 # Osukuuni Practice Questions

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 Osukuuni Practice Questions CHEM 1211 Percent Composition/Limiting reactants/Percent Yield 1. Calculate the percent composition for each of the following compounds: a) NO b) KClO4 c) NaH2PO4 d) C3H4O2 Divide element’s mass (i.e. atomic mass x number of atoms) by molar mass of compound and multiply by 100%. a) 46.68% N and 53.32% O b) 28.22% K, 25.59% Cl, and 46.19% O c) 19.16% Na, 1.68% H, 25.81% P, and 53.34% O d) 49.99% C, 5.61% H, and 44.40% O 2. An oxide of osmium (Os) is a pale yellow solid. If 2.89 g of the compound contains 2.16 g of osmium, what is the empirical formula? Mass of Os = 2.16 g Mass of O = 2.89 - 2.16 = 0.73 g Covert mass of each element to moles using atomic mass. Calculate mole ratio (divide each mole by the smaller) and round to the nearest integer. Write integers as subscripts. Empirical formula is OsO4 3. Two compounds have the same composition: 92.25% C and 7.75% H. a) Obtain the empirical formula corresponding to this composition. b) One of the compounds has a molecular mass of 52.03 amu; the other, of 78.05 amu. Obtain the molecular formulas of both compounds. a) Assume 100.0 g sample so that percentages represent grams of the elements. Covert grams of each element to moles using atomic mass. Calculate mole ratio (divide each mole by the smaller) and round to the nearest integer. Write integers as subscripts. Empirical formula is CH b) Divide molecular mass by empirical formula mass and round to the nearest integer. Molecular formulas are C4H4 and C6H6 4. Iron in the form of fine wire burns in oxygen to form iron(III) oxide. Fe(s) + 3O2(g) → 2Fe2O3(s) a) How many moles of O2 are needed to produce 3.91 mol Fe2O3? b) How many grams of Fe2O3 will be produced from 3.91 mol Fe? First make sure equation is balanced a) 3 mol O2 produces 2 mol Fe2O3 moles O2 = (3.91 mol Fe2O3) x (3 mol O2/2 mol Fe2O3) = 5.87 mol O2 b) 4 mol Fe produces 2 mol Fe2O3 moles Fe2O3 = (3.91 mol Fe) x (2 mol Fe2O3/4 mol Fe) = 1.96 mol Fe2O3 mass Fe2O3 = (1.96 mol Fe2O3) x (159.70 g Fe2O3/1 mol Fe2O3) = 313 g Fe2O3 5. Solutions of sodium hypochlorite, NaClO, are sold as bleach (such as Clorox). They are prepared by the reaction of chlorine with sodium hydroxide. 2NaOH(aq) + Cl2(g) → NaCl(aq) + NaClO(aq) + H2O(l) If a chemist has 1.58 mol of NaOH in solution and 1.78 mol of Cl2 gas available to react, which is the limiting reactant? How many moles of NaClO could be obtained? Make sure equation is balanced 2 mol NaOH produces 1 mol NaClO 1.58 mol NaOH produces (1.58 mol NaOH) x (1 mol NaClO/2 mol NaOH) = 0.790 mol NaClO 1 mol Cl2 produces 1 mol NaClO 1.78 mol Cl2 would produce 1.78 mol NaClO NaOH produces less amount of NaClO Implies NaOH is the limiting reactant and Cl2 is the excess reactant Use mol limiting reactant to determine mol product Moles NaClO = 0.790 mol NaClO 6. Methyl salicylate (oil of wintergreen) is prepared by heating salicylic acid, C7H6O3, with methanol, CH3OH. C7H6O3 + CH3OH → C8H8O3 + H2O In an experiment, 1.50 g of salicylic acid is reacted with 11.20 g of methanol. The yield of methyl salicylate, C8H8O3, is 1.31 g. What is the percent yield? Make sure equation is balanced Mol C7H6O3 = (1.50 g C7H6O3) x (1 mol C7H6O3/138.13 g C7H6O3) = 0.0109 mol 1 mol C7H6O3 produces 1 mol C8H8O3 Mol C8H8O3 = 0.0109 mol Mol CH3OH = (11.20 g CH3OH) x (1 mol CH3OH/32.05 g CH3OH) = 0.3495 mol 1 mol CH3OH produces 1 mol C8H8O3 Mol C8H8O3 = 0.3495 mol Implies limiting reactant is C7H6O3 (produces less mol product) Use mol limiting reactant to determine mol product Mol C8H8O3 = 0.0109 mol Mass C8H8O3 = (0.0109 mol C8H8O3) x (152.16 g C8H8O3/1 mol C8H8O3) = 1.66 g C8H8O3 % yield = (1.31 g/1.66 g) x 100% = 78.9 % 7. Lithium reacts with nitrogen according to the following reaction: 6Li + N2 → 2Li3N A chemist reacts 12.5 g of Li with 34.1 g of N2. Which is the limiting reactant? If the actual yield is 14.7 g Li3N, what is the percent yield? Make sure equation is balanced Mol Li = (12.5 g Li) x (1 mol Li/6.94 g Li) = 1.80 mol 6 mol Li produces 2 mol Li3N Mol Li3N = (1.80 mol Li) x (2 mol Li3N/6 mol Li) = 0.600 mol Li3N Mol N2 = (34.1 g N2) x (1 mol N2/28.02 g N2) = 1.22 mol 1 mol N2 produces 2 mol Li3N Mol Li3N = (1.22 mol N2) x (2 mol Li3N/1 mol N2) = 2.44 mol Li3N Li produces less mol product Implies Li is the limiting reactant and N2 is the excess reactant Use mol limiting reactant to determine mol product Mol Li3N = 0.600 mol Mass Li3N = (0.600 mol Li3N) x (34.83 g Li3N/1 mol Li3N) = 20.9 g Li3N = theoretical yield % yield = (14.7 g/20.9 g) x 100% = 70.3 % 8. What are the percentages by mass of Al, S, and O in Al(SO4)3? Solution is similar to question 1 above: 8.56% Al, 30.52% S, and 60.92% O 9. What is percent by mass of carbon in potassium hydrogen phthalate, KC8H5O4? Solution is similar to question 1 above: 47.04% C 10. What is the percent by mass of hydrogen in ammonium carbonate, (NH4)2CO3? Solution is similar to question 1 above: 8.41% H 11. Analysis of a sample of a covalent compound showed that it contained 14.4% hydrogen and 85.6% carbon by mass. What is the empirical formula for this compound? Assume 100.0 g sample so that percentages represent grams of the elements. Covert grams of each element to moles using atomic mass. Calculate mole ratio (divide each mole by the smaller) and round to the nearest integer. Write integers as subscripts. Empirical formula is CH2 12. What is the empirical formula for a compound containing 68.3% lead, 10.6% sulfur, and the remainder oxygen? Assume 100.0 g sample so that percentages represent grams of the elements. Covert grams of each element to moles using atomic mass. Calculate mole ratio (divide each mole by the smallest) and round to the nearest integer. Write integers as subscripts. Empirical formula is PbSO4 13. A compound contains carbon, hydrogen, and oxygen. Analysis of a sample shows that it contains 4.92% hydrogen and 26.18% oxygen by mass. What is the empirical formula for this compound? Assume 100.0 g sample so that percentages represent grams of the elements. Covert grams of each element to moles using atomic mass. Calculate mole ratio (divide each mole by the smaller) and round to the nearest integer. Mole ratio of C is 3.507 so multiply each ratio by 2 before rounding. Write integers as subscripts. Empirical formula is C7H6O2 14. Analysis of vitamin C indicates that it contains 40.9% carbon, 4.58% hydrogen, and 54.5% oxygen. What is the simplest formula for this compound? Assume 100.0 g sample so that percentages represent grams of the elements. Covert grams of each element to moles using atomic mass. Calculate mole ratio (divide each mole by the smaller) and round to the nearest integer. Mole ratio of H is 1.33 so multiply each ratio by 3 before rounding. Write integers as subscripts. Empirical formula is C3H4O3 15. Analysis of a compound containing nitrogen hydrogen, and oxygen was found to be 1.59% hydrogen and 76.21% oxygen. Determine the empirical formula for this compound. Assume 100.0 g sample so that percentages represent grams of the elements. Covert grams of each element to moles using atomic mass. Calculate mole ratio (divide each mole by the smaller) and round to the nearest integer. Write integers as subscripts. Empirical formula is HNO3Dostları ilə paylaş:

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