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Chapter 8: Chemical composition

Chapter 8: Chemical composition. Chemistry 1020: Interpretive chemistry Andy Aspaas, Instructor. Atomic masses. Atomic mass unit, amu: a very small unit of mass in which masses of atoms and molecules are given A single carbon-12 atom has mass of 12 amu

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Chapter 8: Chemical composition

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  1. Chapter 8: Chemical composition Chemistry 1020: Interpretive chemistry Andy Aspaas, Instructor

  2. Atomic masses • Atomic mass unit, amu: a very small unit of mass in which masses of atoms and molecules are given • A single carbon-12 atom has mass of 12 amu • But, since there are different naturally occurring isotopes of carbon (carbon-12, carbon-13, carbon-14), the average mass of a carbon atom is larger than 12 amu • Average atomic mass (atomic weight) of carbon is 12.01 amu • The weighted average of all isotopes in a natural sample

  3. The mole • 2 objects = 1 pair, 12 objects = 1 dozen • Grouping can allow for large quantities to be more easily counted • Laboratory-sized samples of chemicals contain a very, very large number of atoms or molecules • The “mole” is a unit chemists use to represent very large numbers of particles (atoms, molecules, ions) • Could be used for anything though, just like pair, dozen

  4. What is a mole? • 1 mole (mol) defined as number of atoms in exactly 12 g of carbon-12 • Avogadro’s number, NA= 6.022 x 1023 particles/mole • Just another quantity like • Pair = 2 objects • Dozen = 12 objects • Gross = 144 objects • Ream = 500 objects • Mole = 6.02 x 1023 objects

  5. How big is Avogadro’s number? • 1 mole of chemistry textbooks would cover the surface of the earth to a depth of 300 km • If you won 1 mole of dollars when you were born and spent a billion dollars per second, 99.999% would still be left at 90 years old • A mole of pennies placed side by side would stretch more than a million light years

  6. Converting between particles and moles • Avogadro’s number is the same no matter the type of particles • Use 1 mol = 6.022 x 1023 particles as a conversion factor • Number of particles in 5.000 mol carbon? • (Same as 5.00 mol Fe, H2O, or anything!) • Number of moles in 5.21 x 1024 Al atoms?

  7. Converting between moles and mass • Since 1 mol = number of atoms in 12 g of carbon-12, • 1 mol of natural carbon atoms has a mass of 12.01 g • The mass of 1 mol of any element is equal to its atomic weight (from the periodic table) in grams! • 1 mol Fe = 55.85 g, 1 mol Na = 22.99 g, etc • Use values like this from the periodic table as conversion factors

  8. Molar mass • Idea extends past just atoms • Once can calculate mass of 1 mol of any molecule too! • Just add up atomic weights of all atoms in the molecule, and you get the molar mass of that compound • Molar mass of H2O = 2(1.008) + 16.00 = 18.016 g/mol

  9. Mass percent • Sometimes useful to know composition of a compound in terms of the masses of elements involved • Mass percent: mass of 1 element in 1 mol of the compound divided by mass of 1 mol of the entire compound, times 100% • Percents are just fractions multiplied by 100 • Practice: mass percentages of each element of ethanol, C2H5OH

  10. Empirical and molecular formulas • Empirical formula: formula that describes the simplest ratio of elements in a compound • Molecular formula: formula that describes the actual number of atoms of different elements in a single molecule • Ex. Butane • Molecular formula: C4H10 (actual number of atoms in molecule) • Empirical formula: C2H5 (simplest whole-number ratio)

  11. Calculating empirical formulas • If given masses of elements in a compound, convert each mass to moles • Then divide all mole values by the smallest one to get the ratios of atoms in the compound • If all the ratio values are integers, they become subscripts in the empirical formula • If there are non-integer values, multiply them all by the smallest integer to make them all integers

  12. Empirical formula calculation practice • Determine the empirical formula of the following: • A sample of phosphoric acid contains 0.3086 g hydrogen, 3.161 g phosphorus, and 6.531 g oxygen • A sample of para-dichlorobenzene contains 5.657 g carbon, 0.3165 g hydrogen, and 5.566 g chlorine.

  13. Empirical formula from percent composition • If you’re given percent composition, assume you have a 100-g sample, and convert percentages directly to grams • Ex. 52.5% carbon becomes 52.5 g carbon • All should add up to 100 g since all percentages must add up to 100% • Then, work as before by converting to moles and finding ratios

  14. Calculation of molecular formulas • If you know the molar mass, you can convert an empirical formula into a molecular formula • Calculate empirical formula mass, and divide by molar mass (molar mass) / (empirical formula mass) = n • Multiply n by subscripts in empirical formula to get molecular formula • Ex. empirical formula = CH, molar mass = 78

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