# Chemistry 2202 Final Review

Topics: Stoichiometry, Mole, Atom Pages: 7 (1725 words) Published: November 4, 2012
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 = m/M, n = v/V, n = # particles/6.02 × 10exp23
- where n = # moles (mol) , m = mass given (g), M = molar mass (g/mol), v = volume given (L), V = molar volume (L/mol) 12. Practice Problems
13. Pressure – the force that is exerted on an object, per unit of surface area. STP – (standard temperature and pressure) 0 °C and 101.325 kPa.

14. Molar Volume – the amount of space that is occupied by 1 mol of a substance; equal to 22.4 L for a gas at STP; V = 22.4 L/mol. 15. ni = nf
Vi Vf ; initial mole is always 1.0 mol.
16. Practice Problems

Chapter 3: Chemical Proportions in Compounds

1. Percent Composition – the relative mass of each element in a compound (max. 100%). To Calculate Percent Composition of a Compound
- % Composition = total mass of element × 100%
mass of compound
Ex. Calculate the % composition of water (H20).
- H = 2.02 g/mol, O = 16.00 g/mol = Total 18.02 g/mol
- % H = 2.02/18.02 × 100% = 11.2%, %0 = 16.00/18.02 × 100% = 88.8%
2. Practice Problems
3. Empirical Formula – shows the lowest whole number ratio of atoms of each element in a compound.
Molecular Formula – a formula that gives that gives the actual number of atoms of each element in a molecule.
Molecular = actual, Empirical = simplest Ex. H2O2, HO
4. Calculation of Empirical Formulas
Ex. What is the empirical formula of a compound that is 25.9% N and 74.1% O by mass?
Step 1: Assume 100g of compound
Step 2: Convert percent to mass
N = 25.9g O = 74.1g
Step 3: Find molar mass (M) of each element and divide by their mass.
N = 25.9 g
14.01 g/mol = 184 mol
O = 74.1 g
16.00 g/mol = 4.63 mol
Step 3: Divide by smallest number and if it is not a whole number double all digits.
N = 25.9 g
14.01 g/mol = 1.84 mol
O = 74.1 g1.84 mol = 1 × 2 = 2
16.00 g/mol = 4.63 mol
1.84 mol = 2.5 × 2 = 5
Empirical Formula = N205
5. Practice problems
6. Calculation of Molecular Formulas
Ex. 7.36 g of compound is...