1 / 15

Example 2.1 A Conceptual Example

Example 2.1 A Conceptual Example.

marvin
Télécharger la présentation

Example 2.1 A Conceptual Example

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Example 2.1 A Conceptual Example Jan Baptista van Helmont (1579–1644) first measured the mass of a young willow tree and, separately, the mass of a bucket of soil and then planted the tree in the bucket. After five years, he found that the tree had gained 75 kg in mass even though the soil had lost only 0.057 kg. He had added only water to the bucket, and so he concluded that all the mass gained by the tree had come from the water. Explain and criticize his conclusion. Analysis and Conclusions Van Helmont’s explanation anticipated the law of conservation of mass by dismissing the possibility that the increased mass of the tree could have been created from nothing. His main failure, however, was in not identifying all the substances involved. He did not know about the role of carbon dioxide gas in the growth of plants. In applying the law of conservation of mass, we must focus on all the substances involved in a chemical reaction. Exercise 2.1A A sealed photographic flashbulb containing magnesium and oxygen has a mass of 45.07 g. On firing, a brilliant flash of white light is emitted and a white powder is formed inside the bulb. What should the mass of the bulb be after firing? Explain. Exercise 2.1B Is the mass of the burned match pictured in the photograph the same as, less than, or more than the mass of the burning match? Explain your answer.

  2. Example 2.2 The mass ratio of oxygen to magnesium in the compound magnesium oxide is 0.6583:1. What mass of magnesium oxide will form when 2.000 g of magnesium is completely converted to magnesium oxide by burning in pure oxygen gas? Strategy Solution We usually write mass ratios in a form such as “the ratio of O to Mg is 0.6583:1.” The first number represents the mass of the first element named—in this case, a mass of oxygen, say 0.6583 g oxygen—and the second number represents the mass of the second element named—here a mass of magnesium. Although written only as “1,” we assume that the second number is as precisely known as the first, that is, 1.0000 g magnesium. According to the law of constant composition, the mass of oxygen that combines with the 2.000 g of magnesium must be just the right amount that makes the mass ratio of oxygen to magnesium in the product 0.6583:1. According to the law of conservation of mass, the mass of magnesium oxide must equal the sum of the masses of the magnesium and oxygen that react. We can state the mass ratio in the form of a conversion factor and then determine the required mass of oxygen. The mass of the sole product, magnesium oxide, equals the total of the masses of the substances entering into the reaction.

  3. Example 2.2 continued Assessment As a simple check, compare the calculated mass with the starting mass of magnesium. From the law of conservation of mass, the product of the reaction of magnesium with oxygen must have a mass greater than the starting mass of magnesium; and 3.317 g magnesium oxide is indeed greater than 2.000 g magnesium. Exercise 2.2A What mass of magnesium oxide is formed when 1.500 g of oxygen combines with magnesium? Exercise 2.2B When a strip of magnesium metal was burned in pure oxygen gas, 1.554 g of oxygen was consumed and the only product formed was magnesium oxide. What must have been the masses of magnesium metal burned and magnesium oxide formed?

  4. Example 2.3 How many protons, neutrons, and electrons are present in a 81Br atom? Strategy Solution We can use the identity of the element and several simple relationships between the subatomic particles described above to determine the numbers of these particles. The atomic number of bromine Z = 35 = number of protons is not given here, but we can obtain it from the list of elements on the inside front cover. In the bromine atom, the number Number of protons = number of electrons = 35 of positively charged protons equals the number of negatively charged electrons. From mass number 81 and Number of neutrons = A – Z = 81 – 35 = 46 atomic number 35, we calculate the number of neutrons, using Equation (2.1).

  5. Example 2.3 continued Exercise 2.3A Use the notation to represent the isotope of tin having 66 neutrons. Exercise 2.3B Isobars are atoms with the same mass number but different atomic numbers. Indicate the numbers of protons, neutrons, and electrons and the that is an isobar of tin-116. notation for a cadmium atom

  6. Example 2.4 Use the data cited above to determine the weighted average atomic mass of carbon. Strategy Assessment The contribution each isotope makes to the weighted average atomic mass is given by Equation (2.2). The weighted average atomic mass is the sum of the two contributions. This is the value listed in a table of atomic masses. As expected, the atomic mass of carbon is much closer to 12 u than to 13 u. Solution The contributions are Contribution of carbon-12 = 0.98892 x 12.00000 u = 11.867 u Contribution of carbon-13 = 0.01108 x 13.00335 u = 0.1441 u The weighted average mass is Atomic mass of carbon = 11.867 u + 0.1441 u = 12.011 u Exercise 2.4A There are three naturally occurring isotopes of neon. Their percent abundances and atomic masses are neon-20, 90.51%, 19.99244 u; neon-21, 0.27%, 20.99395 u; neon-22, 9.22%, 21.99138 u. Calculate the weighted average atomic mass of neon. Exercise 2.4B The two naturally occurring isotopes of copper are copper-63, mass 62.9298 u, and copper-65, mass 64.9278 u. What must be the percent natural abundances of the two isotopes if the atomic mass of copper listed in a table of atomic masses is 63.546 u?

  7. Example 2.5 An Estimation Example Indium has two naturally occurring isotopes and a weighted average atomic mass of 114.82 u. One of the isotopes has a mass of 112.9043 u. Which is likely to be the second isotope: 111In, 112In, 114In, or 115In? Analysis and Conclusions The masses of isotopes differ only slightly from whole numbers, which tells us that the isotope with mass 112.9043 u is 113In. To account for the observed weighted average atomic mass of 114.82 u, the second isotope must have a mass number greater than 114. It can be only 115In. Exercise 2.5A The masses of the three naturally occurring isotopes of magnesium are 24Mg, 23.98504 u; 25Mg, 24.98584 u; 26Mg, 25.98259 u. Use the atomic mass given inside the front cover to determine which of the three is the most abundant. Can you determine which is the second most abundant? Explain. Exercise 2.5B For the three magnesium isotopes described in Exercise 2.5A, (a) could the percent natural abundance of 24Mg be 60.00%? (b) What is the smallest possible value for the percent natural abundance of 24Mg?

  8. Example 2.6 Write the molecular formula and name of a compound for which each molecule contains six oxygen atoms and four phosphorus atoms. Strategy We need to write the chemical symbols of the two elements and use the stated numbers of atoms as subscripts following the symbols. We then determine which element to place first in the molecular formula. Solution We represent the six atoms of oxygen as O6 and the four atoms of phosphorus as P4. According to the scheme in Figure 2.9, the element O is followed only by F. Phosphorus therefore comes first in the formula; we write P4O6. The name of a compound with four (tetra-) P atoms and six (hexa-) oxygen atoms in its molecules is tetraphosphorus hexoxide. Exercise 2.6A Write the molecular formula and name of a compound for which each molecule contains four fluorine atoms and two nitrogen atoms. Exercise 2.6B Write the molecular formula and name of a compound the molecules of which each contain one oxygen atom and eight sulfur atoms.

  9. Example 2.7 Write (a) the molecular formula of phosphorus pentachloride and (b) the name of S2F10. Solution (a) Choosing which element symbol goes first. The order of elements shown in the molecular formula must be the same as the order in the name. Writing subscripts. The lack of a prefix on phosphorus signifies one P atom per molecule. The prefix penta- indicates five chlorine atoms. The molecular formula is PCl5. (b) The subscripts indicate two (di-) sulfur atoms and ten (deca-) fluorine atoms. The compound is disulfur decafluoride. Exercise 2.7A Write (a) the molecular formula of tetraphosphorus decoxide and (b) the name of S7O2. Exercise 2.7B Write a plausible molecular formula for a compound that has one sulfur atom, two oxygen atoms, and two fluorine atoms in each of its molecules. Comment on any ambiguity that exists in this case.

  10. Example 2.8 Determine the formula for (a) calcium chloride and (b) magnesium oxide. Solution (a) First we write the symbols for the ions, with the cation first: Ca2+ and Cl–. The simplest combination of these ions that gives an electrically neutral formula unit is one Ca2+ ion for every two Cl– ions. The formula is CaCl2 . Ca2+ + 2 Cl– = CaCl2 (b) Figure 2.10 tells us that the ions are Mg2+ and O2–. The simplest ratio for an electrically neutral formula unit is 1:1. The formula of this binary ionic compound is MgO. Mg2+ + O2– = MgO Exercise 2.8A Give the formula for each of the following ionic compounds: (a) potassium sulfide (b) lithium oxide (c) aluminum fluoride Exercise 2.8B Give the formula for each of the following ionic compounds: (a) chromium(III) oxide (b) iron(II) sulfide (c) lithium nitride

  11. Example 2.9 What are the names of (a) MgS and (b) CrCl3? Solution (a) MgS is made up of Mg2+ and S2– ions. Its name is magnesium sulfide. (b) From Figure 2.10 we see that there are two simple ions of chromium, Cr3+ and Cr2+. Because there are three Cl– ions in the formula unit CrCl3, the cation must have a charge of 3+, that is, it must be Cr3+. Because there are two chromium cations, there are two chlorides, CrCl2 and CrCl3, and we must assign a different name to each. Therefore, the name of our compound cannot be simply chromium chloride; instead, to indicate the 3+ cation, we say the name is chromium(III) chloride. Exercise 2.9A Name the following compounds: (a) CaBr2 (b) Li2S (c) FeBr2 (d) CuI Exercise 2.9B Write the name and formula for each of the following compounds: (a) the sulfide of copper(I) (b) the oxide of cobalt(III) (c) the nitride of magnesium

  12. Example 2.10 Write the formula for (a) sodium sulfite and (b) ammonium sulfate. Solution (a) It is possible to identify the sulfite ion without memorizing all the ions in Table 2.4. If you remember the name and formula of one of the sulfur–oxygen polyatomic anions, you should be able to deduce the names of others. Suppose you remember that sulfate is SO42–. The -ite anion has one fewer oxygen atom, 3 instead of 4, and so it is SO32–. The charges of the two ions in a formula unit must balance, which means the formula unit of sodium sulfite must have Na+ and SO32– in the ratio 2:1. The formula is therefore Na2SO3. (b) The ammonium ion is NH4+, and the sulfate ion is SO42–. A formula unit of ammonium sulfate has two NH4+ ions and one SO42– ion. To represent the two NH4+ ions, we place parentheses around the NH4, followed by a subscript 2, (NH4)2, and thus arrive at the formula (NH4)2SO4. Exercise 2.10A What is the formula for (a) ammonium carbonate, (b) calcium hypochlorite, and (c) chromium(III) sulfate? Exercise 2.10B Write a plausible formula for (a) potassium aluminum sulfate (b) magnesium ammonium phosphate

  13. Example 2.11 What is the name of (a) NaCN and (b) Mg(ClO4)2? Solution (a) The ions in this compound are Na+, sodium ion, and CN–, cyanide ion (see Table 2.4). The name of the compound is sodium cyanide. (b) The ions present are Mg2+, magnesium ion, and ClO4–, perchlorate ion. The name of the compound is magnesium perchlorate. Exercise 2.11A Name each of the following compounds: (a) KHCO3 (b) FePO4 (c) Mg(H2PO4)2 Exercise 2.11B Give a plausible name for the following: (a) Na2SeO4 (b) FeAs (c) Na2HPO3

  14. Cumulative Example Show that the following experiment is consistent with the law of conservation of mass (within the limits of experimental error): A 10.00-g sample of calcium carbonate was dissolved in 100.0 mL of hydrochloric acid solution (d = 1.148 g/mL). The products were 120.40 g of solution (a mixture of hydrochloric acid and calcium chloride) and 2.22 L of carbon dioxide gas (d = 0.0019769 g/mL). Strategy This problem may give the initial impression of being quite formidable because several data are given. Upon examination, it is much less challenging than you might think. We must show that mass is conserved in the experiment. That means that we must compare the mass of the starting materials to the mass of the end products of the chemical change. If mass is conserved, the two masses should be identical, within the limits of experimental error. Our job then is to find the masses of the starting materials and of the end products. Solution Let’s begin by identifying the starting materials and end products. The context of the problem makes it clear that calcium carbonate reacts with a hydrochloric acid solution, and so calcium carbonate and the HCl solution are the starting materials. The end products are another solution and carbon dioxide gas. Starting mass: The mass of calcium carbonate is given. We can use the density and the volume of the HCl solution to find its mass.

  15. Cumulative Example continued Solution continued Assessment Mass of products: This time, the mass of the solution is given. We must use volume and density of carbon dioxide gas to find its mass. However, we must first convert the volume in liters to milliliters because the density is given in grams per milliliter. Then we can add the masses of the two products. Then we can add the masses of the two starting materials. We note that the masses of reactants and products are not exactly the same. However, the 124.8 g of reactants is reported to four significant figures, and so it is precise only to 0.1 g. The difference between the masses of the starting materials and end products is less than 0.1 g. Clearly, the difference in masses is smaller than the uncertainty in the mass of the starting materials. We can therefore conclude that this experiment is consistent with the law of conservation of mass.

More Related