1 / 12

Formulas, Equations, and Stoichiometry

Formulas, Equations, and Stoichiometry. Chemical Equations. Both sides of the chemical equation must be balanced. Can only change the coefficient, not the subscripts, to balance the equation. Example: methane gas burning to produce carbon dioxide and water.

elma
Télécharger la présentation

Formulas, Equations, and Stoichiometry

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. Formulas, Equations, and Stoichiometry

  2. Chemical Equations • Both sides of the chemical equation must be balanced. • Can only change the coefficient, not the subscripts, to balance the equation. Example: methane gas burning to produce carbon dioxide and water. Unbalanced: CH4 + O2 CO2 + H2O Balanced: CH4 + 2O2 CO2 + 2H2O

  3. Balance the following: • Na + H2O  NaOH + H2 • Fe + O2  Fe2O3 • C2H4 + O2  CO2 + H2O 2 2 2 4 3 2 3 2 2

  4. Formula or Molecular Mass • Sum of the atomic weights of each atom in a substance’s chemical formula Formula weight of sucrose, C12H22O11: 12 C atoms = 12 (12.0 amu) = 144.0 amu 22 H atoms = 22 (1.0 amu) = 22.0 amu 11 O atoms = 11 (16.0 amu) = 176.0 amu Formula weight = 342.0 amu

  5. Percent Composition 144.0 amu C / 342.0 amu C12H22O11 = .421 (.421) x 100 = 42.1% 22.0 amu H / 342.0 amu C12H22O11 = .064 (.064) x 100 = 6.4% 176.0 amu O / 342.0 amu C12H22O11 = .515 (.515) x 100 = 51.5%

  6. Example • Calculate percentage of nitrogen, by mass, in Ca(NO3)2

  7. Moles • 1 mol of X= 6.02 x 1023 of X = molar mass of X • 1 g = 6.02 x 1023 amu 1 mole C atoms = 6.02 x 1023 C atoms = 12g C 1 carbon atom = 12amu

  8. Calculate • Number of H atoms in .350 mol of C12H22O11 • Number of O atoms in .25 mol of Ca(NO3)2 • Formula weight or molecular weight of Ca(NO3)2 • Molar mass of Ca(NO3)2 • Moles of Sucrose, C12H22O11, in 5.3 g of sucrose • Molecules of sucrose in 5.3 g of sucrose

  9. Formulas • Empirical formula: chemical formula giving only the relative number of atoms of each type in a molecule. • Molecular formula: chemical formula giving the actual number of atoms of each type in a molecule. • Example: • Empirical formula: HO CH2 • Molecular formula: H2O2 C2H4

  10. Calculate • Ascorbic Acid (vitamin C) contains 40.92% C, 4.58% H and 54.50% O by mass. What is the empirical formula? • If the molar mass of Ascorbic Acid is 176g, what is the molecular formula? C3H4O3 C6H8O6

  11. Stoichiometry and Limiting Reactants

  12. Percent Yield

More Related