Chemistry 101 with Dr. Finnegan Week 7 10/3-10/7 Lectures #17,18,19
Chemistry 101 with Dr. Finnegan Week 7 10/3-10/7 Lectures #17,18,19 101
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This 7 page Class Notes was uploaded by Shannon Dooley on Saturday October 8, 2016. The Class Notes belongs to 101 at Washington State University taught by Michael Finnegan in Fall 2015. Since its upload, it has received 954 views. For similar materials see Chemistry in Chemistry at Washington State University.
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Oxidationreduction (Redox) reactions: reactions that involve the transfer of electrons from one substance to another. o i.e the number of electrons assigned to each atom in the reactants change Oxidation the loss of electrons. (oxidation # goes up) Reduction the gain of electrons. (oxidation # goes down) o HELPFUL HINT: remember O.I.L. R.I.G Oxidation Is Loss, Reduction Is Gain Oxidation number: [oxidation state] a method of counting electrons and assigning them to atoms. Oxidation numbers (i.e. +2, 2) can look similar to charges (i.e. 2+, 2) . o Tell the difference: Charges are written as number then sign 2+, 3 Oxidation #s are written sign then number +2, 3 Oxidation Number (O#) Rules: 1. The oxidation number of an element (in pure form, i.e. Ni, K, Na, etc. by itself) is always 0. Practice examples: o Mg O# = 0 o Kr O# = 0 2. The oxidation number of a monatomic ion (i.e. Lithium ion or sulfur ion or Lead (II) ion) is its charge. Practice examples: o (Li) Lithium ion charge of 1+ O# = +1 o (S) Sulfur ion charge of 2 O# = 2 o (Pb) Lead (II) ion charge of 2+ O# = +2 3. The oxidation numbers in a molecule or polyatomic ion must add up to the overall charge To find the oxidation number of atoms in a molecule or polyatomic ion: 1) Draw the Lewis structure of the molecule or polyatomic ion: 2) Oxidation # (O#) = v – w v = the normal number of valence electrons for that atom (look at the periodic table group i.e. F has 7 valence electrons normally) w = the number of electrons around the atom in the compound or ion. Electrons in bonds belong to the more electronegative atom. (i.e. if two atoms are sharing electrons, you count the electrons as if the more electronegative atom had taken them H—O—H In this example, O is more electronegative, so you count the electrons in the bond (two electrons per bond) as if they are attached only to O. Therefore, Hydrogen, would not be counted as having any electrons in this molecule, because they have been “taken” by oxygen. Therefore In H2O, which has an overall charge of 0 H O# = VW V=1, W=0 O# = 10 O# = +1 O O# = VW V=6. W=8 O# = 68 O# = 2 Because there are 2 H and 1 O, you have +1 +1 – 2 which = 0, the overall charge of H2O. (You’ll remember from above that the O# in a molecule or polyatomic ion must add up to the overall charge of that molecule or ion) Identify the oxidation number for each element in the following substances: NH3, CO2, V2(SO4)3 NH3 Nv=5, w=8 O# = -3 Hv=1, w=0 O# = +1 (-3+1+1+1= 0) CO2 Cv=4, w=0 O# = +4 Ov=6, w=8 O# = -2 (+4 -2 -2 = 0) V2(SO4)3 V (charge 3+) O# = +3 SO4 Ov=6, w=8 O# = -2 Sv=6, w=0 O# = +6 (-2 -2 -2 -2 +6 = -2 overall charge of sulfate ) (+3 +3 -2 -2 -2 = 0) Remember: A reaction is a redox reaction if the oxidation number (O#) of any element changes in the course of the reaction. NOTE oxidizing agents and reducing agents MUST be reactants!!!! They are NEVER the products. Oxidizing agent: gains one or more electrons Causes oxidation Undergoes reduction Becomes more negative (less positive) Reducing agent: loses one or more electrons Causes reduction Undergoes oxidation Becomes more positive (less negative) Practice Examples: The combination reaction of aluminum and bromine 0 0 3+ 2Al + 3Br 22Al Br (s) 3 (I showed charges so that it is easier to see electron transfer) Al goes from 0 to 3+ charge, it lost 3 electrons(O.I.L.) Al is oxidized (the element that is oxidized, or loses electrons is the reducing agent) Al is reducing agent Br goes from 0 to – charge, it gains one electron (3 total for Br ) (R.I.G.) 3r is reduced (the element that is reduced, or gains electrons is the oxidizing agent) Br is the oxidized agent The decomposition of hydrogen peroxide **It is important to note that when you have two of the same elements (i.e. O=O) bonded together, half the electrons go the one, the other half to the other when counting for W. So O=O each has two lone pairs (4 electrons) plus there are two bonds (4 electrons) between them, half the shared electrons goes to one and half goes to the other, so each O gets 2 of the electrons that they share, and therefore the W in that case would be 6. 2H O2 (2 2H O (l)2+ O (g) 2 2H O# (V–W) 1–0 = +1 2O O# (V–W) 6–7= –1 2 2 2H O2 H O# 2 O O# (VW**) 6–6 = –2 O 2 O# = 0 So notice that on both sides, the H has a2O# of +1, therefore it neither oxidized nor reduced. But 2O (2 O# = –1) which was split between 2H O and O gained2electrons i2 H O 2 (OO# = –2) and lost electrons in O (O O# =20) and was therefore both oxidized and reduced. It was reduced in H O beca2se it gained electrons, and it was oxidized in O 2 because it lost electrons. Then again, remembering the rules, the reducing and oxidizing agents MUST be a reactant, therefore H O is the b2th2 he reducing and the oxidizing agent because it is the only reactant. The combustion (burning therefore a reaction with oxygen) of hexane: C H (l) 6 14 C H (l) + O (g) CO (g) + H O (g) 6 14 2 2 2 C 6 (14 C O# = 4–6 AND 4–7 = –2 AND –3** O 2 (is an element) charge of 0 so O# = 0 **observe that in the chain link of carbon, the two carbons on the end take 3 electrons from the hydrogens, while the carbons in the middle only are bonded to two hydrogens so they only take 2 electrons. That is why there are varying O#s for the carbons. CO 2C O# = 4 – 0 = +4 O 2 O# = 6–8 = –2 H 2 H O2 = 1–0 = +1 O O# = 6–8 = –2 C O# goes from –2 and –3 to +4 loses electrons and therefore is oxidized, reducing agent is C 6 14cause it is the reactant that contains C H O# remains +1 throughout reaction O O# goes from 0 to –2 gains electrons and is therefore reduced, oxidizing agent is O 2 because it is the reactant that contains O The reaction of zinc metal with hydrochloric acid. Zn (g) + 2H Cl (aq) H (g) + 2n Cl +2 2–1 All of the oxidation numbers in this equation are derived from the charge of the elements and ions, so no subtraction had to be done. Zn O# goes from 0 to +2 loses electrons and therefore is oxidized, reducing agent is Zn because that is the reactant that contains Zn H O# goes from +1 to 0 gains an electron and therefore is reduced, oxidizing agent is HCl because is the reactant that contains H Cl O# remains the same The reaction of aluminum metal and iron(III) oxide. 2Al (s) + Fe O 2(g) 3 Al O + 2Fe2+3 3–2 0 Al O# goes from 0 to +3 oxidized, reducing agent (2Al (s) the reactant that contains Al) Fe O# goes from +3 to 0 reduced, oxidizing agent (Fe O (g) the re2ct3 t that contains Fe) O O# remains the same Lecture #18 (10/5) & Lecture #19 (10/7) Which sample contains more atoms: 100.0 g of iron or 100.0 g of carbon? **Comparatively, which contains more objects, a 100 lbs of bowling balls or 100 lbs of golf balls? Golf balls obviously they are smaller so it takes more of them to make 100lbs Carbon has a smaller mass than iron so it takes MORE carbon atoms to make 100g than it takes iron atoms to make 100g. So the answer to the question above is carbon. The atomic weights are the average mass of atom of that element given in atomic mass units (amu). 1 amu = 1.66054x10 g this isn’t really important to remember What mass of iron would have the same number of atoms as 100.0 g of carbon? Well if we know that 100g of iron has LESS atoms than 100g of carbon we know that to get the same number of atoms, we will have to RAISE the number of atoms of iron, which will raise the mass. Therefore, we know that the mass should be greater than 100.0g 100.0gC × 55.85gFe =465.0gFe 12.01gC o conversion factor is the mass of Fe over the mass of C (g of C cancel out, leaving g of Fe) o in addition, note that the g of Fe are greater than 100.0 g, which is what was predicted above How many grams of iron have the same number of atoms as 12.01 g of carbon? 55.85g 12.01 is the atomic weight of C so we could do the equation : 55.85gFe 12.01gC× =55.85g Fe but the 12.01 g C just cancel out, so it’s a 12.01 gC lot simpler to just understand that the atomic weight of each element always has the same number of atoms How many atoms is that? 23 Remember its Avagadros number 6.022×10 atoms 23 6.022x10 atoms of any element has a mass equal to that element’s atomic weight in grams! That is a useful number. So useful that we give that number a name: the mole. Why give it a name? “a mole” is a lot easier to say and write than “6.022x1023". In other words One mole of any atom has a mass of that atoms atomic weight in grams. So for carbon 12.01g= 1mol, for Hydrogen 1.008g = 1mol, and so on What is the mass of one mole of magnesium? Lecture #18 (10/5) & Lecture #19 (10/7) 24.30g hint, it’s the atomic weight of that element in grams Another unit for the atomic weight is g/mole. What is the mass of 0.003500 mole of carbon? 0.003500 molC× 12.01 gC =0.042035gC 1mol C 0.04202g C (4 sig figs) Remember back to conversion factors: start with the number you have, in this case 0.003500mol C. Then you know you want to cancel mol C and end up with grams (mass), so you put mol C on the bottom of the fraction(to cancel), and g C on the top(to remain). So how do you know what numbers get assigned to the units of g C/ mol C? Well think about what you know. You know that 1 mol of any element is equal to that elements weight in grams. So 1 mol of C is equal to the weight of C, or, 12.01 g C. Insert the numbers next to the units of your fraction and do the math. The mol C cancel out leaving you with g C which is what the question was asking for (mass). molar mass: the mass of one mole of a compound is equal to the sum of the atomic weights of the component atoms. What is the mass of one mole of carbon monoxide? CO=12.01g+16.00 g=28.01g Add together the atomic weight of C and the atomic weight of O What is the molar mass of iron(III) oxide? Fe 2 =3 (55.85 )g+3 (16.00 =159.70g REMEMBER* use addition rules of sig figs (count the number of digits behind the decimal) when you calculate molar mass most of the time you will have 2 digits after the decimal What mass of carbon monoxide is required to react with 500.0g of iron(III) oxide: Fe2O 3(s) + 3 CO (g) 2 Fe (l) + 3 CO 2(g) 500gFe O × 1molFe O2 3 × 3molCO × 28.01gCO =263.1gCO 2 3 159.7gFe O2 3 1molFe O 2 3 1molCO There are three steps to answering a stoichiometry question: 1) Convert the mass you are given to moles. (by multiplying by the conversion of 1 mol of a substance / atomic weight of substance or vice a versa) 2) From the balanced equation, find the ratio of moles being used in the reaction use the coefficients in front of each element of compound to tell you the number of moles of that element in the reaction (i.e. in example above there is 1 mole of Fe O and 3 moles of CO) 2 3 3)Convert moles to mass (by multiplying by the conversion of 1 mol of a substance / atomic weight of substance or vice a versa) Lecture #18 (10/5) & Lecture #19 (10/7)