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Chemistry 100 Chapter 3

Chemistry 100 Chapter 3. Balancing Chemical Equations and Stoichiometry. The Chemical Reaction. What happens in a chemical reaction? Example 2 H 2 + O 2  2H 2 O Starting materials (reactants) are converted into different chemical substance(s) (the product(s)).

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Chemistry 100 Chapter 3

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  1. Chemistry 100 Chapter 3 Balancing Chemical Equations and Stoichiometry

  2. The Chemical Reaction • What happens in a chemical reaction? • Example • 2 H2 + O2 2H2O • Starting materials (reactants) are converted into different chemical substance(s) (the product(s)). • Described by ‘standard shorthand’ - chemical equation

  3. Balancing Equations • Balanced equations have the same number of atoms of a given element on the LHS and the RHS • Law of conservation of mass. • All the reactants and products must be identified! • Chemical equations report the results of experimentation!

  4. A General Method of Balancing Chemical Equations. • Write the basic, or ‘skeletal’ equation, showing the formula of all reactants and products. • Balance the equation according to the law of conservation of mass. • Balance by adjusting the coefficients in front of the chemical formulas, never by adjusting the subscripts. • Best to start with an atoms that appears only once on the left and right hand side of an equation.

  5. A Balanced Chemical Equation

  6. Chemical Reaction Types • Combustion reaction  can be written for any compound containing C, H, O or C, H, S etc. O2 (g) is present as a reactant and is usually in excess. • We can also have combination and decomposition reactions • Combination reaction  two or more substances combine to form one product 2Mg(s) + O2(g)  2MgO(s)

  7. Reactions Types (II) • Decomposition reaction  one substance breaks down to form two or more different substances • MgCO3(s)  MgO(s) + CO2(g) • NH4Cl(s)  NH3(g) + HCl(g) • Decomposition of sodium azide, NaN3 • 2NaN3(s)  2Na (s) + 3N2(g)

  8. Molecular Masses • How do we use the atomic masses of elements to determine the molecular masses of molecules? • We simply add the masses of the constituent elements in the molecule.

  9. The Mole and Avogadro’s Number • NOTE: 1 amu = 1.6605 x 10-24 g; this is a very small mass!! • It is not a very convenient unit of measurement on the laboratory scale. • In chemistry, we use a special unit (the mole) when dealing with atoms, molecules, and ions.

  10. The Definition of a Mole • 1 mole (SI definition) • the amount of substance that contains as many elementary particles (atoms, molecules, ions) as there are atoms in 12 grams (exactly) of carbon-12. • This number is called Avogadro's number (it is an experimentally determined quantity).

  11. The Molar Mass • 1 mole = 6.022 x 1023 particles. • 1 mole of carbon-12 has a mass of 12.00000 .... g exactly! • The molar mass of any element in g/mole is the same numerically as its atomic mass in amu’s!

  12. The molar mass of a compound is the mass (in grams) of 1 mole of a compound. • Numbers in periodic table • e.g., Cl = 35.453 g/mole = 35.453 amu/atom

  13. Converting Between Moles and Molecules

  14. Masses of Anions and Cations • Electron mass is small • mass of Na+ mass of Na • mass of Cl- mass of Cl • mass of O2- mass of O atom!!

  15. Percent Composition • Percent composition is the percent by mass of each element in a compound. • Q: If we have the percent composition of the compound, can we calculate its empirical formula?

  16. A Schematic for Doing Empirical Formula Calculations

  17. Chemical Analysis • Use a variety of techniques to obtain the identity and the % by mass of each element in a sample. • Combustion analysis  a sample containing C, H, or C,H, and O is combusted (burned). • (C,H,O)  CO2 + H2O • all the C in original compound gets converted to CO2 • all the H in original compound is converted to H2O

  18. Combustion Analysis

  19. Quantitative Information from Chemical Reactions • Example • A + B  C + D • Normally, we start out with certain quantities of reactants. • How much product can we expect? • How much reactant would we need to obtain a specified amount of product?

  20. The Mole Method • Stoichiometric coefficients in the balanced chemical equation represent the number of moles of reactants and products. • certain mass of reactants  ? mass of products • initial volume of reactants  ? Volume of products

  21. Example N2 (g) + 3 H2 (g)  2 NH3 (g) 1 mole N2  3 moles H2 2 moles NH3 the  symbol means stoichiometrically equivalent

  22. A Schematic of the Mole Method

  23. Limiting Reagent • Chemical equations  give the molecular or molar ratios of reactants needed and the products obtained. N2 (g) + 3 H2 (g)  2 NH3 (g) • We assumed 100% complete conversion of of the N2 (g) and H2 (g) to NH3 (g). • Normally, the reactants will react in the proper ratio until one of them is consumed completely.

  24. The Definition of the Limiting Reagent • The amount of product formed is limited by the reactant that is completely consumed  the limiting reagent. • The amount of product obtained assuming complete consumption of the limiting reagent  theoretical yield. • The reactant that is left over  the excess reagent.

  25. Limiting Reagents

  26. The Limiting Reagent Schematic Mass of A and B Obtain Theor. Yield Use Molar Masses of A, B Use amount of Limiting Reagent Determine Limiting Reagent Moles of A, B Use Coefficients from balanced Equation

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