x ions of Na + = 3.32 x 10 8 mol of Na +

4 More examples (# of particles) How many moles of carbon dioxide are if we have 6.00 x 10 24 molecules of carbon dioxide? 6.00 x 10 24 molec of CO 2 6.02 x 10 23 1 molec of CO 2 x mole of CO 2 = 9.97 moles of CO 2 How many formula units of sodium chloride are in 0.25 moles of sodium chloride? 0.25 mol of NaCl 1 6.02 x 10 23 mol of NaCl x Formula units of NaCl = 1.5 x 10 23 form. Units of NaCl

5 Molar Mass and Atomic Mass It is equal to the atomic mass from the periodic table, expressed in grams. Atomic mass: the average mass of all the isotopes of an element considering their percentage of abundance. The atomic mass can be expressed in amu´s (mass of atoms) or in g (mass of mole of atoms). An amu (atomic mass unit) is the mass of 1/12 the mass of one atom of 12 C. Molar Mass: The mass of one mole of substance ( 6.02 x 10 23 particles).

6 The molar mass can be: atomic mass (for atoms), molecular mass (for molecules) or formula mass (for salts), but the term molar mass can be applied to all of them. Examples: What is are the molar masses of Carbon, Sodium Chloride, Carbon Dioxide, Oxygen Atoms and Oxygen Molecules? Molar Mass Remember: 1 Mole = Molar Mass (in grams)

7 The mole and the mass of substances 0.250 mol Na mol Na g Na22.99 1 What is the mass of 0.250 moles of sodium? = x5.75 g Na How many moles of methane (CH 4 ) are in 45.0g of methane? From the Periodic Table, the molar mass of Na is 22.99g 45.0 g CH 4 g CH 4 mol CH 4 1 16.05 Molar mass of CH 4 : 1C= 12.01 4H= 4 x 1.01= 4.04 16.05 + = x 2.80 mol CH 4

8 Relative atomic and molecular mass The relative molecular mass M r is the mass of the molecule relative to carbon 12. It has no units. The relative atomic mass A r is the mass of an atom relative to carbon 12. It has no units. It is important to distinguish between the atomic mass (A or A r ) of elements and the molecular mass (M or M r ) of elements. For example, the atomic mass of chlorine is 35.45 and its molecular mass is 70.90 because it exists as Cl 2

9 Mass of elements Molecules of a compound or formula units of a salt. Mass of a compound Atoms of an element 12g Fe x g Fe mol Fe 1 55.85 0.50 mol Na x mol Na g Na22.99 1 12g NaCl x gNaCl mol NaCl1 58.44 0.64 mol NaOH x mol NaOH g NaOH 1 40.00 3.1 x 10 40 molec.NaF X molec NaF mol1 6.02x10 23 0.53 mol CaCl 2 x mol CaCl 2 molec CaCl 2 1 6.02 x 10 23 3.1 x 10 40 atoms of B X atoms of B mol of B1 6.02 x 10 23 0.72 mol of Ni x mol of Ni atoms of Ni 1 6.02 x 10 23 mol

10 Mass of elements Molecules of a compound or formula units of a salt. Mass of a compound Atoms of an element.. mol Divide by the Molar Mass (of the element) Multiply by the Molar Mass (of the element) Divide by the Molar Mass (of the compound) Multiply by the Molar Mass (of the compound) Divide by 6.02 x 10 23 Multiply by 6.02 x 10 23 Divide by 6.02 x 10 23

11 Formula: % composition of X = mass of X in the compound x 100% mass of the compound Percent composition: The percent by mass of each element in a compound. It is also called percent by mass. Percent composition

12 Example 1: Calculate the percent composition of each element in sulfuric acid. Note: If the formula of the compound is not known, it can be determined experimentally. Example 2: In the laboratory 32.0 g of a compound were analyzed. The compound contained 24.0g of Carbon and 8.00g of hydrogen. What is the percent composition of the of each element in the compound Example 3: Sodium Hydrogen Carbonate, NaHCO 3, also called baking soda is an active ingredient in some antacids used for the relief of indigestion. Determine the percent composition of sodium hydrogen carbonate. Examples

13 Molecular formula: The formula for molecular compounds, indicating total atoms of each element in a compound. (For example H 2 O 2 ). Empirical formula: The formula that represents the smallest whole number ratio of types of atoms in a compound (or formula unit). (For example HO) Empirical and Molecular Formulas The molecular formula can be equal to or a multiple of the empirical formula.

14 If the data is in percentage assume 100 g sample and use the percentages as grams of each element. Calculate the moles of each element Divide by the smallest number of moles obtained If the numbers are whole numbers they are the subscripts in the formula. If the numbers are not whole numbers, multiply them by the smaller number that makes them all whole numbers. The obtained values are the subscripts in the formula. To determine the empirical and molecular formulas:

15 Example – Methyl acetate, is a compound commonly used in some paints, inks, and adhesives. Determine the empirical formula of methyl acetate, with the following chemical analysis: Percent by mass: 48.65% C, 8.16% H 43.20% O

16 To obtain the molecular formula divide the molar mass of the compound (given) by the molar mass of the empirical formula. Multiply the subscripts of the empirical formula by the number obtained in the previous step. To obtain the Molecular formula

17 Example – Methyl acetate, is a compound commonly used in some paints, inks, and adhesives. Determine the empirical formula of methyl acetate, with the following chemical analysis: Percent by mass: 48.65% C, 8.16% H 43.20% O

18 Example: Succinic acid is a substance produce by lichens. Chemical analysis indicates it is composed of 40.68% oarbon, 5.08 % hydrogen and 54.24% oxygen and has a molar mass of 118.1 g/mol. Determine the empirical and molecular formulas for succinic acid.


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