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C HEM 167 F INAL R EVIEW Part 1. E LECTROLYTES VS. N ONELECTROLYTES Eletrolyte: Can break apart into ions Examples: salts – NaCl, acids – HF, bases –

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Presentation on theme: "C HEM 167 F INAL R EVIEW Part 1. E LECTROLYTES VS. N ONELECTROLYTES Eletrolyte: Can break apart into ions Examples: salts – NaCl, acids – HF, bases –"— Presentation transcript:

1 C HEM 167 F INAL R EVIEW Part 1

2 E LECTROLYTES VS. N ONELECTROLYTES Eletrolyte: Can break apart into ions Examples: salts – NaCl, acids – HF, bases – NaOH Types: Strong – complete dissociation into ions Weak – not complete dissociation Nonelectrolyte: does not break apart into ions Examples: ethanol, sugars (sucrose, fructose)

3 M ASS S PECTROMETER VIADD Vaporization: heat, laser, needs to be in gas phase Ionization: electron shower/magnetic plate Acceleration: ions are sped through spectrometer Deflection: Magnetic field sorts larger/smaller ions Detection: how many of which ion are present Gives a Mass:Charge ratio Data displayed on graph with peaks Helps to find average mass of an element

4 F RACTIONAL A BUNDANCE OF AN I SOTOPE Multiply fraction by the mass of the isotope Example: 24 Mg 23.99 amu – 78.99% 25 Mg 24.99 amu – 10.00% 26 Mg 25.99 amu – 11.01% Work: (.7899x23.99) + (.1x24.99) + (.1101x25.99) = 24.31 amu which is value on periodic table Can also be asked to find the fractional abundance

5 F IND THE FRACTIONAL ABUNDANCE Problem: Bromine has two naturally occurring isotopes 79 Br 78.918 amu and 81 Br 80.916 amu. The average mass for bromine is 79.904 amu. What is the fractional abundance of 79 Br?

6 F INDING F RACTIONAL A BUNDANCE Work: X + Y = 1  Y = 1-X  X= 79 Br,Y= 81 Br 78.918(X) + 80.916(1-X) = 79.904 78.918X + 80.916 – 80.916X = 79.904 78.918X-80.916X = 79.904-80.916 -1.998X = -1.012 X = 0.51 = 79 Br

7 P ERCENT BY M ASS /E MPIRICAL F ORMULA Assume a 100g sample %mass  mass of element  moles  mole ratio Problem: A compound is 54.05% Ca, 43.24% O, and 2.71 % H. What is its empirical formula?

8 A NSWER 54.05%  54.05g Ca/40.1g/mol  1.348 mol Ca 43.24%  43.24g O/16g/mol  2.703 mol O 2.71%  2.71g H/1.01g/mol  2.683 mol H Ca has the smallest number of moles 1.348mol/1.348  1Ca 2.703mol/1.348  2O 2.683mol/1.348  2H Answer: CaO 2 H 2  Ca(OH) 2

9 L INE D RAWINGS AND T HEIR F ORMULAS If no element is written assume is C at every joint and end. Fill in H’s until number of bonds is met Examples: drawn on board/paper

10 N AMING C OMPOUNDS Know group charges and specific element charges, such as B, Al, Zn, and Ag Know polyatomic ions Carbon-carbon bonds: single –ane, double –ene, triple –yne Transition metals need oxidation state in parentheses Covalent bonding needs prefixes Examples: a) Fe 2 O 3  iron (III) oxide b) Cl 2 O 7  dichlorine heptoxide c) (NH 4 ) 2 S  ammonium sulfide

11 S OLUBILITY VS. I NSOLUBILITY Review or write down solubility rules Examples: are the following salts soluble or insoluble? a) KBr - soluble b) AgCl - insoluble c) Al(OH) 3 - insoluble d) BaSO 4 - soluble e) Ca 3 (PO 4 ) 2 - insoluble

12 B REAKDOWN OF AN E LEMENT An element contains a specific number of protons, neutrons, and electrons If the element is neutral then the number of electrons and protons is equal The number of protons and neutrons is equal to the mass of the element (A) The number of protons is equal to the atomic number of the element (Z)

13 T YPES OF B ONDING Ionic bonding: between a metal and nonmetal; is a transfer of electrons Covalent bonding: between nonmetals; is a sharing of electrons Metallic bonding: between metals; electrons are free flowing and shared between many nuclei

14 B ALANCING E QUATIONS Matter cannot be created or destroyed, need the same amount of each element on both sides of the arrow. Balance these equations: a) C4H10 + O2  CO2 + H2O b) NaClO3  NaCl + O2 c) CaCl2 + NaOH  NaCl + Ca(OH)2 d) C + SO2  CS2 + CO

15 A NSWERS a) 2,13,8,10 b) 2,2,3 c) 1,2,2,1 d) 5,2,1,4

16 M OLES, M OLARITY, AND D ILUTION Moles = mass/molar mass; this is true for elements and compounds Molarity = moles/liters Dilution = M1V1 = M2V2 Example Dilution problem: What volume (mL) of 6M NaOH must be diluted to create a solution of 200mL and 1.5M?

17 A NSWER Use M1V1=M2V2 Assign M1=6M V1=? M2=1.5M V2=200mL V1=(M2V2)/M1 V1=50mL

18 T OTAL I ONIC, M OLECULAR, N ET I ONIC Total ionic equation: write out all the ions on both sides of arrow Molecular equation: combines ions and includes states; aq, and l Net ionic equation: eliminate spectator ions from the total ionic equation Write the net ionic equation of: AgNO3 + NaCl *look for formed insoluble salts

19 A NSWER AgNO 3 + NaCl  AgCl + NaNO 3 AgCl is insoluble Total ionic: Ag + + NO 3 - + Na + + Cl -  AgCl + Na + + NO 3 - Spectator ions: Na + and NO 3 - Net ionic: Ag + + Cl -  AgCl

20 G RAPHS TO K NOW Coulomb’s Law: repulsion on top half (same ions), attraction on lower half (different ions). As radius (r) increases force (F) decreases. Maxwell-Boltzmann distribution: explains how the speed of molecules in a gas increases with higher temperature, and the speed also increases with smaller mass

21 E XAM T WO M ATERIAL T YPES OF L IGHTS Incandescent bulbs: a wire inside is heated up by electricity, excites the electrons in tungsten (W) which then emits excited electrons to create light. Fluorescent bulbs: an arc/plasma is sent through bulb, excites the electrons of the gas in the bulb (gas varies), then emitted excited electrons interact with phosphorous coating to create light. LEDs (light emitting diodes): promote electron from ground to excited state, emits light to fall back to ground state. Has a narrow wavelength and frequency distribution, single color, high efficiency Laser (light amplification by stimulated emission of radiation): emits light same as LED. Need population inversion. All emitted photons are in same phase (coherent light). Single color with narrow wavelength and frequency distribution

22 A TOMIC M ODELS AND P RINCIPLES Bohr’s: states that 1) electrons exist in well defined orbits 2) each orbit has specific energy associated with it and 3) energy is released or absorbed when electrons change energy levels Heisenberg uncertainty principle: proves Bohr’s model wrong because it states that it is impossible to pin down where an electron is at exactly. Another thing to prove Bohr’s model wrong is that electrons diffract through slits so they can also behave like waves

23 O RBITALS Orbitals have a set of quantum numbers that describe them: n, l, m l, and m s. N: principle quantum level, gives number of energy shell L: secondary quantum number; determines shape of orbital. Is equal to n-1or less. 0=s, 1=p, 2=d, 3=f M l : magnetic quantum number; tells how many orbitals are in a subshell. Is equal to any number from –l to +l M s : spin quantum number. Can only be +1/2 or -1/2. Each orbital can only have 2 electrons and they spin in opposite directions. Pauli exclusion principle: each electron in an atom has a unique set of quantum numbers Hund’s rule: single electrons first fill empty orbitals before they pair up in an occupied one.

24 S HAPE OF AN ORBITAL S orbital: shaped like a sphere, groups 1&2 P orbial: has two lobes, groups 3-8 D orbital: has 4 lobes, transition metals F orbital: has 6 lobes, lanthanides &actinides Orbitals also have nodes (areas where no electrons exist) the number of nodes is equal to n- 1. Nodes can be planar or spherical. Examples: Which orbital is this? n=2, l=1, m l =0, m s =+1/2 Which quantum numbers describe a 3p orbital?

25 A NSWER 2p because n=2 and l=1  p n= 3, l=1, m l =-1,0,1, m s = +1/2 or -1/2

26 P HOTOELECTRIC E FFECT Energy cost to remove one electron from a surface E = h*v Energy cost to remove one electron using a certain wavelength E=(h*c)/ and c= v Example: What is the wavelength in meters of a wave with a photon energy of 4.5*10^-28J? Know photoelectric graphs, there are 4

27 A NSWER Use E=(h*c)/   h*c)/E h= Planck’s constant = 6.626*10^-34J*s c= speed of light = 3*10^8m/s = 442m

28 E LECTRON C ONFIGURATIONS Fill in the orbitals across the periods of the periodic table There are full electron configurations and condensed configurations that use noble gases Examples: a) Write the full configuration of Ga. b) Write the condensed state of Ga.

29 A NSWER 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 1 [Ar] 4s 2 3d 10 4p 1

30 G ROUND S TATE AND E XCITED S TATE Electrons can either be in their ground state or their excited state. This can be demonstrated in electron configurations. Examples: 1) Write the stable configuration of nitrogen (N) 2) Which is an excited state? a) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 b) 1s 2 2s 2 2p 6 4s 1 c) 1s 2 2s 2 2p 6 3s 1

31 A NSWER 1s 2 2s 2 2p 3 b

32 P ERIODIC P ROPERTIES Z eff : increases as move left to right because Z eff = Z-S. S= shielding (core) and remains the same, Z=atomic number and it increases. Radius, diameter, and volume: increase as move left and down towards francium (Fr). Ionization energy: increases as move right and up towards fluorine (F). Is a positive value Electron affinity: increases in magnitude as move right and up towards fluorine (F). Is a negative value. Noble gases are extremely positive and do not follow trend.

33 L ATTICE E NERGY The energy cost to break an ionic species into cations and anions Larger energy is characterized by larger ionic charges and smaller distance apart (smaller r) Example: a) Which has the smallest radius? O 2-, F -, Ne, Na +, Mg 2+ b) Which will have the largest lattice energy? NaCl, MgCl 2, MgO, Na 2 O In general cations are smaller than anions because of the missing electron(s) which occupy a lot of space by repulsion.

34 G AS L AWS P1V1=P2V2, P1/T1=P2/T2, V1/T1=V2/T2 Combines into ideal gas law PV=nRT where R=0.08206 L*atm/mol*K Conditions where any gas behaves most ideally are at high temperatures and low pressures Van der Waals equation helps account for not ideal conditions (P+(n 2 /V 2 )a) (V-nb) = nRT Example: What is the V occupied by 2 moles of N 2 at P=1atm and T=273K?

35 A NSWER Use ideal gas law V=(nRT)/P V=(2*0.08206*273)/1=44.8L

36 D ALTON ’ S L AW AND M OLE F RACTION Dalton’s Law: total pressure is equal to the sum of the partial pressures. Mole fraction: equal to partial moles divided by total moles is symbolized by X i Partial pressure can be found by applying mole fraction and Dalton’s law. P i =X i * P T

37 C ALCULATING THE EXTENT OF A REACTION Need to make sure the equation is balanced and then use the stoichiometry to decide if there is a limiting reactant. Example: a) N 2 + H 2  NH 3 *balance it What mass of N 2 is needed to fully react with 5.11g of H 2 ? b) O 2 + 2H 2  2H 2 O What is the yield of H 2 O if 7.9g O 2 and 2.9g H 2 react to completion?

38 A NSWERS N 2 + 3H 2  2NH 3 5.11g H 2 /2.02  2.53 mol H 2 H 2 and N 2 in 1:3 ratio; 2.53mol/3  0.843 mol N 2 0.843mol * 28g/mol = 23.6g N 2 7.9g O 2 /32  0.247mol O 2 2.9g H 2 /2.02  1.435mol H 2 Limiting reactant is O 2 because of mole ratio 0.247mol O 2  0.494mol H 2 O from stoichiometry 0.494mol * 18g/mol = 8.89g H 2 O

39 R EACTIONS WITH SOLIDS AND GASES Can use gaseous reactants in reactions to calculate product yields by using the ideal gas law. Examples: a) O 2 (g) + 2H 2 (g)  2H 2 O(g) What is the V of H 2 needed to make 36g H 2 O at T=30C and P=0.995 atm? b) CaO(s) + CO 2 (g)  CaCO 3 (s) What V of CO 2 can be reacted exactly with 112g CaO at T=800C and P=1.25 atm?

40 A NSWERS First find moles of H2O created 36g/18g/mol  2mol H2O  2mol H2 by reaction stoichiometry Use number of moles in PV=nRT  V=(nRT)/P V=(2*0.08206*303)/0.995 = 50L First find moles of CaO used 112g/56.1g/mol  2mol CaO  2mol CO2 Use number of moles in V=(nRT)/P V=(2*0.08206*1073)/1.25 = 141L

41 P RESSURE MEASURING DEVICES Capacitance manometer: uses a flexible diaphragm and measures the distance change (capacitance) between the diaphragm and a metal plate. Ionization gauge: measures the current created by ions, an increase in P makes the current increase Thermocouple gauge: as pressure decreases the measured temperature of the filament increases Mass spectrometer: measures partial pressures of the sample

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