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Chapter six DAY 1: 6.1-6.3 •H O W M U C H S O D I U M ? •C O U N T I N G N A I L S B Y T H E P O U N D •C O U N T I N G A T O M S B Y T H E G R A M Why Is Knowledge of Composition Important? 2 Some Applications: The amount of sodium in sodium chloride for diet. The amount of iron in iron ore for steel production. The amount of hydrogen in water for hydrogen fuel. The amount of chlorine in freon to estimate ozone depletion. Tro's "Introductory Chemistry", Chapter 6 Counting Nails by the Pound 3 A hardware store customer buys 2.60 pounds of nails. A dozen nails has a mass of 0.150 pounds. How many nails did the customer buy? 1 dozen nails = 0.150 lbs. 12 nails = 1 dozen nails Solution map: Tro's "Introductory Chemistry", Chapter 6 Counting Nails by the Pound, Continued 4 1 doz. nails 12 nails 2.60 lbs. 208 nails 0.150 lbs. 1 doz. The customer bought 2.60 lbs of nails and received 208 nails. He counted the nails by weighing them! Tro's "Introductory Chemistry", Chapter 6 Counting Atoms by Moles 5 If we can find the mass of a particular number of atoms, we can convert the mass of an element sample to the number of atoms in the sample. The conversion factor we will use is the MOLE 1 mole = 6.022 x 1023 things. Like 1 dozen = 12 things. Avogadro’s number. Tro's "Introductory Chemistry", Chapter 6 Moles- as a conversion factor 6 Mole = Number of things equal to the number of atoms in 12 g of C-12. 1 atom of C-12 weighs exactly 12 amu. 1 mole of C-12 weighs exactly 12 g. In 12 g of C-12 there are 6.022 x1023 C-12 atoms. 6.022 10 atoms 1 mole 23 Tro's "Introductory Chemistry", Chapter 6 1 mole 6.022 10 23 atoms Conversions When converting between moles and number of atoms Use the conversion factor: 1 mole = 6.02 x1023 atoms Practice: A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring? Conversions continued When converting from grams to moles of an element Use the conversion factor: Molar mass (atomic mass in grams) = 1 mole Practice: Calculate the number of moles of sulfur in 57.8 g of sulfur. Conversions, continued When converting between grams and number of atoms Use two conversion factors molar mass= 1 mole and then 1 mole = 6.022 x1023 atoms Practice: How many aluminum atoms are in an aluminum can with a mass of 16.2 g? Day 2: 6.4-6.5 •C O U N T I N G M O L E C U L E S B Y T H E G R A M •C H E M I C A L F O R M U L A S A S C O N V E R S I O N FACTORS Molar Mass of Compounds 12 The relative weights of molecules can be calculated from atomic weights. Formula mass = 1 molecule of H2O = 2(1.01 amu H) + 16.00 amu O = 18.02 amu. Since 1 mole of H2O contains 2 moles of H and 1 mole of O. Molar mass = 1 mole H2O = 2(1.01 g H) + 16.00 g O = 18.02 g. Tro's "Introductory Chemistry", Chapter 6 Conversions When converting from grams to moles of a compound Use the conversion factor: Formula mass (in grams) = 1 mole Practice: Calculate the number of moles of Hydrogen in 57.8 g of water. Conversions, continued When converting between grams and number of molecules Use two conversion factors formula mass= 1 mole and then 1 mole = 6.022 x1023 molecules Practice: How many carbon dioxide molecules are in 45.3g of dry ice (carbon dioxide)? Chemical Formulas as Conversion Factors 15 1 spider 8 legs. 1 chair 4 legs. 1 H2O molecule 2 H atoms 1 O atom. Tro's "Introductory Chemistry", Chapter 6 Counting Parts 16 We can use the total number to determine the number of each part: When all the desks in the room have 4 legs, if there are 30 desks in the room, there will be 120 legs (4 x 30). Since every H2O molecule has 2 H atoms, in 100 H2O molecules, there are 200 H atoms. In 1 mole of H2O molecules, there are 2 moles of H atoms. Tro's "Introductory Chemistry", Chapter 6 Mole Relationships in Chemical Formulas 17 Since we count atoms and molecules in mole units, we can find the number of moles of a constituent element if we know the number of moles of the compound. Moles of compound 1 mol NaCl 1 mol H2O 1 mol CaCO3 Moles of constituents 1 mol Na, 1 mol Cl 2 mol H, 1 mol O 1 mol Ca, 1 mol C, 3 mol O 1 mol C6H12O6 6 mol C, 12 mol H, 6 mol O Tro's "Introductory Chemistry", Chapter 6 Converting from moles of compound to moles of (constituent) part Calculate the Moles of Oxygen in 1.7 Moles of CaCO3 1.7 moles of CaCO3 x 3 mole O = 5.1 mol O 1 mole CaCO3 Practice: How many grams of Cl are in CF3Cl? How many grams of sodium are in 24g of sodium phosphate? Day 3: QUIZ 6.1-6.5 SECTION: 6.6 & 6.7 Mass Percent This is the % of the compound that is composed of one of its elements. Mass % = mass of one element x 100 mass of compound sample Mass percent is a conversion factor too! 100g of the compound/ mass % of element Practice What is the mass percent of calcium in calcium nitrate? The atomic mass of calcium is 40.08g The formula mass of calcium nitrate is 164g 40.08g x 1oo = 24.4% 164 g Using %mass as a conversion factor Silver Chloride, used in silver plating, contains 75.27% Ag. Calculate the mass of silver chloride in grams, required to make 4.8 g of silver plating. 4.8 g Ag x 100 g AgCl = 6.38g AgCl 75.27 g Ag Mass % in a compound Mass of single element/ formula mass of compound Practice: What is the mass % of Cl in: C2Cl4F2 Day 4: BAGGIE LAB & WRITE UP Day 5 6.8-6.9 EMPIRICAL FORMULAS Empirical Formulas 28 The simplest, whole-number ratio of atoms in a molecule is called the empirical formula. Can be determined from percent composition or combining masses. The molecular formula is a multiple of the empirical formula. %A 100g mass A (g) Molar MassA moles A moles A moles B %B 100g mass B (g) Tro's "Introductory Chemistry", Chapter 6 Molar MassB moles B Empirical Formulas, Continued 29 Hydrogen Peroxide Molecular formula = H2O2 Empirical formula = HO Glucose Molecular formula = C6H12O6 Empirical formula = CH2O Benzene Molecular formula = C6H6 Empirical formula = CH Finding an Empirical Formula 30 1. Convert the percentages to grams. a. Skip if already grams. 2. Convert grams to moles. a. Use molar mass of each element. 3. Write a pseudoformula using moles as subscripts. 4. Divide all by smallest number of moles. 5. Multiply all mole ratios by number to make all whole numbers, if necessary. a. If ratio ?.5, multiply all by 2; if ratio ?.33 or ?.67, multiply all by 3, etc. b. Skip if already whole numbers after Step 4. Tro's "Introductory Chemistry", Chapter 6 Empirical Formula from Data Empirical formula is a ratio of moles not mass Convert any data in grams into moles (use molar mass) Look at the number of moles and divide the number of moles by the smallest in the formula If the number you are left with is a decimal, multiply by a whole number (see previous slide) to determine subscript. Find the empirical formula of aspirin with the given mass percent composition. 32 A laboratory analysis of aspirin determined the following mass percent composition: Carbon= 60.00%, hydrogen= 4.48%, and oxygen = 35.53% (Remember % mass= g of element/100g compound) gC gH gO Information: Given:60.00 g C, 4.48 g H, 35.53 g O Find: empirical formula, CxHyOz Conversion Factors: 1 mol C = 12.01 g; 1 mol H = 1.01 g; 1 mol O = 16.00 g mol C mol H mol O Tro's "Introductory Chemistry", Chapter 6 pseudoformula mole ratio whole number ratio empirical formula Example: Find the empirical formula of aspirin with the given mass percent composition. 33 Information: Given: 60.00 g C, 4.48 g H, 35.53 gO Find: empirical formula, CxHyOz Conversion Factors: 1 mol C = 12.01 g; 1 mol H = 1.01 g; 1 mol O = 16.00 g Solution Map: g C,H,O mol C,H,O mol ratio empirical formula Apply the solution map: Calculate the moles of each element. 1 mol C 4.996 mol C 12.01 g C 1 mol H 4.48 g H 4.44 mol H 1.01 g H 1 mol O 35.53 g O 2.221 mol O 16.00 g O 60.00 g C Tro's "Introductory Chemistry", Chapter 6 Practice—Determine the Empirical Formula of Stannous Fluoride, which Contains 75.7% Sn (118.70) and the Rest Fluorine (19.00) 34 Given: 75.7% Sn, (100 – 75.3) = 24.3% F in 100 g stannous fluoride there are 75.7 g Sn and 24.3 g F. Find: SnxFy Conversion Factors: 1 mol Sn = 118.70 g; 1 mol F = 19.00 g Solution Map: g Sn gF mol Sn mol F Tro's "Introductory Chemistry", Chapter 6 pseudoformula whole mole number empirical ratio ratio formula Calculating the molecular formula If you are given the empirical formula you can figure out the molecular formula. Empirical formula x n = molecular formula Molar mass of compound . =n Formula mass of empirical formula Practice The empirical formula for Butane is C2H5, and has a molar mass of 58.12 g/mol. What is the molecular formula? 58.12 g = 2 29.07 g C2H5 x 2 = C4H10 Day 6 EMPIRICAL LAB Day 7:Lab write up Day 8: Review Day 9: Chapter 5/6 Test