The Mole and Molar Mass 1 Molar mass is the mass of one mole of a substance. Molar mass is numerically equal to atomic mass, molecular mass, or formula mass. However the units of molar mass are g/mol. Examples: 1 atom Na = 22.99 u 1 mol Na = 22.99 g 1 molecule CO 2 = 44.01 u 1 formula unit KCl = 74.56 u 1 mol CO 2 = 44.01 g 1 mol KCl = 74.56 g 2 1
3 We can read formulas in terms of moles of atoms or ions. Molecular Masses and Formula Masses 4 Molecular mass: sum of the masses of the atoms represented in a molecular formula. Simply put: the mass of a molecule. Molecular mass is specifically for molecules. Ionic compounds don t exist as molecules; for them we use Formula mass: sum of the masses of the atoms or ions present in a formula unit. 2
Determining the Formula/Molar Mass of Ammonium Sulfate 5 The molar mass of ammonium sulfate is g mol -1 The molar mass of acetic acid (CH 3 COOH) is g mol -1 Conversion Factors involving Mass, Moles, and Number of Atoms/Molecules 6 1 mol CH 3 COOH = 6.022 10 23 CH 3 COOH molecules 60.05 g CH 3 COOH We can use these equalities to construct conversion factors: molec. 6.022 10 23 mol (CH 3 COOH) 60.05 g mol CH 3 COOH 1 mol 6.022 10 23 molec. (CH 3 COOH) 1 60.05 mol g CH 3 COOH 3
Mass Percent Composition from Chemical Formulas The mass percent composition of a compound refers to the mass proportion of the constituent elements: g element % element = 100 g compound OR X g element X % element = 100 g compound #grams of the element per 100 grams of the compound. 7 8 Percentage Composition of Butane g/mol g/mol 4
9 אנליזת יסודות Elemental Analysis is one method of determining empirical formulas in the laboratory. This method is used primarily for simple organic compounds (that contain carbon, hydrogen, oxygen). The organic compound is burned in oxygen. The products of combustion (usually CO 2 and H 2 O) are weighed. The amount of each element is determined from the mass of products. 10 Elemental Analysis Setup H 2 O, which is absorbed by MgClO 4, and The sample is burned in a stream of oxygen gas, producing CO 2, which is absorbed by NaOH. 5
11 Elemental Analysis (cont d) If our sample were CH 3 OH, every two molecules of CH 3 OH would give two molecules of CO 2 and four molecules of H 2 O. 12 Elemental analysis of butane (cooking gas) gave the following mass composition: C 82.66%, H 17.34%. a) Determine the empirical formula of butane. b) The molar mass of butane is 58.12 g/mol; what is the molecular formula? Determine the empirical formula of phenol (a general disinfectant) from its elemental analysis: C 76.57% H 6.43% H. 6
Writing Chemical Equations 13 A chemical equation is a shorthand description of a chemical reaction, using symbols and formulas to represent the elements and compounds involved. Writing Chemical Equations 14 Sometimes additional information about the reaction is conveyed in the equation. 7
איזון משוואות Balancing chemical equations 15 A balanced equation must conform to the law of חוק שימור המסה mass! conservation of To balance an equation we need to adjust the מקדמים סטויכיומטריים stoichiometric coefficients Balance the following equations: H 2 + O 2 H 2 O C 2 H 6 + O 2 CO 2 + H 2 O Fe + O 2 Fe 2 O 3 Guidelines for Balancing Chemical Equations 16 If an element is present in just one compound on each side of the equation, try balancing that element first. Balance any reactants or products that exist as the free element last. In some reactions, certain groupings of atoms (such as polyatomic ions) remain unchanged. In such cases, treat these groupings as a unit. At times, an equation can be balanced by first using a fractional coefficient(s). The fraction is then cleared by multiplying each coefficient by a common factor. 8
Stoichiometric Equivalence and Reaction Stoichiometry 17 A stoichiometric factor or mole ratio is a conversion factor obtained from the stoichiometric coefficients in יחס סטויכיומטרי equation. a chemical In the equation: CO(g) + 2 H 2 (g) CH 3 OH(l) 1 mol CO 1 mol CO 1 mol CO 2 mol H reacted per 2 1 mol CH 3 OH 2 mol H 2 1 mol CO was consumed per 1 mol CH 3 OH that evolved 2 mol H 2 1 mol CO 2 mol H 2 reacted per 1 mol CO 1 mol CH 3 OH 2 mol H 2 1 mol CH 3 OH was formed for each 2 mol of H 2 that reacted 18 Outline of Simple Reaction Stoichiometry Stoichiometric factor (or: mole ratio) 9
19 When 0.105 mol propane is burned in an excess of oxygen, how many moles of oxygen are consumed? The reaction is C 3 H 8 + 5 O 2 3 CO 2 + 4 H 2 O Outline of Stoichiometry Involving Mass 20 To our simple stoichiometry scheme and a conversion to mass at the end. we ve added a conversion from mass at the beginning Substances A and B may be two reactants, two products, or reactant and product. Think: If we are given moles of substance A initially, do we need to convert A to grams? 10
The final step in the production of nitric acid involves the reaction of nitrogen dioxide with water; nitrogen monoxide is also produced. How many grams of nitric acid are produced for every 5 kg of nitrogen dioxide that reacts? 21 22 Summary of Concepts Mass percent composition can be used to calculate the empirical formula. Additional information on the molecular weight of the material allows the calculation of the molecular formula A balanced equation must conform to the law of conservation of mass. Calculations involving reactions use stoichiometric factors based on stoichiometric coefficients in the balanced equation. 11