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Chemical Composition

Chemical Composition
Chapter 6 Chemical Composition www.ThesisScientist.comWhy Is Knowledge of Composition Important • All matter is either chemically or physically combined into substances. • Knowing the fraction of material you have can tell you:  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. www.ThesisScientist.comHow much seed do you plant • In a garden you count the seeds by hand. How many seeds would you know to plant in a field www.ThesisScientist.comCounting by Weighing • Building a house requires a lot of nails. • If you know that a single nail weighs .0122 g, than 100 nails weigh 1.22 g, a 1000 nails weigh 12.2 g and so on. • Analogy: You want to make 100 lbs of Al O , how much aluminum do 2 3 you use www.ThesisScientist.comCounting Nails by the Pound, Continued 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: www.ThesisScientist.comCounting Nails by the Pound, Continued 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 www.ThesisScientist.comCounting Nails by the Pound, Continued • What if he bought a different size nail Would the mass of a dozen be 0.150 lbs Would there still be 12 nails in a dozen Would there be 208 nails in 2.60 lbs How would this effect the conversion factors www.ThesisScientist.comCounting Atoms by Moles • If we can find the mass of a particular number of atoms, we can use this information to convert the mass of an element sample to the number of atoms in the sample. 23 • The number of atoms we will use is 6.022 x 10 and we call this a mole. 23 1 mole = 6.022 x 10 things. Like 1 dozen = 12 things. Avogadro’s number. 100 Like a kilo = 1000 or a Google = 1×10 www.ThesisScientist.comChemical Packages—Moles • Mole = Number of carbon atoms ―in‖ 12 g of C12. 1 mole protons or 1 mole of neutrons = 1 amu C12 exactly 6 protons and 6 neutrons since 1 mole × 1 amu = 1 g. 1 mole of C12 (which is 12 amu) weighs exactly 12 g. 23 • In 12 g of C12 there are 6.022 x10 C12 atoms. 23 1 mole 6.02210 atoms 23 1 mole 6.02210 atoms www.ThesisScientist.comExample 6.1: 22 • A silver ring contains 1.1 x 10 silver atoms. How many moles of silver are in the ring www.ThesisScientist.comExample: A silver ring contains 1.1 22 x 10 silver atoms. How many moles of silver are in the ring • Write down the given quantity and its units. 22 Given: 1.1 x 10 Ag atoms www.ThesisScientist.comInformation: Example: 22 Given: 1.1 x 10 Ag atoms A silver ring contains 1.1 22 x 10 silver atoms. How many moles of silver are in the ring • Write down the quantity to find and/or its units. Find: moles www.ThesisScientist.comInformation: Example: 22 Given: 1.1 x 10 Ag atoms A silver ring contains 1.1 22 Find: moles x 10 silver atoms. How many moles of silver are in the ring • Collect needed conversion factors: 23 1 mole Ag atoms = 6.022 x 10 Ag atoms. www.ThesisScientist.comInformation: Example: 22 Given: 1.1 x 10 Ag atoms A silver ring contains 1.1 Find: moles 22 x 10 silver atoms. How Conversion Factor: many moles of silver are 23 1 mole = 6.022 x 10 in the ring • Write a solution map for converting the units: atoms Ag moles Ag 1 mole Ag 23 6.02210 Ag atoms www.ThesisScientist.comInformation: Example: 22 Given: 1.1 x 10 Ag atoms A silver ring contains 1.1 Find: moles 22 x 10 silver atoms. How Conversion Factor: many moles of silver are 23 1 mole = 6.022 x 10 in the ring Solution Map: atoms  mole • Apply the solution map: 1 mole Ag 22 1.110 Ag atoms moles 23 6.02210 Ag atoms 2 = 1.8266 x 10 moles Ag • Significant figures and round: 2 = 1.8 x 10 moles Ag www.ThesisScientist.comInformation: Example: 22 Given: 1.1 x 10 Ag atoms A silver ring contains 1.1 Find: moles 22 x 10 silver atoms. How Conversion Factor: many moles of silver are 23 1 mole = 6.022 x 10 in the ring Solution Map: atoms  mole • Check the solution: 22 2 1.1 x 10 Ag atoms = 1.8 x 10 moles Ag The units of the answer, moles, are correct. The magnitude of the answer makes sense 22 since 1.1 x 10 is less than 1 mole. www.ThesisScientist.comPractice—Calculate the Number of Atoms in 2.45 Mol of Copper. www.ThesisScientist.comPractice—Calculate the Number of Atoms in 2.45 Mol of Copper, Continued. Given: 2.45 mol Cu Find: atoms Cu Solution Map: mol Cu atoms Cu 23 6.02210 atoms 1 mol 23 Relationships: 1 mol = 6.022 x 10 atoms 23 Solution: 6.02210 atoms 2.45 mol Cu 1 mol 24  1.4810 atoms Cu Check: Since atoms are small, the large number of atoms makes sense. www.ThesisScientist.comRelationship Between Moles and Mass • The mass of one mole of atoms is called the molar mass. • The molar mass of an element, in grams, is numerically equal to the element’s atomic mass, in amu. • The lighter the atom, the less a mole weighs. • The lighter the atom, the more atoms there are in 1 g. www.ThesisScientist.comMole and Mass Relationships Substance Pieces in 1 mole Weight of 1 mole 23 Hydrogen 6.022 x 10 atoms 1.008 g 23 Carbon 6.022 x 10 atoms 12.01 g 23 Oxygen 6.022 x 10 atoms 16.00 g 23 Sulfur 6.022 x 10 atoms 32.06 g 23 Calcium 6.022 x 10 atoms 40.08 g 23 Chlorine 6.022 x 10 atoms 35.45 g 23 Copper 6.022 x 10 atoms 63.55 g 1 mole 1 mole carbon sulfur 12.01 g 32.06 g www.ThesisScientist.comExample 6.2—Calculate the Moles of Sulfur in 57.8 g of Sulfur. Given: 57.8 g S Find: mol S Solution Map: g S mol S 1 mol S 32.07 g Relationships: 1 mol S = 32.07 g 1 mol Solution: 57.8 g S 32.07 g  1.80 mol S Check: Since the given amount is much less than 1 mol S, the number makes sense. www.ThesisScientist.comPractice—Calculate the Moles of Carbon in 0.0265 g of Pencil Lead. www.ThesisScientist.comPractice—Calculate the Moles of Carbon in 0.0265 g of Pencil Lead, Continued. Given: 0.0265 g C Find: mol C Solution Map: g C mol C 1 mol 12.01 g Relationships: 1 mol C = 12.01 g 1 mol Solution: 0.0265 g C 12.01 g 3  2.2110 mol C Check: Since the given amount is much less than 1 mol C, the number makes sense. www.ThesisScientist.comExample 6.3: • How many aluminum atoms are in an aluminum can with a mass of 16.2 g www.ThesisScientist.comExample 6.3—How Many Aluminum Atoms Are in a Can Weighing 16.2 g Given: 16.2 g Al Find: atoms Al Solution Map: g Al mol Al atoms Al 23 1 mol 6.02210 atoms 26.98 g 1 mol 23 Relationships: 1 mol Al = 26.98 g, 1 mol = 6.022 x 10 23 Solution: 1 mol Al 6.02210 atoms 16.2 g Al 26.98 g Al 1 mol 23  3.6210 atoms Al Check: Since the given amount is much less than 1 mol Cu, the number makes sense. www.ThesisScientist.comMolar Mass of Compounds • The relative weights of molecules can be calculated from atomic weights. Formula mass = 1 molecule of H O 2 = 2(1.01 amu H) + 16.00 amu O = 18.02 amu. • Since 1 mole of H O contains 2 moles of H and 1 2 mole of O. Molar mass = 1 mole H O 2 = 2(1.01 g H) + 16.00 g O = 18.02 g. www.ThesisScientist.comExample 6.4—Calculate the Mass of 1.75 Mol of H O. 2 Given: 1.75 mol H O 2 Find: g H O 2 Solution Map: mol H O g H O 2 2 18.02 g H 2 1.01 amu 1 mol H O 2 O 1 16.00 amu H O 18.02 amu 2 Relationships: 1 mol H O = 18.02 g 2 Solution: 18.02 g 1.75 mol H O 2 1 mol  31.535 g  31.5 g H O 2 Check: Since the given amount is more than 1 mol, the mass being 18 g makes sense. www.ThesisScientist.comPractice—How Many Moles Are in 50.0 g of PbO (Pb = 207.2, O = 16.00) 2 www.ThesisScientist.comPractice—How Many Moles Are in 50.0 g of PbO 2 (Pb = 207.2, O = 16.00), Continued Given: 50.0 g mol PbO 2 Find: moles PbO 2 Solution Map: g PbO mol PbO 2 2 1 mol PbO 2 Pb 1 209.2 amu 239.2 g O 2 16.00 amu Relationships: 1 mol PbO = 239.2 g 2 PbO 239.2 amu 2 1 mol Solution: 50.0 g PbO 2 239.2 g  0.20903 mol  0.209 mol PbO 2 Check: Since the given amount is less than 239.2 g, the moles being 1 makes sense. www.ThesisScientist.com24 Example 6.5—What Is the Mass of 4.78 x 10 NO Molecules 2 24 Given: 4.78 x 10 NO molecules 2 Find: g NO 2 Solution Map: molecules mol NO g NO 2 2 46.01 g 1 mol NO 2 N 1 14.01 amu 23 6.02210 molec 1 mol O 2 16.00 amu 23 Relationships: 1 mol NO = 46.01 g, 1 mol = 6.022NO x 10 46.01 amu 2 2 Solution: 1 mol 46.01 g 24 4.7810 molec NO 2 23 6.02210 molec 1 mol NO 2  365 g NO 2 Check: Since the given amount is more than Avogadro’s number, the mass 46 g makes sense. www.ThesisScientist.comCounting and ratio’s It takes me .2 gal of gas to get to IVC. It is a very simple ratio: = What if I only had .1 gal 200 2X4’s, 3 sinks, 2 showers, you can make a house with 3 bathrooms and 3 bedrooms. What if you had 12 sinks…how many houses could you make. 200 3 2 = 1 3 3 www.ThesisScientist.comChemical Formulas as Conversion Factors • 1 spider  8 legs. • 1 chair  4 legs. • 1 H O molecule  2 H atoms  1 O atom. 2 www.ThesisScientist.comMole Relationships in Chemical Formulas • 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 Moles of constituents 1 mol NaCl 1 mol Na, 1 mol Cl 1 mol H O 2 mol H, 1 mol O 2 1 mol CaCO 1 mol Ca, 1 mol C, 3 mol O 3 1 mol C H O 6 mol C, 12 mol H, 6 mol O 6 12 6 www.ThesisScientist.comExample 6.6—Calculate the Moles of Oxygen in 1.7 Moles of CaCO . 3 Given: 1.7 mol CaCO 3 Find: mol O Solution Map: mol CaCO mol O 3 3 mol O 1 mol CaCO 3 Relationships: 1 mol CaCO = 3 mol O 3 Solution: 3 mol O 1.7 mol CaCO 3 1 mol CaCO 3  5.1 mol O Check: Since the given amount is much less than 1 mol S, the number makes sense. www.ThesisScientist.comExample 6.7: • Carvone (C H O) is the main component in spearmint 10 14 oil. It has a pleasant odor and mint flavor. It is often added to chewing gum, liqueurs, soaps, and perfumes. Find the mass of carbon in 55.4 g of carvone. www.ThesisScientist.comExample: Find the mass of carbon in 55.4 g of carvone, (C H O). 10 14 • Write down the given quantity and its units. Given: 55.4 g C H O 10 14 www.ThesisScientist.comInformation: Example: Given: 55.4 g C H O Find the mass of carbon in 10 14 55.4 g of carvone, (C H O). 10 14 • Write down the quantity to find and/or its units. Find: g C www.ThesisScientist.comInformation: Example: Given: 55.4 g C H O Find the mass of carbon in 10 14 Find: g C 55.4 g of carvone, (C H O). 10 14 • Collect needed conversion factors: Molar mass C H O = 10(atomic mass C) + 14(atomic mass H) + 10 14 1(atomic mass O) = 10(12.01) + 14(1.01) + (16.00) = 150.2 g/mol 1 mole C H O = 150.2 g C H O 10 14 10 14 1 mole C H O  10 mol C 10 14 1 mole C = 12.01 g C www.ThesisScientist.comInformation: Example: Given: 55.4 g C H O 10 14 Find the mass of carbon in Find: g C 55.4 g of carvone, Conversion Factors: (C H O). 10 14 1 mol C H O = 150.2 g 10 14 1 mol C H O  10 mol C 10 14 1 mol C = 12.01 g • Write a solution map for converting the units: g mol mol g C H O C H O C C 10 14 10 14 10 mol C 12.01 g C 1 mol C H O 10 14 1 mol C H O 1 mol C 150.2 g C H O 10 14 10 14 www.ThesisScientist.comInformation: Example: Given: 55.4 g C H O 10 14 Find the mass of carbon in Find: g C Conversion Factors: 55.4 g of carvone, 1 mol C H O = 150.2 g 10 14 (C H O). 10 14 1 mol C H O  10 mol C 10 14 1 mol C = 12.01 g Solution Map: g C H O  mol C H O  10 14 10 14 mol C  g C • Apply the solution map: 1 mole C H O 10 mol C 12.01 g C 10 14 55.4 g C H O 10 14 150.2 g C H O 1 mol C H O 1 mole C 10 14 10 14 = 44.2979 g C • Significant figures and round: = 44.3 g C www.ThesisScientist.comPercent Composition • Percentage of each element in a compound.  By mass. • Can be determined from:  The formula of the compound.  The experimental mass analysis of the compound. • The percentages may not always total to 100 due to rounding. mass of element X in 1 mol part Percentage100 Percentage100 mass of 1 mol of the compound whole www.ThesisScientist.comExample 6.9—Find the Mass Percent of Cl in C Cl F . 2 4 2 Given: C Cl F 2 4 2 Find: Cl by mass 4 molar mass Cl Solution Map: Mass Cl 100 molar mass C Cl F 2 4 2 mass element X in 1 mol Mass element X 100 Relationships: mass 1 mol of compound 4 molar mass Cl  4(35.45 g/mol) 141.8 g/mol Solution: molar mass C Cl F  2(12.01) 4(35.45) 2(19.00) 203.8 g/mol 2 4 2 141.8 g/mol Mass Cl  100 69.58 203.8 g/mol Check: Since the percentage is less than 100 and Cl is much heavier than the other atoms, the number makes www.ThesisScientist.co se mnse.Practice—Determine the Mass Percent Composition of the Following: CaCl (Ca = 40.08, Cl = 35.45) 2 www.ThesisScientist.comPractice—Determine the Percent Composition of the Following, Continued: molar mass Ca Mass Ca 100 CaCl 2 molar mass CaCl 2 2 molar mass Cl Mass Cl 100 molar mass CaCl 2 2 molar mass Cl  2(35.45 g/mol)  70.90 g/mol molar mass CaCl 1(40.08) 2(35.45)110.98 g/mol 2 40.08 g/mol Mass Ca  100 36.11 110.98 g/mol 70.90 g/mol Mass Cl  100 63.88 110.98 g/mol www.ThesisScientist.comMass Percent as a Conversion Factor • The mass percent tells you the mass of a constituent element in 100 g of the compound. The fact that NaCl is 39 Na by mass means that 100 g of NaCl contains 39 g Na. • This can be used as a conversion factor. 100 g NaCl  39 g Na 39 g Na 100 g NaCl g NaCl  g Na g Na  g NaCl 100 g NaCl 39 g Na www.ThesisScientist.comEmpirical Formulas • The simplest, wholenumber 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. www.ThesisScientist.comEmpirical (CH O) 2 vs Molecular Formula molecular formula Formaldehyde CH O 1 30.03 g/mol 2 Acetic Acid C H O 2 60.06 g/mol 2 4 2 Lactic Acid C H O 3 90.09 g/mol 3 6 3 Erythrose C H O 4 120.12 g/mol 4 8 4 Ribose C H O 5 150.15 g/mol 5 10 5 Glucose C H O 6 180.18 g/mol 6 12 6 www.ThesisScientist.comEmpirical Formulas, Continued Hydrogen Peroxide Molecular formula = H O 2 2 Empirical formula = HO Benzene Molecular formula = C H 6 6 Empirical formula = CH Glucose Molecular formula = C H O 6 12 6 Empirical formula = CH O 2 www.ThesisScientist.comExample 6.11—Finding an Empirical Formula from Experimental Data www.ThesisScientist.comExample: • A laboratory analysis of aspirin determined the following mass percent composition. Find the empirical formula. C = 60.00 H = 4.48 O = 35.53 www.ThesisScientist.comExample: Find the empirical formula of aspirin with the given mass percent composition. • Write down the given quantity and its units. Given: C = 60.00 H = 4.48 O = 35.53 Therefore, in 100 g of aspirin there are 60.00 g C, 4.48 g H, and 35.53 g O. www.ThesisScientist.comInformation: Example: Given: 60.00 g C, 4.48 g H, 35.53 g O Find the empirical formula of aspirin with the given mass percent composition. • Write down the quantity to find and/or its units. Find: empirical formula, C H O x y z www.ThesisScientist.comInformation: Example: Given: 60.00 g C, 4.48 g H, 35.53 g O Find the empirical Find: empirical formula, C H O formula of aspirin with x y z the given mass percent composition. • Collect needed conversion factors: 1 mole C = 12.01 g C 1 mole H = 1.01 g H 1 mole O = 16.00 g O www.ThesisScientist.comInformation: Example: Given: 60.00 g C, 4.48 g H, 35.53 g O Find the empirical Find: empirical formula, C H O x y z formula of aspirin with Conversion Factors: the given mass percent 1 mol C = 12.01 g; 1 mol H = 1.01 g; composition. 1 mol O = 16.00 g • Write a solution map: whole g C mol C mole number empirical pseudo ratio ratio g H mol H formula formula g O mol O www.ThesisScientist.comInformation: Example: Given: 60.00 g C, 4.48 g H, 35.53 g O Find the empirical Find: empirical formula, C H O x y z formula of aspirin with Conversion Factors: the given mass percent 1 mol C = 12.01 g; 1 mol H = 1.01 g; 1 mol O = 16.00 g composition. 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 60.00 g 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 www.ThesisScientist.comInformation: Example: Given: 4.996 mol C, 4.44 mol H, Find the empirical 2.221 mol O formula of aspirin with Find: empirical formula, C H O x y z Conversion Factors: the given mass percent 1 mol C = 12.01 g; composition. 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:  Write a pseudoformula. C H O 4.996 4.44 2.221 www.ThesisScientist.comInformation: Example: Given: C H O 4.996 4.44 2.221 Find the empirical Find: empirical formula, C H O x y z formula of aspirin with Conversion Factors: the given mass percent 1 mol C = 12.01 g; 1 mol H = 1.01 g; 1 mol O = 16.00 g composition. Solution Map: g C,H,O  mol C,H,O  mol ratio  empirical formula • Apply the solution map:  Find the mole ratio by dividing by the smallest number of moles. C H O 4.996 4.44 2.221 2.221 2.221 2.221 C H O 2.25 2 1 www.ThesisScientist.comInformation: Example: Given: C H O 2.25 2 1 Find the empirical Find: empirical formula, C H O x y z formula of aspirin with Conversion Factors: the given mass percent 1 mol C = 12.01 g; 1 mol H = 1.01 g; 1 mol O = 16.00 g composition. Solution Map: g C,H,O  mol C,H,O  mol ratio  empirical formula • Apply the solution map:  Multiply subscripts by factor to give whole number. C H O x 4 2.25 2 1 C H O 9 8 4 www.ThesisScientist.comExample 6.12—Finding an Empirical Formula from Experimental Data www.ThesisScientist.comExample: • A 3.24g sample of titanium reacts with oxygen to form 5.40 g of the metal oxide. What is the formula of the oxide www.ThesisScientist.comExample: Find the empirical formula of oxide of titanium with the given elemental analysis. • Write down the given quantity and its units. Given: Ti = 3.24 g compound = 5.40 g www.ThesisScientist.comInformation: Example: Given: 3.24 g Ti, 5.40 g compound Find the empirical formula of oxide of titanium with the given elemental analysis. • Write down the quantity to find and/or its units. Find: empirical formula, Ti O x y www.ThesisScientist.comInformation: Example: Given: 3.24 g Ti, 5.40 g compound Find the empirical Find: empirical formula, Ti O formula of oxide of x y titanium with the given elemental analysis. • Collect needed conversion factors: 1 mole Ti = 47.88 g Ti 1 mole O = 16.00 g O www.ThesisScientist.comExample: Information: Find the empirical Given: 3.24 g Ti, 5.40 g compound formula of oxide of Find: empirical formula, Ti O x y titanium with the Conversion Factors: given elemental 1 mol Ti = 47.88g;1 mol O = 16.00g analysis. • Write a solution map: whole g Ti mol Ti mole number empirical pseudo ratio ratio formula formula g O mol O www.ThesisScientist.comInformation: Example: Given: 3.24 g Ti, 5.40 g compound Find the empirical Find: empirical formula, Ti O x y formula of oxide of Conversion Factors: titanium with the 1 mol Ti= 47.88g;1 mol O= 16.00g given elemental Solution Map: g Ti,O  mol Ti,O  analysis. mol ratio  empirical formula • Apply the solution map:  Calculate the mass of each element. 5.40 g compound − 3.24 g Ti = 2.16 g O www.ThesisScientist.comInformation: Example: Given: 3.24 g Ti, 2.16 g O Find the empirical Find: empirical formula, Ti O x y formula of oxide of Conversion Factors: titanium with the 1 mol Ti= 47.88g;1 mol O= 16.00g given elemental Solution Map: g Ti,O  mol Ti,O  analysis. mol ratio  empirical formula • Apply the solution map:  Calculate the moles of each element. 1 mol Ti 3.24 g Ti 0.0677 mol Ti 47.88 g Ti 1 mol O 2.16 g O 0.135 mol O 16.00 g O www.ThesisScientist.comInformation: Example: Given: 0.0677 mol Ti, 0.135 mol O Find the empirical Find: empirical formula, Ti O x y formula of oxide Conversion Factors: of titanium with 1 mol Ti= 47.88g;1 mol O= 16.00g the given Solution Map: g Ti,O  mol Ti,O  elemental analysis. mol ratio  empirical formula • Apply the solution map:  Write a pseudoformula. Ti O 0.0677 0.135 www.ThesisScientist.comInformation: Example: Given: 0.0677 mol Ti, 0.135 mol O Find the empirical Find: empirical formula, Ti O x y formula of oxide of Conversion Factors: titanium with the 1 mol Ti= 47.88g;1 mol O= 16.00g given elemental Solution Map: g Ti,O  mol Ti,O  analysis. mol ratio  empirical formula • Apply the solution map:  Find the mole ratio by dividing by the smallest number of moles. Ti O 0.0677 0.135 0.0677 0.0677 Ti O 1 2 www.ThesisScientist.comPractice—Determine the Empirical Formula of Tin (II) Fluoride, which Contains 75.7 Sn (118.70) and the Rest Fluorine (19.00). www.ThesisScientist.comPractice—Determine the Empirical Formula of Tin (II) Fluoride, which Contains 75.7 Sn (118.70) and the Rest Fluorine (19.00), Continued. Given: 75.7 Sn, (100 – 75.3) = 24.3 F  in 100 g Tin (II) fluoride there are 75.7 g Sn and 24.3 g F. Find: Sn F x y Conversion Factors: 1 mol Sn = 118.70 g; 1 mol F = 19.00 g Solution Map: whole mole number g Sn mol Sn empirical pseudo ratio ratio formula formula g F mol F www.ThesisScientist.comPractice—Determine the Empirical Formula of Tin (II) Fluoride, which Contains 75.7 Sn (118.70) and the Rest Fluorine (19.00), Continued. Apply solution map: 1 mol Sn 75.7 g Sn 0.638 mol Sn 118.70 g Sn F 0.638 1.28 1 mol F 24.3 g F 1.28 mol F 19.00 g Sn F  Sn F 0.638 1.28 1 2 0.638 0.638 SnF 2 www.ThesisScientist.comPractice—Determine the Empirical Formula of Hematite, which Contains 72.4 Fe (55.85) and the Rest Oxygen (16.00). www.ThesisScientist.comPractice—Determine the Empirical Formula of Hematite, which Contains 72.4 Fe (55.85) and the Rest Oxygen (16.00), Continued. Given: 72.4 Fe, (100 – 72.4) = 27.6 O  in 100 g hematite there are 72.4 g Fe and 27.6 g O. Find: Fe O x y Conversion Factors: 1 mol Fe = 55.85 g; 1 mol O = 16.00 g Solution Map: whole mole number g Fe mol Fe pseudo empirical ratio ratio formula formula g O mol O www.ThesisScientist.comPractice—Determine the Empirical Formula of Hematite, which Contains 72.4 Fe (55.85) and the Rest Oxygen (16.00), Continued. Apply solution map: 1 mol Fe 72.4 g Fe 1.30 mol Fe 55.85 g Fe O 1.30 1.73 1 mol O 26.7 g O 1.73 mol O 16.00 g Fe O  Fe O 1.30 1.73 1 1.33 1.30 1.30 Fe O3 Fe O 1 1.33 3 4 www.ThesisScientist.comAll These Molecules Have the Same Empirical Formula. How Are the Molecules Different Name Molecular Empirical Formula Formula Glyceraldehyde C H O CH O 3 6 3 2 Erythrose C H O CH O 4 8 4 2 Arabinose C H O CH O 5 10 5 2 Glucose C H O CH O 6 12 6 2 www.ThesisScientist.comAll These Molecules Have the Same Empirical Formula. How Are the Molecules Different, Continued Name Molecular Empirical Molar Formula Formula Mass, g Glyceraldehyde C H O CH O (30) 90 3 6 3 2 Erythrose C H O CH O 120 4 8 4 2 Arabinose C H O CH O 150 5 10 5 2 Glucose C H O CH O 180 6 12 6 2 www.ThesisScientist.comMolecular Formulas • The molecular formula is a multiple of the empirical formula. • To determine the molecular formula, you need to know the empirical formula and the molar mass of the compound. Molar mass real formula = Factor used to multiply subscripts Molar mass empirical formula www.ThesisScientist.comExample—Determine the Molecular Formula of Cadinene if it has a Molar Mass of 204 g and an Empirical Formula of C H . 5 8 1. Determine the empirical formula.  May need to calculate it as previous. C H 5 8 2. Determine the molar mass of the empirical formula. 5 C = 60.05, 8 H = 8.064 C H = 68.11 g/mol 5 8 www.ThesisScientist.comExample—Determine the Molecular Formula of Cadinene if it has a Molar Mass of 204 g and an Empirical Formula of C H , 5 8 Continued. 3. Divide the given molar mass of the compound by the molar mass of the empirical formula.  Round to the nearest whole number. 204 g/mol  3 68.11 g/mol www.ThesisScientist.comExample—Determine the Molecular Formula of Cadinene if it has a Molar Mass of 204 g and an Empirical Formula of C H , 5 8 Continued. 4. Multiply the empirical formula by the factor above to give the molecular formula. (C H ) = C H 5 8 3 15 24 www.ThesisScientist.comPractice—Benzopyrene has a Molar Mass of 252 g and an Empirical Formula of C H . What is its 5 3 Molecular Formula (C = 12.01, H=1.01) www.ThesisScientist.comPractice—Benzopyrene has a Molar Mass of 252 g and an Empirical Formula of C H . What is its 5 3 Molecular Formula (C = 12.01, H=1.01), Continued C = 5(12.01 g) = 60.05 g 5 H = 3(1.01 g) = 3.03 g 3 C H = 63.08 g 5 3 252 g/mol n 4 63.08 g/mol Molecular formula = C H x 4 = C H 5 3 20 12 www.ThesisScientist.comPractice—Determine the Molecular Formula of Nicotine, which has a Molar Mass of 162 g and is 74.0 C, 8.7 H, and the Rest N. (C=12.01, H=1.01, N=14.01) www.ThesisScientist.comPractice—Determine the Molecular Formula of Nicotine, which has a Molar Mass of 162 g and is 74.0 C, 8.7 H, and the Rest N, Continued Given: 74.0 C, 8.7 H, 100 – (74.0+8.7) = 17.3 N  in 100 g nicotine there are 74.0 g C, 8.7 g H, and 17.3 g N. Find: C H N x y z Conversion Factors: 1 mol C = 12.01 g; 1 mol H = 1.01 g; 1 mol N = 14.01 g Solution Map: whole g C mol C mole number pseudo empirical ratio ratio g H mol H formula formula g N mol N www.ThesisScientist.comPractice—Determine the Molecular Formula of Nicotine, which has a Molar Mass of 162 g and is 74.0 C, 8.7 H, and the Rest N, Continued. Apply solution map: 1 mol C C = 5(12.01 g) = 60.05 g 5 74.0 g C 6.16 mol C 12.01 g N = 1(14.01 g) = 14.01 g 1 1 mol H H = 7(1.01 g) = 7.07 g 7 8.7 g H 8.6 mol H C H N = 81.13 g 1.01 g 5 7 1 mol N 17.3 g N 1.23 mol N 14.01 g mol. mass nicotine 162 g  2 C H N mol. mass emp. form. 81.13 g 6.16 8.6 1.23 C H N  C H N 6.16 8.6 1.23 5 7 C H N x 2 = C H N 5 7 10 14 2 1.23 1.23 1.23 www.ThesisScientist.com
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