Mole and Empirical Formulas

Avogadro's number (mole) - number of atoms in 12g of pure carbon-12  

• equal to 6.02*10²³
• molar mass - numerically equal to the element's atomic mass (1 atom of carbon-12 = 12 amu, 1 mol of carbon-12 = 12 grams)
• use dimensional analysis to convert between masses, moles, and numbers of particles

moles to number of representative particles  

• multiply number of moles by 6.02 x 10²³
• How many carbon particles in four moles of carbon? 4 x 6.02 x 10²³ = 2.408 x 10²⁴

number of representative particles to moles  

• divide number of representative particles by 6.02 x 10²³
• How many moles in 6.02 x 10²³ atoms? (6.02 x 10²³)/(6.02 x 10²³) = 1

mass (grams) to mole  

• divide the mass by the representative particle's molar mass
• How many moles are in 24 grams of carbon-12? 24/12 = 2

moles to mass (grams)  

• multiply number of moles by the representative particle's molar mass
• How many grams are in 23 moles of iron? 23 x 55.845 = 1.3 x 10³

number of representative particles to mass (grams)  

• divide number of representative particles by 6.02 x 10²³ and then multiply that by the representative particle's molar mass
• What's the mass of 5.234 x 10³⁰ silicon atoms? (5.234 x 10³⁰) / (6.02 x 10²³) x 28.0855 = 2.441 x 10⁸

mass (grams) to number of representative particles  

• divide the mass by the representative particle's molar mass and then multiply it by 6.02 x 10²³
• How many particles of carbon-12 are found in 12 grams of it? 12 g / 12/g/mol x 6.02 x 10²³ = 6.02 x 10²³

empirical formula - tells relative number of atoms of each element in a compound  

• finding the ratios of elements in a compound gives the empirical formula (from percentage composition)
  • C₆H₁₂O₂ (molecular formula) >> CH₂O (empirical formula)
• subscripts in the molecular formula can be found by multiplying the subscripts in the empirical formula by a whole number
• whole number multiple can be found by dividing the molecular weight by empirical formula weight
• combustion analysis - all the C goes into CO₂ and all the H goes into H₂O in combustion; moles of C/H in original compound can be found from the masses of CO₂/H₂O in the product
• Find empirical/molecular formula of caffeine, which contains 49.5% carbon, 5.15% hydrogen, 28.9% nitrogen, 16.5% oxygen. It has a molar mass of 195 g.
  • 49.5 / 12 = 4.125 mol carbon
  • 5.15 / 1 = 5.15 mol hydrogen
  • 28.9 / 14 = 2.1 mol nitrogen
  • 16.5 / 16 = 1.0 mol oxygen
  • C : H : N : O = 4:5:2:1
  • empirical formula is C₄H₅N₂O
  • 49.5% x 195 / 12 = 8.0 mol carbon (2x amount in empirical calculations)
  • molecular formula is C₈H₁₀N₄O₂
• Find empirical/molecular formula for nicotine (contains carbon, hydrogen, nitrogen) if 5.250 mg nicotine combusted to form 14.242 mg CO₂ and 4.083 mg H₂O. It has a molar mass of 160 g.
  • 14.242 + 4.083 - 5.250 = 13.075 mg O₂ used
  • 12 / (12 + 2 x 16) = 27% C in CO₂
  • 27% x 14.242 = 3.85 mg C
  • 2 / (2 + 16) = 11% H in H₂O
  • 11% x 4.083 = 0.45 mg H
  • 5.250 - 0.45 - 3.85 = 0.95 mg N
  • 0.95 / 5.250 = 18% N; 0.45 / 5.250 = 8.6 % H; 3.85 / 5.250 = 74% C
  • 18 / 14 = 1.3 mol N; 8.6 / 1 = 8.6 mol H; 74 / 12 = 6.2 mol C
  • C : H : N = 5 : 7 : 1
  • empirical formula is C₅H₇N
  • 18% x 160 / 14 = 2.1 mol nitrogen (about 2x amount in empirical calculations)
  • molecular formula is C₁₀H₁₄N₂