OWL Assignment #2 Study Sheet

Transcription

1 OWL Assignment #2 Study Sheet Binary Acid Nomenclature Binary compounds are composed of two elements. When one of the elements is a binary acid can be formed. Examples of this are HCl or H 2 S. When put into water, binary acids give H + (aq), which makes the solution acidic. There are two ways to name these compounds: 1. Molecular Name : Hydrogen + nonmetal with ide ending 2. Acid Name : Hydro ic acid. Used for aqueous solutions of the compound. Formul a HCl H 2 S Molecular Name chloride sulfide Acid Name hydrochloric acid hydrosulfuric acid There is one additional substance that is named like a binary compound although it is composed of three elements instead of two. HCN = cyanide or hydrocyanic acid. Below is a list of the acids you should be able to name. Formul a Molecular Name Acid Name Strong Binary Acids HCl HBr chloride bromide hydrochloric acid hydrobromic acid HI iodide hydriodic acid

2 HF fluoride hydrofluoric acid Weak Binary Acids H 2 S sulfide hydrosulfuric acid HCN cyanide hydrocyanic acid Oxo Acid Nomenclature Oxo acids are composed of H + and an oxygen-containing anion. For example: HNO 3 or H 2SO 4. If you learn your polyatomic anions, you will be able to name these acids. Recall that oxygen-containing anions end in either ate or ite. If you learn the ate anions, all the rest follow. ate anions form ic acids. HNO 3 = nitric acid ite anions form ous acids. HNO 2 = nitrous acid Below is a list of the oxo acids you should be able to name. Anion Name Anion Formula Acid Name Acid Formula Kind nitrate NO 3 nitric acid HNO 3 perchlorate ClO 4 perchloric acid HClO 4 strong acids ate anions sulfate SO 4 2 sulfuric acid H 2SO 4 acetate CH 3COO acetic acid CH 3COOH carbonate CO 3 2 carbonic acid H 2CO 3 phosphate PO 4 3 phosphoric acid H 3PO 4 nitrite NO 2 nitrous acid HNO 2 weak acids ite anions hypochlorite ClO hypochlorous acid HClO sulfite SO 3 2 sulfurous acid H 2SO 3 phosphite PO 3 3 phosphorous acid H 3PO 3

3 Finding Formulas From Percent Composition silicon = Si Molar mass = g / mol Si chlorine = Cl Molar mass = g / mol Cl bromine = Br Molar mass = g / mol Br QUESTION 1: Find the empirical formula for this compound. The EMPIRICAL FORMULA gives the smallest whole number ratio of moles of elements in the compound. Percents by weight: % Si, % Cl, % Br In 100 grams of the compound there are: g Si 1 mol Si = mol g Si Si g Cl 1 mol Cl = mol g Cl Cl g Br 1 mol Br = mol g Br Br Divide by the smallest number of moles calculated above to get the lowest mole ratio moles silicon / = moles Si moles chlorine / = moles Cl moles bromine / = moles Br These numbers are close enough to integer values to be rounded to integers mol Si = 1 mol Si

4 2.000 mol Cl = 2 mol Cl mol Br = 2 mol Br The EMPIRICAL FORMULA is SiCl 2Br 2. QUESTION 2 Using the empirical formula and the molecular weight, find the molecular formula for the compound. The MOLECULAR FORMULA is a simple multiple of the empirical formula based on the molecular weight. The molecular weight for this compound is given as g/mol. Empirical weight for SiCl 2Br 2 = 1 mol Si g g g g Si + mol Cl + mol Br = 258. SiCl 2Br 2/empi mol Si Cl mol Cl Br mol Br 8 rical unit To find the number of empirical units per molecule, divide the molecular weight by the empirical weight: Molecular Weight / Empirical Weight = g / g = Therefore, there is 1 SiCl 2Br 2 unit per molecule. MOLECULAR FORMULA = (SiCl 2Br 2) 1 = SiCl 2Br 2 Summary Table Element Grams Moles Mole Formula Ratio Ratio Empirical Formula Empirical Weight Number of Units Molecular Formula Si Cl Br SiCl 2Br SiCl 2Br 2

5 Finding Moles From Moles According to the following reaction, how many moles of water are necessary to form moles hydrochloric acid? chlorine (g) + water (l) hydrochloric acid (aq) + chloric acid (HClO 3) (aq) 1. Write a balanced chemical equation for the above reaction: 3 Cl 2 (g) + 3 H 2O (l) 5 HCl (aq) + HClO 3 (aq) 2. Determine the number of moles of water needed: GIVEN: mol HCl...WANTED: mol H 2O PER:...3 mol H 2O / 5 mol HCl PATH:... mol HCl mol H 2O mol HCl mol H 2O...= mol H2O 5 mol HCl Finding Grams From Moles C 3H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2O (g) GIVEN: mol CO 2...WANTED:... g O 2 PER:...5 mol O 2 / 3 mol CO g O 2 / mol O 2 PATH:... mol CO mol O g O mol CO mol O g O 2...= g O 2 3 mol CO 2 1 mol O 2

6 Finding The Limiting Reagent 1. Write a balanced chemical equation for the above reaction: H 2SO 4 (aq) + Zn(OH) 2 (s) ZnSO 4 (aq) + 2 H 2O (l) 2. Determine the number of moles of zinc sulfate that can be formed from the available quantity of sulfuric acid: mol ZnSO 4 = mol H 2SO 4 1 mol ZnSO 4 = mol ZnSO4 1 mol H2SO 4 3. Determine the number of moles of zinc sulfate that can be formed from the available quantity of zinc hydroxide. mol ZnSO 4 = mol Zn(OH) 2 1 mol ZnSO 4 = mol ZnSO4 1 mol Zn(OH)2 4. It is possible to form moles of zinc sulfate from the available zinc hydroxide, but only moles of zinc sulfate can be formed from the available sulfuric acid. The limiting reagent is sulfuric acid. The maximum amount of zinc sulfate that can be formed is moles.