Chemistry Review Mr. Halfen June 2011

Substances Pure substance - a substance composed on only one type of element or molecule Examples - diamond, oxygen, water Mixture - a physical combination of more than one pure substance Examples - salad, steel, soda Compound - a chemical combination of two or more pure substances

Mixtures Heterogeneous Mixture - substances are not uniformly mixed together Examples - salad, granite, chocolate chip cookie Homogeneous mixture - substances are evenly mixed; each part is identical to all others, even under a light microscope Examples - steel, soda, air

Atomic Structure 3 main particles - protons, neutrons & electrons nucleus - center of atom; contains neutrons & protons electron cloud - electrons found in region outside of nucleus in defined energy levels valence electrons - electrons in outer shell (highest energy level)

Elements mass number = #protons + #neutrons 41 Ca has 20 p + and 21 n 0 isotopes - same #p + & different #n 0

Periodic Table Families, natural phase, etc. Trends - pp in Holt

Periodic Table Using the text and your periodic table, list at least 10 trends or patterns in the periodic table.

Mass Calculations molar mass = atomic weight in grams/mole Avogadro’s number = x atoms/mole mass of atom = molar mass/A.N. #moles = measured mass/molar mass

Molar Mass Glucose - C 6 H 12 O 6 C - 6 x = g H - 12 x = g O - 6 x = g = g Glucose molecule = g/6.022 x 10 23

Mass Ratios mass of 1 element:mass of 2nd element CO - carbon monoxide : :1.332

Mass Ratio - 2 Glucose - C 6 H 12 O 6 C - 6 x = H - 12 x = O - 6 x = : : : 1 : 7.94

Formulas Structural formula - chemical formula showing the ratios of different elements in a chemical and the arrangement of the atoms Empirical formula - chemical formula showing the relative ratios of the elements in a molecule

Formulas - Practice StructuralEmpirical Acetic Acid - C 2 H (HC 2 H ) CH 2 O Formaldehyde - CH 2 OCH 2 O Glucose - C 6 H 12 O 6 CH 2 O

Calculation of Percent Composition from Chemical Formulas Calculate the percent composition for Mg(CN) Find the molar mass. Mg - 1 x g = g C - 2 x g = g N - 2 x g = g molar mass = g

Percent composition Find the % of each element % Mg = / = 31.84% % C = / = 31.47% % N = / = 36.70%

Calculation of Empirical Formulas from Percent Compositions What is the empirical formula for a liquid with a composition of 18.0% C, 2.26% H and 79.7% Cl? Assume 100 g of substance. Convert the % to grams. So there is 18.0 g of C, etc.

Empirical formula -2 Calculate moles of each substance. mole-C = 18.0g/(12.01g/mole) = 1.50 mole-H = 2.26g/(1.01g/mole) = 2.24 mole-Cl = 79.7g/( g/mole) = 2.24

Divide moles of each by the smallest subscript. mole-C = 1.50/1.50 = 1 mole-H = 2.2/1.50 = 1.5 mole-Cl = 2.24/1.50 = 1.5 By inspection, C 2 H 3 Cl 3 Structurally, it could be ClH 2 C 2 HCl 2 or H 3 C 2 Cl 3.

Chemical Reactions - Evidence release or absorption of heat production of light (or flames) production of sound release or absorption of electricity formation of a gas formation of a precipitate change in color change in odor

Chemical Reactions - Types Combustion Ex: C 3 H 8 + 5O 2 --> 3CO 2 + 4H 2 O Synthesis Ex: 2H 2 + O 2 --> 2H 2 O

Decomposition Ex.: CH 4 --> C + 2H 2 Displacement Ex.: Cu + FeO --> CuO + Fe Double Displacement Ex.: H 2 SO 4 + Ca(OH) 2 --> 2H 2 O + CaSO 4

Balancing Chemical Reactions In the examples of types of chemical reactions, all of the equations are balanced. Balancing chemical reactions is simply applying the Law of the Conservation of Mass. That is, if you start with 2 atoms (or moles) of Hydrogen, you have to end up with 2 atoms of Hydrogen.

Balancing Chemical Reactions worksheet work thru the first few together

Gas Laws - Ideal Gas Law PV = nRT Others derived this by combining the other gas laws. Einstein derived it from a mathematical description of the kinetic theory of gases. I suspect you would rather not see this derivation right now.... R is the “gas constant.” R = L kPa/(mol K), or R = L atm/(mol K)

Gas Laws - Boyle’s Law PV = nRT If the amount (i.e., moles) of the gas are constant (k 1 ) and the temperature is held constant (k 2 ), then the equation becomes... PV = k 1 R k 2 Let k 1 R k 2 = k B and PV = k B

Boyle’s Law - 2 If we start with conditions 1, then P 1 V 1 = k B If we change to conditions 2, then P 2 V 2 = k B Since the k B ’s are equal, P 1 V 1 = P 2 V 2

Gas Laws - Charles’ Law PV = nRT If the amount (i.e., moles) of the gas are constant (k 1 ) and the pressure is held constant (k 3 ), then the equation becomes... k 3 V = k 1 R T and let k 1 R/k 3 = k C and we get V = k C T or V/T = k C

Charles’ Law - 2 If we start with conditions 1, then V 1 /T 1 = k C If we change to conditions 2, then V 2 /T 2 = k C Since the k C ’s are equal, V 1 /T 1 = V 2 /T 2

Gas Laws - Gay-Lussac’s Law PV = nRT If the amount (i.e., moles) of the gas are constant (k 1 ) and the volume is held constant (k 4 ), then the equation becomes... P k 4 = k 1 R T Let k 1 R /k 4 = k G and P = k G T

Gay-Lussac’s Law - 2 If we start with conditions 1, then P 1 /T 1 = k C If we change to conditions 2, then P 2 /T 2 = k C Since the k C ’s are equal, P 1 /T 1 = P 2 /T 2

Gas Laws - Avogadro’s Law PV = nRT If the pressure of the gas is constant (k 5 ) and the temperature is held constant (k 6 ), then the equation becomes... k 5 V= n R k 6 Let k 6 R /k 5 = k A and V = k A n

Gas Law Problems Boyle’s: p.425 #1 & #3 Charles’: p.428 #1 & #3 G-L’s: p. 431 #1 & #3 Section Review: #5, #7 & #9 Ideal Gas Law: p.435 #1 & #3