Slide 1 / 29 Slide 2 / 29 The Mole Unit 3 - Presentation A Recall - PDF document

Slide 1 / 29 Slide 2 / 29 The Mole Unit 3 - Presentation A Recall that 1 mole is defined as 6.022 x 10 23 units of a given substance. The Mole, Empirical, and Molecular Formulas 1 mol of electrons = 6.022 x 10 23 electrons 1 mol of H 2 O molecules = 6.022 x 10 23 molecules of water 1 mol of NaCl formula units = 6.022 x 10 23 formula units NaCl 1 mol of K atoms = 6.022 x 10 23 atoms of K The molecular formula for nicotine is C 10 H 14 N 2 Slide 3 / 29 Slide 4 / 29 The Mole The Mole Within 1 mole of a compound, there are often differing moles of Example: How many O atoms are present in 2.0 moles of aluminum nitrate? each element In 1 mole of Al(NO 3 ) 3 2.0 mol Al(NO 3 ) 3 x 9 mol O = 18.0 mol O move for answer = 1 mol of Al 3+ ions 1 mol Al(NO 3 ) 3 = 3 mol of NO 3- ions 18.0 mol O x 6.022 x 10 23 atoms O = 1.08 x 10 25 atoms O = 3 mol of N atoms 18.0 mol O 1 mol O = 9 mol of O atoms Slide 5 / 29 Slide 6 / 29 Molar Mass and Volume Molar Mass and Volume Recall that the mass of 1 mol of a substance is called the molar Example: What is the volume occupied @STP by 88 grams of mass and is measured in g/mol. This can be found on the carbon dioxide? periodic table. Molar mass of CaCl 2 = 110 g/mol 88 g CO 2 x 1 mol CO 2 x 22.4 L = 44.8 L Molar Mass of Ag = 108 g/mol 44 g CO 2 1 mol move for answer Recall also that 1 mol of any gaseous substance will occupy 22.4 L of space at STP. molar mass molar volume 1 mol of Ar(g) = 22.4 L @STP 1 mol of H 2 (g) = 22.4 L @STP

Slide 7 / 29 Slide 7 (Answer) / 29 1 How many hydroxide ions are present in 1.2 x 10 24 1 How many hydroxide ions are present in 1.2 x 10 24 formula units of magnesium hydroxide: Mg(OH) 2 ? formula units of magnesium hydroxide: Mg(OH) 2 ? 2.4 x 10 24 hydroxide ions Answer Answer [This object is a pull tab] Slide 8 / 29 Slide 8 (Answer) / 29 2 How many mL of methane gas (CH 4 ) are present @STP 2 How many mL of methane gas (CH 4 ) are present @STP in a 100 gram sample of a gas that is 32% methane by in a 100 gram sample of a gas that is 32% methane by mass? mass? Answer Answer 44,800 mL Natural gas (methane) pipeline. Natural gas (methane) pipeline. [This object is a pull tab] Slide 9 / 29 Slide 9 (Answer) / 29 3 Which of the following contains the most atoms of H? 3 Which of the following contains the most atoms of H? A 2 grams of H 2 gas A 2 grams of H 2 gas B 16 grams of methane (CH 4 ) B 16 grams of methane (CH 4 ) Answer Answer B C 22.4 L of H 2 gas C 22.4 L of H 2 gas D 9 grams of water (H 2 O) D 9 grams of water (H 2 O) E They all contain the same # of H atoms E They all contain the same # of H atoms [This object is a pull tab]

Slide 10 / 29 Slide 10 (Answer) / 29 4 How many moles of fluoride ions (F-) are present in a 79 4 How many moles of fluoride ions (F-) are present in a 79 gram sample of SnF 2 ? gram sample of SnF 2 ? A 0.5 moles A 0.5 moles B 78.5 moles B 78.5 moles Answer Answer C 1 mole C 1 mole C D 1.5 moles D 1.5 moles E 2 moles E 2 moles [This object is a pull tab] Slide 11 / 29 Slide 12 / 29 Chemical Formulas Empirical and Molecular Formulas A chemical formula provides the ratio of atoms or moles of each An empirical formula provides the simplest whole number ratio of element in a compound. atoms or moles of each element in a compound. H 2 O = 2 atoms H or 2 mol H Examples: H 2 O, NaCl, C 3 H 5 O 1 atom O 1 mol O A molecular formula represents the actual number of atoms or moles of each element in a compound. Al(NO 3 ) 3 = 1 Al 3+ ion or 1 mol Al 3+ ions 3 NO 3- ions 3 mol NO 3- ions Examples: H 2 O, C 3 H 5 O, C 6 H 12 O 6 Slide 13 / 29 Slide 14 / 29 Empirical and Molecular Formulas Calculating an Empirical Formula To find an empirical formula: There are two reasons for determining an empirical formula. 1. Determine the moles of each element within the compound Reason 1: then..... Many compounds are really a gigantic molecule composed of Compound "X" consists of 1.2 g C, 0.2 g H, and 1.6 g O trillions upon trillions of atoms. No one would want to write the actual formula of an NaCl crystal as... = 0.1 mol C, 0.2 mol H, and 0.1 mol O Na 1.8 x 1024 Cl 1.8 x 1024 so we just write the empirical formula instead (NaCl) 2. Find the whole number ratio of these moles by dividing by smallest mole value! Reason 2: 0.1 mol C = 1 C 0.2 mol H = 2 H 0.1 mol O = 1 O In order to determine the molecular formula, we must first calculate the empirical formula anyway. 0.1 mol 0.1 mol 0.1 mol Empirical formula = CH 2 O

Slide 15 / 29 Slide 16 / 29 Calculating Empirical and Calculating a Molecular Formula Molecular Formulas Determining the molecular formula of a compound is easy once the empirical formula and the molecular weight of the compound Class Example: Given the following data, calculate the empirical are known. formula of phosphine gas. Phosphine gas is created by reacting solid phosphorus with H 2 (g). 1. Determine the ratio of the molecular weight to the empirical formula weight. Mass of P(s) initial Mass of P(s) unreacted MW of Compound "X" = 60 u 1.45 g 1.03 g Empirical formula weight of CH 2 O = 30 u Mass of H 2 (g) initial Mass of H 2 (g) unreacted Ratio = 60/30 = 2/1. 0.041 g 0.000 g The molecule is twice as heavy as the empirical formula. 0.42 g P reacted = 0.0135 mol P reacted 0.041 g H2 reacted = 0.0202 mol H2 = 0.0405 mol H move for answer 2. Multiply each subscript of empirical formula by the ratio determined in step 1 0.0405/0.0135 = 3 H 0.0135/0.0135 = 1 P PH 3 = empirical formula CH 2 O x 2 = C 2 H 4 O 2 = Molecular Formula Slide 17 / 29 Slide 18 / 29 Calculating Empirical and Calculating Empirical and Molecular Formulas Molecular Formulas Class example 2: Black iron oxide (aka magnetite) is used as a Class example 3: Lactic acid is produced when our muscle cells contrast agent in MRI scans of human soft tissue. To determine run out of oxygen. the empirical formula, a student reacted solid iron with O 2 (g). Fe(s) reacted Mass of iron oxide obtained. Fe(s) reacted Mass of iron oxide obtained. 3.05 g 4.22 g 3.05 g 4.22 g What is the empirical formula? What is the empirical formula? 3.05 g Fe = 0.055 mol Fe 3.05 g Fe = 0.055 mol Fe 4.22 - 3.05 = 1.17 g O 2 (g) = 0.037 mol O 2 = 0.073 mol O 4.22 - 3.05 = 1.17 g O 2 (g) = 0.037 mol O 2 = 0.073 mol O move for answer move for answer 0.055/0.055 = 1 Fe 0.073/0.055 = 1.33 mol O 0.055/0.055 = 1 Fe 0.073/0.055 = 1.33 mol O x each by 3 to get a whole number ratio ... x each by 3 to get a whole number ratio ... Fe 3 O 4 Fe 3 O 4 Slide 19 / 29 Slide 20 / 29 Calculating Empirical and 5 Which of the following is NOT an empirical formula? Molecular Formulas A Fe 2 O 3 Class Example 4: Butane gas can be produced when solid carbon is reacted with hydrogen gas. If 0.45 grams of carbon were found B H 2 NNH 2 to react with 1.05 L of H 2 gas @STP, what is the molecular formula Answer of butane given it has a molar mass of 58 g/mol. C CH 3 OH 0.45 g C = 0.0375 mol C D CH 3 CH 2 Cl 1.05 L H 2 (g) = 0.0469 mol H 2 (g) = 0.0938 mol H E All are empirical formulas 0.0375/0.0375 = 1 C x 2 = C 2 0.0938/0.0375 = 2.5 H x 2 = H 5 move for answer Empirical Formula = C 2 H 5 58/29 = 2 Molecular Formula = C 4 H 10

Slide 20 (Answer) / 29 Slide 21 / 29 5 Which of the following is NOT an empirical formula? 6 A compound used in airbags degrades into sodium and nitrogen gas (N 2 ) when ignited. If 3.36 L of N 2 (g) was produced @STP from an initial mass of the compound of A Fe 2 O 3 6.50 grams, what is the empirical formula? B H 2 NNH 2 A Na 2 N 3 Answer C CH 3 OH B Answer B Na 3 N D CH 3 CH 2 Cl C NaN 3 E All are empirical formulas D NaN [This object is a pull tab] E None of these Slide 21 (Answer) / 29 Slide 22 / 29 6 A compound used in airbags degrades into sodium and 7 A compound containing carbon, hydrogen, and chlorine is nitrogen gas (N 2 ) when ignited. If 3.36 L of N 2 (g) was 14.1% carbon by mass, 83.5% Cl, with the rest being produced @STP from an initial mass of the compound of hydrogen. What is the empirical formula? 6.50 grams, what is the empirical formula? A C 2 H 5 Cl A Na 2 N 3 B CH 2 Cl 2 Answer Answer B Na 3 N C C C 2 H 6 Cl C NaN 3 D CH 3 Cl D NaN E None of these [This object is a pull tab] E None of these Slide 22 (Answer) / 29 Slide 23 / 29 7 A compound containing carbon, hydrogen, and chlorine is 8 Hydrazine is a component of rocket fuel. It consists of 14.1% carbon by mass, 83.5% Cl, with the rest being 87.5% N with the rest being hydrogen by mass. If the hydrogen. What is the empirical formula? molecular weight of the compound is 32 grams/mol, what is the molecular formula? A C 2 H 5 Cl A NH 2 Answer B CH 2 Cl 2 Answer B NH 3 B C C 2 H 6 Cl C N 2 H 6 D CH 3 Cl D N 2 H 4 E None of these [This object is a pull tab] E None of these