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Stoichiometry: Chemical Equations and Balancing

This chapter explores the concepts of stoichiometry, chemical equations, and balancing. Learn how to write and balance chemical equations, predict products, and calculate formula weights and percent composition.

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Stoichiometry: Chemical Equations and Balancing

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  1. Chapter 3 – Stoichiometry: Calculations with Chemical Formulas and Equations Jennie L. Borders

  2. Section 3.1 – Chemical Equations Stoichiometry is the area of study that examines the quantities of substances consumed and produced in chemical reactions. Remember: Atoms are neither created nor destroyed during any chemical reaction or physical process.

  3. Chemical Equations Below are the parts of a chemical equation: The arrow is called the yield sign.

  4. Balancing Equations Since atoms are neither created nor destroyed in a chemical reaction, a chemical equation must have an equal number of atoms of each element on each side of the arrow.

  5. Balancing Equations When balancing equations the coefficients must be the lowest whole-numbers possible. There are some exceptions if a reaction is particularly describing the reaction of 1 mole of a substance.

  6. Rules/Hints for Balancing Equations To balance equations, only coefficients can be added, subscripts can never be changed or added. Balance H and O last. If a polyatomic ion stays together on both sides of the arrow, then keep it together when balancing. If you have an odd number of a substance and you want an even number, try doubling the coefficient.

  7. Sample Exercise 3.1 The following diagram represents a chemical reaction in which the red spheres are oxygen atoms and the blue spheres are nitrogen atoms. a. Write the chemical formulas for the reactants and products.

  8. Sample Exercise 3.1 b. Write a balanced equation for the reaction. c. Is the diagram consistent with the law of conservation of mass?

  9. Practice Exercise In the following diagram, the white spheres represent hydrogen atoms, and the blue sphere represent nitrogen atoms. To be consistent with the law of conservation of mass, how many NH3 molecules should be shown in the right box?

  10. States of Matter We use the symbols (g) for gas, (l) for liquid, (s) for solid, and (aq) for aqueous to identify the states of matter of the reactants and products. Symbols listed above the arrow indicate catalysts. When heat is a catalyst the Delta symbol is used, D.

  11. Sample Exercise 3.2 Balance this equation: Na(s) + H2O(l) NaOH(aq) + H2(g)

  12. Practice Exercise Balance the following equations: a. Fe(s) + O2(g) Fe2O3(s) b. C2H4(g) + O2(g)  CO2(g) + H2O(g) c. Al(s) + HCl(aq)  AlCl3(aq) + H2(g)

  13. Section 3.2 – Some Simple Patterns of Chemical Reactivity In synthesis/combination reactions two or more substances react to form one product. Generic Reaction: A + B  AB Real Reaction: 2Mg + O2  2MgO

  14. Decomposition Reactions In a decomposition reaction one substance undergoes a reaction to produce two or more products. Generic Reaction: AB  A + B Real Reaction: CaCO3  CaO + CO2

  15. Writing Equations Remember to balance charges if you are predicting the products of a reaction. Ionic compounds (a metal and a nonmetal) tend to be solids and acids tend to be aqueous in reaction.

  16. Sample Exercise 3.3 Write the balanced equations for the following reactions: a. The synthesis reaction that occurs when lithium metal and fluorine gas react. b. The decomposition reaction that occurs when solid barium carbonate is heated.

  17. Hint for Predicting Products Like in the previous exercise, carbonate ions tend to produce CO2 when they break down.

  18. Practice Exercise Write the balance equation for the following reactions: a. Solid mercury (II) sulfide decomposes into its elements when heated. b. The surface of aluminum metal undergoes a combination reaction with oxygen in the air.

  19. Combustion Combustion reactions are rapid reactions that produce a flame. They require air (O2) as a reactant. When hydrocarbons are combusted completely, the products are CO2 and H2O. Generic Reaction: CxHy + O2 CO2 + H2O Real Reaction: C3H8 + 5O2  3CO2 + 4H2O

  20. Sample Exercise 3.4 Write the balanced equation for the reaction that occurs when methanol, CH3OH(l), is burned in air.

  21. Practice Exercise Write the balance equation for the reaction that occurs when ethanol, C2H5OH(l), is burned in air.

  22. Formula Weights The formula weight is a substance is the sum of the atomic weights of each atom in its chemical formula. Ex: H2SO4 = 2(1 amu) + 32 amu + 4(16 amu) = 98 amu If the chemical formula is that of the molecule, then the formula weight is also called the molecular weight.

  23. Sample Exercise 3.5 Calculate the formula weight of a. sucrose (C12H22O11) b. calcium nitrate

  24. Practice Exercise Calculate the formula weight of a. aluminum hydroxide b. methanol (CH3OH)

  25. Percent Composition Percent Composition is the percentage by mass contributed by each element in a substance. Percent Composition = mass of element x 100 mass of compound

  26. Sample Exercise 3.6 Calculate the percent of carbon, hydrogen, and oxygen (by mass) in C12H22O11.

  27. Practice Exercise Calculate the percentage of nitrogen, by mass, in calcium nitrate.

  28. Section 3.4 – Avogadro’s Number and the Mole In Chemistry the unit for dealing with the number of atoms, ions, or molecules in a common-sized sample is the mole, abbreviated mol. 1 mol = 6.02 x 1023 representative particles

  29. Sample Exercise 3.7 Arrange the following samples in order of increasing numbers of carbon atoms: 12g C12, 1 mol C2H2, 9 x 1023 molecules of CO2.

  30. Practice Exercise Arrange the following samples in order of increasing number of O atoms: 1 mol H2O, 1 mol CO2, 3 x 1023 molecules O3.

  31. Sample Exercise 3.8 Calculate the number of H atoms in 0.350 mol of C6H12O6.

  32. Practice Exercise How many oxygen atoms are in a. 0.25 mol of calcium nitrate? b. 1.50 mol of sodium carbonate?

  33. Molar Mass A mole is always the same number (6.02 x 1023), but 1 mole sample of different substances will have different masses. The mass of a single atom of an element (in amu) is numerically equal to the mass (in grams) of 1 mole of that element. The mass in grams of 1 mole of a substance is called the molar mass of the substance.

  34. Sample Exercise 3.9 What is the mass in grams of 1.000 mol of glucose, C6H12O6?

  35. Practice Exercise Calculate the molar mass of calcium nitrate.

  36. Sample Exercise 3.10 Calculate the number of moles of glucose in 5.380g of glucose.

  37. Practice Exercise How many moles of sodium bicarbonate are in 508g of sodium bicarbonate (also known as sodium hydrogen carbonate)?

  38. Sample Exercise 3.11 Calculate the mass, in grams. Of 0.433 mol of calcium nitrate.

  39. Practice Exercise What is the mass, in grams, of a. 6.33 mol sodium bicarbonate? b. 3.0 x 10-5 mol of sulfuric acid?

  40. Sample Exercise 3.12 a. How many molecules are in 5.23g of glucose? b. How many oxygen atoms are in this sample?

  41. Practice Exercise How many molecules are in 4.20g of nitric acid? How many O atoms are in this sample?

  42. Section 3.5 – Empirical Formulas from Analyses An empirical formula is the lowest whole number ratio of elements in chemical formula. The ratio of the number of moles of each element in a compound gives the subscripts in the compound’s empirical formula.

  43. Empirical Formula 1. When given percentages, assume 100g so that the percentages can be grams. 2. Divide the mass of each element by its molar mass. 3. Divide all answers by the lowest number to get subscripts. 4. If you get a .5, then multiply all numbers by 2. If you get a .33 or .66, then multiply all numbers by 3.

  44. Sample Exercise 3.13 Ascorbic acid (vitamin C) contains 40.92% C, 4.58% H, and 54.5% O by mass. What is the empirical formula of ascorbic acid?

  45. Practice Exercise A 5.325g sample of methyl benzoate, a compound used in perfumes, contains 3.758g C, 0.316g H, and 1.251g of O. What is the empirical formula of this substance?

  46. Molecular Formula The molecular formula is the actual formula of a molecule. It can be the same as the empirical formula. When the molecular formula is not the same, it is a whole number multiple of the empirical formula.

  47. Molecular Formula If you know the empirical formula and the mass of the molecular formula, then you can calculate the molecular formula. Whole # = molecular weight empirical weight

  48. Sample Exercise 3.14 Mesitylene, a hydrocarbon in crude oil, has an empirical formula of C3H4. The molecular weight of the substance is 121 amu. What is the molecular formula of mesitylene?

  49. Practice Exercise Ethylene glycol, the substance used in antifreeze, is composed of 38.7%C, 9.7% H, and 51.6% O by mass. Its molar mass is 62.1 g/mol. a. What is the empirical formula of ethylene glycol?

  50. Practice Exercise b. What is the molecular formula for the previous question?

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