V05s9QE8

1
2MC #2 (Ch. 2 and 3)
3Due: 11:59pm on Tuesday, October 1, 2019
4You will receive no credit for items you complete after the assignment is due. Grading Policy
5± Law of Definite Proportions and Atomic Theory
6Using mass measurements, early chemists were able to determine that for pure substances, the mass proportions of the composite elements were the same, no matter how much of the substance was analyzed. This became known as the law of definite proportions, or the law of constant composition. Dalton interpreted that this constant mass composition arose from constant composition of atoms. Working with this concept will help you to better understand the particulate nature of matter. To answer the questions, refer to the structures of water and ammonia shown here.A molecule of water consists of an oxygen atom and two hydrogen atoms attached. A molecule of ammonia consists of a nitrogen atom and three hydrogen atoms attached.
7Part A
8What is the ratio of hydrogen atoms (H) to oxygen atoms (O) in 2 L of water?  Enter the simplest whole number ratio in order of hydrogen to oxygen, respectively.
9Express your answer as two integers, separated by a comma (e.g., 3,4).
10Hint 1. Find the atom ratio for a single molecule
11What is the ratio of hydrogen atoms (H) to oxygen atoms (O) in one water molecule?   Enter the simplest whole number ratio in order of hydrogen to oxygen, respectively.
12Express your answer as two integers, separated by a comma (e.g., 3,4).
13ANSWER:
142,1
15Hint 2. Find the atom ratio for four molecules
16What is the atom ratio of hydrogen to oxygen in this sample of four water molecules?Four water molecules. Each molecule consists of an oxygen atom and two hydrogen atoms attached.Enter the simplest whole number ratio in order of hydrogen to oxygen, respectively.
17Express your answer as two integers, separated by a comma (e.g., 3,4).
18ANSWER:
192,1
20ANSWER:
212,1
22Correct
23Your answer illustrates how Dalton's atomic theory explains the law of definite proportions. In any sample of water, no matter how large or small, the ratio of hydrogen atoms to oxygen atoms will always be 2:1 because that is the ratio within one single molecule.
24Part B
25How many atoms of hydrogen (H)  are present in 200 molecules of ammonia (NH3)?
26Express your answer numerically.
27Hint 1. Determine the hydrogen atom to ammonia molecule ratio
28What is the ratio of hydrogen atoms to ammonia molecules?  Enter the simplest whole number ratio in order of hydrogen to ammonia, respectively.
29Express your answer as two integers, separated by a comma (e.g., 3,4).
30Hint 1. Determine the number of hydrogen atoms in a single molecule
31How many hydrogen atoms are in a single ammonia molecule? Refer the image in the introduction, if needed.
32Express your answer numerically as an integer.
33ANSWER:
343
35 H atoms
36ANSWER:
373,1
38ANSWER:
39600
40 H atoms 
41Correct
42Since each ammonia molecule contains three hydrogen atoms, there will always be three times as many H atoms as there are NH3 molecules.
43Part C
44A solution of ammonia and water contains 1.60×1025 water molecules and 6.80×1024 ammonia molecules. How many total hydrogen atoms are in this solution?
45Enter your answer numerically.
46Hint 1. How to approach the problem
47First, determine the total number of hydrogen atoms in the water molecules.  Second, determine the total number of hydrogen atoms in the ammonia molecules. Finally, add these two numbers together.
48Hint 2. Calculate the number of H atoms from water
49How many hydrogen atoms are in 1.60×1025 water molecules?
50Express your answer numerically.
51Hint 1. The relation between hydrogen atoms and water molecules
52Recall that each molecule of water contains two hydrogen atoms.
53ANSWER:
543.20×1025
55 H atoms 
56Hint 3. Calculate the number of H atoms from ammonia
57How many hydrogen atoms are in 6.80×1024 ammonia molecules?
58Express your answer numerically.
59Hint 1. The relation between hydrogen atoms and ammonia molecules
60Recall that each molecule of ammonia contains three hydrogen atoms.
61ANSWER:
622.04×1025
63 H atoms 
64ANSWER:
655.24×1025
66 H atoms 
67Correct
68Exercise 2.32 - Enhanced - with Feedback and Hints
69Upon decomposition, one sample of magnesium fluoride produced 1.65 kg of magnesium and 2.58 kg of fluorine. A second sample produced 1.28 kg of magnesium.
70Part A
71How much fluorine (in grams) did the second sample produce?
72Express your answer in grams to three significant figures.
73Hint 1. Calculate the mass ratio
74In accordance with the law of definite proportions, all samples of a given compound, regardless of their source or method of preparation, have the same proportions of their constituent elements. Calculate the mass of fluorine per 1.00 kg of magnesium.
75Express your answer in kilograms to three significant figures.
76ANSWER:
771.56
78  kgfluorine  
79Correct
80Divide the mass of fluorine by the mass of magnesium produced during the single decomposition to calculate the mass of fluorine per 1.00 kg of magnesium:
81
82mass ratio===kgfluorinekgmagnesium2.58kgfluorine1.65kgmagnesium1.56kgfluorine1.00kgmagnesium
83
84Now, use this ratio to calculate the mass of fluorine produced by the second sample.
85
86ANSWER:
87m = 	
882000
89  g  
90Correct
91The decomposition reaction is consistent with the law of definite proportions. So, the mass ratio of fluorine to magnesium for the first sample is
92
93mass ratio===kgfluorinekgmagnesium2.58fluorine1.65magnesium1.56kgfluorine1.00kgmagnesium
94
95As a result, the mass of fluorine (in grams) after the decomposition of the second sample is
96
971.28kgmagnesium×1.56kgfluorine1.00kgmagnesium×1000g1kg=2000gfluorine
98
99± Neutrons and Isotopes
100Periodic table
101
102The periodic table lists two main numbers for each element. The atomic number is an integer that equals the number of protons. The number of neutrons is not given in the periodic table because it will vary with different isotopes. The mass number of an element is the sum of the protons and neutrons. To find the number of neutrons, the atomic number must be subtracted from the mass number.
103
104Elemental symbol
105
106When writing the symbol for an element, a superscript indicates the mass number and a subscript indicates the atomic number. For example, 14 6C has a mass number of 14 and an atomic number of 6. This isotope can also be written as just 14C. You do not need to put in the 6 because the atomic number of carbon is always 6 regardless of which isotope you are dealing with.
107
108Part A
109How many neutrons are found in one atom of 16N?
110Express your answer numerically.
111Hint 1. How to approach the problem
112The superscript number in front of an elemental symbol, in this case 16, is called the mass number. It is the sum of the numbers of protons and neutrons in that atom.
113Hint 2. Determine the number of protons
114How many protons are in an atom of 16N?
115Express your answer numerically.
116Hint 1. Atomic number
117The atomic number for an element is the integer given next to that element in the periodic table. The atomic number is equal to the number of protons in an atom of that element.
118ANSWER:
1197
120ANSWER:
1219
122Correct
123Atomic mass
124Since the number of neutrons varies in the periodic table, neither then number of neutrons nor the mass number (number of neutrons plus protons) is shown. What we do see is the atomic mass, or a weighted average of all the isotope masses. This is the noninteger listed with an element; it is the number with several decimal places. An element's atomic mass is the weighted average of the isotope masses. In other words, it is an average that takes into account the percentage of each isotope. To find the weighted average, multiply each isotopic mass by its relative abundance and find the sum for each isotope of an element.
125Part B
126A certain element X has four isotopes.
1275.845% of X has a mass of 53.93961 amu.
12891.75% of X has a mass of 55.93494 amu.
1292.123% of X has a mass of 56.93539 amu.
1300.2820% of X has a mass of 57.93328 amu.
131What is the average atomic mass of element X?
132Express your answer numerically to four significant figures.  
133Hint 1. How to approach the problem
134The atomic mass of X is found by calculating the weighted average of the four given masses.
135Hint 2. What's a weighted average?
136You are probably familiar with the process of averaging. For example, the average of 6 and 8 is (6+8)/2=7. However, in this process one assumes that both numbers (6 and 8) have equal weight. In other words, 50% of 6 plus 50% of 8 is equal to 7. In a weighted average, the numbers to be averaged have varying percentage weights. For example, if 6 were 30% and 8 were 70%, the weighted average would be
137
138(0.30)(6)+(0.70)(8)=7.4
139
140This is closer to 8 because it has the greater percentage.
141
142ANSWER:
14355.85
144  amu  
145Correct
146Part C
147What is the identity of element X from Part B?
148Express your answer as a chemical symbol.
149ANSWER:
150Fe
151Correct
152The atomic mass of iron, 55.85 amu, is very close to 55.93494 amu because that is the isotope with the greatest percent abundance.
153Part D
154The two isotopes of chlorine are 3517Cl and 3717Cl. Which isotope is the most abundant?
155Hint 1. How to approach the problem
156By comparing the atomic mass of chlorine to the mass of each isotope of chlorine, you will be able to tell which isotope contributes more toward the weighted average without doing any calculations.
157Hint 2. Find the atomic mass
158What is the atomic mass of chlorine?
159Express your answer numerically to three decimal places.
160Hint 1. How to find atomic mass
161The atomic mass for each element is given in the periodic table.
162ANSWER:
16335.5
164  amu  
165Hint 3. Mass of each isotope
166Even though you were not given the mass of each isotope, you can safely assume that the mass of any isotope is very close to its mass number. For example, the mass of 40K is 39.974 amu, which is very close to 40. Notice that the individual masses are not simply the sum of the protons and neutrons because of nuclear binding energy effects, which slightly alter the experimentally observed mass.
167ANSWER:
168
1693517Cl 
1703717Cl
171Correct
172Since the atomic mass of chlorine, 35.45 amu, is closer to 35 than it is to 37, 35Cl must be the most abundant isotope.
17335Cl does in fact account for 75.53% of Earth's chlorine. The full calculation is
174
175(0.7553)(34.96885)+(0.2447)(36.96590)=35.45amu
176
177Subatomic Particles
178Learning Goal:
179To determine the number of protons, neutrons, and electrons in a given isotope or ion.
180Atoms are composed of three fundamental particles. Protons are positively charged, neutrons are neutral, and electrons are negatively charged. Protons and neutrons are clustered into a dense core called the nucleus, whereas electrons are found outside of the nucleus at a relatively large distance. Elements differ from one another by how many protons their atoms they contain. The number of protons is called the atomic number (Z) of the element. Since protons and neutrons make up most of the mass of an atom, the sum of the protons and neutrons is its mass number (A). In neutral atoms, the numbers of protons and electrons are equal. In ions, the numbers of electrons and protons are not equal.
181Part A
182Specify the number of protons, neutrons, and electrons in the neutral atom copper-64.
183Enter your answers as integers separated by commas.
184Hint 1. How to approach the problem
185All copper-64 atoms have the same number of protons, which can be found by looking at the periodic table. The number 64 is the mass of this particular isotope, which can be used with the atomic number to determine the number of neutrons. The number of electrons can be determined by looking at both the charge of the atom and the number of protons.
186Hint 2. Identify a method to determine the number of electrons
187ANSWER:
188In a neutral atom, the number of protons is 
189greater than
190equal to
191less than
192 the number of electrons.
193Hint 3. Identify the particles that make up the mass number
194Copper-64 has a mass number of 64. Which of the following are included in the mass number of an atom?
195Check all that apply.
196ANSWER:
197
198electrons
199protons
200neutrons
201ANSWER:
20229,35,29
203 protons, neutrons, electrons
204Correct
205All copper atoms have 29 protons. A neutral Cu atom also has 29 electrons. The sum of the protons and neutrons gives the mass number, 29+35=64 .
206Part B
207Complete the sentence indicating the number of protons and electrons in the given ion.
208Match the numbers in the left column to the appropriate blanks in the sentences on the right.
209Hint 1. How to approach the problem
210Even though N3− is an ion, it still has the same number of protons as a neutral atom of nitrogen. Ions are formed when the atom gains or loses electrons. Use the periodic table to determine the number of protons. Then determine how many electrons would be needed to create a −3 charge overall.
211Hint 2. Identify how the charge impacts the number of electrons
212Which statement is true for an ion with a −3 charge?
213ANSWER:
214
215There are three less protons than electrons.
216There is one more proton than there are electrons.
217There is one less proton than there are electrons.
218There are three more protons than electrons.
219The number of protons equals the number of electrons.
220Hint 3. Find the atomic number of nitrogen
221What is the atomic number of nitrogen?
222ANSWER:
223
2245
2256
2267
2278
22814
22915
23016
231ANSWER:
232ResetHelp
233The ion N3− has 7The ion N 3 − has 7 protons and 10 electrons. protons and 10The ion N 3 − has 7 protons and 10 electrons. electrons.
234Correct
235Regardless of how many electrons are present, every nitrogen nucleus contains 7 protons. The identity of an element is determined by the number of protons, not the number of electrons. When the element is neutral, the number of positively charged protons and negatively charged electrons will be equal. When there are more electrons than protons, the ion will be negative, as in this example. When there are more protons than electrons, the ion will be positive.
236Part C
237What isotope has 14 protons and 15 neutrons?
238Enter the name of the element followed by a hyphen and the mass number (e.g., uranium-234).
239Hint 1. How to approach the problem
240The number of protons determines the identity of the element and the mass number is determined by the total number of protons and neutrons.
241Hint 2. Determine the mass number of the atom
242What is the mass number of an atom that has 14 protons and 15 neutrons?
243ANSWER:
244
24528
24614
24715
2481
24929
250Hint 3. Identify the element symbol
251The element symbol is prominent in the center of a cell in the periodic table.What is the element symbol of the element with 14 protons?
252Express your answer as the element symbol.
253Hint 1. How to use the atomic number
254Recall that the atomic number of an element represents the number of protons. Identify the element that has atomic number 14. Be careful not to answer with the element that has a mass number close to 14.
255ANSWER:
256element symbol: 
257Si
258ANSWER:
259Silicon-29
260Correct
261Silicon has 14 protons in its nucleus. It is the only element that has 14 protons. If it had more or fewer protons, it would not be silicon. However, the number of neutrons can vary, which is why the mass is written in the name.
262Part D
263Which element does X represent in the following expression: 5525X?
264Enter the chemical symbol of the element.
265Hint 1. Classify the superscript
266ANSWER:
267For the isotope 5525X, the number 55 refers to the 
268atomic number.
269number of protons.
270mass number.
271number of electrons.
272Hint 2. Classify the subscript
273ANSWER:
274For the element 5525X, the number 25 refers to the 
275mass number.
276atomic number.
277number of neutrons.
278ANSWER:
279X is the element symbol 
280Mn
281 .
282Correct
283In this example we only needed the number of protons to identify the symbol of the element. If you were asked to name the isotope, the mass number would be needed as well. In that case the correct answer would be manganese-55.
284Exercise 2.101 - Enhanced - with Feedback and Hints
285Antimony has only two naturally occurring isotopes. The mass of antimony-121 is 120.90382 u, and the mass of antimony-123 is 122.90421 u.
286Part A
287Calculate the relative abundance of antimony-121.
288Express your answer to three decimal places.
289Hint 1. Express the fraction of antimony-123
290The atomic mass is calculated according to the following equation, where the fractions of each isotope are the percents of natural abundances converted to their decimal values:
291
292Atomic mass=∑n(fraction of isotope n)×(mass of isotope n)
293
294Let x equal the fraction of antimony-121. What is the fraction of antimony-123 in terms of x?
295
296Express your answer in terms of x.
297ANSWER:
298Fraction of antimony-123 = 	
2991−x
300ANSWER:
30157.199
302  %  
303Correct
304The atomic mass is calculated according to the following equation, where the fractions of each isotope are the percents of natural abundances converted to their decimal values:
305
306Atomic mass=∑n(fraction of isotope n)×(mass of isotope n)
307
308Let x equal the fraction of antimony-121; then 1−x is the fraction of antimony-123. Substituting the given atomic mass and masses of isotopes into the equation above gives
309
310121.760u=(x)(120.90382u)+(1−x)(122.90421u)
311
312Rearrange this equation to solve for x:
313
314−1.14421x==−2.00039x0.57199
315
316Thus, the relative abundance of antimony-121 is 57.199 %.
317
318Part B
319Calculate the relative abundance of antimony-123.
320Express your answer to three decimal places.
321Hint 1. Express the fraction of antimony-121
322The atomic mass is calculated according to the following equation, where the fractions of each isotope are the percents of natural abundances converted to their decimal values:
323
324Atomic mass=∑n(fraction of isotope n)×(mass of isotope n)
325
326Let x equal the fraction of antimony-123. What is the fraction of antimony-121 in terms of x?
327
328Express your answer in terms of x.
329ANSWER:
330Fraction of antimony-121 = 	
3311−x
332ANSWER:
33342.801
334  %  
335Correct
336The atomic mass is calculated according to the following equation, where the fractions of each isotope are the percents of natural abundances converted to their decimal values:
337
338Atomic mass=∑n(fraction of isotope n)×(mass of isotope n)
339
340Let x equal the fraction of antimony-123; then 1−x is the fraction of antimony-121. Substituting the given atomic mass and masses of isotopes into the equation above gives
341
342121.760u=(1−x)(120.90382u)+(x)(122.90421u)
343
344Rearrange this equation to solve for x:
345
3460.85618x==2.00039x0.42801
347
348Thus, the relative abundance of antimony-123 is 42.801 %.
349
350Exercise 2.46
351Determine the number of protons and neutrons in each of the following isotopes.
352Part A
3534420Ca
354Enter your answers numerically separated by a comma.
355ANSWER:
356Np, Nn = 	
35720,24
358 protons, neutrons 
359Correct
360Part B
36122287Fr
362Enter your answers numerically separated by a comma.
363ANSWER:
364Np, Nn = 	
36587,135
366 protons, neutrons 
367Correct
368Part C
3699943Tc
370Enter your answers numerically separated by a comma.
371ANSWER:
372Np, Nn = 	
37343,56
374 protons, neutrons 
375Correct
376Part D
3773216S
378Enter your answers numerically separated by a comma.
379ANSWER:
380Np, Nn = 	
38116,16
382 protons, neutrons 
383Correct
384Exercise 2.103 - Enhanced - with Feedback and Hints
385A 67.2 g sample of a gold and palladium alloy contains 3.08×1023 atoms.
386Part A
387What is the mass percentage of the gold in the alloy?
388Express your answer in percents to three significant figures.
389Hint 1. Calculate the mass of gold in the alloy
390Calculate the mass of Au in a 67.2-g sample of a gold and palladium alloy containing 3.08×1023 atoms. Assume that X=gAu and 67.2−X=gPd, set up an equation relating the number of atoms of each type, and solve for X.
391Express your answer in grams to three significant figures.
392ANSWER:
39327.8
394  gAu  
395ANSWER:
39641.3
397  %Au  
398Answer Requested
399Let X=gAu and 67.2−X=gPd. Express the number of Au and Pd atoms in terms of X using the molar mass as a conversion factor between grams and moles and Avogadro's number as a conversion factor between moles and the number of atoms:
400
401XgAu×1molAu196.97gAu×6.022×1023atomsAu1molAu=3.057×1021XatomsAu
402
403(67.2−X)gPd×1molPd106.42gPd×6.022×1023atomsPd1molPd=(3.803×1023−5.659×1021X)atomsPd
404
405The sum of the numbers of Au and Pd atoms equals the given number of atoms. Therefore,
406
4073.057×1021X+(3.803×1023−5.659×1021X)2.601×1021XX===3.08×10237.227×102227.8
408
409Finally, use the calculated mass of gold and the given mass of alloy to calculate the mass percentage of gold:
410
411%Au==27.8g67.2g×100%41.3%Au
412
413Part B
414What is the mass percentage of the palladium in the alloy?
415Express your answer in percents to three significant figures.
416Hint 1. Calculate the mass of palladium in the alloy
417Calculate the mass of Pd in a 67.2-g sample of a gold and palladium alloy containing 3.08×1023 atoms. Assume that X=gPd and 67.2−X=gAu, set up an equation relating the number of atoms of each type, and solve for X.
418Express your answer in grams to three significant figures.
419ANSWER:
42039.4
421  gPd  
422ANSWER:
42358.7
424  %Pd  
425Correct
426Let X=gPd and 67.2−X=gAu. Express the number of Pd and Au atoms in terms of X using the molar mass as a conversion factor between grams and moles and Avogadro's number as a conversion factor between moles and number of atoms:
427
428XgPd×1molPd106.42gPd×6.022×1023atomsPd1molPd=5.659×1021XatomsPd
429
430(67.2−X)gAu×1molAu196.97gAu×6.022×1023atomsAu1molAu=(2.055×1023−3.057×1021X)atomsAu
431
432The sum of the numbers of Pd and Au atoms equals the given number of atoms. Therefore,
433
4345.659×1021X+(2.055×1023−3.057×1021X)2.601×1021XX===3.08×10231.025×102339.4
435
436Finally, use the calculated mass of palladium and the given mass of alloy to calculate the mass percentage of palladium:
437
438%Pd==39.4g67.2g×100%58.7%Pd 
439
440Exercise 2.55
441The atomic mass of aluminum is 26.98 u, and its mass spectrum shows a large peak at this mass. The atomic mass of copper is 63.55 u, yet the mass spectrum of copper does not show a peak at this mass.
442Part A
443Explain the difference.
444ANSWER:
445
446Aluminum exists only as Al−27, so the mass spectrum of aluminum exhibits just one line at 26.98 u. Copper has two isotopes, Cu−63 and Cu−65, and the mass of 63.55 u is the weighted average of these two isotopes, so there is no line at 63.55 u.
447Aluminum hasn't d-orbital, so the mass spectrum of aluminum exhibits just one line at 26.98 u. Copper has d-orbital, so there is no line at 63.55 u.
448Aluminum is light element, so the mass spectrum of aluminum exhibits just one line at 26.98 u. Copper is heavy element, so there is no line at 63.55 u.
449Correct
450Exercise 2.98
451A pure titanium cube has an edge length of 2.85 in .
452Part A
453How many titanium atoms does it contain? Titanium has a density of 4.50g/cm3.
454ANSWER:
455N = 	
4562.15×1025
457  atoms  
458Correct
459Exercise 2.63
460What is the amount, in moles, of each of the following substances?
461Part A
46212.8 g Ar
463ANSWER:
4640.320
465 mol 
466Correct
467Part B
4683.60 g Zn
469ANSWER:
4705.51×10−2
471 mol 
472Correct
473Part C
47426.0 g Ta
475ANSWER:
4760.144
477 mol 
478Correct
479Part D
4800.213 g Li
481ANSWER:
4823.07×10−2
483 mol 
484Correct
485Exercise 2.64
486What is the mass, in grams, of each of the following?
487Part A
4882.6×10−3 mol Sb
489Express your answer using two significant figures.
490ANSWER:
491m = 	
4920.32
493  g  
494Correct
495Part B
4963.51×10−2 mol Ba
497Express your answer using three significant figures.
498ANSWER:
499m = 	
5004.82
501  g  
502Correct
503Part C
50444.1 mol Xe
505Express your answer using three significant figures.
506ANSWER:
507m = 	
5085790
509  g  
510Correct
511Part D
5121.5 mol W
513Express your answer using two significant figures.
514ANSWER:
515m = 	
5162.8×102
517  g  
518Correct
519Ions and the Periodic Table
520A main-group metal tends to lose electrons, forming a cation with the same number of electrons as the nearest noble gas in the periodic table. A main-group nonmetal tends to gain electrons, forming an anion with the same number of electrons as the nearest noble gas. The various groups gain or lose electrons as summarized in the following table:
521Group	Tendency	Charge
5221	Lose one electron	1+
5232	Lose two electrons	2+
52413	Lose three electrons	3+
52515	Gain three electrons	3–
52616	Gain two electrons	2–
52717	Gain one electron	1–
52818	Rarely gain or lose electrons	0
529Part A
530If the following elements were to form ions, they would attain the same number of electrons as which noble gas?
531Drag the appropriate elements to their respective bins.
532Hint 1. Locating the appropriate noble-gas configuration
533Metals and nonmetals differ in how they attain an octet of electrons.
534Metals tend to lose the number of electrons required to attain the configuration of the noble gas preceding them in the periodic table.
535Nonmetals tend to gain the number of electrons required to attain the configuration of the noble gas following them in the periodic table.
536Hint 2. Identify the metals and nonmetals
537Classify each element as a metal or a nonmetal.
538Drag the appropriate elements to their respective bins.
539ANSWER:
540ResetHelp
541       
542Metal
543BeMgCaSr
544Nonmetal
545FSBr
546Hint 3. Identify the preceding noble gas for a metal
547Beryllium, Be, is located in group 2 and in period 2 of the periodic table. Identify the preceding noble gas.
548ANSWER:
549
550helium
551neon
552argon
553Hint 4. Identify the succeeding noble gas for a nonmetal
554Bromine, Br, is located in group 17 in period 4 of the periodic table. Identify the next noble gas succeeding Br in the periodic table?
555ANSWER:
556
557argon
558krypton
559xenon
560ANSWER:
561ResetHelp
562       
563He
564Be
565Ne
566MgF
567Ar
568SCa
569Kr
570BrSr
571All attempts used; correct answer displayed
572The Be2+ ion has two electrons, just like He. The ions Mg2+ and F− each have ten electrons like Ne. The S2− and Ca2+ each have 18 electrons like Ar. The ions Br− and Sr2+ each have 36 electrons like Kr.
573Part B
574If the following elements were to form ions, what is the expectant charge?
575Drag the appropriate elements to their respective bins.
576Hint 1. How to approach the problem
577First determine the noble gas electron configuration an element will obtain just as completed in Part A. Metals will lose electrons, and nonmetals will gain electrons to reach the closest noble gas electron configuration. Next, count the number of electrons that must be gained or lost to obtain that electron configuration.
578
579For example, the metal aluminum would lose electrons. The closest noble gas is neon, and you must move over three groups, groups 1, 2, and 13, to reach this noble gas. You also have to remove one p electron and two s electrons to reach this configuration. Therefore, the charge on this ion will be 3+.
580
581Hint 2. Determine the electron difference for an element
582Calcium, Ca, is located in group 2. Determine how many electrons will be lost or gained by an element in group 2 to reach the nearest noble-gas electron configuration.
583ANSWER:
584
585Three electrons will be gained.
586Two electrons will be gained.
587One electron will be gained.
588One electron will be lost.
589Two electrons will be lost.
590Three electrons will be lost.
591ANSWER:
592ResetHelp
593     
594-3
595P
596-2
597S
598-1
599F
600+1
601K
602+2
603Ca
604+3
605Correct
606Part C
607A certain element forms an ion with 36 electrons and a charge of 2+. Identify the element.
608Express your answer as a chemical symbol.
609Hint 1. Determine which group this element belongs to
610Which group in the periodic table tends to form ions with a 2+ charge?
611ANSWER:
612
613Group 1
614Group 2
615Group 13
616Group 15
617Group 16
618Group 17
619Group 18
620Hint 2. Determine whether this element has gained or lost electrons
621How many electrons does the 2+ ion have compared to the neutral atom?
622ANSWER:
623The 2+ ion has 
624more electrons
625fewer electrons
626 than the neutral atom.
627ANSWER:
628Sr
629Correct
630Exercise 2.83
631Predict the charge of the ion formed by each of the following elements.
632Part A
633Se
634Express your answer as an ion.
635ANSWER:
636Se2−
637Correct
638Part B
639K
640Express your answer as an ion.
641ANSWER:
642K+
643Correct
644Part C
645Al
646Express your answer as an ion.
647ANSWER:
648Al3+
649Correct
650Part D
651Cs
652Express your answer as an ion.
653ANSWER:
654Cs+
655Correct
656Exercise 2.79
657Part A
658Classify each of the following elements as an alkali metal, alkaline earth metal, halogen, or noble gas.
659Drag the elements into the appropriate bins.
660ANSWER:
661ResetHelp
662     
663alkali metal
664sodium
665alkaline earth metal
666calciumbarium
667halogen
668iodine
669noble gas
670xenon
671Correct
672Exercise 3.156 - Enhanced - with Feedback and Hints
673A 2.52-g sample of a compound containing only carbon, hydrogen, nitrogen, oxygen, and sulfur was burned in excess O to yield 4.36 g of CO2 and 0.892 g of H2O as the only carbon- and hydrogen-containing products, respectively.
674Another sample of the same compound, of mass 4.14 g, yielded 2.60 g of SO3 as the only sulfur-containing product.
675
676A third sample, of mass 5.66 g, was burned under different conditions to yield 2.80 g of HNO3 as the only nitrogen-containing product.
677Part A
678Calculate the empirical formula of the compound.
679Express your answer as a chemical formula. Enter the elements in the order C, H, S, N, O.
680Hint 1. Calculate the mass of oxygen in the 2.52-g sample
681The first step in the determination of an empirical formula is to find the masses of all elements that initially were present in the sample. Since the masses of C, H, S, and N can be found directly from the masses of CO2, H2O, SO3, and HNO3, finding the mass of O in the initial sample is more complicated. Calculate the mass of O in the 2.52-g sample.
682Express your answer in grams to two significant figures.
683ANSWER:
6840.32
685  gO  
686Correct
687During combustion, all carbon goes to form CO2 and all hydrogen goes to form H2O. To calculate the masses of carbon and hydrogen in the sample, first convert the mass of the product to moles using the molar mass conversion factor. Then use the chemical formula to convert between moles of the compound and moles of the constituent element. Finally, use the molar mass of the corresponding element to find the mass of the element in the sample.
688
689massCinsample1==4.36gCO2×1molCO244.01gCO2×1molC1molCO2×12.01gCmolC1.19gC
690
691massHinsample1==0.892gH2O×1molH2O18.02gH2O×2molH1molH2O×1.008gH1molH0.0998gH
692
693The masses of SO3 and HNO3 are obtained from samples of different masses. Use the ratio of the mass of SO2 to the mass of the sample and the ratio of the mass of HNO3 to the mass of the sample as conversion factors to calculate the mass of SO2 and HNO3 that would be produced when the 2.52-g sample is burned. Then make the same conversions as above.
694
695massSinsample1==2.52gsample×2.60gSO34.14gsample×1molSO380.07gSO3×1molS1molSO3×32.07gS1molS0.634gS
696
697massNinsample1==2.52gsample×2.80gHNO35.66gsample×1molHNO363.02gHNO3×1molN1molHNO3×14.01gN1molN0.277gN
698
699Finally, calculate the mass of oxygen in the 2.52-g sample as the difference between the mass of the sample and the sum of masses of the remaining elements:
700
701massOinsample1==2.52g−(1.19g+0.0998g+0.634g+0.277g)0.32g
702
703Now, convert all the calculated masses to moles and divide them to the smallest one to obtain the subscripts for a chemical formula.
704
705ANSWER:
706C5H5SNO
707Correct
708According to the hint for this part, the 2.52-g sample contains 1.19 gC (9.91×10−2mol), 0.0998 gH (9.90×10−2mol), 0.634 gS (1.98×10−2mol), 0.277 gN (1.98×10−2mol), and 0.32 gO (2.0×10−2mol). Now that you know the number of moles of each element in the sample, write a pseudoformula for this compound:
709
710C9.91×10−2H9.90×10−2O2.0×10−2N1.98×10−2S1.98×10−2
711
712Then, divide the numbers of moles by the smallest one (that is, 1.98×10−2) and, if the coefficients are not whole numbers, multiply them by the smallest whole number to get whole-number subscripts in the empirical formula:
713
714C9.91×10−21.98×10−2H9.90×10−21.98×10−2O2.0×10−21.98×10−2N1.98×10−21.98×10−2S1.98×10−21.98×10−2→C5H5O1N1S1−→−−−−−−−−omitsubscripts1C5H5ONS
715
716Thus, the empirical formula of the compound is C5H5ONS.
717
718± Percent Composition
719Percent composition refers to the mass percent of each element in a compound:
720
721mass percent=mass of elementmass of compound×100%
722
723For example, the percent composition of water, H2O, is 11.2% hydrogen and 88.8% oxygen. Therefore, a 100-g sample of water contains 11.2 g of hydrogen atoms and 88.8 g of oxygen atoms.
724
725The periodic table will be useful when doing this problem. You can access a periodic table by clicking the "Tools" link in the upper right corner of this page.
726
727Part A
728A hydrocarbon is a compound that contains mostly carbon and hydrogen. Calculate the percent composition (by mass) of the following hydrocarbon: C7H16.
729Enter the percentages of carbon and hydrogen numerically to four significant figures, separated by commas.
730Hint 1. How to approach the problem
731Start by assuming that you have exactly 1 mol of the compound. Then calculate the mass of carbon and the mass of hydrogen in this sample. Finally, make a percentage out of the mass of each element and the mass of the compound.
732Hint 2. Calculate the molar mass of the compound
733What is the molar mass of C7H16?
734Express your answer numerically in grams per mole using four significant figures.
735Hint 1. Definition of molar mass
736The molar mass of C7H16is the sum of the masses of 7 mol of C and 16 mol of H.
737ANSWER:
738100.2
739  g/mol  
740Correct
741Hint 3. Calculate the mass of carbon
742Calculate the mass of carbon in exactly 1 mol of C7H16.
743Express your answer numerically in grams using four significant figures.
744Hint 1. How to approach the problem
745There are 7 mol C in 1 mol C7H16. Each mole of C has a mass of 12.01 g.
746ANSWER:
74784.07
748  g  
749Incorrect; Try Again; 5 attempts remaining
750Hint 4. Calculate the mass of hydrogen
751Calculate the mass of hydrogen in exactly 1 mol of C7H16.
752Express your answer numerically in grams using four significant figures.
753ANSWER:
75416.13
755  g  
756ANSWER:
757carbon, hydrogen = 	
75883.90,16.10
759  %  
760Correct
761Notice that the sum of the percentages is 100%.
762Part B
763A certain metal hydroxide, M(OH)2, contains 32.8% oxygen by mass. What is the identity of the metal M?
764Enter the full name of the element.
765Hint 1. How to approach the problem
766Assume that you have exactly 1 mol of M(OH)2 and calculate the mass of oxygen present. Then use the mass percent formula to calculate the molar mass of the compound. Once you know the molar mass of the compound, you can figure out the molar mass of the metal by subtracting the mass of the hydroxide ions.
767Hint 2. Calculate the molar mass of the metal hydroxide
768Calculate the molar mass of M(OH)2 using the fact that it contains 32.8% oxygen by mass.
769Express your answer in grams per mole using four significant figures.
770Hint 1. How to rearrange the mass percent formula
771If you assume 1 mol of compound, the formula from the introduction becomes
772
773mass percent of O= mass of O in 1 mol of compoundmolar mass of compound×100%
774
775This can be rearranged to solve for the molar mass of the compound:
776
777molar mass of compound=mass of O in 1 mol of compoundmass percent of O×100%
778
779Hint 2. Calculate the mass of oxygen in one mole of the compound
780How many grams of oxygen are in 1 mol of M(OH)2?
781Express your answer numerically in grams.
782Hint 1. How to approach the problem
783First determine the number of moles of O in 1 mol of M(OH)2. Use the periodic table to determine the mass of each mole of O (the molar mass).
784Hint 2. Determine the number of moles of oxygen
785How many moles of oxygen are in 1 mol of M(OH)2?
786Express your answer as an integer.
787ANSWER:
7882
789Hint 3. Calculate the mass of one mole of oxygen
790What is the mass of 1 mol of O?
791Express your answer in grams to four significant figures.
792ANSWER:
79316.00
794  g  
795ANSWER:
796oxygen mass = 	
79732.0
798  g  
799ANSWER:
800molar mass = 	
80197.56
802  g/mol  
803Hint 3. Calculate the molar mass of the metal
804A certain metal hydroxide, M(OH)2, has a molar mass of 97.6 g/mol. Find the molar mass of the metal M.
805Express your answer in grams per mole using four significant figures.
806Hint 1. How to approach the problem
807You have already calculated the molar mass of the compound M(OH)2. Now determine the molar mass of (OH)2. The difference between the two numbers represents the molar mass of M:
808
809molar mass of M=[molar mass of M(OH)2]−[molar mass of (OH)2]
810
811ANSWER:
81263.54
813  g/mol  
814ANSWER:
815Copper
816Correct
817The hydroxide formula is Cu(OH)2.
818± Hydrates
819Hydrates are solid compounds that contain water molecules. When hydrates are heated, the water molecules evaporate, producing the anhydrous form of the compound.The formula of a hydrate may be written in two different ways. For example, aluminum chloride hexahydrate can be written as [Al(H2O)6]Cl3 or as AlCl3⋅6H2O. The first formula more accurately shows the bonding arrangement, whereas the second formula makes it easier to see that the anhydrous form of the compound is aluminum chloride.
820Part A
821Name the hydrate MgSO4⋅7H2O.
822Hint 1. Name the anhydrous salt
823What is the name of MgSO4?
824Enter the name with correct spelling.
825ANSWER:
826magnesium sulfate
827Hint 2. Identify the prefix for seven
828When naming compounds, Greek prefixes such as mono, di, and tri are often used. What Greek prefix is used to indicate seven?
829Enter the prefix.
830ANSWER:
831hepta
832ANSWER:
833magnesium sulfate heptahydrate
834Correct
835Magnesium sulfate heptahydrate, MgSO4⋅7H2O, is also called epsom salt. Epsom salt is often used in bath salts as well as in gardening.
836Part B
837If 19.0 g of MgSO4⋅7H2O is thoroughly heated, what mass of anhydrous magnesium sulfate will remain?
838Express your answer to three significant figures and include the appropriate units.
839Hint 1. How to approach the problem
840First, determine the molar mass of MgSO4⋅7H2O. Then determine what percentage of that mass is attributed to water, and what percentage is attributed to magnesium sulfate. This will allow you to determine the mass of just the magnesium sulfate in any sample of MgSO4⋅7H2O.
841Hint 2. Calculate the molar mass of the hydrate
842Using the periodic table, determine the molar mass of MgSO4⋅7H2O.
843Express your answer to four significant figures and include the appropriate units.
844ANSWER:
845246.5 gmol
846Hint 3. Calculate the molar mass of magnesium sulfate
847What is the molar mass of  MgSO4?
848Express your answer to four significant figures and include the appropriate units.
849ANSWER:
850120.4 gmol
851Hint 4. Calculate the percent composition of the hydrate
852What is the percent mass of magnesium sulfate in MgSO4⋅7H2O?
853Express your answer as a percent to four significant figures.
854ANSWER:
85548.84
856  %  
857ANSWER:
8589.28 g
859Correct
860Percent Composition and Formulas
861One of the inherently satisfying features of chemistry is that chemical materials form and interact in a rational and predictable manner. For example, one can predict a great deal about a particular kind of molecule by experimentally determining the percentage composition of the elements in that compound. This gives us the relative proportions of the elements in the molecule. For a molecule made up of elements A, B, and C, the proportions might be A:B:C2, meaning that there are two atoms of C for each atom of A and each atom of B. This may not be the actual description of the molecule (which might actually be A2B2C4), but it is the "reduced" version of that formula, called the empirical formula. The actual formula is some multiple of the empirical formula. To know the actual formula we need to know both the empirical formula and the molecular mass of the compound. This provides us with the multiplier value in whole units that must be applied to the empirical formula to get the actual formula.
862A compound is 40.0% C, 6.70% H, and 53.3% O by mass. Assume that we have a 100.-g sample of this compound.
863Part A
864What are the subscripts in the empirical formula of this compound?
865Enter the subscripts for C, H, and O, respectively, separated by commas (e.g., 5,6,7).
866Hint 1. Find how many moles of each element are involved
867How many moles of each element are present in 100 g of this compound?
868Enter the number of moles of C, H, and O, respectively, separated by commas (e.g. 5, 6, 7).
869Hint 1. How to find the masses
870To obtain the mass of each component, multiply the percentage of each component by the mass of the sample. For a 100-g sample, there are 40.0 g of C, 6.70 g of H, and 53.3 g of O.
871Hint 2. How to convert grams to moles
872To convert grams to moles, divide the mass by the molar mass.
873ANSWER:
874number of moles of C,H, O = 	
8753.33,6.65,3.33
876  mol  
877Hint 2. How to convert the mole ratio to integers
878To convert the ratio of moles of  C:H:O to the integers needed to determine the empirical formula, divide each number in the ratio by the smallest number in the ratio.
879ANSWER:
8801,2,1
881Correct
882The empirical formula of this compound is CH2O.
883Part B
884The molecular formula mass of this compound is 210 amu . What are the subscripts in the actual molecular formula?
885Enter the subscripts for C, H, and O, respectively, separated by commas (e.g., 5,6,7).
886Hint 1. How to approach the problem
887Based on the molecular mass of the compound, determine how many units of the empirical formula mass there are per actual mass of the molecule. The number of units is a multiplication factor (n) that can be multiplied by each subscript in the empirical formula to give the molecular formula.
888Hint 2. Calculate the empirical formula mass
889What is the mass of the empirical formula, CH2O?
890Express your answer in atomic mass units using four significant figures.
891ANSWER:
89230.03
893  amu  
894Correct
895Hint 3. Equation for the multiplication factor n
896Let n be the number by which the empirical formula mass is a factor of the molecular formula mass.  Find n using the following equation:
897
898n=molecularformulamassempiricalformulamass
899
900Express your answer as an integer.
901ANSWER:
902n = 	
9037
904ANSWER:
9057,14,7
906Correct
907Naming Binary Molecular Compounds
908Molecular compounds are usually composed solely of nonmetals.  A binary molecular compound is one in which the compound contains only two elements (regardless of how many atoms are present of each).  When naming binary molecular compounds, prefixes are used to specify the number of atoms of each element.  Take a moment to review some of the prefixes shown here.
909
910Prefix	Number
911mono	one
912di	two
913tri	three
914tetra	four
915penta	five
916hexa	six
917hepta	seven
918octa	eight
919nona	nine
920deca	ten
921For example, SF6 is named sulfur hexafluoride. Note that the prefix mono is not used in naming the first element. Also note that the second element in the name should end with the suffix ide.
922
923Part A
924What is the name for the compound IF7?
925Spell out the full name of the compound.
926Hint 1. Identify the number of atoms
927How many iodine atoms are in a molecule of IF7? How many fluorine atoms are there?
928Enter the number of iodine atoms followed by the number of fluorine atoms separated by a comma.
929ANSWER:
930Number of iodine atoms, fluorine atoms 
9311,7
932Hint 2. Identify the prefix for fluoride
933In naming this compound, what prefix would be used to specify the number of fluorine atoms per molecule?
934Enter the prefix.
935ANSWER:
936hepta
937ANSWER:
938iodine heptafluoride
939Correct
940The prefix mono- is not included on the first element, however if there were two or more iodine atoms, then the appropriate prefix would be necessary.
941Part B
942What is the name for the compound N2H4?
943Spell out the full name of the compound.
944Hint 1. Identify the number of atoms
945How many nitrogen atoms are in a molecule of N2H4? How many hydrogen atoms are there?
946Enter the number of nitrogen atoms first followed by a comma and then the number of hydrogen atoms.
947ANSWER:
9482,4
949Hint 2. Identify the prefix for nitrogen
950In naming this compound, what prefix would be used to specify the number of nitrogen atoms per molecule?
951Enter the prefix.
952ANSWER:
953di
954Hint 3. Identify the prefix for H
955In naming this compound, what prefix would be used to specify the number of hydrogen atoms per molecule?
956Enter the prefix.
957ANSWER:
958tetra
959ANSWER:
960dinitrogen  tetrahydride
961Correct
962Since there are 2 atoms of nitrogen in the molecule, the prefix di is necessary.
963± The Mass of One Sugar Molecule
964Matter is made up of atoms, but in a real chemistry lab, we measure substances in grams. Therefore, you need to be able to convert from atoms to grams, and vice versa.
965Sucrose, the sugar extracted from sugarcane and sugar beets, is an organic compound that contains 12 carbon atoms, 22 hydrogen atoms, and 11 oxygen atoms per molecule.
966Part A
967What is the chemical formula for sucrose?
968Express your answer as a chemical formula.
969ANSWER:
970C12H22O11
971Correct
972Sucrose has the chemical formula: C12H22O11.
973Part B
974How much does one sucrose molecule weigh in grams?
975Express your answer with the appropriate units.
976Hint 1. How to approach the problem
977Start by determining how many molecules are in one mole of sucrose. Then, using the molar mass of sucrose convert the number of moles of sucrose to the number of grams of sucrose.
978Hint 2. Determine the number of molecules per mole
979How many molecules are in one mole of sucrose?
980Express your answer numerically in molecules using four significant figures.
981Hint 1. Avogadro's number and the mole
982The mole is defined as the number of atoms in exactly 12 g of carbon-12. The mole equals 6.022×1023 items of whatever you are counting. This number is also called Avogadro's number, after Amadeo Avogadro (1776–1856).
983ANSWER:
9846.022×1023
985molecules
986Incorrect; Try Again; 4 attempts remaining
987Hint 3. Calculate the molar mass of sucrose
988What is the molar mass of sucrose?
989Express your answer with the appropriate units.
990ANSWER:
991342 gmol
992ANSWER:
993mass of one molecule of sucrose = 	
9945.68×10−22 g
995Correct
996Simple Organic Compounds
997Organic molecules generally consist of a chain of carbon atoms surrounded by hydrogen atoms or functional groups. The number of carbon atoms in the main chain is specified by using a prefix, the most common of which are listed in the table.
998Prefix	Meaning
999meth	1
1000eth	2
1001prop	3
1002but	4
1003pent	5
1004hex	6
1005hept	7
1006oct	8
1007non	9
1008A carbon atom will form four bonds to complete its octet, while oxygen will form two bonds and hydrogen will form one bond to complete its duplet.
1009For this problem, draw all hydrogen atoms explicitly.
1010Part A
1011Draw a heptane molecule.
1012Draw the molecule on the canvas by choosing buttons from the Tools (for bonds), Atoms, and Advanced Template toolbars. The single bond is active by default. Include all hydrogen atoms.
1013Hint 1. Determine the number of carbon atoms in heptane
1014How many carbon atoms are in a molecule of heptane?
1015Express your answer numerically as an integer.
1016ANSWER:
10177
1018Hint 2. Determine the number of bonds for each carbon atom
1019How many bonds does a carbon atom form in organic molecules?
1020Express your answer numerically as an integer.
1021ANSWER:
10224
1023Hint 3. Draw propane as an example
1024As an example, draw propane, an alkane with three carbon atoms.
1025CH3CH2CH3
1026
1027Draw the molecule on the canvas by choosing buttons from the Tools (for bonds), Atoms, and Advanced Template toolbars. The single bond is active by default. Include all hydrogen atoms.
1028ANSWER:
1029
1030ANSWER:
1031
1032Correct
1033The formula for heptane is C7H16. Heptane is classified as an alkane.
1034Part B
1035Draw a molecule of 2-pentanol.
1036Draw the molecule on the canvas by choosing buttons from the Tools (for bonds), Atoms, and Advanced Template toolbars. The single bond is active by default. Include all hydrogen atoms.
1037Hint 1. Determine the number of carbon atoms in pentanol
1038How many carbon atoms are in a molecule of pentanol?
1039Express your answer numerically as an integer.
1040ANSWER:
10415
1042Hint 2. Determine the meaning of the ol ending
1043What does the ending ol indicate?
1044ANSWER:
1045
1046the presence of a ring structure
1047the presence of a double bond
1048the presence of a triple bond
1049the presence of an −OH group
1050Hint 3. Determine the meaning of the number "2"
1051What does the number "2" indicate in 2-pentanol?
1052ANSWER:
1053
1054There are two pentanol molecules.
1055There are two carbon atoms.
1056There is an −O−H group on the middle carbon atom.
1057There are two −O−H groups.
1058There is an −O−H group on the second carbon atom.
1059Hint 4. How to place an −O−H group
1060Hydrogen atoms only form one bond whereas oxygen atoms can form two bonds. Therefore, you must attach the oxygen atom to the carbon chain rather than the hydrogen atom.
1061Correct:Incorrect:−C−O−H−C−H−O
1062
1063You must also explicitly show the bond between the oxygen atom and the hydrogen atom so that the system will be able to interpret your answer correctly.
1064Correct:Incorrect:−C−O−H−C−OH
1065
1066Hint 5. Classify butanol structures by the position of the −O−H group
1067To help you understand how the position of the −O−H group relates to the name of the compound, consider these structures of butanol. Match each structure to its name.
1068Drag the appropriate items to their respective bins.
1069ANSWER:
1070ResetHelp
1071    
10721-butanol
1073CH2CH2CH2CH3, with an OH group attached to the first (from left to right) carbon atom.CH3CH2CH2CH2, with an OH group attached to the fourth (from left to right) carbon atom.
10742-butanol
1075CH3CHCH2CH3, with an OH group attached to the second (from left to right) carbon atom.CH3CH2CHCH3, with an OH group attached to the third (from left to right) carbon atom.
1076ANSWER:
1077
1078Correct
1079The formula for pentanol is C5H12O or C5H11OH. Pentanol is classified as an alcohol.
1080Part C
1081What can be said about the relationship between the number of hydrogen atoms and the number of carbon atoms in an alkane or alcohol with only single bonds?
1082Hint 1. How to approach the problem
1083Notice in the structures from Parts A and B that the carbon atoms at each end of the chain are bonded to three hydrogen atoms each. However, the inner carbon atoms are either directly or indirectly bonded to two hydrogen atoms each.
1084Think about how you could predict the number of hydrogen atoms for a carbon chain of any length. Each carbon atom is bonded to at least two hydrogen atoms, and each of the end carbon atoms has one extra hydrogen atom.
1085
1086Hint 2. Count the hydrogen atoms in heptane
1087Look at the heptane molecule from Part A. There are seven C atoms. How many hydrogen atoms are there?
1088Express your answer numerically as an integer.
1089ANSWER:
109016
1091  H atoms
1092Hint 3. Count the hydrogen atoms in pentanol
1093Look at the pentanol molecule from Part B. There are five C atoms. How many hydrogen atoms are there?
1094Express your answer numerically as an integer.
1095ANSWER:
109612
1097  H atoms
1098ANSWER:
1099To arrive at the number of hydrogen atoms, 
1100triple the number of carbon atoms, then subtract five.
1101double the number of carbon atoms, then add two.
1102triple the number of carbon atoms, then subtract three.
1103there is no consistent formula.
1104double the number of carbon atoms.
1105triple the number of carbon atoms.
1106Correct
1107For this reason, the general formula for alkanes is expressed as CnH2n+2. Similarly, the general formula for alcohols could be expressed as CnH2n+2O or CnH2n+1OH.
1108Empirical Formula from Experimental Data
1109In a student experiment, the empirical formula of a copper halide was found by adding aluminum metal to an aqueous solution of the halide, displacing copper metal. The copper metal was filtered, washed with distilled water, dried, and weighed; three separate determinations were performed. The copper halide solution contained 42.62 g of copper chloride per liter. The student recorded the following experimental data.
1110
1111Trial	Volume of copper chloride solution
1112(mL)	Mass of filter paper
1113(g)	Mass of filter paper with copper
1114(g)
1115A	49.6	0.908	1.694
1116B	48.3	0.922	1.693
1117C	42.2	0.919	1.588
1118Part A
1119Write the empirical formula of copper chloride based on the experimental data.
1120Express your answer as a chemical formula.
1121Hint 1. Determine the mass of copper in each trial
1122What mass of copper was collected in each trial?
1123Express the masses of copper in grams from trial A, trial B, and trial C, respectively, separated by commas.
1124Hint 1. How to find the mass of copper
1125The mass of copper is the difference between the mass of the filter paper with copper and the mass of the filter paper.
1126ANSWER:
1127mass of copper in trials A, B, C = 	
11280.786,0.771,0.669
1129  g  
1130Hint 2. Determine the mass of copper chloride in each trial
1131The concentration of the copper chloride solution was 42.62 g/L. Based on this concentration and the volume from each trial, what mass of copper chloride was used in each trial?
1132Express the masses of copper chloride in grams used in trial A, trial B, and trial C, respectively, separated by commas.
1133Hint 1. Identify how to find mass from concentration and volume
1134If C is the concentration of the solution in grams per liter (g/L) and V is the volume of solution in liters (L), which operation will give units of grams?
1135ANSWER:
1136
1137C/V
1138V/C
1139CV
1140Hint 2. Convert the volumes to liters
1141Convert these volumes from milliliters to liters.
1142
1143Trial	Volume of copper chloride solution
1144(mL)
1145A	49.6
1146B	48.3
1147C	42.2
1148Express the volumes from trials A, B, and C, respectively, in liters separated by commas.
1149ANSWER:
11504.96×10−2,4.83×10−2,4.22×10−2
1151  L  
1152ANSWER:
1153mass of copper chloride in trials A, B, C = 	
11542.11,2.06,1.80
1155  g  
1156Hint 3. Determine the mass of chloride in each trial
1157The mass of a compound is the sum of the masses of its constituent elements. A sample of copper chloride, for example, has a mass equal to the sum of the masses of the copper and chloride in that sample:
1158
1159 
1160(mass of copper chloride) = (mass of copper) + (mass of chloride)
1161
1162Given the masses of copper and of copper chloride you calculated for each trial, what is the mass of chloride for each trial?
1163
1164Express the masses of chloride in trial A, trial B, and trial C, respectively, separated by commas.
1165ANSWER:
1166mass of chloride in trials A, B, C = 	
11671.33,1.29,1.13
1168  g  
1169Hint 4. Determine the average number of moles of copper
1170Convert the average mass of copper to moles.
1171Express your answer numerically in moles.
1172Hint 1. Calculate the average mass of copper
1173What is the average mass of copper across the three trials?
1174
1175Trial	Mass of copper
1176(g)
1177A	0.786
1178B	0.771
1179C	0.669
1180Express your answer numerically in grams.
1181ANSWER:
11820.742
1183  g  
1184Hint 2. Find the molar mass of copper
1185According to the periodic table, what is the molar mass of copper, Cu?
1186Express your answer numerically in grams per mole.
1187ANSWER:
118863.5
1189  g/mol  
1190ANSWER:
11911.17×10−2
1192  mol  
1193Hint 5. Determine the average number of moles of chloride
1194Convert the average mass of chloride to moles.
1195Express your answer numerically in moles.
1196Hint 1. Calculate the average mass of chloride
1197What is the average mass of chloride across the three trials?
1198
1199Trial	Mass of chloride
1200(g)
1201A	1.33
1202B	1.29
1203C	1.13
1204Express your answer numerically in grams.
1205ANSWER:
12061.25
1207  g  
1208Hint 2. Find the molar mass of chloride
1209According to the periodic table, what is the molar mass of Cl?
1210Express the molar mass numerically in grams per mole.
1211ANSWER:
121235.5
1213  g/mol  
1214ANSWER:
12153.52×10−2
1216  mol  
1217ANSWER:
1218CuCl3
1219Correct
1220Part B
1221Is the formula CuCl3 reasonable?
1222Hint 1. Determine the charge of chlorine
1223What is the charge of the chloride ion?
1224Express the charge numerically (e.g., +1).
1225ANSWER:
1226-1
1227Hint 2. Determine the charge of copper
1228The charge of the chloride ion is given in the first hint for this part (Part B). According to the data, three chloride ions combine with one copper ion to form a neutral compound CuCl3. Given this information, what is the charge of the copper ion in CuCl3?
1229Express the charge numerically (e.g., +1).
1230ANSWER:
1231+3
1232ANSWER:
1233
1234Yes, CuCl3 is reasonable because copper usually has a +3 charge.
1235Yes, CuCl3 is reasonable because copper usually has a +1 or +2 charge.
1236No, CuCl3 is not reasonable because copper usually has a +1 or +2 charge.
1237No, CuCl3 is not reasonable because copper usually has a +3 charge.
1238Correct
1239± Empirical Formula by Combustion Analysis
1240Learning Goal:
1241To use combustion analysis data to determine an empirical formula
1242A molecular formula expresses the number of each kind of atom in a molecule. For example, the molecular formula for propene, C3H6, indicates three carbon atoms and six hydrogen atoms per molecule. This also means that one mole of propene contains three moles of carbon and six moles of hydrogen.An empirical formula expresses the mole ratio of the elements. The empirical formula for propene is CH2, indicating twice as much hydrogen as carbon. When analyzing unknown compounds in a lab, it is often possible to identify the mole ratios, and thus the empirical formula, but not the molecular formula.Notice that the molecular mass of propene, 3(12)+6(1)=42amu, is a multiple of the empirical formula mass, 1(12)+2(1)=14amu .
1243An unknown compound contains only carbon, hydrogen, and oxygen (CxHyOz). Combustion of 2.50 g of this compound produced 3.67 g of carbon dioxide and 1.50 g of water.
1244Part A
1245How many moles of carbon, C, were in the original sample?
1246Express your answer to three significant figures and include the appropriate units.
1247Hint 1. How to approach the problem
1248All of the carbon in carbon dioxide came from the original sample. So first convert the mass of CO2 to moles using the molar mass. Then determine the number of moles of C in that amount of CO2.
1249Hint 2. Calculate the number of moles of carbon dioxide
1250Convert the mass of a sample of carbon dioxide, 3.67 g , to the number of moles of carbon dioxide.
1251Express your answer to three significant figures and include the appropriate units.
1252Hint 1. Determine the molar mass of carbon dioxide
1253What is the molar mass of CO2?
1254Express your answer to three significant figures and include the appropriate units.
1255Hint 1. Definition of molar mass
1256The molar mass of CO2 is the sum of the masses of one mole of C and two moles of O.
1257Hint 2. Determine the molar mass of C
1258According to the periodic table, what is the molar mass of C?
1259Express your answer to two decimal places and include the appropriate units.
1260ANSWER:
126112.01 gmol
1262ANSWER:
126344.0 gmol
1264ANSWER:
12658.33×10−2 mol
1266ANSWER:
12678.33×10−2 mol
1268Correct
1269Since there is one mole of carbon for every mole of carbon dioxide, the number of moles of each are equivalent.
1270Part B
1271How many moles of hydrogen, H, were in the original sample?
1272Express your answer to three significant figures and include the appropriate units.
1273Hint 1. How to approach the problem
1274All of the hydrogen in water came from the original sample. So first convert the mass of H2O to moles using the molar mass. Then determine the number of moles of H in that amount of H2O.
1275Hint 2. Calculate the number of moles of water
1276Convert the mass of a sample of water, 1.50 g , to the number of moles of water.
1277Express your answer to three significant figures and include the appropriate units.
1278Hint 1. Determine the molar mass of water
1279What is the molar mass of water?
1280Express your answer to three significant figures and include the appropriate units.
1281Hint 1. Definition of molar mass
1282The molar mass of H2O is the sum of the masses of two moles of H and one mole of O.
1283Hint 2. Determine the molar mass of H
1284According to the periodic table, what is the molar mass of H?
1285Express your answer to three significant figures and include the appropriate units.
1286ANSWER:
12871.01 gmol
1288ANSWER:
128918.0 gmol
1290ANSWER:
12918.33×10−2 mol
1292ANSWER:
12930.167 mol
1294Correct
1295Since there are two moles of hydrogen for every mole of water, the number of moles of hydrogen is twice the number of moles of water.
1296Part C
1297If 2.50 g of the unknown compound contained 8.33×10−2 mol of C and 0.167 mol of H, how many moles of oxygen, O, were in the sample?
1298Express your answer to three significant figures and include the appropriate units.
1299Hint 1. How to approach the problem
1300Because O2 is a reactant in combustion, we can't say that all the oxygen in CO2 and H2O came from the unknown. But we do know that the sum of all the elements in the unknown is equal to the total mass of the unknown. In other words,(mass of C in sample)+(mass of H in sample)+(mass of O in sample)= 2.50 g You determined the number of moles of carbon and hydrogen in Parts A and B. Next, use these values to calculate the masses of carbon and hydrogen in the original sample. Finally, solve for the mass of O in the sample.
1301Hint 2. Calculate the mass of the hydrogen
1302The unknown sample contained 0.167 mol of H. Convert this to grams.
1303Express your answer to three significant figures and include the appropriate units.
1304Hint 1. Determine the molar mass of H
1305What is the molar mass of H?
1306Express your answer to three significant figures and include the appropriate units.
1307ANSWER:
13081.01 gmol
1309ANSWER:
13100.168 g
1311Hint 3. Calculate the mass of carbon
1312The unknown sample contained 8.33×10−2 mol of C. Convert this to grams.
1313Express your answer to three significant figures and include the appropriate units.
1314ANSWER:
1315mass of carbon = 	
13161.00 g
1317Hint 4. Calculate the mass of oxygen
1318Given the sample mass of 2.50 g , as well as the masses of carbon, 1.00 g , and hydrogen, 0.168 g , in the sample, calculate the mass of oxygen in the sample.
1319Express your answer to three significant figures and include the appropriate units.
1320ANSWER:
1321mass of oxygen in sample = 	
13221.33 g
1323ANSWER:
13248.32×10−2 mol
1325Correct
1326Part D
1327What is the empirical formula of a substance that contains 8.33×10−2 mol of carbon, 0.167 mol of hydrogen, and 8.32×10−2 mol of oxygen?
1328Express your answer as a chemical formula.
1329Hint 1. Definition of empirical formula
1330We now know that there is a 8.33×10−2:0.167:8.32×10−2 mole ratio of carbon:oxygen:hydrogen in the unknown. However it would not be proper to express the empirical formula as C 8.33×10−2 H 0.167 O 8.32×10−2 , or even as C 83.3 H 167 O 83.2 . Instead, we must find the most reduced set of three whole numbers that have the same ratio as these values.
1331Hint 2. Calculate the minimum molar ratios of the elements in the compound
1332What is the smallest whole-number ratio that expresses the relative moles of carbon, oxygen, and hydrogen in the unknown sample?
1333Express the relative moles of carbon, oxygen, and hydrogen as whole numbers separated by commas (e.g., 1,2,3).
1334Hint 1. How to approach the problem
1335First, identify which of these three values (8.33×10−2, 0.167, or 8.32×10−2) is the smallest. Then, divide each of the three values by the smallest one.
1336ANSWER:
13371.00,2.00,0.999
1338ANSWER:
1339CH2O
1340Correct
1341Part E
1342Determine the molecular formula for the unknown if the molecular mass is 60.0 amu and the empirical formula is CH2O.
1343Express your answer as a chemical formula.
1344Hint 1. Relating the empirical formula and molecular mass
1345The molecular formula is an integer multiple of the empirical formula. This integer is calculated by dividing the molecular mass to the empirical formula mass.
1346Hint 2. Calculate the empirical formula mass
1347Given the empirical formula of CH2O, calculate the empirical formula mass.
1348Express your answer to three significant figures and include the appropriate units.
1349Hint 1. Finding the formula mass
1350To calculate the formula mass, add the atomic masses of all the atoms in the formula.
1351ANSWER:
1352empirical mass = 	
135330.0 amu
1354Hint 3. Calculate the ratio of the molecular mass to empirical mass
1355How many times greater is the molecular mass (60.0 amu) than the empirical formula mass (30.0 amu)?
1356Express your answer numerically as an integer.
1357ANSWER:
13582
1359ANSWER:
1360C2H4O2
1361Correct
1362Acetic acid, a component in vinegar, is one compound with the molecular formula C2H4O2.
1363A Formula for Formulas
1364Learning Goal:
1365To understand the strategy for writing chemical formulas.
1366Chemistry can be thought of as a language, and, like any language, there are rules that govern usage. In chemistry these rules are particularly strict and invariant, being governed by the properties of the elements that combine into the chemical words or formulas.
1367
1368In any chemical formula, the subscripts indicate the number of each type of atom in one molecule or formula unit of the compound. For example, one molecule of C3H6O3 contains three carbon atoms, six hydrogen atoms, and three oxygen atoms. One NF3 molecule contains one nitrogen atom and three fluorine atoms. One formula unit of Na2S contains two sodium atoms and one sulfur atom.
1369
1370There are three main types of chemical formulas: organic, molecular inorganic, and ionic.
1371
1372C3H6O3 is an organic compound because it contains predominantly carbon and hydrogen atoms.
1373NF3 is a molecular inorganic compound because it does not contain mainly carbon and it contains only nonmetal atoms.
1374Na2S is an ionic compound because it is made up of both metal and nonmetal atoms.
1375
1376
1377Organic compounds
1378Organic formulas follow the convention of placing C first in the formula, followed by H, followed by the remaining symbols in alphabetical order.
1379Part A
1380Which of the following could be the structure of C3H6O3?
1381There are 3 structures. Structure A is CH3CH2C with an OH group single-bonded to the third (from left to right) carbon and an O atom double-bonded to the third carbon. Structure B is CH3OCH2C with an OH group single-bonded to the third (from left to right) carbon and an O atom double-bonded to the third carbon. Structure C is CH2OCH2 with an OH group attached to each carbon.
1382ANSWER:
1383
1384a
1385b
1386c
1387Correct
1388Part B
1389Nitrobenzene contains 5 hydrogen atoms, 6 carbon atoms, 2 oxygen atoms, and 1 nitrogen atom. What is the chemical formula for nitrobenzene?
1390Express your answer as a chemical formula.
1391Hint 1. Choose a convention
1392Which formula convention should be used for nitrobenzene?
1393ANSWER:
1394
1395ionic
1396organic
1397inorganic
1398ANSWER:
1399C6H5NO2
1400Correct
1401Molecular inorganic compounds
1402Molecular inorganic formulas generally follow the convention of placing elements of groups 13 to 15 first (in that order), followed by the rest of the symbols starting with those furthest to the left in the periodic table. Elements in the same column are listed alphabetically.
1403Part C
1404A certain compound is made up of two chlorine atoms, one carbon atom, and one oxygen atom. What is the chemical formula of this compound?
1405Express your answer as a chemical formula.
1406Hint 1. Choose a convention
1407Which formula convention should be used for this compound?
1408ANSWER:
1409
1410ionic
1411organic
1412inorganic
1413ANSWER:
1414COCl2
1415Correct
1416Ionic compounds
1417Ionic compounds are always made up of a cation (positive ion) and an anion (negative ion). The cation is always listed first in the formula.
1418
1419The hard part about writing ionic formulas is determining how many of each ion is needed to make a neutral compound. In Na2S, the cation is Na+ and the anion is S2−. Na forms a cation with a +1 charge because it must lose an electron to achieve 10 electrons like Ne, the nearest noble gas to Na in atomic number. S forms an anion with a −2 charge because it must gain two electrons to have 18 electrons like Ar, the nearest noble gas to S in atomic number.
1420Na+ and S2− must now be combined in such a way that the compound is neutral. Two Na+ ions have a total charge of +2 and one S2− ion has a total charge of −2. Therefore, Na2S is a neutral compound because the total charge sums to zero.
1421
1422Part D
1423Aluminum fluoride contains only aluminum and fluorine. What is the formula for this compound?
1424Express your answer as a chemical formula.
1425Hint 1. Choose a convention
1426Which formula convention should be used for aluminum fluoride?
1427ANSWER:
1428
1429ionic
1430organic
1431inorganic
1432Hint 2. Determine the ionic charges
1433What are the charges on the aluminum ion and the fluoride ion?
1434Enter the charge on aluminum followed by the charge on fluorine separated by a comma (e.g., -5,+6).
1435Hint 1. How to find the charge
1436The atomic number of F is 9. The closest noble gas to F in atomic number is Ne (whose atomic number is 10). The atomic number of Al is 13. Therefore, Ne is also the closest noble gas to aluminum. To find the charges on F and Al, determine how many electrons each one must gain or lose to have 10 electrons like Ne.
1437ANSWER:
1438+3,-1
1439Hint 3. Symbol order in ionic compounds
1440By convention, ionic formulas are written with the cation first, followed by the anion.
1441ANSWER:
1442AlF3
1443Correct
1444Score Summary:
1445Your score on this assignment is 88.6%.
1446You received 21.27 out of a possible total of 24 points.