Fig. 3.11

3 general classes of chemical reactions: Precipitation reactions Ex: geology, heavy metal analysis Solid formation from ionic compounds 2) Acid/base reactions Ex: many biochemical reactions Proton transfer in polar covalent compounds 3) Oxidation/Reduction (“redox”) reactions Ex: batteries, metabolic energy production Electron transfer in ionic & molecular compounds

Classifying Chemical Reactions Fundamental rxn. similarities Precipitation (Ppt) Acid/base (A/B) Redox (R/O) Textbook Similarities in written equations Combination reactions (CR) often redox Decomposition reactions (DR) Single replacement (SR) Double displacement (DD) Acid/base, precipitation, redox Lab

Reactions Involving Ions: Molecular vs. Ionic Equations Chemical Reaction can be expressed by: Molecular Equation (balanced chemical equation) Complete Ionic Equation (showing all ions in reaction) Net Ionic Equation (showing only those ions directly involved in reaction) Consider Copper (III) sulfate reacts with sodium hydroxide to form copper (III) hydroxide and sodium sulfate (all in water). Express reaction in molecular, complete ionic, and net ionic equations

The Solubility of Ionic Compounds in Water The solubility of ionic compounds in water depends upon the relative strengths of the electrostatic forces between ions in the ionic compound and the attractive forces between the ions and water molecules in the solvent. There is a tremendous range in the solubility of ionic compounds in water! The solubility of so called “insoluble” compounds may be several orders of magnitude less than ones that are called “soluble” in water, for example: Solubility of NaCl in water at 20oC = 365 g/L Solubility of MgCl2 in water at 20oC = 542.5 g/L Solubility of AlCl3 in water at 20oC = 699 g/L Solubility of PbCl2 in water at 20oC = 9.9 g/L Solubility of AgCl in water at 20oC = 0.009 g/L Solubility of CuCl in water at 20oC = 0.0062 g/L

Precipitation Reactions: Will a Precipitate Form? If we add a solution containing potassium chloride to a solution containing ammonium nitrate, will we get a precipitate? KCl(aq) + NH4NO3 (aq) = K+(aq) + Cl-(aq) + NH4+(aq) + NO3-(aq) By exchanging cations and anions we see that we could have potassium chloride and ammonium nitrate, or potassium nitrate and ammonium chloride. In looking at the solubility table it shows all possible products as soluble, so there is no net reaction! KCl(aq) + NH4NO3 (aq) = No Reaction! If we mix a solution of sodium sulfate with a solution of barium nitrate, will we get a precipitate? From the solubility table it shows that barium sulfate is insoluble, therefore we will get a precipitate! Na2SO4 (aq) + Ba(NO3)2 (aq) BaSO4 (s) + 2 NaNO3 (aq)

Solubility Soluble = ability to dissolve in a liquid Insoluble = inability to dissolve in a liquid Not all Ionic Compounds are water soluble Not all molecular compounds are insoluble!

8 Simple Rules For Common Ionic Compounds: Predicting Precipitation

AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq) Molecular Equation A chemical equation in which the reactants and products are written as if they were molecular substances, even though they may actually exist in solution as ions. State symbols are include: (s), (l), (g), (aq). For example: AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq) Although AgNO3, NaCl, and NaNO3 exist as ions in aqueous solutions, they are written as compounds in the molecular equation.

AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq) Complete Ionic Equation A chemical equation in which strong electrolytes are written as separate ions in the solution. Other reactants and products are written in molecular form. State symbols are included: (s), (l), (g), (aq). For example: AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq) In ionic form: Ag+(aq) + NO3-(aq) + Na+(aq)Cl-(aq)  AgCl(s) + Na+(aq) + NO3-(aq)

Spectator Ion An ion in an ionic equation that does not take part in the reaction. It appears as both a reactant and a product.

Ag+(aq) + Cl-(aq)  AgCl(s) Net Ionic Equation A chemical equation in which spectator ions are omitted. It shows the reaction that actually occurs at the ionic level. For example: Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq)  AgCl(s) + Na+(aq) + NO3-(aq) In net ionic form: Ag+(aq) + Cl-(aq)  AgCl(s)

Figure 4. 6: Reaction of magnesium chloride and silver nitrate Figure 4.6: Reaction of magnesium chloride and silver nitrate. Photo courtesy of American Color. Write molecular and ionic equations for this reaction. Ionic equation: Ag+(aq) + Cl-(aq) AgCl(s)

Decide whether the following reaction occurs Decide whether the following reaction occurs. If it does, write the molecular, ionic, and net ionic equations. KBr + MgSO4  Determine the product formulas: K+ and SO42- make K2SO4 Mg2+ and Br- make MgBr2 Determine whether the products are soluble: K2SO4 is soluble MgBr2 is soluble KBr + MgSO4  no reaction

Decide whether the following reaction occurs Decide whether the following reaction occurs. If it does, write the molecular, ionic, and net ionic equations. NaOH + MgCl2  Determine the product formulas: Na+ and Cl- make NaCl Mg2+ and OH- make Mg(OH)2 Determine whether the products are soluble: NaCl is soluble Mg(OH)2 is insoluble

2OH-(aq) + Mg2+(aq)  Mg(OH)2(s) Molecular Equation (Balance the reaction and include state symbols) 2NaOH(aq) + MgCl2(aq)  2NaCl(aq) + Mg(OH)2(s) Ionic Equation 2Na+(aq) + 2OH-(aq) + Mg2+(aq) + 2Cl-(aq)  2Na+(aq) + 2Cl-(aq) + Mg(OH)2(s) Net Ionic Equation 2OH-(aq) + Mg2+(aq)  Mg(OH)2(s)

Decide whether the following reaction occurs Decide whether the following reaction occurs. If it does, write the molecular, ionic, and net ionic equations. K3PO4 + CaCl2  Determine the product formulas: K+ and Cl- make KCl Ca2+ and PO43- make Ca3(PO4)2 Determine whether the products are soluble: KCl is soluble Ca3(PO4)2 is insoluble

2PO43-(aq) + 3Ca2+(aq)  Ca3(PO4)2(s) Molecular Equation (Balance the reaction and include state symbols) 2K3PO4(aq) + 3CaCl2(aq)  6KCl(aq) + Ca3(PO4)2(s) Ionic Equation 6K+(aq) + 2PO43-(aq) + 3Ca2+(aq) + 6Cl-(aq)  6K+(aq) + 6Cl-(aq) + Ca3(PO4)2(s) Net Ionic Equation 2PO43-(aq) + 3Ca2+(aq)  Ca3(PO4)2(s)

Mass - Mole Relationships of a Compound For an Element For a Compound Mass (g) of Element Mass (g) of compound Moles of Element Amount (mol) of compound Amount (mol) of compound Molecules (or formula units of compound) Atoms of Element

Calculating the Number of Moles and Atoms in a Given Mass of Element Problem: Tungsten (W) is the element used as the filament in light bulbs, and has the highest melting point of any element 3680oC. How many moles of tungsten, and atoms of the element are contained in a 35.0 mg sample of the metal? Plan: Convert mass into moles by dividing the mass by the atomic weight of the metal, then calculate the number of atoms by multiplying by Avogadro’s number! Solution: Converting from mass of W to moles: Moles of W = 35.0 mg W x = 0.00019032 mol 1.90 x 10 - 4 mol NO. of W atoms = 1.90 x 10 - 4 mol W x = = 1.15 x 1020 atoms of Tungsten 1 mol W 183.9 g W 6.022 x 1023 atoms 1 mole of W

Calcite is a mineral composed of calcium carbonate, CaCO3 Calcite is a mineral composed of calcium carbonate, CaCO3. A sample of calcite composed of pure calcium carbonate weighs 23.6 g. How many moles of calcium carbonate is this? First, find the molar mass of HNO3: 1 Ca 1(40.08) = 40.08 1 C 1(12.01) = 12.01 3 O 3(16.00) = 48.00 100.09 2 decimal places 100.09 g/mol 23

Next, find the number of moles in 23.6 g: 24

The daily requirement of chromium in the human diet is 1. 0 × 10-6 g The daily requirement of chromium in the human diet is 1.0 × 10-6 g. How many atoms of chromium does this represent? 25

(2 significant figures) First, find the molar mass of Cr: 1 Cr 1(51.996) = 51.996 Now, convert 1.0 x 10-6 grams to moles: =1.157781368 x 1016 atoms 1.2 x 1016 atoms (2 significant figures) 26

Chemical Formulas Empirical Formula - Shows the relative number of atoms of each element in the compound. It is the simplest formula, and is derived from masses of the elements. Molecular Formula - Shows the actual number of atoms of each element in the molecule of the compound. Structural Formula - Shows the actual number of atoms, and the bonds between them ; that is, the arrangement of atoms in the molecule. 27

Percentage Composition The mass percentage of each element in the compound The composition is determined by experiment, often by combustion. When a compound is burned, its component elements form oxides—for example, CO2 and H2O. The CO2 and H2O are captured and weighed to determine the amount of C and H in the original compound. 28

Lead(II) chromate, PbCrO4, is used as a paint pigment (chrome yellow) Lead(II) chromate, PbCrO4, is used as a paint pigment (chrome yellow). What is the percentage composition of lead(II) chromate? First, find the molar mass of PbCrO4: 1 Pb 1(207.2) = 207.2 1 Cr 1(51.996) = 51.996 4 O 4(16.00) = 64.00 (1 decimal place) 323.2 g/mol 323.196 29

Now, convert each to percent composition: Check: 64.11 + 16.09 + 19.80 = 100.00 30

Flow Chart of Mass Percentage Calculation Moles of X in one mole of Compound M (g / mol) of X Mass (g) of X in one mole of compound Divide by mass (g) of one mole of compound Mass fraction of X Multiply by 100 Mass % of X

Calculating Mass Percentage and Masses of Elements in a Sample of a Compound - I Problem: Sucrose (C12H22O11) is common table sugar. ( a) What is the mass percent of each element in sucrose? ( b) How many grams of carbon are in 24.35 g of sucrose? (a) Determining the mass percent of each element: mass of C = 12 x 12.01 g C/mol = 144.12 g C/mol mass of H = 22 x 1.008 g H/mol = 22.176 g H/mol mass of O = 11 x 16.00 g O/mol = 176.00 g O/mol 342.296 g/mol Finding the mass fraction of C in Sucrose & % C : Total mass of C 144.12 g C mass of 1 mole of sucrose 342.30 g Cpd Mass Fraction of C = = = 0.421046 To find mass % of C = 0.421046 x 100% = 42.105%

Calculating Mass Percents and Masses of Elements in a Sample of Compound - II (a) continued Mass % of H = x 100% = x 100% = 6.479% H Mass % of O = x 100% = x 100% = 51.417% O (b) Determining the mass of carbon: Mass (g) of C = mass of sucrose X( mass fraction of C in sucrose) Mass (g) of C = 24.35 g sucrose X = 10.25 g C mol H x M of H 22 x 1.008 g H mass of 1 mol sucrose 342.30 g mol O x M of O 11 x 16.00 g O mass of 1 mol sucrose 342.30 g 0.421046 g C 1 g sucrose

Empirical and Molecular Formulas Empirical Formula - The simplest formula for a compound that agrees with the elemental analysis! The smallest set of whole numbers of atoms. Molecular Formula - The formula of the compound as it exists, it may be a multiple of the Empirical formula.

Steps to Determine Empirical Formulas Mass (g) of Element M (g/mol ) Moles of Element use no. of moles as subscripts Preliminary Formula change to integer subscripts Empirical Formula

Empirical Formula (Simplest Formula) The formula of a substance written with the smallest integer subscripts For example: The empirical formula for N2O4 is NO2. The empirical formula for H2O2 is HO 36

Some Examples of Compounds with the Same Elemental Ratio’s Empirical Formula Molecular Formula CH2(unsaturated Hydrocarbons) C2H4 , C3H6 , C4H8 OH or HO H2O2 S S8 P P4 Cl Cl2 CH2O (carbohydrates) C6H12O6

Determining the Empirical Formula Beginning with percent composition: Assume exactly 100 g so percentages convert directly to grams. Convert grams to moles for each element. Manipulate the resulting mole ratios to obtain whole numbers. 38

Manipulating the ratios: Divide each mole amount by the smallest mole amount. If the result is not a whole number: Multiply each mole amount by a factor. For example: If the decimal portion is 0.5, multiply by 2. If the decimal portion is 0.33 or 0.67, multiply by 3. If the decimal portion is 0.25 or 0.75, multiply by 4. 39

Benzene is composed of 92. 3% carbon and 7. 7% hydrogen Benzene is composed of 92.3% carbon and 7.7% hydrogen. What is the empirical formula of benzene? Empirical formula: CH 40

Molecular Formula A formula for a molecule in which the subscripts are whole-number multiples of the subscripts in the empirical formula 41

To determine the molecular formula: Compute the empirical formula weight. Find the ration of the molecular weight to the empirical formula weight. Multiply each subscript of the empirical formula by n. 42

Combustion Train for the Determination of the Chemical Composition of Organic Compounds. 2 m 2 CnHm + (n+ ) O2 = n CO2(g) + H2O(g) Fig. 3.4

Determine % composition by combustion Benzene is a liquid compound composed of carbon and hydrogen; it is used in the preparation of polystyrene plastic. A sample of benzene weighing 342 mg is burned in oxygen and forms 1158 mg of carbon dioxide. What is the percentage composition of benzene? 44

1. Use the mass of CO2 to find the mass of carbon from the benzene. Strategy 1. Use the mass of CO2 to find the mass of carbon from the benzene. 2. Use the mass of benzene and the mass of carbon to find the mass of hydrogen. 3. Use these two masses to find the percent composition. 45

First, find the mass of C in 1156 mg of CO2: = 315.5 mg C 46

Now, we can find the percentage composition: Next, find the mass of H in the benzene sample: 342 mg benzene -315.5 mg C 26.5 mg H (the decimal is not significant) Now, we can find the percentage composition: 47

Determining a Chemical Formula from Combustion Analysis - I Problem: Erthrose (M = 120 g/mol) is an important chemical compound as a starting material in chemical synthesis, and contains Carbon Hydrogen, and Oxygen. Combustion analysis of a 700.0 mg sample yielded 1.027 g CO2 and 0.4194 g H2O. Plan: We find the masses of Hydrogen and Carbon using the mass fractions of H in H2O, and C in CO2. The mass of Carbon and Hydrogen are subtracted from the sample mass to get the mass of Oxygen. We then calculate moles, and construct the empirical formula, and from the given molar mass we can calculate the molecular formula.

Determining a Chemical Formula from Combustion Analysis - II Calculating the mass fractions of the elements: Mass fraction of C in CO2 = = = = 0.2729 g C / 1 g CO2 Mass fraction of H in H2O = = = = 0.1119 g H / 1 g H2O Calculating masses of C and H Mass of Element = mass of compound x mass fraction of element mol C x M of C mass of 1 mol CO2 1 mol C x 12.01 g C/ 1 mol C 44.01 g CO2 mol H x M of H mass of 1 mol H2O 2 mol H x 1.008 g H / 1 mol H 18.02 g H2O

Determining a Chemical Formula from Combustion Analysis - III 0.2729 g C 1 g CO2 Mass (g) of C = 1.027 g CO2 x = 0.2803 g C Mass (g) of H = 0.4194 g H2O x = 0.04693 g H Calculating the mass of O: Mass (g) of O = Sample mass -( mass of C + mass of H ) = 0.700 g - 0.2803 g C - 0.04693 g H = 0.37277 g O Calculating moles of each element: C = 0.2803 g C / 12.01 g C/ mol C = 0.02334 mol C H = 0.04693 g H / 1.008 g H / mol H = 0.04656 mol H O = 0.37277 g O / 16.00 g O / mol O = 0.02330 mol O C0.02334H0.04656O0.02330 = CH2O formula weight = 30 g / formula 120 g /mol / 30 g / formula = 4 formula units / cpd = C4H8O4 0.1119 g H 1 g H2O

Some Compounds with Empirical Formula CH2O (Composition by Mass 40.0% C, 6.71% H, 53.3%O) Molecular M Formula (g/mol) Name Use or Function CH2O 30.03 Formaldehyde Disinfectant; Biological preservative C2H4O2 60.05 Acetic acid Acetate polymers; vinegar ( 5% solution) C3H6O3 90.08 Lactic acid Causes milk to sour; forms in muscle during exercise C4H8O4 120.10 Erythrose Forms during sugar metabolism C5H10O5 150.13 Ribose Component of many nucleic acids and vitamin B2 C6H12O6 180.16 Glucose Major nutrient for energy in cells

Ascorbic Acid ( Vitamin C ) - I Contains C , H , and O Upon combustion in excess oxygen, a 6.49 mg sample yielded 9.74 mg CO2 and 2.64 mg H2O Calculate its Empirical formula! C: 9.74 x10-3g CO2 x(12.01 g C/44.01 g CO2) = 2.65 x 10-3 g C H: 2.64 x10-3g H2O x (2.016 g H2/18.02 gH2O) = 2.92 x 10-4 g H Mass Oxygen = 6.49 mg - 2.65 mg - 0.30 mg = 3.54 mg O

Vitamin C Combustion - II C = 2.65 x 10-3 g C / ( 12.01 g C / mol C ) = = 2.21 x 10-4 mol C H = 0.295 x 10-3 g H / ( 1.008 g H / mol H ) = = 2.92 x 10-4 mol H O = 3.54 x 10-3 g O / ( 16.00 g O / mol O ) = = 2.21 x 10-4 mol O Divide each by 2.21 x 10-4 C = 1.00 Multiply each by 3 = 3.00 = 3.0 H = 1.32 = 3.96 = 4.0 O = 1.00 = 3.00 = 3.0 C3H4O3

H H H C6H8O6

Atoms Molecules Moles Molecular Formula Avogadro’s Number 6.022 x 1023

Chemical Equations Reactants Products 2 H2 (g) + O2 (g) 2 H2O (g) Qualitative Information: Reactants Products States of Matter: (s) solid (l) liquid (g) gaseous (aq) aqueous 2 H2 (g) + O2 (g) 2 H2O (g)

Chemical Equation Calculation - I Atoms (Molecules) Avogadro’s Number 6.02 x 1023 Molecules Reactants Products

Chemical Equation Calculation - II Mass Atoms (Molecules) Molecular Weight Avogadro’s Number g/mol 6.02 x 1023 Molecules Reactants Products Moles

Interpreting a Chemical Equation Stoichiometry The calculation of the quantities of reactants and products involved in a chemical reaction Interpreting a Chemical Equation The coefficients of the balanced chemical equation may be interpreted in terms of either (1) numbers of molecules (or ions or formula units) or (2) numbers of moles, depending on your needs. 59

Information Contained in a Balanced Equation Viewed in Reactants Products terms of: 2 C2H6 (g) + 7 O2 (g) = 4 CO2 (g) + 6 H2O(g) + Energy Molecules 2 molecules of C2H6 + 7 molecules of O2 = 4 molecules of CO2 + 6 molecules of H2O Amount (mol) 2 mol C2H6 + 7 mol O2 = 4 mol CO2 + 6 mol H2O Mass (amu) 60.14 amu C2H6 + 224.00 amu O2 = 176.04 amu CO2 + 108.10 amu H2O Mass (g) 60.14 g C2H6 + 224.00 g O2 = 176.04 g CO2 + 108.10 g H2O Total Mass (g) 284.14g = 284.14g

To find the amount of B (one reactant or product) given the amount of A (another reactant or product): 1. Convert grams of A to moles of A  Using the molar mass of A 2. Convert moles of A to moles of B  Using the coefficients of the balanced chemical equation 3. Convert moles of B to grams of B  Using the molar mass of B 61

C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) Propane, C3H8, is normally a gas, but it is sold as a fuel compressed as a liquid in steel cylinders. The gas burns according to the following equation: C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) How many grams of CO2 are produced when 20.0 g of propane is burned? 62

(3 significant figures) Molar masses C3H8: 3(12.01) + 8(1.008) = 44.094 g CO2: 1(12.01) + 2(16.00) = 44.01 g 59.9 g CO2 (3 significant figures) 63

C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) Propane, C3H8, is normally a gas, but it is sold as a fuel compressed as a liquid in steel cylinders. The gas burns according to the following equation: C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) How many grams of O2 are required to burn 20.0 g of propane? 64

(3 significant figures) Molar masses: O2 2(16.00) = 32.00 g C3H8 3(12.01) + 8(1.008) = 44.094 g 72.6 g O2 (3 significant figures) 65

Once one reactant has been completely consumed, the reaction stops. Limiting Reactant The reactant that is entirely consumed when a reaction goes to completion Once one reactant has been completely consumed, the reaction stops. Any problem giving the starting amount for more than one reactant is a limiting reactant problem. 66

How can we determine the limiting reactant? All amounts produced and reacted are determined by the limiting reactant. How can we determine the limiting reactant? Use each given amount to calculate the amount of product produced. The limiting reactant will produce the lesser or least amount of product. 67

ZrCl4(g) + 2Mg(s)  2MgCl2(s) + Zr(s) Magnesium metal is used to prepare zirconium metal, which is used to make the container for nuclear fuel (the nuclear fuel rods): ZrCl4(g) + 2Mg(s)  2MgCl2(s) + Zr(s) How many moles of zirconium metal can be produced from a reaction mixture containing 0.20 mol ZrCl4 and 0.50 mol Mg? 68

ZrCl4 is the limiting reactant. 0.20 mol Zr will be produced. 69

Urea, CH4N2O, is used as a nitrogen fertilizer Urea, CH4N2O, is used as a nitrogen fertilizer. It is manufactured from ammonia and carbon dioxide at high pressure and high temperature: 2NH3 + CO2(g)  CH4N2O + H2O In a laboratory experiment, 10.0 g NH3 and 10.0 g CO2 were added to a reaction vessel. What is the maximum quantity (in grams) of urea that can be obtained? How many grams of the excess reactant are left at the end of the reactions? 70

CO2 is the limiting reactant. Molar masses NH3 1(14.01) + 3(1.008) = 17.02 g CO2 1(12.01) + 2(16.00) = 44.01 g CH4N2O 1(12.01) + 4(1.008) + 2(14.01) + 1(16.00) = 60.06 g CO2 is the limiting reactant. 13.6 g CH4N2O will be produced. 71

To find the excess NH3, we find how much NH3 reacted: Now subtract the amount reacted from the starting amount: 10.0 at start -7.73 reacted 2.27 g remains 2.3 g NH3 is left unreacted. (1 decimal place) 72

Theoretical Yield The maximum amount of product that can be obtained by a reaction from given amounts of reactants. This is a calculated amount. 73

Actual Yield The amount of product that is actually obtained. This is a measured amount. Percentage Yield 74

(2 significant figures) 2NH3 + CO2(g)  CH4N2O + H2O When 10.0 g NH3 and 10.0 g CO2 are added to a reaction vessel, the limiting reactant is CO2. The theoretical yield is 13.6 of urea. When this reaction was carried out, 9.3 g of urea was obtained. What is the percent yield? Theoretical yield = 13.6 g Actual yield = 9.3 g = 68% yield (2 significant figures) 75

Other Resources Visit the student website at college.hmco.com/pic/ebbing9e 76

The chemical name of table sugar is sucrose, C12H22O11 The chemical name of table sugar is sucrose, C12H22O11. How many grams of carbon are in 68.1 g of sucrose. First, find the molar mass of C12H22O11: 12 C 12(12.01) = 144.12 11 O 11(16.00) = 176.00 22 H 22(1.008) = 22.176 (2 decimal places) 342.30 g/mol 342.296 77

Now, find the mass of carbon in 61.8 g sucrose: 78

Sodium pyrophosphate is used in detergent preparations Sodium pyrophosphate is used in detergent preparations. It is composed of 34.5% Na, 23.3% P, and 42.1% O. What is its empirical formula? Empirical formula Na4P2O7 79

Hexamethylene is one of the materials used to produce a type of nylon Hexamethylene is one of the materials used to produce a type of nylon. It is composed of 62.1% C, 13.8% H, and 24.1% N. Its molecular weight is 116 amu. What is its molecular formula? Empirical formula C3H8N 80

Sodium pyrophosphate is used in detergent preparations Sodium pyrophosphate is used in detergent preparations. It is composed of 34.5% Na, 23.3% P, and 42.1% O. What is its empirical formula? Empirical formula Na4P2O7 81

Hexamethylene is one of the materials used to produce a type of nylon Hexamethylene is one of the materials used to produce a type of nylon. It is composed of 62.1% C, 13.8% H, and 24.1% N. Its molecular weight is 116 amu. What is its molecular formula? Empirical formula C3H8N 82

Molecular formula: C6H16N2 The empirical formula is C3H8N. Find the empirical formula weight: 3(12.01) + 8(1.008) + 1(14.01) = 58.104 amu Molecular formula: C6H16N2 83

C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) Propane, C3H8, is normally a gas, but it is sold as a fuel compressed as a liquid in steel cylinders. The gas burns according to the following equation: C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) How many grams of CO2 are produced when 20.0 g of propane is burned? 84

(3 significant figures) Molar masses C3H8: 3(12.01) + 8(1.008) = 44.094 g CO2: 1(12.01) + 2(16.00) = 44.01 g 59.9 g CO2 (3 significant figures) 85

C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) Propane, C3H8, is normally a gas, but it is sold as a fuel compressed as a liquid in steel cylinders. The gas burns according to the following equation: C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) How many grams of O2 are required to burn 20.0 g of propane? 86

(3 significant figures) Molar masses: O2 2(16.00) = 32.00 g C3H8 3(12.01) + 8(1.008) = 44.094 g 72.6 g O2 (3 significant figures) 87

Converting a Concentrated Solution to a Dilute Solution