Calculations with Chemical Formulas and Equations (Basic crap)

Patterns of Reactivity:

  • Combination reaction: Two things make one (i.e. 2Na + Cl2 —> 2NaCl)
  • Decomposition reaction: One thing divides into two (i.e. MgCO3 —> MgO + CO2)
  • Combustion reaction: Usually involves oxygen; often from air reacting with hydrocarbons or organic molecules with carbon, hydrogen, and oxygen to make CO2 AND H2O (i.e. CH3CH2OH + 3O2 —> 2CO2 + 3H2O)

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  • Avogadro’s Number: 6.02 x 10^23
  • molar mass: grams/ mole; add atomic masses of all atoms in formula
  • grams to moles= divide by molar mass
  • moles to particles= multiply by avog’s #
  • percentage composition: {100 x [(mol mass of element X subscript for element) / molar mass of substance]}

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Calculating Empirical Formula from % Comp:

What is the emp form of a compound with 68.4% chromium and 31.6% oxygen?

1. Convert percentages to grams. 

68.4g Cr

31.6g O

2. Convert grams to moles.

68.4g Cr/ 52g Cr = 1.315g Cr

31.6g O/ 16g O = 1.975g O 

3. Divide by the smaller mole

1.315 Cr/ 1.315 Cr = 1 Cr

1.975g O/ 1.315 Cr = 1.5 O

4. OPTIONAL: Multiply the previous answers into a small number that will turn them into whole numbers

1 Cr x 2 = 2 Cr

1.5 O x 2 = 3 O

Cr2O3 is empirical formula

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Molecular Formula from Empirical Formulas:

Cr2O3 is emp form with molar mass of 304 g/mol

1. Divide the known molar mass by the mass of one mole of emp form.

(304g/mol) / [(52x2)+(16x3)]g/mol)= 2

2. Multiply answer by the subscripts of emp form

Cr(2x2)O(3x2)

Cr4O6 is molecular formula

3 years ago     39 notes    
AP Chem   Molecular Formula   Empirical Formula   Reactivity   I hate this crap.   
  1. irllylikeboyz reblogged this from destrucshann and added:
    Fuckin yes
  2. destrucshann posted this