Chemistry ReviewMr. HalfenJune 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

• 41Ca has 20 p+ and 21 n0

• isotopes - same #p+ & different #n0

Periodic Table

• Families, natural phase, etc.

• www.chemicool.com

• Trends - pp. 132-141 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 = 6.022 x 1023

atoms/mole

• mass of atom = molar mass/A.N.

• #moles = measured mass/molar mass

Molar Mass

• Glucose - C6H12O6

• C - 6 x 12.0107 = 72.0642 g

• H - 12 x 1.00794 = 12.09528 g

• O - 6 x 15.9994 = 95.9964 g

• 72.0642 + 12.09528 + 95.9964 = 180.1559g

• Glucose molecule = 180.1559g/6.022 x 1023

Mass Ratios

• mass of 1 element:mass of 2nd element

• CO - carbon monoxide

• 12.0107:15.9994

• 1:1.332

Mass Ratio - 2

• Glucose - C6H12O6

• C - 6 x 12.0107 = 72.0642

• H - 12 x 1.00794 = 12.09528

• O - 6 x 15.9994 = 95.9964

• 72.0642 : 12.09528 : 95.9964

• 5.96 : 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

Structural Empirical

Acetic Acid - C2H402(HC2H302)

CH2O

Formaldehyde - CH2O CH2O

Glucose - C6H12O6 CH2O

Calculation of Percent Composition from Chemical

Formulas• Calculate the percent composition for

Mg(CN)2.

• 1 - Find the molar mass.

• Mg - 1 x 24.3050g = 24.3050g

• C - 2 x 12.0107g = 24.0214g

• N - 2 x 14.00674g = 28.01348g

• molar mass = 76.33988g

Percent composition - 2

• 2 - Find the % of each element

• % Mg = 24.3050/76.33988 = 31.84%

• % C = 24.0214/76.33988 = 31.47%

• % N = 28.01348/76.33988 = 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/(35.4527g/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, C2H3Cl3

• Structurally, it could be ClH2C2HCl2 or H3C2Cl3.

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: C3H8 + 5O2 --> 3CO2 + 4H2O

• Synthesis

• Ex: 2H2 + O2 --> 2H2O

• Decomposition

• Ex.: CH4 --> C + 2H2

• Displacement

• Ex.: Cu + FeO --> CuO + Fe

• Double Displacement

• Ex.: H2SO4 + Ca(OH)2 --> 2H2O + CaSO4

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 = 8.314 L kPa/(mol K), or

• R = 0.0821 L atm/(mol K)

Gas Laws - Boyle’s Law

• PV = nRT

• If the amount (i.e., moles) of the gas are constant (k1) and the temperature is held constant (k2), then the equation becomes...

• PV = k1 R k2

• Let k1 R k2 = kB and PV = kB

Boyle’s Law - 2

• If we start with conditions 1, then

• P1V1 = kB

• If we change to conditions 2, then

• P2V2 = kB

• Since the kB’s are equal, P1V1 = P2V2

Gas Laws - Charles’ Law

• PV = nRT

• If the amount (i.e., moles) of the gas are constant (k1) and the pressure is held constant (k3), then the equation becomes...

• k3 V = k1 R T and

• let k1 R/k3 = kC and we get V = kC T or V/T = kC

Charles’ Law - 2

• If we start with conditions 1, then

• V1/T1 = kC

• If we change to conditions 2, then

• V2/T2 = kC

• Since the kC’s are equal, V1/T1 = V2/T2

Gas Laws - Gay-Lussac’s Law

• PV = nRT

• If the amount (i.e., moles) of the gas are constant (k1) and the volume is held constant (k4), then the equation becomes...

• P k4 = k1 R T

• Let k1 R /k4 = kG and P = kGT

Gay-Lussac’s Law - 2

• If we start with conditions 1, then

• P1/T1 = kC

• If we change to conditions 2, then

• P2/T2 = kC

• Since the kC’s are equal, P1/T1 = P2/T2

Gas Laws - Avogadro’s Law

• PV = nRT

• If the pressure of the gas is constant (k5) and the temperature is held constant (k6), then the equation becomes...

• k5 V= n R k6

• Let k6 R /k5 = kA and V = kA 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