1 / 28

Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations

Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations. CHEMISTRY The Central Science 9th Edition. Chemical Changes . Chemical properties describes the reaction a substance undergo to form new substances. The study of chemical changes is at the heart of chemistry.

zavad
Download Presentation

Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 3Stoichiometry: Calculations with Chemical Formulas and Equations CHEMISTRY The Central Science 9th Edition

  2. Chemical Changes • Chemical properties describes the reaction a substance undergo to form new substances. • The study of chemical changes is at the heart of chemistry. • Some chemical changes or simple and some or complex. • For example, changes that occur in your brain and eyes allow you to see and think. • In this chapter, chemical changes will be used to explore the quantity of substances consumed and produced in a chemical reaction. • Chemical Equations are used to describe chemical reactions.

  3. Defining Chemistry • Lavoisier: mass is conserved in a chemical reaction. • His careful measurements turned chemistry into a science (Father of Chemistry) • Investigate the reaction of hydrogen with oxygen to produce water (2H2 + O2 2H2O).

  4. Chemical Equations • The chemical equation for the formation of water can be visualized as two hydrogen molecules reacting with one oxygen molecule to form two water molecules: • 2H2 + O2 2H2O Products Reactants

  5. Reading Chemical Equations • The plus sign (+) means “react” and the arrow points towards the substance produce in the reaction. • The chemical formulas on the right side of the equation are called reactants and after the arrow are called product. • The numbers in front of the formulas are called stoichiometric coefficients. • 2Na + 2H2O 2NaOH + H2 • Stoichiometric coefficients: numbers in front of the chemical formulas; give ratio of reactants and products. Coefficient Reactants Products

  6. Understanding Chemical Equations Coefficients and subscripts included in the chemical formula have different effects on the composition.

  7. Balancing Chemical Equations • Law of conservation of mass: matter cannot be lost in any chemical reactions. O

  8. Class Practice Problem • Balance the following equations: • (a) Na(s) + H2O(l) NaOH(aq) + H2(g) • (b) Al(s) + HCl(aq) AlCl3(aq) + H2(g) • (c) C2H4(g) + O2(g) CO2(g) + H2O(l)

  9. In Combination reactions two or more substances react to form products: 2Mg(s) + O2(g)  2MgO(s) The Mg has combined with O2 to form MgO (ionic compounds). Decomposition reactions is when one substance undergoes a reaction to produce two or more substances: 2NaN3(s)  2Na(s) + 3N2(g) (the reaction that occurs in an air bag) The NaN3 has decomposed into Na and N2 gas. Combination and Decomposition Reactions

  10. Summarization of Reactions

  11. Patterns in Chemical Reactivity • The periodic table can be used to predict how elements will react in a combination reaction: • 2K(s) + 2H2O(l)  2KOH(aq) + H2(g) • All alkali metals will react with water to form the hydroxide compound and hydrogen. • Thus, if let M represent the alkali metal, we able to write: • 2M(s) + 2H2O(l)  2MOH(aq) + H2(g) • Alkali metal + water  Metal hydroxide + hydrogen

  12. Formula and Molecular Weights Formula weights (FW) is the sum of the atomic weights of each atom in the chemical formula. FW (H2SO4) = 2AW(H) + AW(S) + 4AW(O) = 2(1.0 amu) + (32.0 amu) + 4(16.0) = 98.0 amu If the chemical formula is also its molecular formula then the weight is called the molecular weight (MW). MW(C6H12O6) = 6(12.0 amu) + 12(1.0 amu) + 6(16.0 amu) Formula Weights

  13. Formula Weights • Percentage Composition from Formulas • Percent composition is the atomic weight for each element divided by the formula weight of the compound multiplied by 100:

  14. Class Practice Problem • Calculate the FW of C12H22O11. • Calculate the percent composition of H2O. • 12C = 12 x 12.01 = 144.12 • 22H = 22 x 1.01 = 22.22 • 11O = 11 x 16.00 = 176.0 • C12H22O11 = 342.34 amu • % H = 11.21% • % O = 88.79%

  15. Molar Mass • Molar mass: mass in grams of 1 mole of substance (units g/mol, g.mol-1). • Experimentally, 1 mole of 12C = 12 g, which can be written as 12g/mol.

  16. The Mole • The unit we use to express the quantity of atoms, ions, and molecules that an object contains is called mole. • Mole: convenient measure chemical quantities. • The actual number of atoms, ions, or molecules in 1 mole of something = 6.0221367  1023(Advogadro’s number) of that thing. • Thus, • 1 mole of 12C atoms = 6.02 x 102312C atoms • 1 mole of H2O molecules = 6.02 x 1023 molecules • 1 mole of NO3- ions = 6.02 x 1023 ions

  17. Visualizing The Mole Concept Different Units

  18. Class Practice Problem • How many C atoms are in 0.350 mol of C6H12O6? • C atoms = [0.350 mol C6H12O6 (6.02 x 1023 molecules/1 mole C6H12O6)(6 C atoms/1 molecule)] = 1.26 x 1024 C atoms

  19. Class Practice Problem • Converting moles to mass • Calculate the number of moles of glucose C6H12O6 in 5.380 g of C6H12O6. • Moles of C6H12O6 = [5.380 g C6H12O6 (1 mole C6H12O6/ 180 g C6H12O6)] = 0.02989 mol C6H12O6

  20. Class Practice Problem • Converting mass to particles • Calculate the number of atoms of Cuin 3 g of Cu? • Atoms of Cu = [3 g Cu (1 mole Cu/180 g Cu)(6.02 x 1023 atoms Cu/1 mole Cu)] = 3 x 1023Cu atoms

  21. Empirical Formulas from Analyses • Start with mass % of elements (i.e. empirical data) and calculate a formula, or • Start with the formula and calculate the mass % elements. • For example: • Ascorbic acid contains 40.92 percent C, 4.58 percent H, and 54.50 percent O by mass. What is the empirical Formula? The experimentally determined molecular weight is 176 amu.

  22. Empirical Formulas from Analyses • Molecular Formula from Empirical Formula • Once we know the empirical formula, we need the MW to find the molecular formula. • Example 3.14, page 98 • Subscripts in the molecular formula are always whole-number multiples of subscripts in the empirical formula

  23. Quantitative Information from Balanced Equations • Balanced chemical equation gives number of molecules that react to form products. • Interpretation: ratio of number of moles of reactant required to give the ratio of number of moles of product. • These ratios are called stoichiometric ratios. • NB: Stoichiometric ratios are ideal proportions • Real ratios of reactants and products in the laboratory need to be measured (in grams and converted to moles).

  24. Limiting Reactants • If the reactants are not present in stoichiometric amounts, at end of reaction some reactants are still present (in excess). • Limiting Reactant: one reactant that is consumed

  25. Limiting Reactants

  26. Limiting Reactants • Theoretical Yields • The amount of product predicted from stoichiometry taking into account limiting reagents is called the theoretical yield. • The percent yield relates the actual yield (amount of material recovered in the laboratory) to the theoretical yield:

  27. The End of Chapter 3The test will cover Chapters 1-3, Scheduled for September 28, 2005Chemistry Feud will be held on Monday, September 26, 2005Homework: 3.9, 3.11, 3.15, 3.17, 3.19, 3.21, 3.25, 3.27, 3.31, 3.33, 3.43, 3.47

More Related