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AP Chemistry Summer Assignment Please read the following sheets of information they include notes and practice problems that will be covered the first week of school. The worksheets include work from the first three chapters and should be considered a review of information from PreAP chemistry. If they are not a review, don't worry! But try to attempt the problems so that you know what to expect your first week in AP Chemistry. We will be going over these worksheets the first couple of days when you come back to school. You will have: 1. A quiz on the most common polyatomic ions after the firstclass meeting and frequent quizzes on nomenclature and netionic equations every Monday thereafter. Correctly writing chemical equations is an essential part of learning chemistry and should be a focus of study early in the course. It is beneficial to start learning these polyatomic ions and review balancing equations before school starts in August. 2. A test on the first three chapters within the second week of school AP CHEMISTRY SUMMER REVIEW PROBLEMS Answers in [ ] after problem. I am NOT interested in the correct answer, rather the path you take to get to the answer. This is a sample of what you will be responsible for knowing on the first test (the second week of school). At the end of the problems, you will find notes in case you get stuck! Good luck, and I will see you in August. 1. Boron has two naturally occurring isotopes, boron10 and boron11. The relative abundance of boron10 is 19.9%. The atomic mass of boron10 is 10.01 amu and the atomic mass of boron11 is 11.01 amu. What is the atomic weight of boron? [10.81 amu] 2. CALCULATE the number of moles present: A. 159 grams of aluminum sulfate [.465 mole] B. 3.00 kg of calcium carbonate [30.0 mole] 20 – 4 C. 1.12 x 10 formula units of sodium hydroxide [1.86 x 10 mole] 3. CALCULATE the number of grams present: A. 8.20 moles calcium carbonate [821 g] 15 7 B. 3.01 x 10 formula units of sodium hydroxide [2.00 x 10 g] 4. CALCULATE the number of ions/atoms present: A. number of hydrogen atom is in 5.00 molecules in isopropyl alcohol, C H O [40 atoms] 3 8 24 B. number of ions in 5.00 moles of ammonium sulfide [9.03 x 10 ions] 5. NAME the following compounds. A. CrF D. Mg N 3 3 2 B. Hg CO E. Zn(C H O ) 2 3 2 3 2 2 C. Pb(C H O ) F. BaO 2 3 24 2 6. WRITE the following compounds into symbols. A. hydrochloric acid D. sodium oxide B. phosphoric acid E. lead (II) chromate C. magnesium nitrate F. aluminum arsenate 7. DETERMINE the percent composition of the following compounds: 1. NH Cl [26.20% N; 7.55% H; 66.26% Cl] 4 2. CH O [39.99% C; 6.73% H; 53.28% O] 2 3. SnI [18.95% Sn; 81.05% I] 4 8. You have 125 g of FeO determine the amount of iron (Fe) in the compound. [97.16 g iron] 9. Determine the percent composition of sodium carbonate decahydrate. (Na CO ・10 H O) 2 3 2 [37.06% Na CO ; 62.94% H O] 2 3 2 10. Determine the formula of a hydrate. You have 1.023g of CoSO hydrate. After heating and driving 4 off all the water, the mass of the anhydride is 0.603 g. Determine the hydrates chemical formula (number of waters attached). [CoSO ∙ 6 H O] 4 2 11. Determine the empirical formula given 76.0% iodine and the other element in the compound is oxygen. [I O ] 2 5 12. You have 245.0 grams of ammonium nitrate. How many grams of nitrogen do you have? [85.8 g nitrogen] 13. Calcium oxide is prepared by heating and decomposing calcium carbonate, how much of the carbonate would be required to produce 15.0 grams o the oxide? [26.8 g] 14. 15.0 grams of calcium oxide reacts with water, what amount of product would be produced? [19.8 g] 15. How many grams of aluminum sulfide are formed if 9.00 grams of aluminum react with 8.00 grams of sulfur? How many grams remain unreacted? [12.5 g aluminum sulfide; 4.51 g of excess left unreacted] Write and balance the following chemical reactions. 16. iron (III) chloride + hydrochloric acid > 17. barium + magnesium bromide > 18. chlorine + magnesium bromide > 19. iron (III) hydroxide > 20. silver + sulfur > 21. aluminum sulfate + calcium hydroxide > 22. potassium + water > 23. sodium chlorate > 24. magnesium carbonate > 25. calcium hydroxide + phosphoric acid > 26. sodium carbonate + nitric acid > 27. aluminum hydroxide + sulfuric acid > 28. magnesium + nitric acid > 29. zinc + lead (II) acetate > 30. ethyl alcohol (C H OH) + oxygen > 2 5 31. Chlorine has two isotopes, Cl – 35 and Cl – 37. Their abundances are 75.54% and 24.4%, respectively. Assume that the only hydrogen isotope present is H – 1. A. How many different HCl molecules are possible? [2 molecules] B. What is the sum of the mass numbers of the two atoms in each molecule? HCl35 = 36; HCl37 = 38] 32. Hexachlorophene, a compound made up of atoms of carbon, hydrogen, chlorine, and oxygen, is an ingredient in germicidal soaps. Combustion of a 1.000 g sample yields 1.407 g of carbon dioxide, 0.134 g of water and 0.523 g of chlorine gas. What are the mass percents of carbon, hydrogen, oxygen and chlorine in hexachlorophene? [38.40% c; 1.50% h; 7.8% O; 52.3% Cl] 33. Nickel reacts with sulfur to form a sulfide. If 2.986 g of nickel reacts with enough sulfur to form 5.433 g of nickel sulfide, what is the simplest formula for the sulfide? Name the sulfide. [nickel (III) sulfide] 34. Methyl salicylate is a common “active ingredient” in liniments such as BenGay. It is also known as oil of wintergreen. It is made up of carbon, hydrogen and oxygen atoms. When a sample of methyl salicylate weighing 5.287 g is burned in excess oxygen, 12.24 g of carbon dioxide and 2.505 g of water are formed. What is the simplest formula for oil of wintergreen? [C H O ] 8 8 3 35. Riboflavin is one of the B vitamins. It is also known as vitamin B and is made up of carbon, 6 hydrogen, nitrogen, and oxygen atoms. When 10.00 g of vitamin B is burned in oxygen, 19.88 g of 6 CO and 4.79 g of H O are obtained. Another experiment shows that vitamin B is made up of 2 2 6 14.89% N. What is the simplest formula for vitamin B ? [C H N O ] 6 17 20 4 6 36. Dimethylhydrazine, the fuel used in the Apollo lunar descent module, has a molar mass of 60.10 g/mole. It is made up of carbon, hydrogen, and nitrogen atoms. The combustion of 2.859 g of the fuel in excess oxygen yields 4.190 g of carbon dioxide and 3.428 g of water. What are the simplest and molecular formulas for dimethylhydrazine? [CH N; C H N ] 4 2 8 2 37. Phosphine gas reacts with oxygen according to the following equations : 4PH → P 3 (g) + 8 O 2 (g) 4O 10 (g) + 6 H 2O. Calculate: A. the mass of tetraphosphorus decaoxide produced from 12.43 mole phosphine. [882.2g] B. the mass of PH required to form 0.739 mol of water. [16.75 g] 3 C. the mass of oxygen gas that yields 1.00g of water. [2.368 g] D. the mass of oxygen required to react with 20.50 g of phosphine. [38.6 g] 38. Aluminum reacts with sulfur gas to form aluminum sulfide. Initially, 1.18 mole of aluminum and 2.25 mole of sulfur are combined. A. Write a balanced equation for the reaction. B. What is the limiting reactant? [aluminum] C. What is the theoretical yield of aluminum sulfide in moles? [.59 mole] D. How many moles of excess reactant remain uncreated? [.38 moles] 39. When iron and steam react at high temperatures, the following reaction takes place. 3 Fe (s) + 4 H O (g) → Fe O (s) + 4 H (g) How much iron must react with excess steam to 2 3 4 2 form 879 g of Fe O if the reaction yield is 69.0%? [940.7g] 3 4 How to setup problems for AP Chem and other information. The following four (4) step set up is recommended for credit to be given on homework, quizzes, labs or tests. Calculations must also be in PENCIL to ensure the work is legible. This is a science class and the answer is not the most important part of this class; I am testing the chemistry concepts, not how well you push buttons on a calculator. I am very rigid on showing your work, and all numbers must have units to recieve credit. 1 . Label the given information in the problem. [ What numbers are given with a symbol and what you are determining in the problem with a symbol ?] (See example BELOW.) 2 . Write the mathematical equation in the correct form or write and balance the reaction . That form is the unknown variable on one side of the equal marks and all the other variables on the other side o f the equal marks. [Isolate the unknown ] (See example. BELOW) 3. Substitution of the numbers and units into the equation written in step 2. [ Substitute .] (See example BELOW.) 4. Final answer with units . I don't want to see your little calculations to the final answer that is why you use the calculator. ) [Evaluate.] (See example BELOW) EXAMPLE : Mercury has a density of 13.6 g/mL, and a volume of 91.0 mL. What is the mass of the sample of mercury? step 1 step 2 step 3 step 4 3 D = 13.6 g/mL m = D x V m = (13.6 g/mL) ( 91.0 mL) m = 1.24 x 10 g V = 91.0 mL m = ? POLYATOMIC IONS (Memorize) 1 + 1 + 1. ammonium NH 4 1+ 3. hydronium H O 3 1 1 4. acetate C H O 2 3 2 1 5. chlorate ClO 3 1 6. chlorite ClO 2 1 7. cyanide CN 1 8. dihydrogen phosphate H PO 2 4 1 9. hydrogen carbonate HCO 3 1 10. hydrogen sulfate HS0 4 1 11. hydroxide OH 1 12. hypochlorite ClO 1 13. iodate IO 3 1 14. nitrate NO 3 1 15. nitrite NO 2 1 16. perchlorate ClO 4 1 17. permanganate MnO 4 1 18. thiocyanate SCN 2 + 2 + 2. mercury (I) Hg 2 2 2 19. carbonate CO 3 2 20. chromate CrO 4 2 21. dichromate Cr O 2 7 2 22. monohydrogen phosphate HPO 4 2 23. oxalate C O 2 4 2 24. peroxide O 2 2 25. silicate SiO 3 2 26. sulfate SO 4 2 27. sulfite SO 3 3 3 28. arsenate AsO 4 3 29. phosphate PO 4 3 – 30. phosphite PO 3 1. SYNTHESIS 1. element + element > compound (Main one to know except the ones mentioned below.) A + B > AB 2H + O > 2H O 2 2 2 2. compound + element > compound (See decomposition of chlorates.) 1. metallic chloride + oxygen > metallic chlorate 2 KCl + 3 O > 2 KClO 2 3 3. Compound + compound > compound (See decomposition of carbonates, hydroxides and ternary/oxyacids.) 1. metallic oxide + carbon dioxide > metallic carbonate CaO + CO > CaCO 2 3 2. metallic oxide + water > metallic hydroxide (a base) CaO + H O > Ca(OH) 2 2 4. Compound + compound > compound (See decomposition of carbonates, hydroxides and ternary/oxyacids.) 3. nonmetallic oxide + water > ternary/oxyacid The oxidation number of the nonmetal other than oxygen determines the polyatomic ion in the acid. N O + H O > 2 HNO 2 5 2 3 4. metallic oxide + nonmetallic oxide > salt (composed of a polyatomic ion). The oxidation number of the nonmetal in the polyatomic ion is the oxidation number in the nonmetallic oxide. CaO + N O > Ca(NO ) 2 5 3 2 2. DECOMPOSITION 1. compound > element + element AB > A + B 2NaCl > 2Na + Cl 2 Must MEMORIZE the following special decomposition reactions. 1. metallic chlorate > metallic chloride + oxygen 2KClO > 2KCl + 3O 3 2 2. metallic carbonate > metallic oxide + carbon dioxide CaCO > CaO + CO 3 2 3. metallic hydroxide > metallic oxide + water Ba(OH) > BaO + H O 2 2 (base) (basic anhydride metallic oxice) 4. Acids (ternary/oxyacids) > nonmetallic oxide + water (acid anhydride – nonmetallic oxide) Must know that when sulfurous acid and carbonic acid are made as products; they immediately decomposed into a nonmetallic oxide and water. H SO > SO + H O; H CO > CO + H O 2 3 2 2 2 3 2 2 3. SINGLE DISPLACEMENT 1. element + compound > element + compound (These products must be different elements and compounds than the element and compound as reactant to be a reaction. If they aren't different then NO reaction takes place and write NR.) M + QN > MN + Q or N + MR > MN + R 1+ 1 2K + 2H O > 2KOH + H (Think of water when in a SD to be H and OH ) 2 2 Cl + 2NaBr > 2NaCl + Br 2 2 If the element is a metal (on the reactant side) then it may displace only the metal in the compound. The metal (element) doing the displacing must be more active than the "metal" in the compound. To determine activity you must use the activity (electromotive) series. If the element is a nonmetal (on the reactant side) then it may displace only the nonmetal in the compound. The nonmetal (element) doing the displacing must be more active than the "nonmetal" in the compound. To determine activity you use Group VII A on the Periodic table for the nonmetal activity. 4. DOUBLE DISPLACEMENT 1. Compound + compound > compound + compound (These are different compounds from than the reactants and products.) AB + CD > AD + CB [The metals A & C switch places from the reactant side to the product side.] NaCl + AgNO > NaNO + AgCl (↓) 3 3 These reactions are carried out in water. For the reaction to occur one or more of the products must leave the chemical reaction environment by forming: 1. a gas You must identify that a gas was formed as a product with a symbol in the reaction. 2. a molecular species > For general chemistry this is water. 3. a precipitate (ppt) must form > You need to use the solubility rules below or if given a solubility chart. The ppt must be identified as a product with a symbol in the reaction. Solubility Rules MEMORIZE at least the first four. You will have a quiz the third day, along with the polyatomic ions. You need to memorize the following solubility rules for determining your ppt (precipitate). 1. Common sodium ions, potassium ions, ammonium ions, and hydrogen ion compounds are soluble in water. (If soluble in water they will not precipitate out of the chemical environment.) 2. Common nitrates, acetates and chlorates are soluble in water. 3. Common chlorides, bromides and iodides are soluble in water EXCEPT silver, mercury (I), lead (II) and thallium (I). [Lead (II) chloride is soluble only in HOT water]. The except means these compounds will precipitate out of a water solution. 4. Common sulfates are soluble in water EXCEPT calcium, barium, strontium, silver and lead (II) ions will precipitate. 5. Common carbonates, phosphates and silicates are INSOLUBLE in water (these precipitate), EXCEPT ammonium ions, and Group IA but not lithium ions these are soluble in water. 6. Common sulfides are insoluble in water (these precipitate) EXCEPT Group IA, barium, strontium and ammonium ions are soluble. 7. Common chromates are soluble (these don't precipitate) EXCEPT barium, strontium lead (II) and silver ions will precipitate. 8. Common hydroxides are insoluble in water (these precipitate) EXCEPT Group IA, barium, strontium and ammonium ions are soluble in water. Notes on Double Displacement Rxs for Net ionic form . 1. Since these are solution, ionic compounds dissociate in water. That means when they dissolve the water molecules pull the positive ions and negative ions apart. It is written as follows: 1+ 1 1+ 1 1+ 1 Na (aq) + Cl (aq) + Ag (aq) + NO (aq) + NO ↓ ) 3 (aq) > Na 3 (aq) + AgCl ( 2. Spectator ions are the ones written the same on the product and reactant side. These ions are crossed out when writing the net ionic reaction is written with what species are left on the reactant and product side: 1 1+ Cl (aq) + Ag (aq) > AgCl ( ↓ ) 5. COMBUSTION of HYDROCARBONS (contain carbon, hydrogen and sometimes oxygen). Always use complete combustion. These reactions you may need to add oxygen as a reactant. The only products are carbon dioxide and water. C H + 4 O → 2 CO + 4 H O 2 8 2 2 2 COMBUSTION of other compounds yield oxides of the each element in the compound. CS + 2 O → CO + SO 2 2 2 2 To balance chemical reactions do the following: 1. First write the skeleton reaction. That means to write all elements and compounds correctly. 2. Use coefficients to give the same number of atoms of each element on the product and reactant side. Formulas of Ionic Compounds The total positive and negative charge must equal zero. + Math equation: F = k (q )(q ) → Use this equation to determine force of 2 d attraction in the ionic bond (Coulombic Attraction) Crystal lattice → formula unit → electrolytes → ionic compounds → salts (metals) Naming Compounds NAMING Binary Acids PREFIX STEM SUFFIX ACID hydro ic Acid The stem comes from the other chemical combined with hydrogen. NAMING Ternary (Oxyacids) Acids The easy way to name ternary acids is to name the polyatomic ion. If the polyatomic ion ends in "ate" drop the ending and add "ic" and the word Acid. If the polyatomic ion ends in "ite" drop the ending and add "ous" and add the word Acid. Ternary acids DON'T MENTION HYDROGEN IN THE NAME. Hydrogen is taken care of by the word acid. Remember all acids start the symbols with hydrogen's symbol. STEPS to WRITE the COMPOUND in SYMBOLS 1. Write the positive ions symbol, and its = oxidation number right ABOVE the symbol. 2. Write the negative ions symbol, and its = oxidation number right ABOVE the symbol 3. Check to see if the oxidation n umbers add out to zero. If the addition equals zero then DO NOT add any subscripts. The compound is written correctly. If the addition does not equal zero then do step 6. 4 . If the oxidation numbers are not multiples of each other then cross the numbers only and add them as subscripts; if the oxidation numbers are multiples then divide the smaller one into the larger oxidation number and the answer is the small ones subscript. 5. Check and see if your overall charge is equal to zero. NOMENCLATURE for BINARY and TERNARY COMPOUNDS 1. Name the positive ion first, name unchanged. 2. Check to see if the positive ion needs a Roman Numeral (RN). A. If the positive ion needs a RN then put parentheses after the positive ions name. B. If the positive ion does not need a RN then go and do JUST step # 3 3. Next name the negative ion, changing the ending to "ide". Except if it is a polyatomic ion then DO NOT CHANGE ITS NAME. 4. If you put parentheses in step 2 then go back and figure out the positive ion's oxidation number (that's the RN you write). Remember you always know the negative ions oxidation ROMAN NUMERALS WHEN to use Roman Numerals (RN): when the positive ion has the possibility of more than one Ox #. HOW to use RN: in writing the names in words use the RN in parenthesis after the positive ion indicating the positive ions oxidation number. NEVER USE A RN with Group IA (1), Group II A (2), Group III A (13) (metals), Zn, Cd, or Ag always have the oxidation numbers indicated by their group A number or the last of the numeric group number. PREFIXES (You WILL NOT NAME any compound BY THIS METHOD EXCEPT a few common compounds by this method.) WHEN to use prefixes: when you have a covalent bond. HOW to use prefixes: to indicate the elements subscript. prefix + element name + ∙ ∙ ∙ Examples: diphosphorus pentaoxide, carbon monoxide, carbon dioxide P O CO CO 2 5 2 SEVEN RULES FOR ASSIGNING OXIDATION NUMBERS (Oxidation Number = OX #) 1. Ox # of an atom of a free element is zero. 2. Ox # of a monatomic ion is equal to its charge. 3. Algebraic sum of the oxidation numbers of the atoms in a compound is zero. 4. Ox # of hydrogen is 1 +, except in a metallic hydride it is 1. 5. Ox # of oxygen is 2, except in peroxides it is a 1. 6. Algebraic sum of oxidation numbers of the atoms in a polyatomic ion is equal to its charge. 7. Combination of nonmetals, the less electronegative element is the positive oxidation number the more electronegative element is the negative oxidation number. EXAMPLES: Determine the atomic weight of Cu – 63 with percent abundance of 69.17%, and the atomic mass of 62.94 amu; Cu – 65 has atomic mass of 64.93 amu. (Atomic mass) (% abundance) = % (62.94 amu) (0.6917) = 43.54 amu (64.93 amu) (0.3083) = + 20.02 amu 63.56 amu → atomic weight (at. wt.) OR If not given atomic masses you use the mass number instead of atomic mass: (mass #) (% abundance) = (63 amu) (0.6917) = 43.58 amu (65 amu) (0.3083) = + 20.04 amu 63.62 amu 23 Masses of Individual Atoms – Avogadro's Number (N ) = 6.02 x 10 A Mole Molar Mass (MM) – (in g/mole) is numerically equal to the sum of the atomic weights (from the Periodic Table carried out to 2 decimals) of the atoms in the formula. EMPIRICAL FORMULA Definition B it shows the atoms (elements) in the SIMPLEST WHOLE NUMBER RATIO. What is a ratio? A comparison What are the only things Chemist compare? The mole These above definitions and questions give us the following steps to solve for the empirical formula. 1. If given % composition then cross out the % sign and substitute the unit grams in that position. 2. Convert the grams of all the elements listed in the problem to moles. (Use the A Road Map @ ) 3. Setup the MOLAR RATIO: This is done according to the definition of empirical formula divide the smallest number of moles determined in step 2 into all the moles in step 2. This will give us the ratios. Don = t round the number of moles or ratios calculated. If the ratio number has a decimal of .0, .1 or .2 drop the decimal. If one of the ratio numbers has a decimal of .8 or .9 then round the number up. If one of the ratio numbers are between .3 up to .7 NONE OF THE NUMBERS CAN BE ROUNDED OR DROPPED. You must multiply all the ratios by first 2 to get all the ratios number to .0, .1, .2, .8 or .9. If multiplying by 2 didn't work then erase you must multiple all the ratios by 3. For this class that should be the highest we have to go, but the next number to try would be 5 then 7. These ratio numbers you now have are the subscripts of the elements. 4. Write the symbols of the elements with their appropriate subscripts. Put the elements in the order given in the problem . 25.9% N other element is oxygen 25.9 g N 1 mole N = 1.85 mole N ⁄ 1.85 mole = 1.00 x 2 = 2 14.01 g N 74.1 g O 1 mole O = 4.63 mole O ⁄ 1.85 mole = 2.50 x 2 = 5 N O 2 5 16.00 g O HYDRATES: These are compounds that have water attached to the salt, but they aren't chemically bonded. The waters are shown after the compound=s symbols after a raised dot. This raised dot doesn't mean to multiply it means surrounding by the following number of water molecules. Two typed of problems we do with hydrates. Need to know definitions of: 1. Hydrate: compound with attached water molecules 2. Anhydride: compound with the attached water molecules removed (just the plain compound). Example: Determine the percent composition of sodium carbonate decahydrate. (Na CO ∙ 10 H O) 2 3 2 Na CO = 106.10 g Part 2 3 H O = 10(18.02 g) = 180.02 g 2 Molar mass = 286.30 g Total % = P art x 100% Total % = 106.10 g x 100% = 37.06 % anhydride 286.30 g % = 180.20 g x 100% = 62.94% water 286.30 g To determine the formula of a hydrate you want to figure out the number of water molecules attached to the salt. The way to figure out these coefficients use the following steps: 1. Convert the grams of the compound to moles. 2. Convert the grams of water to moles. 3. Set up molar ratio by dividing the number of moles of water into the number of moles of compound and into the number of moles of water to get the two ratio numbers. 4. The ratio numbers in step 3 are the coefficients for the compound and water. So write the formula using the appropriate coefficients. Example: Determine the formula of a hydrate. You have 1.023g of CoSO hydrate. After heating and 4 driving off all the water, the mass of the anhydride is 0.603 g. Determine the hydrate’s chemical formula (number of waters attached). 1.023g compound 0.603 g CoSO 4 0.420 g H O 2 0.603 g CoSO 1 mole CoSO = 0.00389 mole CoSO / 0.00389 mole = 1 4 4 4 155.00 g CoSO 4 0.420 g H O 1 mole H O = 0.0233 mole H O/ 0.00389 mole = 5.989 = 6 2 2 2 18.02 g H O 2 CoSO ∙ 6 H O 4 2 New way of using % composition. If determining the empirical formula of a hydrocarbon (may have carbon, hydrogen and oxygen), then you must determine the number of grams of carbon by using carbon dioxide's percent composition with respect to carbon. You must also determine the number of grams of hydrogen by using water's percent composition with respect to hydrogen. Grams given of CO 12.01 g C = Grams of carbon 2 44.02 g of CO 2 Grams given of H O 2.02 g of H = grams of hydrogen 2 18.02 g H O 2 Example: The compound that gives vinegar its sour taste is acetic acid, which contains the elements carbon, hydrogen, and oxygen. When 5.00 g of acetic acid is burned in air, it produces 7.33 g of CO and 3.00 g 2 H O. What is the simplest formula of acetic acid? 2 7.33 g CO 12.01 g C = 2.00 g C 2 44.01 g CO 2 3.00 g H O 2.02 g H = 0.336 g H 2 18.02 g H O 2 Mass of O = 5.00g of acetic acid – ( 2.00 g C + 0.336 g H) = 2.66 g O 2.00g C 1mole C = 0.167 mole C/ 0.166 mole = 1 12.01 g C 0 .336 g H 1mole H = 0.333 mole H/0.166 mole = 2 1.01 g H 2.00 g O 1mole O = 0.166 mole O/ 0.166 mole = 1 16.00 g O Empirical Formula is CH O 2 Molecular Formula You need two pieces of information to determine the molecular formula, they are: 1. the empirical formula OR the percentages to have you determine the empirical formula 2. Molar Mass (MM) You determine the formula mass of the empirical formula given or the one you calculated. Divide that mass (empirical formal) into the given molar mass. The number you get tells you the number of times that unit by mass is in the molecular formula. Example Empirical Formula CH Molar Mass 26.02 g/mole 26.02 g/mole = 2 Molecular Formula C H 2 2 13.02 g/mole Empirical Formula CH O 180.0 g/mole = 6 Molecular Formula C H O 2 6 12 6 MM = 180.0 g/mole 30.06 g/mole Mass Relations from Reactions STOICHIOMETRY “STOIC” 4 STEPS ( You know this is a STOIC problem is you have a chemical reaction, given quantities and looking to solve for quantity (s). 1. Write and balance the chemical reaction. (Put give information in the problem above the substances in the chemical reaction.) 2. Convert the given information into moles. 3. Setup a molar ratio. [This is the only place and time coefficients are used. Your ratio will be the unknown substances coefficient on top and the given number of moles coefficient from step 2 on the bottom. 4. Convert moles of step 3 to ∙ ∙ ∙ LIMITING REAGENT (REACTANTS) LR This is a “Stoic” problem. You repeat steps 2 thru 4 for each reactant with given values. After you have completed step 4 for the last reactant, now circle the smallest answer. That’s the answer for a LR problem. Your limiting reactant is the substance in the dominator or the molar ratio DETERMINE AMOUNT OF EXCESS USED IN REACTION : given amount of LR 1 mole LR Coefficient of the Excess = amount of excess substance …………………………..MM (g) of LR Coefficient of LR used up in moles Amount of excess substance given at the beginning of the problem Amount of excess substance used up Amount of excess substance left unreacted (used up THEORETICALLY) PERCENT YIELD Percent yield = experimental yield ∙ (100) Theoretical yield Given as the actual amount made in the lab. Stoichiometry Problem HYDRATES These are compounds that have water attached to the salt, but they aren't chemically bonded. The waters are shown after the compound’s symbols after a raised dot. This raised dot doesn't mean to multiply it means surrounding by the following number of water molecules. To determine the formula of a hydrate you want to figure out the number of water molecules attached to the salt. The way to figure out these coefficients use the following steps: 1. Convert the grams of the compound to moles. 2. Convert the grams of the water to moles. 3. Set up molar ratio by dividing the number of moles of water into the number of moles of compound and into the number of moles of water to get the 2 ratio numbers. 4. The ratio numbers in step 3 are the coefficients for the compound and water. So write the formula using the appropriate coefficients .
