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Chemistry 2202 Final Review

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Chemistry 2202 Final Review
Chemistry 2202 Review Final Exam

Chapter 2: The Mole

1. Isotope – atoms of an element that have the same number of protons but different numbers of neutrons. Ex. The three forms of oxygen are called oxygen-16, oxygen-17, and oxygen-18. They all have 8 electrons and are written as 16/8 O (8 protons + 8 neutrons), 17/8 O (8 protons + 9 neutrons), and 18/8 O (8 protons + 10 neutrons). 2. 3. Mass Number – the sum of the protons and neutrons in the nucleus of one atom of a particular element.
Atomic Mass Unit (u) – a unit of mass that is 1/12 of the mass of carbon-12 atom; equal to 1.66 × 10-24 g.
Isotopic Abundance – the relative amount of an isotope of an element; expressed as a percent or a decimal fraction.
Average Atomic Mass – the average of the masses of all naturally occurring isotopes of an element weighted according to their abundance.
Ex. Aam = (% abundance × mass) + (% abundance × mass)
100
4. Practice Problems 5. Mole – the amount of substance that contains as many particles (atoms, molecules, or formula units) as exactly 12g of carbon-12.
- one mole (1 mol) of a substance contains 6.02 × 10exp23 particles of the substance. It is called Avogadro’s constant without the units mol, and avogardro’s number with the mol units. 6. Converting Moles to Number of Particles n = # of particles/ Avogadro’s Number (6.02 × 10exp23) 7. Practice Problems 8. Molar Mass (M) – the mass of one mole of a substance, numerically equal to the element’s average atomic mass; expressed in g/mol.
- to find the molar mass of a compound add up all individual compounds. Ex. Na2 Cl2 = Na = 2 × 22.99 g/mol = Cl = 2 × 35.45 g/mol 116.88 g/mol 9. Convert from Moles to Mass and Mass to Moles where m = equals mass (g), n = # of moles (mol), M = molar mass (g/mol) m = nM n = m/M 10. Practice Problems 11. Converting Between Moles, Mass. And Number of Particles n =

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