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Molecules, Compounds, and Chemical Equations

Molecules, Compounds, and Chemical Equations. Chapter 3 Chapter 4.5-4.9 And Chapter 18.2. Molecular View of Elements and Compounds. Practice problems Tro – 3.27-3.32. P 4 , S 8 , and Se 8 are the polyatomic elements. Combining Elements to make Compounds. 2 H 2 + O 2  2 H 2 O

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Molecules, Compounds, and Chemical Equations

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  1. Molecules, Compounds, and Chemical Equations Chapter 3 Chapter 4.5-4.9 And Chapter 18.2

  2. Molecular View of Elements and Compounds Practice problems Tro – 3.27-3.32

  3. P4, S8, and Se8 are the polyatomic elements

  4. Combining Elements to make Compounds 2 H2 + O2 2 H2O • the properties of the compound are totally different from the constituent elements

  5. Formation of Water from Its Elements

  6. Chemical Bonds • Forces of attraction holding two or more atoms together • ionic bonds result when electrons are transferred from one atom to another, resulting in oppositely charged ions that are held together by electrostatic attractions • generally found when metal atoms bonded to nonmetal atoms • covalent bonds result when two atoms share some of their electrons • generally found when nonmetal atoms bonded together

  7. Ionic vs. Molecular Compounds Propane – contains individual C3H8 molecules Table salt – contains an array of Na+ ions and Cl- ions

  8. Chemical Formulas • compounds are generally represented by a chemical formula • the amount of information about the structure of the compound varies with the type of formula

  9. Examples of different molecule representations

  10. Molar Mass • The sum of the atomic masses for all the atoms represented in the chemical formula of a compound. • AKA formula mass, molecular mass, molecular weight Practice problems Tro – 3.57-3.58

  11. Molecular Mass Determination • Nitrogen fixation in the root nodules of peas and other legumes occurs with a reaction involving a molybdenum containing enzyme named nitrogenase. This enzyme contains two Mo atoms per molecule and is 0.0872% Mo by mass. What is the molar mass of the enzyme? Practice problems Tro – 3.67-3.72, 3.113-3.116, 3.127, 3.134, 3.135

  12. Empirical Formulas • Simplest formula for a compound. • Can be determined from percent composition data and combustion analysis.

  13. Problem 3.81a • Calculate the empirical formula for nicotine given the following mass percent composition: • C 58.80% • H 8.70% • N 17.27% Practice problems Tro – 3.79-3.84, 3.126

  14. A compound of Ca, C, N, and S was subjected to quantitative analysis and formula mass determination, and the following data were obtained. A 0.250 g sample was mixed with NaCO3 to convert all of the Ca to 0.160 g of CaCO3. A 0.115 g sample of the compound was carried through a series of reactions until all of its S was changed to 0.344g of BaSO4. A 0.712 g sample was processed to liberate all of its N as NH3, and 0.155 g NH3 was obtained. The formula mass was found to be 156. Determine the empirical and molecular formulas of this compound. Practice problems Tro – 3.85-3.90, 3.117-3.120, 3.124

  15. Some compounds can be decomposed quantitatively with water or acid to give known compounds. Suppose you have a 0.643 g sample of a compound known to be composed of C, H, Al, and Cl. Furthermore, you know that it is composed of some number of CH3 groups and chlorine atoms per aluminum atom. The formula could be written as (CH3)xAlCly. To find x and y you decompose the sample with acid in water. The CH3 portion is evolved as methane gas, CH4, and the chloride ions remain in the water. The chloride ions are precipitated as AgCl by adding AgNO3 to the solution. The data collected in the experiment are given here. What are the values of x and y?

  16. (CH3) xAlCly x CH4(g) + Al3+(aq) + y Cl-(aq) • 0.643g 0.222g  • AgNO3 •  • AgCl(s) • 0.996g Practice problems Tro – 3.121-3.122, 3.130

  17. Chemical Reactions • Chemical reactions are processes in which one set of chemicals are converted to a new set of chemicals • Chemical reactions are described by chemical equations.

  18. Chemical Equations 2 C8H18(l) + 25 O2 (g)16 CO2 (g) +18 H2O(g) • must be balanced to satisfy Law of conservation of mass • state designations • (g) gas • (l) liquid • (s) solid • (aq) aqueous

  19. Pure silicon, which is needed in the manufacturing of electronic components, may be prepared by heating silicon dioxide (sand) with carbon at high temperatures, releasing carbon monoxide gas. Write the balanced chemical equation for this process.

  20. Carbon tetrachloride was widely used for many years as a solvent until its harmful properties became well established. Carbon tetrachloride may be prepared by the reaction of natural gas (methane, CH4) and elemental chlorine gas in the presence of ultraviolet light. Write a balanced chemical equation for this process.

  21. Crude gunpowders often contain a mixture of potassium nitrate and charcoal (carbon). When such a mixture is heated until reaction occurs, a solid residue of potassium carbonate is produced. The explosive force of the gunpowder comes from the fact that two gases are also produced (carbon monoxide and nitrogen), which increase in volume with great force and speed. Write the balanced chemical equation for this reaction. Practice problems Tro – 3.91-3.100

  22. Classifying Reactions by Type of Chemistry

  23. Classifying Reactions by Type of Chemistry • Precipitation AX + BZ  AZ + BX • Acid Base HX +BOH  BX +H2O • Gas Evolution • H2X + BCO3 H2O + CO2(g) + BX • H2X + BSO3 H2O + SO2(g) + BX • NH4X + BOH  H2O + NH3(g) + BX • Oxidation Reduction A+2 + B  A + B+2 • Combustion CxHxOx+O2CO2 + H2O

  24. Classifying Reactions by what Atoms Do

  25. Classifying Reactions by what Atoms Do • Combination/Synthesis A + Z  AZ • Decomposition AZ  A + Z • Single Displacement A + BZ  AZ + B • Double displacement AX + BZ  AZ + BX • Neutralization HX +BOH  BX +H2O

  26. REDOX Reactions • Oxidation number - In order to keep track of electrons in chemical reactions, chemists assign an oxidation number to each element.

  27. Determining Oxidation States • Oxidation number of an element in its native state is zero. • Alkali metals have an oxidation number of +1 • Alkaline earth elements have an oxidation number of +2 • The oxidation number of monatomic ions is the same as the charge.

  28. Determining Oxidation States • Fluorine is 1 except for F2. • Cl, Br, and I are 1 in binary compounds. • O is usually 2 (except for peroxides O22 and superoxides O21). • H is usually +1 (except for hydrides H1) • The sum of the oxidation numbers equals the charge on ion or molecule.

  29. Try some • Ga2O3 Fe2(CrO4)3 • K2MnO4 Hg2(BrO3)2 • H2PO4- KClO4 Practice problems Tro – 4.83-4.86

  30. Oxidation – process in which an element loses one or more electrons with an increase in the oxidation number. • Reduction – process in which an element gains one or more electrons with a decrease in oxidation number.

  31. Oxidizing agent – Substance that causes another substance to be oxidized. The oxidizing agent is always reduced. • Reducing agent – Substance that causes another substance to be reduced. The reducing agent is always oxidized.

  32. WO3(s) + 3 H2(g)  W(s) + 3 H2O(l) hydrogen tungsten WO3 H2 Element oxidized – Element reduced – Oxidizing agent – Reducing agent –

  33. SnO2(s) + 2 C(s)  Sn(l) + 2 CO(g) Carbon Tin SnO2(s) C(s) Element oxidized – Element reduced – Oxidizing agent – Reducing agent – Practice problems Tro – 4.87-4.88

  34. Half Reaction Method of Balancing Redox Reactions • Write skeleton ionic reaction. (Usually a given.) • Split into 2 half reactions, one for oxidation and one for reduction. (Determine what is oxidized and what is reduced by calculating oxidation numbers. Remember LEO says GER Loses Electrons Oxidation, Gains Electrons Reduction)

  35. Half Reaction Method of Balancing Redox Reactions • Balance each half reaction. • balance all but H and O. • balance O by adding H2O. • balance H by adding H+. • balance charge by adding electrons.

  36. Half Reaction Method of Balancing Redox Reactions • Add half reactions together after multiplying by a factor to be sure electrons cancel.

  37. Half Reaction Method of Balancing Redox Reactions • This method provides an equation for a reaction occurring in acid. To change to a balanced basic reaction add • H+ + OH H2O • to the reaction to cancel out all H+’s. Practice problems Tro – 18.37-18.42

  38. Balance in acid Cr2O72-(aq) + Cl-1(aq)  Cr3+(aq) + Cl2(g) 14H+(aq) + Cr2O72-(aq) + 6Cl-1(aq)  2Cr3+(aq) + 3Cl2(g) + 7 H2O(l)

  39. Balance in acid MnO2(s) + Hg(l) + Cl-1(aq)  Mn+2(aq) + Hg2Cl2(s) 4H+(aq) + MnO2(s) + 2Hg(l) + 2Cl-1(aq)  Mn+2(aq) + Hg2Cl2(s) + 2H2O(l)

  40. Balance in acid Ag(s) + NO3-1(aq)  Ag+1(aq) + NO(g) 4H+(aq) + 3Ag(s) + NO3-1(aq)  3Ag+1(aq) + NO(g) + 2 H2O(l)

  41. Balance in acid H3AsO4(aq) + Zn(s)  AsH3(g) + Zn+2(aq) 8H+(aq) + H3AsO4(aq) + 4Zn(s)  AsH3(g) + 4Zn+2(aq) + 4 H2O(l)

  42. Balance in acid Au+3(aq) + I2(s)  Au(s) + IO3-1(aq) 10Au+3(aq) + 3I2(s) + 18 H2O(l)  36H+(aq) + 10Au(s) + 6IO3-1(aq)

  43. Balance in acid IO3-1(aq) + I-1(aq)  I3-1(aq) 6H+(aq) + IO3-1(aq) + 8I-1(aq)  3I3-1(aq) + 3 H2O(l)

  44. Balance in acid HS2O3-1(aq)  S(s) + HSO4-1(aq) H+(aq) + 3HS2O3-1(aq)  4S(s) + 2HSO4-1(aq) + H2O(l)

  45. Balance in acid O2-2(aq)  O2(g) + H2O(l) 4H+(aq) + 2 O2-2(aq)  O2(g) + 2H2O(l)

  46. Balance in acid Cr2O7-2(aq) + I2(aq)  Cr+3(aq) + IO3-1 34H+(aq) + 5Cr2O7-2(aq) + 3I2(aq)  10Cr+3(aq) + 6IO3-1 + 17H2O(l)

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