Moles and Atoms

Counting Atoms by the Gram

  • Mole
  • this is the SI base unit for the amount of substance
  • by knowing this, you can calculate the amount of particles of a substance, the number of grams of a substance, and calculate the volume of a gas at a given temperature and pressure
  • 1 mole = 6.022 x 10^23 particles
  • avogadro's number
  • two main substances are ionic compound and molecular compounds (you also have elements)
  • You have acids if they have an H in front and these types of substances are types of molecular compounds
  • element - atoms
  • molecular compounds - molecules
  • ionic compounds- formula units
  • Problems
  • Convert 2.85 mol of He to atoms
    • 2.85 mol/1 * 6.022 x 10^23 atoms/1mol = 1.72 x 10^24 atoms of Helium
  • Convert 8.57 x 10^23 atoms of Cu to moles
    • 8.57 x 10^23 atoms/1 * 1 mol/6.022 x 10^22 atoms = 1.42 moles of Copper
  • Convert 4.52 mol of Na to atoms
    • 4.52 mol/1 * 6.022 x 10^23 atoms/1mol = 2.72 x 10^24 atoms of Sodium
  • Convert 7.94 x 10^22 atoms of S to moles
    • 7.94 x 10^22 atoms/1 * 1mol/6.022 x 10^23 atoms = 0.132 moles of Sulfur
  • molar mass
    • the sum of all of the molar masses of the atoms of that particular formula
    • periodic table - round to the second decimal place
    • molar mass = 1 mole = 6.022 x 10^23 particles
  • He
    • 1 mole
    • 4.00g
    • 6.022 x 10^23 atoms
    • CO2
  • 1 mole
    • 44.01g
    • 6.022 x 10^23 molecules
  • CuSO4
    • 1 mole
    • 159.62g
    • 6.022 x 10^23 formula units
  • Problems
    • Convert 43.76 g of Br to moles
      • 43.76g/1 * 1mol/79.90g = .5477 moles of Bromine
    • Convert 3.76 mol of neon to mass (grams)
      • 3.67 mol/1 * 20.18g/1mol = 74.1 g of Neon
    • Convert 2.35 x 10^23 atoms of zinc to mass (grams)
      • 2.35 x 10^23 atoms/1 * 1mol/6.022×10^23 atoms * 65.39g/1mol = 25.5g of Zinc
    • Convert 6.89g of Ca to atoms
      • 6.89g/1 * 1mol/40.08g * 6.022 x 10^23 atoms/1mol = 1.04 x 1023 atoms of Calcium
  • Problems
    • Convert 15.6 moles of H2) to molecules
      • 15.6 mol/1 * 6.022 x 10^23/1mol = 9.39 x 10^24 moles of water
    • Convert 7.83 x 10^22 formula units of Copper(II) chloride to moles
      • 7.83 x 10^22 formula units CuCl2/1 * 1mol CuCl2/6.022 x 10^23 formula units CuCl2 = 0.130 moles of CuCl2
    • Convert 6.72 moles of carbon monoxide to mass(g)
      • 6.72 moles CO/1 * 28.01g/1mol CO = 188 g of CO
      • 25.8g of calcium hydroxide to moles
      • 25.8g Ca(OH)2/1 * 1mol Ca(OH)2/74.10g = 0.348 moles of Ca(OH)2
    • Convert 34.82 g of nitrogen dioxide to molecules
      • 48.82g NO2/1 * 1mole NO2/46.01 gNO2 * 6.022 x 10^23 molecules NO2/1mol = 4.557 x 10^23 molecules of NO2
    • Convert 7.83 x 10^24 formula units of lithium carbonate to grams
      • 7.83 x 10^24 fu Li2CO3/1 * 1mol LI2CO3/6.022×10^23 fu Li2CO3 * 73.89g/1mol Li2CO3 = 961 g of Li2CO3
    • Convert 8.63 x 10^23 molecules of CH3OH convert to mass
      • 8.63 x 10^23 molec CH3OH * 1 mol CH3OH/6.022 x 10^23 molec CH3OH * 32.05g CH3OH/1 mol CH3OH = 45.9 g of CH3OH
    • Convert 23.87 g of silver nitrate to particles
      • 23.87g AgNO3/1 * 1mol AgNO3/169.88 g AgNO3 * 6.022 x 10^23 particles AgNO3/1mol AgNO3 = 8.462 x 10^22 particles of AgNO3
    • Determine the number of moles of O in 8.54 moles of nitrogen trioxide
      • 8.54 mol NO3 * 3 mol O/1mol NO3 = 25.6 mol of O
    • Determine the number of atoms of N in 9.32 x 10^24 formula units iron (II) nitrate
      • 9.34 x 10^24 form units Fe(NO3)2 * 2 atoms N/1 form unit Fe(NO3)2 = 1.87 x 10^25 atoms of N
    • Calculate the grams of C in 45.78 g of C2H5OH
      • 45.78g C2H5OH/1 * 1 mol C2H5OH/46.08 g C2H5OH * 2 mol C/1 mol C2H5OH * 21.01 g C/1 mol C = 23.86 g of C
    • Calculate the grams of N in 52.3 g of nitrogen trichloride
      • 52.3 g NCl3/1 * 1 mol NCl3/120.36 g NCl3 * 1 mol N/1 mol NCl3 * 14.01 g N/1 mol N = 6.09 g N
    • Calculate the number of hydrogen atoms in 15.8 g of CH4
      • 15.8 g CH4/1 * 1 mol CH4/16.05 g CH4 * 4 mol H/1 mol CH4 * 6.022 x 10^23 atoms H /1 mol H = 2.37 x 10^24 atoms of H
  • Mass Percent Composition
    • mass percent composition (mass percent; percent composition)
    • Total mass of the element/total mass of the compound x 100%
    • 2 ways to determine
      • experimental evidence
      • chemical formula
    • Ammonium nitrate - NH4NO3
    • molar mass = 80.05 g/mol
    • 1 mole (80.05g) contains 28.02g of nitrogen
    • Percent N = 2802g/80.05g x 100% = 35.00%
    • Always round percent by mass to the first decimal place
    • Problems
      • What is the percent composition of a compound formed when 2.70g of aluminum combine with oxygen to form 5.10g of aluminum oxide?
      • Al% = 2.70g/5.10g x 100% = 52.9%
      • O% = 47.1%
      • Calculate the percent composition when 13.3g Fe combine completely with 5.7 g of O.
      • Fe% = 13.3g/19g = 70.0%
      • O% = 30.0%
      • Determine the percent composition of magnesium cyanide
      • Mg% = 24.31g/76.35g x 100 = 31.8%
      • C% = 24.02g/76.35g x 100 = 31.5%
      • N% = 36.7%
      • Determine the percent composition of diphosphorus trioxide.
      • P% = 62.94g/109.94 = 56.3%
      • O% = 43.7%
      • The FDA recommends that adults consume less than 2.4g of sodium per day. How many grams of sodium chloride can you consume and still be within the GDA guidelines?
  • 2.4g Na/1 * 1mol Na/22.99 g Na * 1 mol NaCl/1mol Na * 58.44 g NaCl/1mol NaCl = 6.1g NaCl
    • Calculating Empirical Formulas
    • Empirical formula
    • Molecular formula
    • C4H10, C6H12, H2O2, C3H8
    • Steps for solving empirical formula
      • convert to moles
      • divide each amount of moles by the lowest number of moles
      • use the ratio between the moles to develop the empirical formula
      • Steps for solving the molecular formula
      • calculate the empirical formula
      • molecular formula = molar mass/empirical mass = integer
      • use the integer to multiply the subscripts of empirical formula
  • Problems
    • Determine the empirical and molecular formula if a compound contains 71.72% Cl, 16.16% O, and 12.12% C; molar mass = 198g
      • 71.72 g Cl/1 * 1 mol Cl/35.45g Cl = 2.02 mol Cl
      • 16.16 g O/1 * 1 mol O/16.00 g O = 1.01 mol O
      • 12.12 g C/1 * 1 mol C/1201 g C = 1.01 mol C
      • 2.02 mol Cl/1.01 mol O = 2 mol Cl/1 mol O
      • 1.01 mol C/1.01 mol O = 1 mol C/1 mol O
  • COCl2
  • MF = MM/EM = 198g/98.9g = 2
  • C2O2Cl4
  • Determine the empirical and molecular formula if a compound contains 94.1% O and 5.9% H; molar mass = 34g
    • 94.1g O * 1mol O/16 g O = 5.88 mol O
    • 5.9g H * 1mol H/1.01g H = 5.84 mol H
    • 5.84 mol H/5.88 mol O = 0.993 mol H/1 mol O
    • HO
    • MF = MM/EM = 34/17.01 = 2.0
    • H2O2
    • Determine the empirical and molecular formula if a compound contains 40.05% C, 6.6% H, and 53.4% O; molar mass = 120g.
      • 40.05 g C * 1 mol C/12.01 g C = 3.34 mol C
      • 6.6 g H * 1 mol H/1.01 g H = 6.53 mol H
      • 53.4 g O * 1 mol O/16.00 g O = 3.34 mol O
      • 3.34 mol C/3.34 mol O = 1mol C/1mol O
      • 6.53 mol H/3.34 mol O = 1.96 mol H/1mol O
      • H2CO
      • MF = MM/EM = 120/30.03 = 4.0
      • H8C4O4

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