Unit 1: Sigfigs and Nomenclature AP Chem 2017-2018 Mrs.Grein

Matter A. Pure Substance: doesn’t change composition, fixed properties 1. Molecule: covalent bond 2. Compound: ionic bond B. Mixture: two or more pure substances 1. Properties change

C. States of Matter 1. Solid: strong attraction, low kinetic Energy, constant volume and shape 2. Liquid: constant volume, shape changes 3. Gas: no set volume or shape

Energy A. Definition: ability to do work, apply force to make molecules move 1. Kinetic: energy of motion 2. Potential: stored energy due to position 3. ET: total energy is the sum of kinetic and potential B. 1st Law of Thermodynamics 1. Conservation of energy, energy cannot be created or destroyed

II. Significant Figures A. Accuracy vs. Precision 1. Consistency leads to the correct outcome- precision leads to accuracy B. Measurements 1. Two components: value and unit

2. Uncertainty: estimate a.) The last value in any measurement is a guess b.) all non-zero digits are significant c.) zeroes that place the decimal are not significant 1.) 380.41 -5 sig figs 2.) 0.000410 – 3 sig figs 3.) 2000 – 1 sig fig 4.) 1.0000000 – 8 sig figs

d.) If decimal is present, start from the left and move right to first non-zero, all that follow are sig figs e.) If no decimal start right to left to first non-zero, all that follow are sig figs

Rounding rules for operation 1. Addition and Subtraction 14.36 cm + 2.149 cm = 14.509 = 16.51 cm use least number of decimal places, least precise 2. Multiplication and Division 14.36 cm ÷ 2.149 cm = 6.68217757= 6.682 cm use least number of sig figs

3.) If the number next to the place value to be rounded is a 5, round your answer to an even number 6.6825 = 6.682 6.6835 = 6.684

IV. Metric System A. Standard units Mass ~ kg, g Temperature ~ K Energy ~ Joules Length ~ m Pressure ~ Pa Volume~ L Amount of matter ~ mol Current ~ amps Charge ~ coulomb (c)

B. Prefixes Mega (M) 106 milli (m) 10-3 kilo (k) 103 micro (μ) 10-6 deci (d) 10-1 nano (n) 10-9 centi (c) 10-2 pico (p) 10-12

V. Conversions A. Using multiplication of unit factors to cancel unwanted units 1. 1 mile = 5280 ft 5280 ft 1 mile

B. Sample 1. How many millimeters are equal to 1.85 x 10-4 pm? 2. What is the volume of 62.3g of a gas with a density of 1.2 g/L?

VI. Law of Conservation of Mass A. The total mass of substances does not change during a chemical reaction. Mass Reactant 1 + Mass Reactant 2 = Mass Product

1. Calculating the mass of an element in a compound a. How much N is in 4.55kg of NH4NO3? The Formula Mass is: 4 x H = 4 x 1.008 = 4.032 amu 2 x N = 2 x 14.01 = 28.02 amu 3 x O = 3 x 16.00 = 48.00 amu 80.052 amu b. Mass ratio is : 28.02 amu N / 80.05 amu NH4NO3 = 0.3500 (It doesn’t matter how much of the sample you have because the ratio never changes.)

B. The Law of Constant Composition 1. No matter what its source, a particular chemical compound is composed of the same elements in the same parts (fractions) by mass. a. The number of atoms does not equal the amount of contributed mass

VI. Law of Multiple Proportions A. If elements A and B react to form two compounds, the different masses of B that combine with a fixed mass of A can be expressed as a ratio of small whole numbers. 1. Ex. Nitrogen Oxides 1 and 2 Oxide 1. 46.68%N + 53.32%O Oxide 2. 30.45%N + 69.55%O

2. Assume you have 100g of each compound (drop %) Oxide 1. = g O/ g N = 53.32/46.68 = 1.142:1 Oxide 2 = g O/ g N = 69.55/30.45 = 2.284:1 2.284/1.142 = 2:1 Oxide 1 = NO Oxide 2 = NO2

VII. Dalton’s Atomic Theory A. all matter consists of atoms B. atoms of one element cannot be converted into atoms of another element C. atoms of an element are identical in mass and other properties and are different from atoms of any other element D. compounds result from the chemical combination of a specific ratio of atoms of different elements.

E. Today: 1. the atom is the smallest body that retains the identity of an element. 2. elements can only be converted into other elements in nuclear reactions 3. Isotopes of an element differ in the number of neutrons and thus in mass number. A sample of an element is treated as though its atoms have an average mass. 4. Evidence comes from work of many scientists a. J.J. Thomson: Raisin bun model b. Ernest Rutherford: Nuclear Model from gold foil experiment c. Millikan: Oil drop experiment determined charge/mass ratio of an electron

F. Isotopic Symbols A = mass (protons and neutrons) Z = atomic number (protons) X = atomic symbol

G. Average Atomic Mass 1. Measures the mass and relative abundance of an isotope. 2. Uses this formula Avg. Atomic Mass = (isotope1avg.mass x % abundance) + (isotope2avg.mass x %abundance) 100

3. What is the percent abundance of two isotopes of boron if the mass of isotope one is 10.013 amu and isotope two weighs 11.009 amu

IX. Naming and Writing Formulas A. Non-metals 1. replace ending of non-metal with …ide B. Metals 1. All metal names remain intact 2. Transition Metals get Roman Numerals for their different ions

Element Ion Formula Systematic name Common name Copper Cu+1 copper (I) cuprous Cu+2 copper (II) cupric Mercury Hg+1 mercury (I) mercurous Hg+2 mercury (II) mercuric Cobalt Co+2 cobalt (II) Co+3 cobalt (III) Iron Fe+2 iron (II) ferrous Fe+3 iron (III) ferric Manganese Mn+2 manganese (II) manganous Mn+3 manganese (III) Tin Sn+2 tin (II) stannous Sn+4 tin (IV) stannic Lead Pb+2 lead (II) plumbous Pb+4 lead (IV) plumbic **( -ic ending is higher charge than –ous ending)

C. Naming Binary Compounds a. Magnesium and Nitrogen: Mg3N2 ____________________ b. Iodine and Cadmium: CdI2 _______________________

D. Naming and Formulas of Ionic Compounds a. tin (II) fluoride (stannous fluoride) : SnF2 b. CrI3: chromium (III) iodide (chromic iodide) c. Ferric Oxide: Fe2O3 d. CoS: cobalt (II) sulfide (cobaltous sulfide)

E. Naming Polyatomic Oxyions ClO4- per root ate ClO- hypo root ite ClO3- root ate ClO2- root ite

F. Acid Naming: 1. Based on name of anion. a. binary anion: …..ide 1. acid becomes hydro…..ic acid a) HCl: b. oxyion acids 1. …..ate ions become ……ic acids a) H2CO3: 2. …..ite ions become …..ous acids a) H3PO3:

G. Naming Hydrates and Binary Covalent Compounds 1. Prefixes 1- mono 6- hexa 2- di 7- hepta 3- tri 8- octa 4- tetra 9- nona 5- penta 10- deca

H. Alkanes CnH(2n+2) Saturated hydrocarbons--each C is bonded to the maximum number of other atoms (4) Naming alkanes: root + suffix (the ROOT tells the # of carbon atoms): Meth-1 Hex-6 Eth-2 Hept-7 Prop-3 Oct-8 But-4 Non-9 Pent-5 Dec-10

Alkenes: CnH2n 1. Contains at least one C=C bond 2. Unsaturated hydrocarbons

J. Alkynes: CnH2n-2 Contains at least one triple C bond

K. Examples C6H12 C4H6 Octane Nonene Stannic dichromate tetrahydrate Cupric Acetate

L. Formula Types 1. Empirical: lowest possible ratio of atoms 2. Molecular: actual ratio multiple of empirical 3. Structural: shows arrangement a. Used mainly for hydrocarbons B. Categories of hydrocarbons Alkanes: chain of single bonded carbons Alkenes: chain of carbons with a double bond Benzene: cyclohexene 3. Alkynes: chain with a triple bond

c. Prefixes for chains and branches 1) if dealing with a chain add: -ane ; -ene ; -yne 2) if dealing with a branch of a chain add –yl d. Examples 1) 2)

4. Rules for naming hydrocarbons a. Name the longest chain (root) 1) Find the longest continuous chain 2) Select proper root b. Name the compound type (suffix) 1) –ane ; -ene ; -yne 2) If the chain forms a ring, add ‘cyclo’

C. Name the branches 1) Each branch has a prefix and ends in –yl 2) Branches precede the chain name. If 2 or more branches, alphabetize the branches. 3) To specify where a branch occurs, number the carbon atoms starting at the end closer to a branch 4) If a double or triple bond is in the chain, the number of the carbon where the bond occurs precedes the root name