* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Chemical Calculations, Chemical Equations
Chemical thermodynamics wikipedia , lookup
Transition state theory wikipedia , lookup
Nanofluidic circuitry wikipedia , lookup
Chemistry: A Volatile History wikipedia , lookup
Electrochemistry wikipedia , lookup
Computational chemistry wikipedia , lookup
History of chemistry wikipedia , lookup
Chemical reaction wikipedia , lookup
Electrolysis of water wikipedia , lookup
Host–guest chemistry wikipedia , lookup
Size-exclusion chromatography wikipedia , lookup
Implicit solvation wikipedia , lookup
Rate equation wikipedia , lookup
Debye–Hückel equation wikipedia , lookup
Physical organic chemistry wikipedia , lookup
Hydrogen atom wikipedia , lookup
Isotopic labeling wikipedia , lookup
Resonance (chemistry) wikipedia , lookup
Rutherford backscattering spectrometry wikipedia , lookup
Hypervalent molecule wikipedia , lookup
Gas chromatography–mass spectrometry wikipedia , lookup
Photosynthetic reaction centre wikipedia , lookup
Biochemistry wikipedia , lookup
Chemical bond wikipedia , lookup
Molecular scale electronics wikipedia , lookup
Metalloprotein wikipedia , lookup
IUPAC nomenclature of inorganic chemistry 2005 wikipedia , lookup
Stoichiometry wikipedia , lookup
Handout for semester exam #3/ CHM101 © 2006, Dr. Miroslav Rezac, Nomenclature, Chemical Calculations, Chemical Equations Naming binary compounds where one element is metal Both in the formula and in the name, the “positive” part appears first, the “negative” part as the second. The name of the positive part is identical to the name of the element. Naming of the negative part uses the stem of the element’s name and suffix –ide. The table below shows the most commonly encountered monoatomic negative ions. Name of element fluorine bromine chlorine iodine oxygen nitrogen sulfur NaCl CaI2 SrO Li3N Name of ion fluoride bromide chloride iodide oxide nitride sulfide Formula F– Br– Cl– I– O2– N3– S2– sodium chloride calcium iodide strontium oxide lithium nitride Polyatomic ions. Structures, names and charges of polyatomic ions are best memorized: it is very difficult to guess these properties out of MPT. The table below shows the ions you should know. However, the good news is you treat them as a unit, or a “box” with a charge. So the rules for naming and formula writing are identical as with monoatomic ions. Majority of polyatomic ions are negative, the only positive one you should know is ammonium, NH4+. How to generate a formula and name is shown below. Name of ion Formula Name of ion Formula sulfate SO42– NO3– OH– or HO– CO32- hydrogenphosphate HPO42– HCO3– PO43NH4+ nitrate hydroxide Carbonate hydrogencarbonate Phosphate Ammonium Compound formed between sodium and sulfate ions: Na 1+ 2- (SO4) Na2(SO4)1 1 Na2SO4 sodium sulfate Handout for semester exam #3/ CHM101 © 2006, Dr. Miroslav Rezac, Compound formed between aluminum and sulfate ions: Al 3+ 2- Al2(SO4)3 (SO4) aluminum sulfate Atoms with variable ionic behavior. Atoms forming negative ions always generate one, predictable kind (gaining all electrons to bring the s&p orbital sum to 8). However, some atoms can form more than one positively charged ion, having the ability to lose different amount of electrons each time. This behavior is difficult to predict, and you don’t have to know it . Names of ionic compounds containing these elements must contain reference as to which ion is involved. This is done by including a roman numeral in the name, showing the charge of the involved ion. So, in order to generate name, you first have to figure out what the charge is. For that, use the fact that the negative ion is always predictable. Note: In homework or on exam I will let you know that an element displays a variable ionic behavior. Without this disclaimer, you can assume that the element “behaves properly”. CuCl 1 x Cl - Cu+ 1 copper(I) chloride Count the number of negative ions, figure the total negative charge present. This charge is distributed over 1 copper ion, thus giving Cu+. This charge is indicated by roman numeral in name Cr2(SO4)3 - 3 x(SO 4)2 6 6 per 2 x Cr Cr3+ chromium(III) sulfate 3 sulfate ions carry charge of 3x(-2) = -6. Equal but opposite charge of +6 is distributed per 2 chromium ions, thus each chromium has 3+, making it Cr3+. This is indicated by (III) in name. Binary compounds where both elements are non-metals The only difference here is that we note how many of each atom there are. For example: N2O4 dinitrogen tetroxide SF4 sulfur tetrafluoride NO2 nitrogen dioxide PCl5 phosphorus pentachloride Number Prefix 1 mono 2 di 3 tri Note: the prefix mono- is frequently omitted. Number 4 5 6 2 Prefix tetra penta hexa Handout for semester exam #3/ CHM101 © 2006, Dr. Miroslav Rezac, Formula mass Remember that just like we weigh things in the supermarket in pounds, in context of atom we are using atomic mass unit (amu). This unit is defined as 1/12 fraction of mass of isotope of carbon 12C. Thus, atomic mass of 12C is 12.000000000 with as many zeros as you want!!! 1 atomic mass unit = 1 12.0000000 of mass of 12 6 C Formula mass is the sum of average atomic masses of all atoms shown in the formula. You can view this as the mass of the molecule expressed in amu’s. Example below shows formula mass for calcium carbonate. All atomic masses are average atomic masses and are taken directly from MPT. calcium carbonate: CaCO3 1 atom Ca.......... 1*40.08 = 40.08 1 atom C............ 1*12.01 = 12.01 3 atoms O.......... 3*16.00 = 48.00 Formula mass FW = 100.09 Formula mass tells you that one molecule of calcium carbonate “weighs” 100.09 atomic mass units. Problem: 12C has atomic mass of 12.0000 amu. One can guess that 1 amu is a very, very small mass . What if – what if we borrow the number 12.0000, but use a gram as the unit. You would get 12.0000 grams of carbon, and that would have to contain a very “large number” of atom carbons, since mass of 1 carbon atom is so negligible, right? Mole – The Chemist’s “Dozen” So what is the “large number” of carbon atoms? Is it 100; 10,000 or else? This puzzle was eventually solved, and it is now known that 12.0000 grams of 12C contains 6.023*1023, or the so-called Avogadro’s number. Definition of "mole" 0.000 g keep adding atom by atom until... atoms of 12C 12.000 g How many atoms on the balance? 1 mole of course!!! One mole of any OBJECTS always contains the same number of units. So, 1 mole of eggs, carbon atoms, sugar molecules, sodium ions all contain 6.023*1023 of their respective objects. Just like a dozen of anything contains 12 pieces of that anything. 3 Handout for semester exam #3/ CHM101 dozen © 2006, Dr. Miroslav Rezac, Group count Objects 1 dozen of eggs 12 eggs 1 mole of eggs ~ 6x1023 eggs 1 mole of O2 ~ 6x1023 molecules of O2 0.1 mole of O2 ~ 6x1022 molecules of O2 mole Note: You know that your conting "object" is a molecule because O2 is "2 or more atoms bound by chemical bonds", thus a "molecule" MOLAR MASS = mass of 1 mole of objects. For chemical substances, molar mass is the same number as formula mass, except the unit is “gram”. Glucose: C6H12O6 Formula mass: 6 atoms C = 6x12 12 atoms H = 12x 1 6 atoms O = 6x16 Molar mass: MW = 180 grams FW = 180 AMU Note: Chemical formula can “stand in” for three different things: One molecule One mole Number of grams corresponding to 1 mole EXAMPLE: CO2 may represent - 1 molecule of CO2 - 1 mole of molecules CO2 (molecule is the “object”) - 44 g of CO2 (formula mass of CO2 is 44) This “visualization” is especially important for calculations. Looking at formula of CO2 you may see the following: One molecule of CO2 contains one atom of C and 2 atoms of oxygen One mole of CO2 contains one mole of C and 2 moles of oxygen 44 g of CO2 (weight of 1 mole) contains 12 g of C (mass of 1 mole of C) and 32 g of O (mass of 2 moles of oxygen atoms) CALCULATIONS WITH MOLE Below you see how mole relates to grams and # of molecules. Each of these equivalencies gives 2 conversion factors. Those will be used in the same fashion you used them for unit conversions. 4 Handout for semester exam #3/ CHM101 © 2006, Dr. Miroslav Rezac, MOLES MOLES GRAMS # OF MOLECULES 1 mole ~ 6x1023 molecules 1 mole ~ MW grams MW grams 1 mole 6x1023 molecules 6x1023 molecules 1 mole MW grams 1 mole 1 mole Example 1: How many molecules of C5H10 are present in 1.245 g? Solution: From the available conversion factors you can convert directly between grams and moles, and between moles and # of objects. 1) Calculate molar mass: MW = 5x12 + 10 x 1 = 70 g 2) Plan a route: grams moles # molecules 3) Select appropriate conversion factors and perform conversion: grams 1.245 g * moles 1 mole # molecules 6.023x1023 molecules * 70 grams 1 mole = 1.067x1022 molecules Example 2: What is the mass of 1.25x1022 atoms of gold? Solution: 1) Calculate molar mass of gold: Au – single atoms; MW = 197 g 2) Plan a route: moles # atoms grams 3) Select conversion factors and perform conversion: # atoms 1.25x1022 atoms * moles 1 mole 23 6.023x10 molecules grams 197 g * = 4.088 g 1 mole CALCULATING AMOUNT OF ELEMENT PRESENT IN AN AMOUNT OF COMPOUND Concept of mole will help us to solve practical problems. For example, you are own certain amount of Au2O3 and you are wondering how many grams of gold it contains. 5 Handout for semester exam #3/ CHM101 © 2006, Dr. Miroslav Rezac, To solve these types of problems, we have to make additional considerations. As shown below on line A, one molecule of Au2O3 will contain 2 atoms of Au. If one molecule contains 2 atoms, then larger number of molecules will contain larger number of atoms. Line B shows this for 6x1023 molecules. Line C simply points out that 6x1023 objects is just a codeword for 1 mole. Line D shows that moles of object do have a mass – MW. Frame on the bottom shows a conversion factor linking grams of gold to grams of Au2O3. A) 2 atoms Au 1 molecule Au2O3 B) 6x1023 molecules Au2O3 2 * (6x1023 atoms Au) C) 1 mole of Au2O3 2 moles of Au D) 442 g of Au2O3 2*197= 394g Au 394 g Au 442 g Au2O3 Conversion Factor Armed with the above conversion factor we can convert any number of grams of Au2O3. Example 1: How many grams of gold are present in 4.875 g of Au2O3? Solution: Perform conversion with the above conversion factor. 394 g Au 4.875g Au2O3 * 442 g Au2O3 = 4.346 g Au Example 2: How many grams of C are present in 10.450 g of C3H6O? Solution: 1) Calculate molar mass of C3H6O: MW = 58. 2) Derive proper gram – gram conversion factor and perform conversion. 3 moles of C 1 mole of C3H6O 3*12= 36g C 58 g of C3H6O 36 g C 10.450 g C3H6O * 36 g C 58 g C3H6O = 6.486 g of C 58 g C3H6O PERCENT COMPOSITION OF COMPOUNDS When calculating molar mass, you essentially are figuring out how many grams of each element that amount of compound contains. For example, for NaCl, the molar mass is 23 + 35 = 58g. That 6 Handout for semester exam #3/ CHM101 © 2006, Dr. Miroslav Rezac, essentially tells you that in 58g of sodium chloride there is 23g of sodium and 35g of chlorine to be found. What is the percentage composition? By using proportion, you have to figure out how much of each 100g of NaCl would contain. You figure out what “fraction” of the whole (58g of NaCl) each element represents. Multiplying by 100% you get the percentage: 58 g NaCl amount of 1 mole amount corresponting to % scale 100g NaCl x= 23 g Na 35g Cl x y 23 * 100% = 39.65 % 58 y= 35 * 100% = 60.34 % 58 In this way, we calculate that NaCl contains 39.65% of sodium and 60.34% of chlorine. Obtaining a % composition from an experiment is even easier: you are told how much element is contained in what amount of compound, so you don’t have to worry about molar mass! For example, an experiment finds that 3.8 g of Na reacts with oxygen (O) to give 5.1g of compound. What is the percent composition? First, let’s figure out how much oxygen we have in: 5.1g – 3.8 g = 1.3g of bound oxygen % composition: 5.1 g of oxide 3.8 g Na 1.3g O 100g of oxide x y x= 3.8 * 100% = 74.5 % 5.1 y= 1.3 * 100% = 25.5 % 5.1 Thus, the unknown compound consists of 74.5% of sodium and 25.5% oxygen. EMPIRICAL FORMULA Empirical formula lists the smallest possible ratio of all atoms present. It is precisely this formula we get from experimental data. Let us use the above example of compound of sodium and oxygen. The percentage composition we calculated above tells us how many grams of each element is present in 100g of the compound (definition of a percent!). 7 Handout for semester exam #3/ CHM101 © 2006, Dr. Miroslav Rezac, However, we do not care about ratio of grams, but ratio of atoms… However, anything what holds true for atoms / molecules holds true also for moles! So what if we calculate the ratio of moles of each element present in 100g of compound? It will tell us directly in what ratio are atoms present. Oxide: 74.5 % Na 25.5% O 100g of oxide 74.5 g Na 25.5 g O : 16 (molar mass of O) : 23 (molar mass of Na) 100g of oxide Divide each by the smallest # 3.2391 mol of Na 1.5938 mol O 3.2391 = 2.03 1.5938 1.5938 = 1.00 1.5938 Ratio of # atoms of Na to # atoms of O: 2.03 : 1.00 = 2 : 1 EMPIRICAL FORMULA: Na2O MOLECULAR FORMULA Molecular formula gives truthfully the actual # of atoms present in the molecule, rather than their simplest ratio. One empirical formula is common to an infinite number of molecular formulas: multiply the empirical formula by a whole # and you get a possible molecular formula. Thus, for the compound above you get: Multiply by Molecular formula 1 Na2O 2 Na4O2 3 Na6O3 4 Na8O4 and so on. How do we decide which molecular formula belongs to our compound? We must know the molar mass. Then we can calculate molar mass for all possible molecular formula spawning from the empirical formula and – obviously – the one, which has molar mass identical to the unknown compound, represents its molecular formula. Our compound has molar mass of 56 g. Now let’s have a look at possible molecular formulas: Molecular formula Molar mass Na2O 56 Na4O2 116 Na6O3 168 Na8O4 224 and so on etc. Examining the molar masses of the possible formulas we see that the first one, shown in bold, has molar mass identical to the molar mass of our compound determined by experiment. Thus, our compound has molecular formula of Na2O. Fun, huh? 8 Handout for semester exam #3/ CHM101 © 2006, Dr. Miroslav Rezac, BALANCING EQUATIONS Chemical Equation: description of a chemical process C+D A+B Reactants "turn into" Description of a process Products sum of masses of all reactants = sum of masses of all products sum of atoms of reactants = sum of atoms of products The "equation part" For example: Zn + CuSO4 Cu + ZnSO4 Two requirements imposed on chemical equations: Chemical process must be able to occur Reactants and products must be used in the form they actually exist While the first requirement is pretty self-explanatory (if the process does not happen, you should not write chemical equation describing it…). The second is best illustrated below: Below you find example of bad equation, which does not take into account proper chemical forms H2O 2H + O The following equations pays attention to hydrogen and oxygen existing as diatomic molecules 2H2 + O2 2H2O The second requirement means that the number of atoms of each kind on the left side must be the same as the number of atoms of the same kind on the right side… That means, not only the grand total number of atoms stays the same, but the total for each kind as well. This rule will serve us to balance equations. In the example below you see an unbalanced equation for oxidation of ethanol. C2H6O + O2 CO2 + H2O In this case, the most complex molecule is ethanol. The unique elements are carbon and hydrogen: they appear on the left only in ethanol, and no other reactant; on the right carbon is found only in one product (CO2) and so is hydrogen (H2O). Let us use carbon as a starting point. Let us pick one molecule of ethanol. 9 Handout for semester exam #3/ CHM101 1C2H6O + 1C2H6O + © 2006, Dr. Miroslav Rezac, O2 CO2 + 2 CO2 O2 H2O + H2O 1 ethanol = 2 carbons => need 2 CO2 to account for 2 carbons on the right Now the equation is balanced for carbon. We will do the same for hydrogen. 1C2H6O + O2 2CO2 + 3H2O 1 ethanol = 6 hydrogens => as on the right each molecule of water accounts for 2 hydrogen atoms, we need 3 molecules of water to account for 6 hydrogen atoms (3 molecules * 2 atoms per molecule = 6 atoms) To figure out oxygens, the following will apply. By choosing one molecule of ethanol we determined we get 2 CO2 and 3 H2O. Thus we must find how many atoms of oxygen are contained in the products. The total count is 7. Next, realize that the 7 oxygens must come from ethanol and O2. Ethanol, since we chose 1 molecule, will bring in 1 atom of oxygen. The O2 must provide the rest, that is 7 – 1 = 6 atoms. Since each O2 molecule provides 2 atoms, we will need 3 molecules of O2. 1C2H6O + ? O2 2CO2 + we need 7-1=6 atoms 3H2O 7 oxygen atoms total 1 oxygen atom And so the balanced equation appears below. C2H6O + 3 O2 2CO2 + 3H2O In the next example, oxidation of propanol, we follow the same scheme. The only difference is that the fractions of molecules, which we get in the primordial equation, will be eliminated by doubling the amount of all components of the equation. C3H8O + 1C3H8O O2 + 1C3H8O + O2 O2 CO2 + H2O CO2 + H2O 3 CO2 + H2O 1 propanol = 3 carbons => need 3 CO2 to account for 3 carbons on the right 10 Handout for semester exam #3/ CHM101 1C3H8O + © 2006, Dr. Miroslav Rezac, O2 3CO2 + 4H2O 1 propanol = 8 hydrogens => as on the right each molecule of water accounts for 2 hydrogen atoms, we need 4 molecules of water to account for 8 hydrogen atoms (4 molecules * 2 atoms per molecule = 8 atoms) ? 1C3H8O + O2 3CO2 + we need 10-1=9 atoms 4H2O 10 oxygen atoms total 1 oxygen atom Balanced primordial equation contains fractions: 1C3H8O + 9 O2 2 3CO2 + 4H2O *2 Final balanced equation does not contain fractions: 2C3H8O + 9 O2 6CO2 + 8H2O TYPES OF CHEMICAL REACTIONS Reactions can be formally subdivided into four types. For a given reaction, be able to recognize which type it belongs to: 1) Combination This reaction is characterized by more than 1 reactant and only one product. C + O2 CO2 2Na + Cl2 2NaCl 2) Decomposition Reaction characterized by only one reactant and more than one product 2HgO 2Hg + O2 2NI3 N2 + 3) Single replacement 11 3I2 Handout for semester exam #3/ CHM101 © 2006, Dr. Miroslav Rezac, B B A C A C CuSO4 + Mg Cu 2AgNO3 + Zn 2Ag + MgSO4 + Zn(NO3)3 4) Double replacement D B A C D A B C CuF + NaI CuI CuSO4 + BaCl2 CuCl2 + + NaF BaSO4 The most important representative of single displacement reactions is metal/hydrogen displacement. All metals and hydrogen can be ranked by their reactivity, where more reactive metal (as element) will displace less reactive metal from its compound (or, hydrogen H2 from an acid). K > Ca > Na > Mg > Al > Zn > Fe > Ni > Sn > Pb > H > Cu > Ag > Hg > Au You do not have to memorize the reactivity ranking except that Zn > H > Cu and that for metals in-group Ia and IIa of Mendeleev table the reactivity is increasing top to bottom Fr > Cs > Rb > K > Na > Li > H Ra > Ba > Sr > Ca > Mg > Be However, given the “reactivity chart” you must be able to determine if a reaction occurs or not. Example Decide if the following reactions will occur or not: 12 Handout for semester exam #3/ CHM101 Mg + © 2006, Dr. Miroslav Rezac, CuSO4 Cu + MgSO4 Fe + Pb(NO3)2 H2 + AgSO4 Yes, Mg more reactive than Cu Pb + Fe(NO3)2 No, Pb is less reactive than Fe Ag + H2SO4 No, Ag is less reactive than H Neutralization The most notable example of double replacement reactions is neutralization, a reaction between acids and bases (for our purposes, base is a compound containing a hydroxide ion). During neutralization, a salt and water are produced. HNO3 + acid NaOH NaNO3 base salt + H2O water Each H from an acid reacts with one OH of the base to give 1 molecule of water. Whatever is left will “make up” the salt. Example: H2SO4 + 1 OH Present 2H 1 H2SO4 1*2=2 + NaOH 2H and 2 OH will give 2 molecules of water 1 OH per molecule: need 2 molecules 2NaOH 2 H2O + Na2SO4 2 Na, 1 SO4 left: form one molecule In more complicated cases the neutralization can be balanced the same way as any other equation would: 13 Handout for semester exam #3/ CHM101 H3PO4 + Ca(OH)2 2 H3PO4 + 3 Ca(OH)2 © 2006, Dr. Miroslav Rezac, 6 H2O + 3Ca, 2(PO4) 2*3 = 6 2 H3PO4 "leftovers give one molecule of calcium phosphate + 3 Ca(OH)2 6 H2O + Ca3(PO4)2 HEAT IN CHEMICAL CHANGES The amount of heat can be included in a chemical reaction as a reactant or product. Reaction heat is the heat released by reaction. If heat is shown as a product, the reaction heat is positive, if heat is shown as a reactant, the reaction heat is negative! H2 + Cl2 N2 + O2 2 HCl + 181 kJ exothermic reaction: reaction heat = +185 kJ 185 kJ 2 NO Reaction profile for exothermic reaction R + endothermic reaction: reaction heat = -181kJ Reaction profile for endothermic reaction activation energy P activation energy reaction heat P reaction heat R Know that: Exothermic reaction releases heat Endothermic reaction absorbs heat Amount of heat energy as reactant = reaction is endothermic Amount of heat energy included as product = reaction is exothermic Be able to recognize graphs for exothermic and endothermic reactions 14