{"id":3512,"date":"2018-05-14T16:04:35","date_gmt":"2018-05-14T20:04:35","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/?post_type=chapter&#038;p=3512"},"modified":"2019-05-13T20:12:15","modified_gmt":"2019-05-14T00:12:15","slug":"5-6-end-of-chapter-problems","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/chapter\/5-6-end-of-chapter-problems\/","title":{"raw":"5.5 End of Chapter Problems","rendered":"5.5 End of Chapter Problems"},"content":{"raw":"1. How many gidgets are present in 4.23 mol of gidgets?\r\n\r\n2. How many mol are represented by 5.2 x 10<sup>72<\/sup>atoms of Fe?\r\n\r\n3. a) What is the mass (in grams) of 9.01 mol of Ca?\r\n\r\nb) How many mol are present in 21.2 g of Zn?\r\n\r\n4. a) How many mol are present in 10.421 g of Fe?\r\nb) What is the mass (in kilograms) of 905.25 mol of lead?\r\n\r\n5. a) How many atoms are present in a 25.0 g sample of Na?\r\nb) What is the mass (in grams) of 5.00 x 10<sup>25<\/sup>atoms of Cr?\r\n\r\n6.\u00a0How many atoms are present in 21.2 mg of Ag?\r\n\r\n7.\u00a0How many atoms are in a piece of gold measuring 1 cm x 2 cm x 0.5 cm?\r\n(Density of gold is 19.32 g\/mL)\r\n\r\n8.\u00a0The density of mercury (Hg) is 13.53 g\/mL. How many litres will 4.2 x 10<sup>21<\/sup>atoms of Hg occupy?\r\n\r\n9.\u00a0What is the molar mass of Cu(SO<sub>4<\/sub>)<sub>2<\/sub>?\r\n\r\n10.\u00a0What is the molar mass of:\u00a0\u00a0a)\u00a0 H<sub>2<\/sub>O\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 b)\u00a0Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 c)\u00a0C<sub>2<\/sub>H<sub>5<\/sub>OH\r\n\r\n11.\u00a0How many mol of O are present in 4.32 x 10<sup>32<\/sup> molecules of Cu(SO<sub>4<\/sub>)<sub>2<\/sub>?\r\n\r\n12.\u00a0How many atoms of H are present in 4.2 mol of H<sub>2<\/sub>O?\r\n\r\n13.\u00a0How many grams of S are present in 32.1 grams of Cu(SO<sub>4<\/sub>)<sub>2<\/sub>?\r\n\r\n14.\u00a0How many grams of Ni are present in 50.0 g of NiNO<sub>3<\/sub>?\r\n\r\n15. Complete the following conversions:\r\n\r\n9.024 x 10<sup>23\u00a0<\/sup>Hg atoms = ____ g Hg\r\n\r\n96.35 g NO = ____ molecules NO\r\n\r\n253.52 g CO<sub>2<\/sub>= ____ TOTAL atoms\r\n\r\n521.2 g (NH<sub>4<\/sub>)<sub>2<\/sub>Cr<sub>2<\/sub>O<sub>7<\/sub> \u00a0=\u00a0 _____ g H\r\n\r\n1.371 g C<sub>2<\/sub>H<sub>5<\/sub>OH = ____ atoms of H\r\n\r\n16. The hemoglobin content of blood is about 15.5 g\/100 mL of blood. The molar mass of hemoglobin is about 64,500 g\/mol and there are 4 iron atoms in a hemoglobin molecule. Approximately how many Fe atoms are present in 6 L of blood in a typical adult?\r\n\r\n17. List the conversion factors, or combination of conversion factors that you would use in the following problem situations (no calculations required):\r\n\r\na) You need to go from g H<sub>2<\/sub>O to mol H<sub>2<\/sub>O\r\nb) You need to go from mol H<sub>2<\/sub>SO<sub>4\u00a0<\/sub>to molecules H<sub>2<\/sub>SO<sub>4<\/sub>\r\nc)\u00a0 From g HCl to molecules HCl\r\nd)\u00a0 From mol KMnO<sub>4<\/sub> to atoms O\r\ne)\u00a0 From g KCN to g C\r\nf)\u00a0 From mL CH<sub>3<\/sub>OH\u00a0to mol CH<sub>3<\/sub>OH\r\ng)\u00a0 From molecules NaCl to atoms Cl\r\nh)\u00a0 From kg KBr to number of Br-ions\r\ni)\u00a0 From mol H<sub>2<\/sub>SO<sub>4<\/sub> to mL of H<sub>2<\/sub>SO<sub>4<\/sub>\r\nj)\u00a0 From mol MgO to g O\r\nk)\u00a0 From mm<sup>3<\/sup> NaOH to mol NaOH\r\nl)\u00a0 From molecules HCl to g H\r\nm)\u00a0 From mol C in C<sub>6<\/sub>H<sub>6<\/sub> to g C<sub>6<\/sub>H<sub>6<\/sub>\r\n\r\n18. For all the situations in #17, calculate the final answer based on a starting unit of 5.\r\n\r\nE.g.: \u00a0for (a)\u00a0 How many mol H<sub>2<\/sub>O are in 5 g of H<sub>2<\/sub>O?\r\nfor (b)\u00a0 How many molecules of H<sub>2<\/sub>SO<sub>4\u00a0<\/sub>are in 5 mol H<sub>2<\/sub>SO<sub>4<\/sub>?\u00a0\u00a0\u00a0etc\u2026\r\n\r\nNOTE:\u00a0 for f, i and k, the following densities are needed:\r\nCH<sub>3<\/sub>OH\u00a0 d = 0.791 g\/mL\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 H<sub>2<\/sub>SO<sub>4<\/sub>\u00a0\u00a0 d = 1.45 g\/mL \u00a0 \u00a0 \u00a0 \u00a0NaOH\u00a0\u00a0 d = 1.22 g\/mL\r\n\r\n19. Calculate the following:\r\na) Total number of ions in 38.1 g of CaF<sub>2\u00a0<\/sub>(hint\u2026 how many ions TOTAL are in one molecule of CaF<sub>2<\/sub>?)\r\nb) Mass in g of 2.04 x 10<sup>21<\/sup><sub>\u00a0<\/sub>molecules of N<sub>2<\/sub>O<sub>5<\/sub>\r\nc)\u00a0 Mass in mg of 3.58 mol CuCl<sub>2<\/sub>\r\nd)\u00a0 Mass in g of 9.64 x 10<sup>24<\/sup><sub>\u00a0<\/sub>molecules of Cl<sub>2<\/sub>O<sub>7<\/sub>\r\n\r\n20. Oxygen is required for metabolic combustion of foods. Calculate the number of atoms in 38.0 g of oxygen gas, the amount absorbed from the lungs at rest in about 15 minutes.\r\n\r\n21. An imperial quart of oil is spilled on a lake. If the molecules were to spread out in an film one molecule deep, what would be the area of the oil slick, in square miles? The density of oil = 0.8g\/mL ; area covered by one molecule = 0.5 nm<sup>2<\/sup>. Although oil is really a mixture of compounds of C &amp; H, assume the oil is simply C<sub>16<\/sub>H<sub>34<\/sub> and 1 qt = 1.1L and 1 mile = 1.6 km.\r\n\r\n22. Why would it be necessary to clarify what you mean when you say \u201c1 mole of nitrogen\u201d or \u201c1 mole of hydrogen\u201d? Why is this clarification not necessary when referring to \u201c1 mole of lead\u201d or \u201c1 mole of water\u201d?\r\n\r\n23. How many magnesium ions and how many nitride ions are in 4.75 mol of Mg<sub>3<\/sub>N<sub>2<\/sub>?\r\n\r\n24. What mass of O<sub>2<\/sub> contains the same number of moles of molecules as 52.0 g of N<sub>2<\/sub>?\r\n\r\n25. What mass of sodium hydroxide (NaOH) contains the same number of moles as 126 g of nitric acid (HNO<sub>3<\/sub>)?\r\n\r\n26. Ringer\u2019s lactate is an aqueous (in water) physiological solution used for intraveneous fluid therapy. A 1.00\u00a0L sample of the solution contains 5.96 g of NaCl, 3.1 g of NaC<sub>3<\/sub>H<sub>5<\/sub>O<sub>3<\/sub>(sodium lactate), 0.3 g of KCl and 0.2 g CaCl<sub>2<\/sub>.\r\n\r\na) How many moles of each compound are in 1.00 L of solution?\r\n\r\nb) The ingredients dissolve to form Na<sup>+<\/sup>, K<sup>+<\/sup>, Ca<sup>2+<\/sup>, Cl<sup>-<\/sup> and C<sub>3<\/sub>H<sub>5<\/sub>O<sub>3<\/sub><sup>-<\/sup>\u00a0ions. \u00a0How many moles of each ion are in 1.00 L of solution. \u00a0Note: there are several sources of Na<sup>+<\/sup>and Cl<sup>-<\/sup>.\r\n\r\n27. The density of liquid benzene, C<sub>6<\/sub>H<sub>6\u00a0<\/sub>is 0.879 g\/mL at 15<sup>o<\/sup>C. What is the volume in milliliters of 1.00 mol of benzene at this temperature?\r\n\r\n28. A sample of 0.370 mol of a metal oxide (M<sub>2<\/sub>O<sub>3<\/sub>) weighs 55.4 g.\r\na) How many moles of O are in the sample?\r\nb) How many grams of M are in the sample?\r\nc)\u00a0 What element is represented by M?\r\n\r\n29. a) How many grams of H<sub>2<\/sub>S are there in 0.400 moles of H<sub>2<\/sub>S?\r\nb) How many grams of hydrogen and sulfur are contained in 0.400 moles of H<sub>2<\/sub>S?\r\nc)\u00a0 How many molecules of H<sub>2<\/sub>S are contained in 0.400 moles of H<sub>2<\/sub>S?\r\nd)\u00a0 How many atoms of H and S are contained in 0.400 moles of H<sub>2<\/sub>S?\r\n\r\n30. a) How many moles of SO<sub>2<\/sub> are represented by 9.54 g of SO<sub>2<\/sub>?\r\nb) How many atoms of oxygen does this represent?\r\n\r\n31. What is the average mass (in grams) of one oxygen atom?\r\n\r\n32. What is the mass in grams of one molecule of CH<sub>3<\/sub>OH?\r\n\r\n33. The density of gold is approx. 2 x 10<sup>1<\/sup> g\/cm<sup>3<\/sup>. What is the volume of one gold atom?\r\n\r\n34. How many moles are there in one atom?\r\n\r\n35. What is the mass percent of each atom in (NH<sub>4<\/sub>)<sub>2<\/sub>Cr<sub>2<\/sub>O<sub>7<\/sub>?\r\n\r\n36. Calculate the following mass percents:\r\na) Mass % of H in NH<sub>4<\/sub>HCO<sub>3<\/sub>\u00a0 \u00a0 \u00a0 \u00a0b)\u00a0Mass % of Mn in KMnO<sub>4<\/sub>\r\n\r\n37. Calculate the following mass fractions:\r\na) Mass fraction of Cl in CaCl<sub>2<\/sub>\u00a0 \u00a0 \u00a0b)\u00a0Mass fraction of P in P<sub>4<\/sub>O<sub>7<\/sub>\r\n\r\n38. The effectiveness of nitrogen fertilizer is determined mainly by its mass % N. Rank the following in terms of their effectiveness: KNO<sub>3<\/sub>; NH<sub>4<\/sub>NO<sub>3<\/sub>; Co(NH<sub>2<\/sub>)<sub>2<\/sub>; NH<sub>4<\/sub>SO<sub>4<\/sub>. Show your calculations.\r\n\r\n39. A compound of iodine and cesium contains 63.94 g of metal and 61.06 g of nonmetal. How many grams of cesium are in 38.77 g of the compound? How many grams of iodine?\r\n\r\n40. What is the empirical formula and the empirical formula molar mass of each of the following: \u00a0a) C<sub>2<\/sub>H<sub>4<\/sub>\u00a0 \u00a0b) C<sub>2<\/sub>H<sub>6<\/sub>O<sub>2<\/sub>\u00a0 \u00a0c) N<sub>2<\/sub>O<sub>5<\/sub>\u00a0 \u00a0 d) Ba<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>\u00a0 \u00a0 \u00a0e) Te<sub>4<\/sub>I<sub>16<\/sub>\r\n\r\n41. What is the molecular formula of each of the following:\r\na) Empirical formula = CH<sub>2<\/sub>; molecular molar mass = 42.08 g\/mol\r\nb)\u00a0 EF = NH<sub>2<\/sub>; MMM = 32.05 g\/mol\r\nc)\u00a0 EF = NO<sub>2<\/sub>; MMM = 92.02 g\/mol\r\nd)\u00a0 EF = CHN; MMM = 135.14 g\/mol\r\ne)\u00a0 EF = CH;\u00a0 MMM = 78.11 g\/mol\r\nf)\u00a0 EF = C<sub>3<\/sub>H<sub>6<\/sub>O<sub>2<\/sub>; MMM = 74.08 g\/mol\r\ng)\u00a0 EF = C<sub>7<\/sub>H<sub>4<\/sub>O<sub>2<\/sub>; MMM = 240.02 g\/mol\r\n\r\n42. A 3.450 g sample of nitrogen reacts with 1.970 g of oxygen to form a compound. Determine the empirical formula for this compound.\r\n\r\n43. A compound containing carbon, hydrogen and oxygen is found to be 40.00% carbon and 6.700% hydrogen by mass. The molar mass of the compound is between 115 g\/mol and 125 g\/mol. Determine the empirical formula and molecular formula of this compound.\r\n\r\n44. Determine the empirical formula for the following situations:\r\na) 0.0630 mol of chlorine atom combined with 0.220 mol of oxygen atoms\r\nb) 2.45 g silicon combined with 12.4 g of chlorine\r\nc)\u00a0 a compound with 27.3% carbon and 72.7% oxygen by mass\r\nd)\u00a0 a hydrocarbon (containing only C and H) which has 79.9% by mass C\r\n\r\n45. A compound containing only silicon and chlorine contains 79.1% chlorine by mass and has a molar mass of 269 g\/mol. What is the molecular formula?\r\n\r\n46. A sample of 0.600 mole of metal reacts completely with fluorine to form 46.8 g of MF<sub>2<\/sub>.\r\na) How moles of F are in the sample of MF<sub>2<\/sub> that formed?\r\nb) How many grams of M are in this sample of MF<sub>2<\/sub>?\r\nc)\u00a0 What element is represented by M?\r\n\r\n47. A sample of nicotine contains 6.16 mmol of C, 8.56 mmol of H and 1.23 mmol of N. What is its empirical formula? (note: 1000 mmol = 1 mol)\r\n\r\n48. Cortisol (molar mass = 362.47 g\/mol), one of the major steroid hormones, is a key factor in the synthesis of protein. Its profound effect on the reduction of inflammation explains its use in the treatment of rheumatoid arthritis. Cortisol is 69.6% C, 8.34% H and 22.1% O by mass. What is its molecular formula?\r\n\r\n49. 3.00g of molybdenum (Mo) combines with sulfur to produce 5.50 g of a compound. What is the empirical formula of this compound?\r\n\r\n50. A 5.70 g sample of an iron oxide compound was heated in a stream of hydrogen gas, which reacted with the oxygen in the compound to form water. The water vapor was carried away in the stream of gas, leaving only pure metallic iron weighing 4.20g. Calculate the simplest possible formula for the original iron oxide.\r\n\r\n51. When 2.31g of a carboxylic acid compound is burned in O<sub>2<\/sub>, the only products are 1.33g of H<sub>2<\/sub>O and 3.38g of CO<sub>2<\/sub>. Calculate the empirical (simplest possible) formula for this compound.\r\n\r\n52. A compound was analyzed and found to have the following percentage composition:\u00a0aluminum = 15.7%, sulfur = 28.11%, oxygen = 56.12%. The molar mass of the compound is known to be approximately 684 g\/mol. What is the molecular formula?\r\n\r\n&nbsp;\r\n<h2><strong>Answers<\/strong><\/h2>\r\n1. 2.55 x 10<sup>24\u00a0<\/sup>gidgets\r\n\r\n2.\u00a08.6 x 10<sup>48\u00a0<\/sup>mol of Fe\r\n\r\n3. a) 361 g of Ca \u00a0 \u00a0 \u00a0 \u00a0 b) 0.324 mol of Zn\r\n\r\n4.\u00a0<span> <\/span>a) 0.18660 mol of Fe<span>\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span>b) 187.57 kg of Pb\r\n\r\n5. a)\u00a06.55 x 10<sup>23\u00a0<\/sup>atoms of Na \u00a0 \u00a0b)\u00a04.32 x 10<sup>3<\/sup> g of Cr\r\n\r\n6.\u00a01.18 x 10<sup>20\u00a0<\/sup>atoms\r\n\r\n7.\u00a06 x 10<sup>22<\/sup> Au atoms\r\n\r\n8.\u00a01.0 x 10<sup>-4\u00a0<\/sup>L\r\n\r\n9.\u00a0255.673 g\/mol of Cu(SO<sub>4<\/sub>)<sub>2<\/sub>\r\n\r\n10.\u00a0a) 18.0152 g\/mol<span>\u00a0\u00a0\u00a0\u00a0 <\/span>b)<span>\u00a0<\/span>310.177 g\/mol<span>\u00a0\u00a0\u00a0\u00a0 <\/span>c)<span>\u00a0 <\/span>46.069 g\/mol\r\n\r\n11.\u00a05.74 x 10<sup>9\u00a0<\/sup>mol of O\r\n\r\n12.\u00a05.1 x 10<sup>24\u00a0<\/sup>atoms of H\r\n\r\n13.\u00a08.05 g of S\r\n\r\n14.\u00a024.3 g of Ni\r\n<p class=\"Answers\">15.<span>\u00a0\u00a0a) 300.6 g Hg\r\nb) 1.934 x 10<sup>24\u00a0<\/sup>molecules of NO\r\nc) 1.0407 x 10<sup>25\u00a0<\/sup>atoms\r\nd) 16.67 g H\r\ne) 1.075 x 10<sup>23\u00a0<\/sup>atoms <\/span><\/p>\r\n<p class=\"Answers\">16. \u00a03 x 10<sup>22\u00a0<\/sup>iron atoms<\/p>\r\n<p class=\"Answers\">17.<span>\u00a0\u00a0 <\/span>a) g H<sub>2<\/sub>O\u00a0$latex \\longrightarrow$ mol H<sub>2<\/sub>O using molar mass of H<sub>2<\/sub>O<\/p>\r\n<p class=\"AnswersSub\">b) mol H<sub>2<\/sub>SO<sub>4\u00a0<\/sub>$latex \\longrightarrow$\u00a0molecules H<sub>2<\/sub>SO<sub>4\u00a0<\/sub>using Avogadro\u2019s number<\/p>\r\n<p class=\"AnswersSub\">c) g HCl $latex \\longrightarrow$ mol HCl using molar mass HCl $latex \\longrightarrow$ molecules HCl using Avogadro\u2019s number<\/p>\r\n<p class=\"AnswersSub\">d) mol KMnO<sub>4\u00a0<\/sub>$latex \\longrightarrow$<span>\u00a0<\/span>mol O using chemical formula $latex \\longrightarrow$ atoms O using Avogadro\u2019s number<\/p>\r\n<p class=\"AnswersSub\">e) g KCN $latex \\longrightarrow$ mol KCN using molar mass of KCN $latex \\longrightarrow$ mol C using chemical formula $latex \\longrightarrow$ g C using molar mass of C<\/p>\r\n<p class=\"AnswersSub\">f) mL CH<sub>3<\/sub>OH $latex \\longrightarrow$ g CH<sub>3<\/sub>OH using density $latex \\longrightarrow$ mol CH<sub>3<\/sub>OH using molar mass of CH<sub>3<\/sub>OH<\/p>\r\n<p class=\"AnswersSub\">g) molecules NaCl $latex \\longrightarrow$ atoms Cl using formula ratio\r\n<span>\u00a0<\/span>OR\r\nmolecules NaCl $latex \\longrightarrow$ moles NaCl using Avogadro\u2019s number $latex \\longrightarrow$ moles Cl using the chemical formula $latex \\longrightarrow$ atoms Cl using Avogadro\u2019s number<\/p>\r\n<p class=\"AnswersSub\">h) kg KBr $latex \\longrightarrow$ g KBr using metric conversions $latex \\longrightarrow$ mol KBr using the molar mass of KBr $latex \\longrightarrow$ moles Br<sup>-\u00a0<\/sup>ions using chemical formula $latex \\longrightarrow$ number of Br<sup>-\u00a0<\/sup>ions using Avogadro\u2019s number<\/p>\r\n<p class=\"AnswersSub\">i) mol H<sub>2<\/sub>SO<sub>4<\/sub>\u00a0$latex \\longrightarrow$<span>\u00a0<\/span>g H<sub>2<\/sub>SO<sub>4\u00a0<\/sub>using molar mass of H<sub>2<\/sub>SO<sub>4\u00a0<\/sub>$latex \\longrightarrow$<span>\u00a0<\/span>mL\u00a0H<sub>2<\/sub>SO<sub>4\u00a0<\/sub>using density of H<sub>2<\/sub>SO<sub>4<\/sub><\/p>\r\n<p class=\"AnswersSub\">j) mol MgO $latex \\longrightarrow$<span>\u00a0<\/span>mol O using the chemical formula $latex \\longrightarrow$<span>\u00a0<\/span>g O using molar mass of O<\/p>\r\n<p class=\"AnswersSub\">k) mm<sup>3\u00a0<\/sup>NaOH $latex \\longrightarrow$<span>\u00a0<\/span>cm<sup>3\u00a0<\/sup>NaOH using metric conversions $latex \\longrightarrow$ g\u00a0NaOH using the density $latex \\longrightarrow$ mol NaOH using the molar mass of NaOH<\/p>\r\n<p class=\"AnswersSub\">l) molecules HCl $latex \\longrightarrow$ moles HCl using Avogadro\u2019s number $latex \\longrightarrow$ moles H using the chemical formula $latex \\longrightarrow$ g H using the molar mass of H<\/p>\r\n<p class=\"AnswersSub\">m) mol C $latex \\longrightarrow$ mol C<sub>6<\/sub>H<sub>6\u00a0<\/sub>using the chemical formula $latex \\longrightarrow$ g C<sub>6<\/sub>H<sub>6\u00a0<\/sub>using the molar mass of C<sub>6<\/sub>H<sub>6<\/sub><\/p>\r\n<p class=\"Answers\">18.<span>\u00a0\u00a0 <\/span>a) 0.3 mol \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0b) 3 x 10<sup>24\u00a0<\/sup>molecules \u00a0 \u00a0 \u00a0 \u00a0c) 8 x 10<sup>22\u00a0<\/sup>molecules\r\nd) 1 x 10<sup>25\u00a0<\/sup>atoms \u00a0 \u00a0 \u00a0 \u00a0 \u00a0e) 0.9 g \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0f) 0.1 mol\r\ng) 5 ions (or 5 atoms) \u00a0 \u00a0h) 3 x 10<sup>25\u00a0<\/sup>ions \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0i) 3 x 10<sup>2\u00a0<\/sup>mL\r\nj) 8 x 10<sup>1\u00a0<\/sup>g \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0k) 2 x 10<sup>-4\u00a0<\/sup>mol \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0l) 8 x 10<sup>-24 <\/sup>g\r\nm) 7 x 10<sup>1\u00a0<\/sup>g<\/p>\r\n<p class=\"Answers\">19.<span>\u00a0\u00a0 <\/span>a) 8.82 x 10<sup>23\u00a0<\/sup>ions \u00a0 \u00a0 \u00a0 \u00a0\u00a0b) 0.366 g \u00a0 \u00a0 \u00a0 \u00a0c) 4.81 x 10<sup>5\u00a0<\/sup>mg\r\nd)<span>\u00a0 <\/span>2.93 x 10<sup>3\u00a0<\/sup>g<\/p>\r\n<p class=\"Answers\">20.<span>\u00a0\u00a0 <\/span>1.43 x 10<sup>24\u00a0<\/sup>atoms of O (7.15 x 10<sup>23\u00a0<\/sup>molecules O<sub>2<\/sub>)<\/p>\r\n<p class=\"Answers\">21.<span>\u00a0 <\/span>0.5 mile<sup>2<\/sup><\/p>\r\n<p class=\"Answers\">22.<span>\u00a0 <\/span>Because 1 mole of nitrogen may be 1 mole of N or 1 mole of N<sub>2<\/sub>, etc. Nitrogen and hydrogen (but not lead and water) are commonly found as diatomic molecules.<\/p>\r\n<p class=\"Answers\">23<span>\u00a0 \u00a0<\/span>8.58 x 10<sup>24<\/sup>Mg<sup>2+<\/sup>ions and 5.72 x 10<sup>24<\/sup>N<sup>3-<\/sup>ions<\/p>\r\n<p class=\"Answers\">24.<span>\u00a0 <\/span>59.4 g of O<sub>2\u00a0<\/sub>(note: the nitrogen is given as N<sub>2<\/sub>)<\/p>\r\n<p class=\"Answers\"><span lang=\"ES-MX\">25.<span>\u00a0 <\/span>80.0 g of NaOH<\/span><\/p>\r\n<p class=\"Answers\"><span lang=\"ES-MX\">26.<span>\u00a0 <\/span>a) 0.102 mol NaCl, 0.028 mol NaC<sub>3<\/sub>H<sub>5<\/sub>O<sub>3<\/sub>,\u00a00.004mol KCl\r\n<span>\u00a0\u00a0\u00a0 <\/span>and 0.002 mol CaCl<sub>2<\/sub>\r\nb) Na<sup>+\u00a0<\/sup>= 0.130 mol,<span>\u00a0 <\/span>K<sup>+\u00a0<\/sup>= 0.004 mol, Ca<sup>2+\u00a0<\/sup>= 0.002,\r\n<span>\u00a0\u00a0\u00a0 <\/span>Cl<sup>-\u00a0<\/sup>= 0.110 mol; C<sub>3<\/sub>H<sub>5<\/sub>O<sub>3<\/sub><sup>-\u00a0<\/sup>= 0.028 mol<\/span><\/p>\r\n<p class=\"Answers\"><span lang=\"ES-MX\">27.<span>\u00a0 <\/span>88.9 mL<\/span><\/p>\r\n<p class=\"Answers\"><span lang=\"ES-MX\">28.<span>\u00a0 <\/span>a) 1.11 mol O<span>\u00a0\u00a0\u00a0 <\/span>b) 37.6 g M<span>\u00a0\u00a0\u00a0 <\/span>c) V<\/span><\/p>\r\n<p class=\"Answers\">29.<span>\u00a0 <\/span>a) 13.6 g H<sub>2<\/sub>S \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0b) 0.806 g H, 12.8 g S\r\nc) 2.41 x 10<sup>23\u00a0<\/sup>molecules H<sub>2<\/sub>S \u00a0 \u00a0 \u00a0 \u00a0d)<span>\u00a0 <\/span>4.82 x 10<sup>23<\/sup>H atoms and 2.41 x 10<sup>23\u00a0<\/sup>S atoms<\/p>\r\n<p class=\"Answers\">30.<span>\u00a0 <\/span>a) 0.149 mol SO<sub>2 \u00a0 \u00a0 \u00a0\u00a0<\/sub>b) 1.79 x 10<sup>23<\/sup>atoms O<\/p>\r\n<p class=\"Answers\">31. 2.65682 x 10<sup>-23\u00a0<\/sup>g<\/p>\r\n<p class=\"Answers\">32.<span>\u00a0 <\/span>5.3208 x 10<sup>-23\u00a0<\/sup>g<\/p>\r\n<p class=\"Answers\">33.<span>\u00a0 <\/span>2 x 10<sup>-23\u00a0<\/sup>cm<sup>3<\/sup><\/p>\r\n<p class=\"Answers\">34.<span>\u00a0 <\/span>1.6605388 x 10<sup>-24\u00a0<\/sup>mol<\/p>\r\n<p class=\"Answers\">35.<span>\u00a0 <\/span>N= 11.114%, H= 3.1990%, Cr= 41.256%, O= 44.431%<\/p>\r\n<p class=\"Answers\">36.<span>\u00a0 <\/span>a) 6.3749%<span>\u00a0\u00a0\u00a0\u00a0 <\/span>b) 34.7634%<\/p>\r\n<p class=\"Answers\">37.<span>\u00a0 <\/span>a)<span>\u00a0 <\/span>0.638883<span>\u00a0\u00a0\u00a0\u00a0<\/span>b) 0.525222<\/p>\r\n<p class=\"Answers\">38. NH<sub>4<\/sub>NO<sub>3<\/sub>&gt; Co(NH<sub>2<\/sub>)<sub>2<\/sub>&gt; KNO<sub>3 <\/sub>&gt; NH<sub>4<\/sub>SO<sub>4<\/sub><\/p>\r\n<p class=\"Answers\"><span lang=\"ES-MX\">39. 19.83 g Cs, 18.94 g I<\/span><\/p>\r\n<p class=\"Answers\"><span lang=\"ES-MX\">40.<span>\u00a0 <\/span>a) C<sub>2<\/sub>H<sub>4<\/sub>= CH<sub>2<\/sub>; 14.027 g\/mol\r\nb) C<sub>2<\/sub>H<sub>6<\/sub>O<sub>2<\/sub>= CH<sub>3<\/sub>O; 31.034 g\/mol\r\nc) N<sub>2<\/sub>O<sub>5<\/sub>= N<sub>2<\/sub>O<sub>5<\/sub>; 108.0104 g\/mol\r\nd) Ba<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>= Ba<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>; 601.924 g\/mol\r\ne) Te<sub>4<\/sub>I<sub>16<\/sub>= TeI<sub>4<\/sub>; 635.22 g\/mol<\/span><\/p>\r\n<p class=\"Answers\"><span lang=\"ES-MX\">41.<span>\u00a0 <\/span>a) C<sub>3<\/sub>H<sub>6 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0<\/sub>b) N<sub>2<\/sub>H<sub>4 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0<\/sub>c) N<sub>2<\/sub>O<sub>4\r\n<\/sub>d) C<sub>5<\/sub>H<sub>5<\/sub>N<sub>5 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<\/sub>e) C<sub>6<\/sub>H<sub>6 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<\/sub>f) C<sub>3<\/sub>H<sub>6<\/sub>O<sub>2\r\n<\/sub>g) C<sub>14<\/sub>H<sub>8<\/sub>O<sub>4<\/sub><\/span><\/p>\r\n<p class=\"Answers\"><span lang=\"ES-MX\">42. N<sub>2<\/sub>O<\/span><\/p>\r\n<p class=\"Answers\"><span lang=\"ES-MX\">43.<span>\u00a0 <\/span>EF = CH<sub>2<\/sub>O, MF = C<sub>4<\/sub>H<sub>8<\/sub>O<sub>4<\/sub><\/span><\/p>\r\n<p class=\"Answers\"><span lang=\"ES-MX\">44.<span>\u00a0 <\/span>a) Cl<sub>2<\/sub>O<sub>7 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<\/sub>b) SiCl<sub>4 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<\/sub>c) CO<sub>2 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<\/sub>d) CH<sub>3<\/sub><\/span><\/p>\r\n<p class=\"Answers\"><span lang=\"ES-MX\">45.<span>\u00a0 <\/span>Si<sub>2<\/sub>Cl<sub>6<\/sub><\/span><\/p>\r\n<p class=\"Answers\"><span lang=\"ES-MX\">46.<span>\u00a0 <\/span>a) 1.20 mol F<span>\u00a0\u00a0\u00a0\u00a0 <\/span>b) 24.0 g M<span>\u00a0\u00a0\u00a0\u00a0 <\/span>c) Ca<\/span><\/p>\r\n<p class=\"Answers\">47.<span>\u00a0 <\/span>C<sub>5<\/sub>H<sub>7<\/sub>N<\/p>\r\n<p class=\"Answers\">48.<span>\u00a0 <\/span>C<sub>21<\/sub>H<sub>30<\/sub>O<sub>5<\/sub><\/p>\r\n<p class=\"Answers\">49.<span>\u00a0 <\/span>Mo<sub>2<\/sub>S<sub>5<\/sub><span>\u00a0 <\/span><\/p>\r\n<p class=\"Answers\">50.<span>\u00a0 <\/span>Fe<sub>4<\/sub>O<sub>5<\/sub>(Fe<sub>8<\/sub>O<sub>10<\/sub>etc. are also possible, but not the <i>simplest<\/i>)<\/p>\r\n<p class=\"Answers\">51.<span>\u00a0 <\/span>CH<sub>2<\/sub>O<span>\u00a0\u00a0 <\/span>(Note that, based on the products, the unknown compound could only contain C, H and O, and note that all of the C in the original compound ends up in the CO<sub>2<\/sub>and all the H ends up in the H<sub>2<\/sub>O.)<\/p>\r\n<p class=\"Answers\"><span lang=\"PT-BR\">52.<span>\u00a0 <\/span>Al<sub>4<\/sub>S<sub>6<\/sub>O<sub>24<\/sub><\/span><\/p>\r\n&nbsp;\r\n\r\n&nbsp;","rendered":"<p>1. How many gidgets are present in 4.23 mol of gidgets?<\/p>\n<p>2. How many mol are represented by 5.2 x 10<sup>72<\/sup>atoms of Fe?<\/p>\n<p>3. a) What is the mass (in grams) of 9.01 mol of Ca?<\/p>\n<p>b) How many mol are present in 21.2 g of Zn?<\/p>\n<p>4. a) How many mol are present in 10.421 g of Fe?<br \/>\nb) What is the mass (in kilograms) of 905.25 mol of lead?<\/p>\n<p>5. a) How many atoms are present in a 25.0 g sample of Na?<br \/>\nb) What is the mass (in grams) of 5.00 x 10<sup>25<\/sup>atoms of Cr?<\/p>\n<p>6.\u00a0How many atoms are present in 21.2 mg of Ag?<\/p>\n<p>7.\u00a0How many atoms are in a piece of gold measuring 1 cm x 2 cm x 0.5 cm?<br \/>\n(Density of gold is 19.32 g\/mL)<\/p>\n<p>8.\u00a0The density of mercury (Hg) is 13.53 g\/mL. How many litres will 4.2 x 10<sup>21<\/sup>atoms of Hg occupy?<\/p>\n<p>9.\u00a0What is the molar mass of Cu(SO<sub>4<\/sub>)<sub>2<\/sub>?<\/p>\n<p>10.\u00a0What is the molar mass of:\u00a0\u00a0a)\u00a0 H<sub>2<\/sub>O\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 b)\u00a0Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 c)\u00a0C<sub>2<\/sub>H<sub>5<\/sub>OH<\/p>\n<p>11.\u00a0How many mol of O are present in 4.32 x 10<sup>32<\/sup> molecules of Cu(SO<sub>4<\/sub>)<sub>2<\/sub>?<\/p>\n<p>12.\u00a0How many atoms of H are present in 4.2 mol of H<sub>2<\/sub>O?<\/p>\n<p>13.\u00a0How many grams of S are present in 32.1 grams of Cu(SO<sub>4<\/sub>)<sub>2<\/sub>?<\/p>\n<p>14.\u00a0How many grams of Ni are present in 50.0 g of NiNO<sub>3<\/sub>?<\/p>\n<p>15. Complete the following conversions:<\/p>\n<p>9.024 x 10<sup>23\u00a0<\/sup>Hg atoms = ____ g Hg<\/p>\n<p>96.35 g NO = ____ molecules NO<\/p>\n<p>253.52 g CO<sub>2<\/sub>= ____ TOTAL atoms<\/p>\n<p>521.2 g (NH<sub>4<\/sub>)<sub>2<\/sub>Cr<sub>2<\/sub>O<sub>7<\/sub> \u00a0=\u00a0 _____ g H<\/p>\n<p>1.371 g C<sub>2<\/sub>H<sub>5<\/sub>OH = ____ atoms of H<\/p>\n<p>16. The hemoglobin content of blood is about 15.5 g\/100 mL of blood. The molar mass of hemoglobin is about 64,500 g\/mol and there are 4 iron atoms in a hemoglobin molecule. Approximately how many Fe atoms are present in 6 L of blood in a typical adult?<\/p>\n<p>17. List the conversion factors, or combination of conversion factors that you would use in the following problem situations (no calculations required):<\/p>\n<p>a) You need to go from g H<sub>2<\/sub>O to mol H<sub>2<\/sub>O<br \/>\nb) You need to go from mol H<sub>2<\/sub>SO<sub>4\u00a0<\/sub>to molecules H<sub>2<\/sub>SO<sub>4<\/sub><br \/>\nc)\u00a0 From g HCl to molecules HCl<br \/>\nd)\u00a0 From mol KMnO<sub>4<\/sub> to atoms O<br \/>\ne)\u00a0 From g KCN to g C<br \/>\nf)\u00a0 From mL CH<sub>3<\/sub>OH\u00a0to mol CH<sub>3<\/sub>OH<br \/>\ng)\u00a0 From molecules NaCl to atoms Cl<br \/>\nh)\u00a0 From kg KBr to number of Br-ions<br \/>\ni)\u00a0 From mol H<sub>2<\/sub>SO<sub>4<\/sub> to mL of H<sub>2<\/sub>SO<sub>4<\/sub><br \/>\nj)\u00a0 From mol MgO to g O<br \/>\nk)\u00a0 From mm<sup>3<\/sup> NaOH to mol NaOH<br \/>\nl)\u00a0 From molecules HCl to g H<br \/>\nm)\u00a0 From mol C in C<sub>6<\/sub>H<sub>6<\/sub> to g C<sub>6<\/sub>H<sub>6<\/sub><\/p>\n<p>18. For all the situations in #17, calculate the final answer based on a starting unit of 5.<\/p>\n<p>E.g.: \u00a0for (a)\u00a0 How many mol H<sub>2<\/sub>O are in 5 g of H<sub>2<\/sub>O?<br \/>\nfor (b)\u00a0 How many molecules of H<sub>2<\/sub>SO<sub>4\u00a0<\/sub>are in 5 mol H<sub>2<\/sub>SO<sub>4<\/sub>?\u00a0\u00a0\u00a0etc\u2026<\/p>\n<p>NOTE:\u00a0 for f, i and k, the following densities are needed:<br \/>\nCH<sub>3<\/sub>OH\u00a0 d = 0.791 g\/mL\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 H<sub>2<\/sub>SO<sub>4<\/sub>\u00a0\u00a0 d = 1.45 g\/mL \u00a0 \u00a0 \u00a0 \u00a0NaOH\u00a0\u00a0 d = 1.22 g\/mL<\/p>\n<p>19. Calculate the following:<br \/>\na) Total number of ions in 38.1 g of CaF<sub>2\u00a0<\/sub>(hint\u2026 how many ions TOTAL are in one molecule of CaF<sub>2<\/sub>?)<br \/>\nb) Mass in g of 2.04 x 10<sup>21<\/sup><sub>\u00a0<\/sub>molecules of N<sub>2<\/sub>O<sub>5<\/sub><br \/>\nc)\u00a0 Mass in mg of 3.58 mol CuCl<sub>2<\/sub><br \/>\nd)\u00a0 Mass in g of 9.64 x 10<sup>24<\/sup><sub>\u00a0<\/sub>molecules of Cl<sub>2<\/sub>O<sub>7<\/sub><\/p>\n<p>20. Oxygen is required for metabolic combustion of foods. Calculate the number of atoms in 38.0 g of oxygen gas, the amount absorbed from the lungs at rest in about 15 minutes.<\/p>\n<p>21. An imperial quart of oil is spilled on a lake. If the molecules were to spread out in an film one molecule deep, what would be the area of the oil slick, in square miles? The density of oil = 0.8g\/mL ; area covered by one molecule = 0.5 nm<sup>2<\/sup>. Although oil is really a mixture of compounds of C &amp; H, assume the oil is simply C<sub>16<\/sub>H<sub>34<\/sub> and 1 qt = 1.1L and 1 mile = 1.6 km.<\/p>\n<p>22. Why would it be necessary to clarify what you mean when you say \u201c1 mole of nitrogen\u201d or \u201c1 mole of hydrogen\u201d? Why is this clarification not necessary when referring to \u201c1 mole of lead\u201d or \u201c1 mole of water\u201d?<\/p>\n<p>23. How many magnesium ions and how many nitride ions are in 4.75 mol of Mg<sub>3<\/sub>N<sub>2<\/sub>?<\/p>\n<p>24. What mass of O<sub>2<\/sub> contains the same number of moles of molecules as 52.0 g of N<sub>2<\/sub>?<\/p>\n<p>25. What mass of sodium hydroxide (NaOH) contains the same number of moles as 126 g of nitric acid (HNO<sub>3<\/sub>)?<\/p>\n<p>26. Ringer\u2019s lactate is an aqueous (in water) physiological solution used for intraveneous fluid therapy. A 1.00\u00a0L sample of the solution contains 5.96 g of NaCl, 3.1 g of NaC<sub>3<\/sub>H<sub>5<\/sub>O<sub>3<\/sub>(sodium lactate), 0.3 g of KCl and 0.2 g CaCl<sub>2<\/sub>.<\/p>\n<p>a) How many moles of each compound are in 1.00 L of solution?<\/p>\n<p>b) The ingredients dissolve to form Na<sup>+<\/sup>, K<sup>+<\/sup>, Ca<sup>2+<\/sup>, Cl<sup>&#8211;<\/sup> and C<sub>3<\/sub>H<sub>5<\/sub>O<sub>3<\/sub><sup>&#8211;<\/sup>\u00a0ions. \u00a0How many moles of each ion are in 1.00 L of solution. \u00a0Note: there are several sources of Na<sup>+<\/sup>and Cl<sup>&#8211;<\/sup>.<\/p>\n<p>27. The density of liquid benzene, C<sub>6<\/sub>H<sub>6\u00a0<\/sub>is 0.879 g\/mL at 15<sup>o<\/sup>C. What is the volume in milliliters of 1.00 mol of benzene at this temperature?<\/p>\n<p>28. A sample of 0.370 mol of a metal oxide (M<sub>2<\/sub>O<sub>3<\/sub>) weighs 55.4 g.<br \/>\na) How many moles of O are in the sample?<br \/>\nb) How many grams of M are in the sample?<br \/>\nc)\u00a0 What element is represented by M?<\/p>\n<p>29. a) How many grams of H<sub>2<\/sub>S are there in 0.400 moles of H<sub>2<\/sub>S?<br \/>\nb) How many grams of hydrogen and sulfur are contained in 0.400 moles of H<sub>2<\/sub>S?<br \/>\nc)\u00a0 How many molecules of H<sub>2<\/sub>S are contained in 0.400 moles of H<sub>2<\/sub>S?<br \/>\nd)\u00a0 How many atoms of H and S are contained in 0.400 moles of H<sub>2<\/sub>S?<\/p>\n<p>30. a) How many moles of SO<sub>2<\/sub> are represented by 9.54 g of SO<sub>2<\/sub>?<br \/>\nb) How many atoms of oxygen does this represent?<\/p>\n<p>31. What is the average mass (in grams) of one oxygen atom?<\/p>\n<p>32. What is the mass in grams of one molecule of CH<sub>3<\/sub>OH?<\/p>\n<p>33. The density of gold is approx. 2 x 10<sup>1<\/sup> g\/cm<sup>3<\/sup>. What is the volume of one gold atom?<\/p>\n<p>34. How many moles are there in one atom?<\/p>\n<p>35. What is the mass percent of each atom in (NH<sub>4<\/sub>)<sub>2<\/sub>Cr<sub>2<\/sub>O<sub>7<\/sub>?<\/p>\n<p>36. Calculate the following mass percents:<br \/>\na) Mass % of H in NH<sub>4<\/sub>HCO<sub>3<\/sub>\u00a0 \u00a0 \u00a0 \u00a0b)\u00a0Mass % of Mn in KMnO<sub>4<\/sub><\/p>\n<p>37. Calculate the following mass fractions:<br \/>\na) Mass fraction of Cl in CaCl<sub>2<\/sub>\u00a0 \u00a0 \u00a0b)\u00a0Mass fraction of P in P<sub>4<\/sub>O<sub>7<\/sub><\/p>\n<p>38. The effectiveness of nitrogen fertilizer is determined mainly by its mass % N. Rank the following in terms of their effectiveness: KNO<sub>3<\/sub>; NH<sub>4<\/sub>NO<sub>3<\/sub>; Co(NH<sub>2<\/sub>)<sub>2<\/sub>; NH<sub>4<\/sub>SO<sub>4<\/sub>. Show your calculations.<\/p>\n<p>39. A compound of iodine and cesium contains 63.94 g of metal and 61.06 g of nonmetal. How many grams of cesium are in 38.77 g of the compound? How many grams of iodine?<\/p>\n<p>40. What is the empirical formula and the empirical formula molar mass of each of the following: \u00a0a) C<sub>2<\/sub>H<sub>4<\/sub>\u00a0 \u00a0b) C<sub>2<\/sub>H<sub>6<\/sub>O<sub>2<\/sub>\u00a0 \u00a0c) N<sub>2<\/sub>O<sub>5<\/sub>\u00a0 \u00a0 d) Ba<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>\u00a0 \u00a0 \u00a0e) Te<sub>4<\/sub>I<sub>16<\/sub><\/p>\n<p>41. What is the molecular formula of each of the following:<br \/>\na) Empirical formula = CH<sub>2<\/sub>; molecular molar mass = 42.08 g\/mol<br \/>\nb)\u00a0 EF = NH<sub>2<\/sub>; MMM = 32.05 g\/mol<br \/>\nc)\u00a0 EF = NO<sub>2<\/sub>; MMM = 92.02 g\/mol<br \/>\nd)\u00a0 EF = CHN; MMM = 135.14 g\/mol<br \/>\ne)\u00a0 EF = CH;\u00a0 MMM = 78.11 g\/mol<br \/>\nf)\u00a0 EF = C<sub>3<\/sub>H<sub>6<\/sub>O<sub>2<\/sub>; MMM = 74.08 g\/mol<br \/>\ng)\u00a0 EF = C<sub>7<\/sub>H<sub>4<\/sub>O<sub>2<\/sub>; MMM = 240.02 g\/mol<\/p>\n<p>42. A 3.450 g sample of nitrogen reacts with 1.970 g of oxygen to form a compound. Determine the empirical formula for this compound.<\/p>\n<p>43. A compound containing carbon, hydrogen and oxygen is found to be 40.00% carbon and 6.700% hydrogen by mass. The molar mass of the compound is between 115 g\/mol and 125 g\/mol. Determine the empirical formula and molecular formula of this compound.<\/p>\n<p>44. Determine the empirical formula for the following situations:<br \/>\na) 0.0630 mol of chlorine atom combined with 0.220 mol of oxygen atoms<br \/>\nb) 2.45 g silicon combined with 12.4 g of chlorine<br \/>\nc)\u00a0 a compound with 27.3% carbon and 72.7% oxygen by mass<br \/>\nd)\u00a0 a hydrocarbon (containing only C and H) which has 79.9% by mass C<\/p>\n<p>45. A compound containing only silicon and chlorine contains 79.1% chlorine by mass and has a molar mass of 269 g\/mol. What is the molecular formula?<\/p>\n<p>46. A sample of 0.600 mole of metal reacts completely with fluorine to form 46.8 g of MF<sub>2<\/sub>.<br \/>\na) How moles of F are in the sample of MF<sub>2<\/sub> that formed?<br \/>\nb) How many grams of M are in this sample of MF<sub>2<\/sub>?<br \/>\nc)\u00a0 What element is represented by M?<\/p>\n<p>47. A sample of nicotine contains 6.16 mmol of C, 8.56 mmol of H and 1.23 mmol of N. What is its empirical formula? (note: 1000 mmol = 1 mol)<\/p>\n<p>48. Cortisol (molar mass = 362.47 g\/mol), one of the major steroid hormones, is a key factor in the synthesis of protein. Its profound effect on the reduction of inflammation explains its use in the treatment of rheumatoid arthritis. Cortisol is 69.6% C, 8.34% H and 22.1% O by mass. What is its molecular formula?<\/p>\n<p>49. 3.00g of molybdenum (Mo) combines with sulfur to produce 5.50 g of a compound. What is the empirical formula of this compound?<\/p>\n<p>50. A 5.70 g sample of an iron oxide compound was heated in a stream of hydrogen gas, which reacted with the oxygen in the compound to form water. The water vapor was carried away in the stream of gas, leaving only pure metallic iron weighing 4.20g. Calculate the simplest possible formula for the original iron oxide.<\/p>\n<p>51. When 2.31g of a carboxylic acid compound is burned in O<sub>2<\/sub>, the only products are 1.33g of H<sub>2<\/sub>O and 3.38g of CO<sub>2<\/sub>. Calculate the empirical (simplest possible) formula for this compound.<\/p>\n<p>52. A compound was analyzed and found to have the following percentage composition:\u00a0aluminum = 15.7%, sulfur = 28.11%, oxygen = 56.12%. The molar mass of the compound is known to be approximately 684 g\/mol. What is the molecular formula?<\/p>\n<p>&nbsp;<\/p>\n<h2><strong>Answers<\/strong><\/h2>\n<p>1. 2.55 x 10<sup>24\u00a0<\/sup>gidgets<\/p>\n<p>2.\u00a08.6 x 10<sup>48\u00a0<\/sup>mol of Fe<\/p>\n<p>3. a) 361 g of Ca \u00a0 \u00a0 \u00a0 \u00a0 b) 0.324 mol of Zn<\/p>\n<p>4.\u00a0<span> <\/span>a) 0.18660 mol of Fe<span>\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span>b) 187.57 kg of Pb<\/p>\n<p>5. a)\u00a06.55 x 10<sup>23\u00a0<\/sup>atoms of Na \u00a0 \u00a0b)\u00a04.32 x 10<sup>3<\/sup> g of Cr<\/p>\n<p>6.\u00a01.18 x 10<sup>20\u00a0<\/sup>atoms<\/p>\n<p>7.\u00a06 x 10<sup>22<\/sup> Au atoms<\/p>\n<p>8.\u00a01.0 x 10<sup>-4\u00a0<\/sup>L<\/p>\n<p>9.\u00a0255.673 g\/mol of Cu(SO<sub>4<\/sub>)<sub>2<\/sub><\/p>\n<p>10.\u00a0a) 18.0152 g\/mol<span>\u00a0\u00a0\u00a0\u00a0 <\/span>b)<span>\u00a0<\/span>310.177 g\/mol<span>\u00a0\u00a0\u00a0\u00a0 <\/span>c)<span>\u00a0 <\/span>46.069 g\/mol<\/p>\n<p>11.\u00a05.74 x 10<sup>9\u00a0<\/sup>mol of O<\/p>\n<p>12.\u00a05.1 x 10<sup>24\u00a0<\/sup>atoms of H<\/p>\n<p>13.\u00a08.05 g of S<\/p>\n<p>14.\u00a024.3 g of Ni<\/p>\n<p class=\"Answers\">15.<span>\u00a0\u00a0a) 300.6 g Hg<br \/>\nb) 1.934 x 10<sup>24\u00a0<\/sup>molecules of NO<br \/>\nc) 1.0407 x 10<sup>25\u00a0<\/sup>atoms<br \/>\nd) 16.67 g H<br \/>\ne) 1.075 x 10<sup>23\u00a0<\/sup>atoms <\/span><\/p>\n<p class=\"Answers\">16. \u00a03 x 10<sup>22\u00a0<\/sup>iron atoms<\/p>\n<p class=\"Answers\">17.<span>\u00a0\u00a0 <\/span>a) g H<sub>2<\/sub>O\u00a0[latex]\\longrightarrow[\/latex] mol H<sub>2<\/sub>O using molar mass of H<sub>2<\/sub>O<\/p>\n<p class=\"AnswersSub\">b) mol H<sub>2<\/sub>SO<sub>4\u00a0<\/sub>[latex]\\longrightarrow[\/latex]\u00a0molecules H<sub>2<\/sub>SO<sub>4\u00a0<\/sub>using Avogadro\u2019s number<\/p>\n<p class=\"AnswersSub\">c) g HCl [latex]\\longrightarrow[\/latex] mol HCl using molar mass HCl [latex]\\longrightarrow[\/latex] molecules HCl using Avogadro\u2019s number<\/p>\n<p class=\"AnswersSub\">d) mol KMnO<sub>4\u00a0<\/sub>[latex]\\longrightarrow[\/latex]<span>\u00a0<\/span>mol O using chemical formula [latex]\\longrightarrow[\/latex] atoms O using Avogadro\u2019s number<\/p>\n<p class=\"AnswersSub\">e) g KCN [latex]\\longrightarrow[\/latex] mol KCN using molar mass of KCN [latex]\\longrightarrow[\/latex] mol C using chemical formula [latex]\\longrightarrow[\/latex] g C using molar mass of C<\/p>\n<p class=\"AnswersSub\">f) mL CH<sub>3<\/sub>OH [latex]\\longrightarrow[\/latex] g CH<sub>3<\/sub>OH using density [latex]\\longrightarrow[\/latex] mol CH<sub>3<\/sub>OH using molar mass of CH<sub>3<\/sub>OH<\/p>\n<p class=\"AnswersSub\">g) molecules NaCl [latex]\\longrightarrow[\/latex] atoms Cl using formula ratio<br \/>\n<span>\u00a0<\/span>OR<br \/>\nmolecules NaCl [latex]\\longrightarrow[\/latex] moles NaCl using Avogadro\u2019s number [latex]\\longrightarrow[\/latex] moles Cl using the chemical formula [latex]\\longrightarrow[\/latex] atoms Cl using Avogadro\u2019s number<\/p>\n<p class=\"AnswersSub\">h) kg KBr [latex]\\longrightarrow[\/latex] g KBr using metric conversions [latex]\\longrightarrow[\/latex] mol KBr using the molar mass of KBr [latex]\\longrightarrow[\/latex] moles Br<sup>&#8211;\u00a0<\/sup>ions using chemical formula [latex]\\longrightarrow[\/latex] number of Br<sup>&#8211;\u00a0<\/sup>ions using Avogadro\u2019s number<\/p>\n<p class=\"AnswersSub\">i) mol H<sub>2<\/sub>SO<sub>4<\/sub>\u00a0[latex]\\longrightarrow[\/latex]<span>\u00a0<\/span>g H<sub>2<\/sub>SO<sub>4\u00a0<\/sub>using molar mass of H<sub>2<\/sub>SO<sub>4\u00a0<\/sub>[latex]\\longrightarrow[\/latex]<span>\u00a0<\/span>mL\u00a0H<sub>2<\/sub>SO<sub>4\u00a0<\/sub>using density of H<sub>2<\/sub>SO<sub>4<\/sub><\/p>\n<p class=\"AnswersSub\">j) mol MgO [latex]\\longrightarrow[\/latex]<span>\u00a0<\/span>mol O using the chemical formula [latex]\\longrightarrow[\/latex]<span>\u00a0<\/span>g O using molar mass of O<\/p>\n<p class=\"AnswersSub\">k) mm<sup>3\u00a0<\/sup>NaOH [latex]\\longrightarrow[\/latex]<span>\u00a0<\/span>cm<sup>3\u00a0<\/sup>NaOH using metric conversions [latex]\\longrightarrow[\/latex] g\u00a0NaOH using the density [latex]\\longrightarrow[\/latex] mol NaOH using the molar mass of NaOH<\/p>\n<p class=\"AnswersSub\">l) molecules HCl [latex]\\longrightarrow[\/latex] moles HCl using Avogadro\u2019s number [latex]\\longrightarrow[\/latex] moles H using the chemical formula [latex]\\longrightarrow[\/latex] g H using the molar mass of H<\/p>\n<p class=\"AnswersSub\">m) mol C [latex]\\longrightarrow[\/latex] mol C<sub>6<\/sub>H<sub>6\u00a0<\/sub>using the chemical formula [latex]\\longrightarrow[\/latex] g C<sub>6<\/sub>H<sub>6\u00a0<\/sub>using the molar mass of C<sub>6<\/sub>H<sub>6<\/sub><\/p>\n<p class=\"Answers\">18.<span>\u00a0\u00a0 <\/span>a) 0.3 mol \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0b) 3 x 10<sup>24\u00a0<\/sup>molecules \u00a0 \u00a0 \u00a0 \u00a0c) 8 x 10<sup>22\u00a0<\/sup>molecules<br \/>\nd) 1 x 10<sup>25\u00a0<\/sup>atoms \u00a0 \u00a0 \u00a0 \u00a0 \u00a0e) 0.9 g \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0f) 0.1 mol<br \/>\ng) 5 ions (or 5 atoms) \u00a0 \u00a0h) 3 x 10<sup>25\u00a0<\/sup>ions \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0i) 3 x 10<sup>2\u00a0<\/sup>mL<br \/>\nj) 8 x 10<sup>1\u00a0<\/sup>g \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0k) 2 x 10<sup>-4\u00a0<\/sup>mol \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0l) 8 x 10<sup>-24 <\/sup>g<br \/>\nm) 7 x 10<sup>1\u00a0<\/sup>g<\/p>\n<p class=\"Answers\">19.<span>\u00a0\u00a0 <\/span>a) 8.82 x 10<sup>23\u00a0<\/sup>ions \u00a0 \u00a0 \u00a0 \u00a0\u00a0b) 0.366 g \u00a0 \u00a0 \u00a0 \u00a0c) 4.81 x 10<sup>5\u00a0<\/sup>mg<br \/>\nd)<span>\u00a0 <\/span>2.93 x 10<sup>3\u00a0<\/sup>g<\/p>\n<p class=\"Answers\">20.<span>\u00a0\u00a0 <\/span>1.43 x 10<sup>24\u00a0<\/sup>atoms of O (7.15 x 10<sup>23\u00a0<\/sup>molecules O<sub>2<\/sub>)<\/p>\n<p class=\"Answers\">21.<span>\u00a0 <\/span>0.5 mile<sup>2<\/sup><\/p>\n<p class=\"Answers\">22.<span>\u00a0 <\/span>Because 1 mole of nitrogen may be 1 mole of N or 1 mole of N<sub>2<\/sub>, etc. Nitrogen and hydrogen (but not lead and water) are commonly found as diatomic molecules.<\/p>\n<p class=\"Answers\">23<span>\u00a0 \u00a0<\/span>8.58 x 10<sup>24<\/sup>Mg<sup>2+<\/sup>ions and 5.72 x 10<sup>24<\/sup>N<sup>3-<\/sup>ions<\/p>\n<p class=\"Answers\">24.<span>\u00a0 <\/span>59.4 g of O<sub>2\u00a0<\/sub>(note: the nitrogen is given as N<sub>2<\/sub>)<\/p>\n<p class=\"Answers\"><span lang=\"ES-MX\">25.<span>\u00a0 <\/span>80.0 g of NaOH<\/span><\/p>\n<p class=\"Answers\"><span lang=\"ES-MX\">26.<span>\u00a0 <\/span>a) 0.102 mol NaCl, 0.028 mol NaC<sub>3<\/sub>H<sub>5<\/sub>O<sub>3<\/sub>,\u00a00.004mol KCl<br \/>\n<span>\u00a0\u00a0\u00a0 <\/span>and 0.002 mol CaCl<sub>2<\/sub><br \/>\nb) Na<sup>+\u00a0<\/sup>= 0.130 mol,<span>\u00a0 <\/span>K<sup>+\u00a0<\/sup>= 0.004 mol, Ca<sup>2+\u00a0<\/sup>= 0.002,<br \/>\n<span>\u00a0\u00a0\u00a0 <\/span>Cl<sup>&#8211;\u00a0<\/sup>= 0.110 mol; C<sub>3<\/sub>H<sub>5<\/sub>O<sub>3<\/sub><sup>&#8211;\u00a0<\/sup>= 0.028 mol<\/span><\/p>\n<p class=\"Answers\"><span lang=\"ES-MX\">27.<span>\u00a0 <\/span>88.9 mL<\/span><\/p>\n<p class=\"Answers\"><span lang=\"ES-MX\">28.<span>\u00a0 <\/span>a) 1.11 mol O<span>\u00a0\u00a0\u00a0 <\/span>b) 37.6 g M<span>\u00a0\u00a0\u00a0 <\/span>c) V<\/span><\/p>\n<p class=\"Answers\">29.<span>\u00a0 <\/span>a) 13.6 g H<sub>2<\/sub>S \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0b) 0.806 g H, 12.8 g S<br \/>\nc) 2.41 x 10<sup>23\u00a0<\/sup>molecules H<sub>2<\/sub>S \u00a0 \u00a0 \u00a0 \u00a0d)<span>\u00a0 <\/span>4.82 x 10<sup>23<\/sup>H atoms and 2.41 x 10<sup>23\u00a0<\/sup>S atoms<\/p>\n<p class=\"Answers\">30.<span>\u00a0 <\/span>a) 0.149 mol SO<sub>2 \u00a0 \u00a0 \u00a0\u00a0<\/sub>b) 1.79 x 10<sup>23<\/sup>atoms O<\/p>\n<p class=\"Answers\">31. 2.65682 x 10<sup>-23\u00a0<\/sup>g<\/p>\n<p class=\"Answers\">32.<span>\u00a0 <\/span>5.3208 x 10<sup>-23\u00a0<\/sup>g<\/p>\n<p class=\"Answers\">33.<span>\u00a0 <\/span>2 x 10<sup>-23\u00a0<\/sup>cm<sup>3<\/sup><\/p>\n<p class=\"Answers\">34.<span>\u00a0 <\/span>1.6605388 x 10<sup>-24\u00a0<\/sup>mol<\/p>\n<p class=\"Answers\">35.<span>\u00a0 <\/span>N= 11.114%, H= 3.1990%, Cr= 41.256%, O= 44.431%<\/p>\n<p class=\"Answers\">36.<span>\u00a0 <\/span>a) 6.3749%<span>\u00a0\u00a0\u00a0\u00a0 <\/span>b) 34.7634%<\/p>\n<p class=\"Answers\">37.<span>\u00a0 <\/span>a)<span>\u00a0 <\/span>0.638883<span>\u00a0\u00a0\u00a0\u00a0<\/span>b) 0.525222<\/p>\n<p class=\"Answers\">38. NH<sub>4<\/sub>NO<sub>3<\/sub>&gt; Co(NH<sub>2<\/sub>)<sub>2<\/sub>&gt; KNO<sub>3 <\/sub>&gt; NH<sub>4<\/sub>SO<sub>4<\/sub><\/p>\n<p class=\"Answers\"><span lang=\"ES-MX\">39. 19.83 g Cs, 18.94 g I<\/span><\/p>\n<p class=\"Answers\"><span lang=\"ES-MX\">40.<span>\u00a0 <\/span>a) C<sub>2<\/sub>H<sub>4<\/sub>= CH<sub>2<\/sub>; 14.027 g\/mol<br \/>\nb) C<sub>2<\/sub>H<sub>6<\/sub>O<sub>2<\/sub>= CH<sub>3<\/sub>O; 31.034 g\/mol<br \/>\nc) N<sub>2<\/sub>O<sub>5<\/sub>= N<sub>2<\/sub>O<sub>5<\/sub>; 108.0104 g\/mol<br \/>\nd) Ba<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>= Ba<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>; 601.924 g\/mol<br \/>\ne) Te<sub>4<\/sub>I<sub>16<\/sub>= TeI<sub>4<\/sub>; 635.22 g\/mol<\/span><\/p>\n<p class=\"Answers\"><span lang=\"ES-MX\">41.<span>\u00a0 <\/span>a) C<sub>3<\/sub>H<sub>6 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0<\/sub>b) N<sub>2<\/sub>H<sub>4 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0<\/sub>c) N<sub>2<\/sub>O<sub>4<br \/>\n<\/sub>d) C<sub>5<\/sub>H<sub>5<\/sub>N<sub>5 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<\/sub>e) C<sub>6<\/sub>H<sub>6 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<\/sub>f) C<sub>3<\/sub>H<sub>6<\/sub>O<sub>2<br \/>\n<\/sub>g) C<sub>14<\/sub>H<sub>8<\/sub>O<sub>4<\/sub><\/span><\/p>\n<p class=\"Answers\"><span lang=\"ES-MX\">42. N<sub>2<\/sub>O<\/span><\/p>\n<p class=\"Answers\"><span lang=\"ES-MX\">43.<span>\u00a0 <\/span>EF = CH<sub>2<\/sub>O, MF = C<sub>4<\/sub>H<sub>8<\/sub>O<sub>4<\/sub><\/span><\/p>\n<p class=\"Answers\"><span lang=\"ES-MX\">44.<span>\u00a0 <\/span>a) Cl<sub>2<\/sub>O<sub>7 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<\/sub>b) SiCl<sub>4 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<\/sub>c) CO<sub>2 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<\/sub>d) CH<sub>3<\/sub><\/span><\/p>\n<p class=\"Answers\"><span lang=\"ES-MX\">45.<span>\u00a0 <\/span>Si<sub>2<\/sub>Cl<sub>6<\/sub><\/span><\/p>\n<p class=\"Answers\"><span lang=\"ES-MX\">46.<span>\u00a0 <\/span>a) 1.20 mol F<span>\u00a0\u00a0\u00a0\u00a0 <\/span>b) 24.0 g M<span>\u00a0\u00a0\u00a0\u00a0 <\/span>c) Ca<\/span><\/p>\n<p class=\"Answers\">47.<span>\u00a0 <\/span>C<sub>5<\/sub>H<sub>7<\/sub>N<\/p>\n<p class=\"Answers\">48.<span>\u00a0 <\/span>C<sub>21<\/sub>H<sub>30<\/sub>O<sub>5<\/sub><\/p>\n<p class=\"Answers\">49.<span>\u00a0 <\/span>Mo<sub>2<\/sub>S<sub>5<\/sub><span>\u00a0 <\/span><\/p>\n<p class=\"Answers\">50.<span>\u00a0 <\/span>Fe<sub>4<\/sub>O<sub>5<\/sub>(Fe<sub>8<\/sub>O<sub>10<\/sub>etc. are also possible, but not the <i>simplest<\/i>)<\/p>\n<p class=\"Answers\">51.<span>\u00a0 <\/span>CH<sub>2<\/sub>O<span>\u00a0\u00a0 <\/span>(Note that, based on the products, the unknown compound could only contain C, H and O, and note that all of the C in the original compound ends up in the CO<sub>2<\/sub>and all the H ends up in the H<sub>2<\/sub>O.)<\/p>\n<p class=\"Answers\"><span lang=\"PT-BR\">52.<span>\u00a0 <\/span>Al<sub>4<\/sub>S<sub>6<\/sub>O<sub>24<\/sub><\/span><\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n","protected":false},"author":330,"menu_order":6,"template":"","meta":{"pb_show_title":"on","pb_short_title":"5.5 End of Chapter Problems","pb_subtitle":"","pb_authors":[],"pb_section_license":"cc-by-nc-sa"},"chapter-type":[],"contributor":[],"license":[54],"class_list":["post-3512","chapter","type-chapter","status-publish","hentry","license-cc-by-nc-sa"],"part":1354,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapters\/3512","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/wp\/v2\/users\/330"}],"version-history":[{"count":13,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapters\/3512\/revisions"}],"predecessor-version":[{"id":4863,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapters\/3512\/revisions\/4863"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/parts\/1354"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapters\/3512\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/wp\/v2\/media?parent=3512"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapter-type?post=3512"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/wp\/v2\/contributor?post=3512"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/wp\/v2\/license?post=3512"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}