JavaScript is disabled on your browser. Enable JavaScript to use this site. Learn more

Chemical Composition

Add to Favourites

Post to:

Description
This ppt cover some of the basic concepts of Chemistry such as what are various compositions of different elements in a compound and what are the various ways to find them.

Comments

Presentation Transcript

Slide 1 : Chapter 6 Chemical Composition

Why Is Knowledge of Composition Important? : Tro's "Introductory Chemistry", Chapter 6 2 Why 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.

How much seed do you plant? : How 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? 3

Counting by Weighing : Tro's "Introductory Chemistry", Chapter 6 4 Counting 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 Al2O3, how much aluminum do you use

Counting Nails by the Pound, Continued : Tro's "Introductory Chemistry", Chapter 6 5 Counting 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:

Counting Nails by the Pound, Continued : Tro's "Introductory Chemistry", Chapter 6 6 Counting Nails by the Pound, Continued The customer bought 2.60 lbs of nails and received 208 nails. He counted the nails by weighing them!

Counting Nails by the Pound, Continued : Tro's "Introductory Chemistry", Chapter 6 7 Counting 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?

Counting Atoms by Moles : Tro's "Introductory Chemistry", Chapter 6 8 Counting 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. The number of atoms we will use is 6.022 x 1023 and we call this a mole. 1 mole = 6.022 x 1023 things. Like 1 dozen = 12 things. Avogadro’s number. Like a kilo = 1000 or a Google = 1×10100

Chemical Packages—Moles : Tro's "Introductory Chemistry", Chapter 6 9 Chemical Packages—Moles Mole = Number of carbon atoms “in” 12 g of C-12. 1 mole protons or 1 mole of neutrons = 1 amu C-12 exactly 6 protons and 6 neutrons since 1 mole × 1 amu = 1 g. 1 mole of C-12 (which is 12 amu) weighs exactly 12 g. In 12 g of C-12 there are 6.022 x1023 C-12 atoms.

Slide 10 : Tro's "Introductory Chemistry", Chapter 6 10 Example 6.1: A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring?

Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring? : Tro's "Introductory Chemistry", Chapter 6 11 Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring? Write down the given quantity and its units. Given: 1.1 x 1022 Ag atoms

Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring? : Tro's "Introductory Chemistry", Chapter 6 12 Write down the quantity to find and/or its units. Find: ? moles Information: Given: 1.1 x 1022 Ag atoms Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring?

Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring? : Tro's "Introductory Chemistry", Chapter 6 13 Collect needed conversion factors: 1 mole Ag atoms = 6.022 x 1023 Ag atoms. Information: Given: 1.1 x 1022 Ag atoms Find: ? moles Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring?

Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring? : Tro's "Introductory Chemistry", Chapter 6 14 Write a solution map for converting the units: Information: Given: 1.1 x 1022 Ag atoms Find: ? moles Conversion Factor: 1 mole = 6.022 x 1023 Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring? atoms Ag moles Ag

Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring? : Tro's "Introductory Chemistry", Chapter 6 15 Apply the solution map: = 1.8266 x 10-2 moles Ag = 1.8 x 10-2 moles Ag Significant figures and round: Information: Given: 1.1 x 1022 Ag atoms Find: ? moles Conversion Factor: 1 mole = 6.022 x 1023 Solution Map: atoms  mole Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring?

Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring? : Tro's "Introductory Chemistry", Chapter 6 16 Check the solution: 1.1 x 1022 Ag atoms = 1.8 x 10-2 moles Ag The units of the answer, moles, are correct. The magnitude of the answer makes sense since 1.1 x 1022 is less than 1 mole. Information: Given: 1.1 x 1022 Ag atoms Find: ? moles Conversion Factor: 1 mole = 6.022 x 1023 Solution Map: atoms  mole Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring?

Practice—Calculate the Number of Atoms in 2.45 Mol of Copper. : Tro's "Introductory Chemistry", Chapter 6 17 Practice—Calculate the Number of Atoms in 2.45 Mol of Copper.

Practice—Calculate the Number of Atoms in 2.45 Mol of Copper, Continued. : Tro's "Introductory Chemistry", Chapter 6 18 Practice—Calculate the Number of Atoms in 2.45 Mol of Copper, Continued. Since atoms are small, the large number of atoms makes sense. 1 mol = 6.022 x 1023 atoms 2.45 mol Cu atoms Cu Check: Solution: Solution Map: Relationships: Given: Find:

Relationship Between Moles and Mass : Tro's "Introductory Chemistry", Chapter 6 19 Relationship 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.

Mole and Mass Relationships : Tro's "Introductory Chemistry", Chapter 6 20 Mole and Mass Relationships 1 mole carbon 12.01 g 1 mole sulfur 32.06 g

Example 6.2—Calculate the Moles of Sulfur in 57.8 g of Sulfur. : Tro's "Introductory Chemistry", Chapter 6 21 Example 6.2—Calculate the Moles of Sulfur in 57.8 g of Sulfur. Since the given amount is much less than 1 mol S, the number makes sense. 1 mol S = 32.07 g 57.8 g S mol S Check: Solution: Solution Map: Relationships: Given: Find:

Practice—Calculate the Moles of Carbon in 0.0265 g of Pencil Lead. : Tro's "Introductory Chemistry", Chapter 6 22 Practice—Calculate the Moles of Carbon in 0.0265 g of Pencil Lead.

Practice—Calculate the Moles of Carbon in 0.0265 g of Pencil Lead, Continued. : Tro's "Introductory Chemistry", Chapter 6 23 Practice—Calculate the Moles of Carbon in 0.0265 g of Pencil Lead, Continued. Since the given amount is much less than 1 mol C, the number makes sense. 1 mol C = 12.01 g 0.0265 g C mol C Check: Solution: Solution Map: Relationships: Given: Find:

Slide 24 : Tro's "Introductory Chemistry", Chapter 6 24 Example 6.3: How many aluminum atoms are in an aluminum can with a mass of 16.2 g?

Example 6.3—How Many Aluminum Atoms Are in a Can Weighing 16.2 g? : Tro's "Introductory Chemistry", Chapter 6 25 Example 6.3—How Many Aluminum Atoms Are in a Can Weighing 16.2 g? Since the given amount is much less than 1 mol Cu, the number makes sense. 1 mol Al = 26.98 g, 1 mol = 6.022 x 1023 16.2 g Al atoms Al Check: Solution: Solution Map: Relationships: Given: Find:

Molar Mass of Compounds : Tro's "Introductory Chemistry", Chapter 6 26 Molar Mass of Compounds 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.

Example 6.4—Calculate the Mass of 1.75 Mol of H2O. : Tro's "Introductory Chemistry", Chapter 6 27 Example 6.4—Calculate the Mass of 1.75 Mol of H2O. Since the given amount is more than 1 mol, the mass being > 18 g makes sense. 1 mol H2O = 18.02 g 1.75 mol H2O g H2O Check: Solution: Solution Map: Relationships: Given: Find:

Practice—How Many Moles Are in 50.0 g of PbO2? (Pb = 207.2, O = 16.00) : Tro's "Introductory Chemistry", Chapter 6 28 Practice—How Many Moles Are in 50.0 g of PbO2? (Pb = 207.2, O = 16.00)

Practice—How Many Moles Are in 50.0 g of PbO2? (Pb = 207.2, O = 16.00), Continued : Tro's "Introductory Chemistry", Chapter 6 29 Practice—How Many Moles Are in 50.0 g of PbO2? (Pb = 207.2, O = 16.00), Continued Since the given amount is less than 239.2 g, the moles being < 1 makes sense. 1 mol PbO2 = 239.2 g 50.0 g mol PbO2 moles PbO2 Check: Solution: Solution Map: Relationships: Given: Find:

Example 6.5—What Is the Mass of 4.78 x 1024 NO2 Molecules? : Tro's "Introductory Chemistry", Chapter 6 30 Example 6.5—What Is the Mass of 4.78 x 1024 NO2 Molecules? Since the given amount is more than Avogadro’s number, the mass > 46 g makes sense. 1 mol NO2 = 46.01 g, 1 mol = 6.022 x 1023 4.78 x 1024 NO2 molecules g NO2 Check: Solution: Solution Map: Relationships: Given: Find:

Counting and ratio’s : Counting 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

Chemical Formulas as Conversion Factors : Tro's "Introductory Chemistry", Chapter 6 32 Chemical Formulas as Conversion Factors 1 spider  8 legs. 1 chair  4 legs. 1 H2O molecule  2 H atoms  1 O atom.

Mole Relationships inChemical Formulas : Tro's "Introductory Chemistry", Chapter 6 33 Mole Relationships inChemical 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.

Example 6.6—Calculate the Moles of Oxygen in 1.7 Moles of CaCO3. : Tro's "Introductory Chemistry", Chapter 6 34 Example 6.6—Calculate the Moles of Oxygen in 1.7 Moles of CaCO3. Since the given amount is much less than 1 mol S, the number makes sense. 1 mol CaCO3 = 3 mol O 1.7 mol CaCO3 mol O Check: Solution: Solution Map: Relationships: Given: Find:

Slide 35 : Tro's "Introductory Chemistry", Chapter 6 35 Example 6.7: Carvone (C10H14O) is the main component in spearmint 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.

Example:Find the mass of carbon in 55.4 g of carvone, (C10H14O). : Tro's "Introductory Chemistry", Chapter 6 36 Example:Find the mass of carbon in 55.4 g of carvone, (C10H14O). Write down the given quantity and its units. Given: 55.4 g C10H14O

Example:Find the mass of carbon in 55.4 g of carvone, (C10H14O). : Tro's "Introductory Chemistry", Chapter 6 37 Write down the quantity to find and/or its units. Find: ? g C Information: Given: 55.4 g C10H14O Example:Find the mass of carbon in 55.4 g of carvone, (C10H14O).

Example:Find the mass of carbon in 55.4 g of carvone, (C10H14O). : Tro's "Introductory Chemistry", Chapter 6 38 Collect needed conversion factors: Molar mass C10H14O = 10(atomic mass C) + 14(atomic mass H) + 1(atomic mass O) = 10(12.01) + 14(1.01) + (16.00) = 150.2 g/mol 1 mole C10H14O = 150.2 g C10H14O 1 mole C10H14O  10 mol C 1 mole C = 12.01 g C Information: Given: 55.4 g C10H14O Find: g C Example:Find the mass of carbon in 55.4 g of carvone, (C10H14O).

Example:Find the mass of carbon in 55.4 g of carvone, (C10H14O). : Tro's "Introductory Chemistry", Chapter 6 39 Write a solution map for converting the units: g C10H14O mol C10H14O mol C Information: Given: 55.4 g C10H14O Find: g C Conversion Factors: 1 mol C10H14O = 150.2 g 1 mol C10H14O  10 mol C 1 mol C = 12.01 g Example:Find the mass of carbon in 55.4 g of carvone, (C10H14O). g C

Example:Find the mass of carbon in 55.4 g of carvone, (C10H14O). : Tro's "Introductory Chemistry", Chapter 6 40 Apply the solution map: = 44.2979 g C = 44.3 g C Significant figures and round: Information: Given: 55.4 g C10H14O Find: g C Conversion Factors: 1 mol C10H14O = 150.2 g 1 mol C10H14O  10 mol C 1 mol C = 12.01 g Solution Map: g C10H14O  mol C10H14O  mol C  g C Example:Find the mass of carbon in 55.4 g of carvone, (C10H14O).

Example:Find the mass of carbon in 55.4 g of carvone, (C10H14O). : Tro's "Introductory Chemistry", Chapter 6 41 Check the solution: 55.4 g C10H14O = 44.3 g C The units of the answer, g C, are correct. The magnitude of the answer makes sense since the amount of C is less than the amount of C10H14O. Information: Given: 55.4 g C10H14O Find: g C Conversion Factors: 1 mol C10H14O = 150.2 g 1 mol C10H14O  10 mol C 1 mol C = 12.01 g Solution Map: g C10H14O  mol C10H14O  mol C  g C Example:Find the mass of carbon in 55.4 g of carvone, (C10H14O).

Percent Composition : Tro's "Introductory Chemistry", Chapter 6 42 Percent 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.

Example 6.9—Find the Mass Percent of Cl in C2Cl4F2. : Example 6.9—Find the Mass Percent of Cl in C2Cl4F2. Since the percentage is less than 100 and Cl is much heavier than the other atoms, the number makes sense. C2Cl4F2 % Cl by mass Check: Solution: Solution Map: Relationships: Given: Find:

Practice—Determine the Mass Percent Composition of the Following: : Tro's "Introductory Chemistry", Chapter 6 44 Practice—Determine the Mass Percent Composition of the Following: CaCl2 (Ca = 40.08, Cl = 35.45)

Practice—Determine the Percent Composition of the Following, Continued: : Tro's "Introductory Chemistry", Chapter 6 45 Practice—Determine the Percent Composition of the Following, Continued: CaCl2

Mass Percent as a Conversion Factor : Tro's "Introductory Chemistry", Chapter 6 46 Mass 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

Empirical Formulas : Tro's "Introductory Chemistry", Chapter 6 47 Empirical Formulas 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.

Empirical (CH2O)-vs- Molecular Formula : 48 molecular formula Formaldehyde CH2O 1 30.03 g/mol Acetic Acid C2H4O2 2 60.06 g/mol Lactic Acid C3H6O3 3 90.09 g/mol Erythrose C4H8O4 4 120.12 g/mol Ribose C5H10O5 5 150.15 g/mol Glucose C6H12O6 6 180.18 g/mol Empirical (CH2O)-vs- Molecular Formula

Empirical Formulas, Continued : 49 Empirical Formulas, Continued Benzene Molecular formula = C6H6 Empirical formula = CH Glucose Molecular formula = C6H12O6 Empirical formula = CH2O Hydrogen Peroxide Molecular formula = H2O2 Empirical formula = HO

Example 6.11—Finding an Empirical Formulafrom Experimental Data : Tro's "Introductory Chemistry", Chapter 6 50 Example 6.11—Finding an Empirical Formulafrom Experimental Data

Slide 51 : Tro's "Introductory Chemistry", Chapter 6 51 Example: A laboratory analysis of aspirin determined the following mass percent composition. Find the empirical formula. C = 60.00% H = 4.48% O = 35.53%

Example:Find the empirical formula of aspirin with the given mass percent composition. : Tro's "Introductory Chemistry", Chapter 6 52 Example: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.

Example:Find the empirical formula of aspirin with the given mass percent composition. : Tro's "Introductory Chemistry", Chapter 6 53 Write down the quantity to find and/or its units. Find: empirical formula, CxHyOz Information: Given: 60.00 g C, 4.48 g H, 35.53 g O Example:Find the empirical formula of aspirin with the given mass percent composition.

Example:Find the empirical formula of aspirin with the given mass percent composition. : Tro's "Introductory Chemistry", Chapter 6 54 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 Information: Given: 60.00 g C, 4.48 g H, 35.53 g O Find: empirical formula, CxHyOz Example:Find the empirical formula of aspirin with the given mass percent composition.

Example:Find the empirical formula of aspirin with the given mass percent composition. : Tro's "Introductory Chemistry", Chapter 6 55 Write a solution map: 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 Example:Find the empirical formula of aspirin with the given mass percent composition. g C mol C g H mol H pseudo- formula empirical formula mole ratio whole number ratio g O mol O

Example:Find the empirical formula of aspirin with the given mass percent composition. : Tro's "Introductory Chemistry", Chapter 6 56 Apply the solution map: Calculate the moles of each element. 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 Solution Map: g C,H,O  mol C,H,O  mol ratio  empirical formula Example:Find the empirical formula of aspirin with the given mass percent composition.

Example:Find the empirical formula of aspirin with the given mass percent composition. : Tro's "Introductory Chemistry", Chapter 6 57 Apply the solution map: Write a pseudoformula. Information: Given: 4.996 mol C, 4.44 mol H, 2.221 mol 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 Solution Map: g C,H,O  mol C,H,O  mol ratio  empirical formula Example:Find the empirical formula of aspirin with the given mass percent composition. C4.996H4.44O2.221

Example:Find the empirical formula of aspirin with the given mass percent composition. : Tro's "Introductory Chemistry", Chapter 6 58 Apply the solution map: Find the mole ratio by dividing by the smallest number of moles. Information: Given: C4.996H4.44O2.221 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 Example:Find the empirical formula of aspirin with the given mass percent composition.

Example:Find the empirical formula of aspirin with the given mass percent composition. : Tro's "Introductory Chemistry", Chapter 6 59 Apply the solution map: Multiply subscripts by factor to give whole number. Information: Given: C2.25H2O1 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 Example:Find the empirical formula of aspirin with the given mass percent composition. { } x 4 C9H8O4

Example 6.12—Finding an Empirical Formulafrom Experimental Data : Tro's "Introductory Chemistry", Chapter 6 60 Example 6.12—Finding an Empirical Formulafrom Experimental Data

Slide 61 : Tro's "Introductory Chemistry", Chapter 6 61 Example: A 3.24-g sample of titanium reacts with oxygen to form 5.40 g of the metal oxide. What is the formula of the oxide?

Example:Find the empirical formula of oxide of titanium with the given elemental analysis. : Tro's "Introductory Chemistry", Chapter 6 62 Example: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

Example: Find the empirical formula of oxide of titanium with the given elemental analysis. : Tro's "Introductory Chemistry", Chapter 6 63 Write down the quantity to find and/or its units. Find: empirical formula, TixOy Information: Given: 3.24 g Ti, 5.40 g compound Example: Find the empirical formula of oxide of titanium with the given elemental analysis.

Example: Find the empirical formula of oxide of titanium with the given elemental analysis. : Tro's "Introductory Chemistry", Chapter 6 64 Collect needed conversion factors: 1 mole Ti = 47.88 g Ti 1 mole O = 16.00 g O Information: Given: 3.24 g Ti, 5.40 g compound Find: empirical formula, TixOy Example: Find the empirical formula of oxide of titanium with the given elemental analysis.

Example:Find the empirical formula of oxide of titanium with the given elemental analysis. : Tro's "Introductory Chemistry", Chapter 6 65 Write a solution map: Information: Given: 3.24 g Ti, 5.40 g compound Find: empirical formula, TixOy Conversion Factors: 1 mol Ti = 47.88g;1 mol O = 16.00g Example:Find the empirical formula of oxide of titanium with the given elemental analysis. g Ti mol Ti pseudo- formula empirical formula mole ratio whole number ratio g O mol O

Example:Find the empirical formula of oxide of titanium with the given elemental analysis. : Tro's "Introductory Chemistry", Chapter 6 66 Apply the solution map: Calculate the mass of each element. Information: Given: 3.24 g Ti, 5.40 g compound Find: empirical formula, TixOy Conversion Factors: 1 mol Ti= 47.88g;1 mol O= 16.00g Solution Map: g Ti,O  mol Ti,O  mol ratio  empirical formula Example:Find the empirical formula of oxide of titanium with the given elemental analysis. 5.40 g compound − 3.24 g Ti = 2.16 g O

Example:Find the empirical formula of oxide of titanium with the given elemental analysis. : Tro's "Introductory Chemistry", Chapter 6 67 Apply the solution map: Calculate the moles of each element. Information: Given: 3.24 g Ti, 2.16 g O Find: empirical formula, TixOy Conversion Factors: 1 mol Ti= 47.88g;1 mol O= 16.00g Solution Map: g Ti,O  mol Ti,O  mol ratio  empirical formula Example:Find the empirical formula of oxide of titanium with the given elemental analysis.

Example:Find the empirical formula of oxide of titanium with the given elemental analysis. : Tro's "Introductory Chemistry", Chapter 6 68 Apply the solution map: Write a pseudoformula. Information: Given: 0.0677 mol Ti, 0.135 mol O Find: empirical formula, TixOy Conversion Factors: 1 mol Ti= 47.88g;1 mol O= 16.00g Solution Map: g Ti,O  mol Ti,O  mol ratio  empirical formula Example:Find the empirical formula of oxide of titanium with the given elemental analysis. Ti0.0677O0.135

Example:Find the empirical formula of oxide of titanium with the given elemental analysis. : Tro's "Introductory Chemistry", Chapter 6 69 Apply the solution map: Find the mole ratio by dividing by the smallest number of moles. Information: Given: 0.0677 mol Ti, 0.135 mol O Find: empirical formula, TixOy Conversion Factors: 1 mol Ti= 47.88g;1 mol O= 16.00g Solution Map: g Ti,O  mol Ti,O  mol ratio  empirical formula Example:Find the empirical formula of oxide of titanium with the given elemental analysis.

Practice—Determine the Empirical Formula of Tin (II) Fluoride, which Contains 75.7% Sn (118.70) and the Rest Fluorine (19.00). : Tro's "Introductory Chemistry", Chapter 6 70 Practice—Determine the Empirical Formula of Tin (II) Fluoride, which Contains 75.7% Sn (118.70) and the Rest Fluorine (19.00).

Practice—Determine the Empirical Formula of Tin (II) Fluoride, which Contains 75.7% Sn (118.70) and the Rest Fluorine (19.00), Continued. : Tro's "Introductory Chemistry", Chapter 6 71 Practice—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: SnxFy Conversion Factors: 1 mol Sn = 118.70 g; 1 mol F = 19.00 g Solution Map: g Sn mol Sn g F mol F pseudo- formula empirical formula mole ratio whole number ratio

Practice—Determine the Empirical Formula of Hematite, which Contains 72.4% Fe (55.85) and the Rest Oxygen (16.00). : Tro's "Introductory Chemistry", Chapter 6 73 Practice—Determine the Empirical Formula of Hematite, which Contains 72.4% Fe (55.85) and the Rest Oxygen (16.00).

Practice—Determine the Empirical Formula of Hematite, which Contains 72.4% Fe (55.85) and the Rest Oxygen (16.00), Continued. : Tro's "Introductory Chemistry", Chapter 6 74 Practice—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: FexOy Conversion Factors: 1 mol Fe = 55.85 g; 1 mol O = 16.00 g Solution Map: g Fe mol Fe g O mol O pseudo- formula empirical formula mole ratio whole number ratio

All These Molecules Have the Same Empirical Formula. How Are the Molecules Different? : Tro's "Introductory Chemistry", Chapter 6 76 All These Molecules Have the Same Empirical Formula. How Are the Molecules Different?

All These Molecules Have the Same Empirical Formula. How Are the Molecules Different?, Continued : Tro's "Introductory Chemistry", Chapter 6 77 All These Molecules Have the Same Empirical Formula. How Are the Molecules Different?, Continued

Molecular Formulas : Tro's "Introductory Chemistry", Chapter 6 78 Molecular 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.

Example—Determine the Molecular Formula of Cadinene if it has a Molar Mass of 204 g and an Empirical Formula of C5H8. : Tro's "Introductory Chemistry", Chapter 6 79 Example—Determine the Molecular Formula of Cadinene if it has a Molar Mass of 204 g and an Empirical Formula of C5H8. Determine the empirical formula. May need to calculate it as previous. C5H8 Determine the molar mass of the empirical formula. 5 C = 60.05, 8 H = 8.064 C5H8 = 68.11 g/mol

Example—Determine the Molecular Formula of Cadinene if it has a Molar Mass of 204 g and an Empirical Formula of C5H8, Continued. : Tro's "Introductory Chemistry", Chapter 6 80 Divide the given molar mass of the compound by the molar mass of the empirical formula. Round to the nearest whole number. Example—Determine the Molecular Formula of Cadinene if it has a Molar Mass of 204 g and an Empirical Formula of C5H8, Continued.

Example—Determine the Molecular Formula of Cadinene if it has a Molar Mass of 204 g and an Empirical Formula of C5H8, Continued. : Tro's "Introductory Chemistry", Chapter 6 81 Multiply the empirical formula by the factor above to give the molecular formula. (C5H8)3 = C15H24 Example—Determine the Molecular Formula of Cadinene if it has a Molar Mass of 204 g and an Empirical Formula of C5H8, Continued.

Practice—Benzopyrene has a Molar Mass of 252 g and an Empirical Formula of C5H3. What is its Molecular Formula? (C = 12.01, H=1.01) : Tro's "Introductory Chemistry", Chapter 6 82 Practice—Benzopyrene has a Molar Mass of 252 g and an Empirical Formula of C5H3. What is its Molecular Formula? (C = 12.01, H=1.01)

Practice—Benzopyrene has a Molar Mass of 252 g and an Empirical Formula of C5H3. What is its Molecular Formula? (C = 12.01, H=1.01), Continued : Tro's "Introductory Chemistry", Chapter 6 83 Practice—Benzopyrene has a Molar Mass of 252 g and an Empirical Formula of C5H3. What is its Molecular Formula? (C = 12.01, H=1.01), Continued C5 = 5(12.01 g) = 60.05 g H3 = 3(1.01 g) = 3.03 g C5H3 = 63.08 g Molecular formula = {C5H3} x 4 = C20H12

Practice—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) : Tro's "Introductory Chemistry", Chapter 6 84 Practice—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)

Practice—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 : Tro's "Introductory Chemistry", Chapter 6 85 Practice—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: CxHyNz Conversion Factors: 1 mol C = 12.01 g; 1 mol H = 1.01 g; 1 mol N = 14.01 g Solution Map: g C mol C g H mol H pseudo- formula empirical formula mole ratio whole number ratio g N mol N

Practice—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. : Tro's "Introductory Chemistry", Chapter 6 86 Practice—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: C6.16H8.6N1.23 C5 = 5(12.01 g) = 60.05 g N1 = 1(14.01 g) = 14.01 g H7 = 7(1.01 g) = 7.07 g C5H7N = 81.13 g {C5H7N} x 2 = C10H14N2