Chapter 3 Homework Packet (1-­‐35; 37-­‐80 odd; 81-­‐135 every other odd) 1. The atomic mass of rhenium is 186.2. Given that 37.1% of natural rhenium is rhenium-­‐185, what is the other stable isotope? When we determine an average atomic mass for an element on the periodic table we create what we call a weighted average. Instead of simply averaging the masses of all of the different isotopes, we multiply the mass of each isotope by the abundance (fraction) of the isotope in nature and then add these values. This assigns an “importance” to each mass in the average—the greater an isotope’s abundance in nature, the more important it is in contributing to the stated mass on the periodic table. In this problem, because we have the final averaged mass (186.2 atomic mass units), and the abundance and mass of one of the isotopes, we can create a formula that solves for the other mass. The weighted mass of 185Re is (.371)(185)—use the fraction rather than the percent—and that of the other isotope would be (.629)(mass of other isotope). The sum of these would be 186.2. Subtract the weighted mass of 185Re from the total and then divide by .629 to get the mass of the other isotope. The number you get will not usually be a whole number exactly, but will be close enough that you can easily round to the correct mass number for the isotope—in this case, 187, or 187Re. (.371)(185) + (.629)(mass of other isotope) = 186.2
mass of other isotope =
186.2 − (.371)(185)
= 186.9 = 187 .629
Answer 187Re 2. Consider the element indium, atomic number 49, atomic mass 114.8 g. The nucleus of an atom of indium-­‐112 contains Protons Neutrons Electrons Remember that when we associate a number with an atom of an element—for example, indium-­‐112, the number represents the mass number, which is the sum of the protons and neutrons. Because we know that the atomic number is the number of protons, the number of protons in this atom is 49. As the sum of neutrons and protons is 112, the number of neutrons is 112-­‐49 =63. For a neutral atom, the number of protons equals the number of electrons, and so, 49. However, the question asks for the number of electrons in the nucleus (be careful!), so, 0. Answer 49-­‐protons 63-­‐neutrons 0-­‐electrons 3. A hypothetical element consists of two isotopes of masses 86.95 amu and 88.95 amu with abundances of 35.5% and 64.5%, respectively. What is the average atomic mass of this element? Copyright © Cengage Learning. All rights reserved.
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When we determine an average atomic mass for an element on the periodic table we create what we call a weighted average. Instead of simply averaging the masses of all of the different isotopes, we multiply the mass of each isotope by the abundance (fraction) of the isotope in nature and then add these values. This assigns an “importance” to each mass in the average—the greater an isotope’s abundance in nature, the more important it is in contributing to the stated mass on the periodic table. In this problem we are given the abundances and masses of both isotopes so we can simply multiply the corresponding abundances and masses and add the results together to get the weighted average mass. 3 SF Answer 88.2amu 4. Naturally occurring copper exists in two isotopic forms: 63Cu and 65Cu. The atomic mass of copper is 63.55 amu. What is the approximate natural abundance of 63Cu? You know that the sum of the weighted mass values for each isotope equals 63.55. Right from the start, because the overall average mass value is so much closer to 63 than to 65 you should immediately see that the 63Cu isotope will be much more abundant. In fact, based on the fact that the value is about ¼ of the difference between the two values, the relative abundance of the carbon 63 atom must be about ¾ while that of the carbon 65 atom must be about ¼ . But, we can get closer to this fairly quickly. If we say the abundance of 63Cu is “x,” then the abundance of 65Cu must be 1-­‐x. We can then create a sum adding the product of the abundance of 63Cu and 63 amu—(x)(63)—and the product of the abundance of 65Cu and 65 amu—(x)(65), equaling 63.55. Then solve for x, which will equal the abundance of 63Cu. (x)(63) + (1− x)(65) = 63.55
63x + 65 − 65x = 63.55
−2x = 63.55 − 65
x = .725
Answer .725 or 72.5% 5. Naturally occurring element X exists in three isotopic forms: X-­‐28 (27.977 amu, 92.23% abundance), X-­‐29 (28.976 amu, 4.67% abundance), and X-­‐30 (29.974 amu, 3.10% abundance). Calculate the atomic weight of X. What is the identity of element X? When we determine an average atomic mass for an element on the periodic table we create what we call a weighted average. Instead of simply averaging the masses of all of the different isotopes, we multiply the mass of each isotope by the abundance Copyright © Cengage Learning. All rights reserved.
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(fraction) of the isotope in nature and then add these values. This assigns an “importance” to each mass in the average—the greater an isotope’s abundance in nature, the more important it is in contributing to the stated mass on the periodic table. In this problem we are given the abundances and masses of all three isotopes so we can simply multiply the corresponding abundances and masses and add the results together to get the weighted average mass. We should then be able to identify the element by finding what element on the periodic table has this mass. 3 SF (remember when adding that the number of decimal places determines the number of sig figs). Answer 28.09 amu (silicon) 7. The average mass of a carbon atom is 12.011. Assuming you were able to pick up only one carbon unit, the chances that you would randomly get one with a mass of 12.011 is a) 12.011% b) greater than 50% c) 0% d) 0.011% e) about 12% Remember that the average mass quoted on the periodic table is a mass that is the weighted average of all isotopes of carbon. Therefore, no atom of carbon has a mass of 12.011. An atom has a mass that is basically a whole number because it has a definite number of protons and neutrons. An atom with a mass that contains a decimal fraction would have fractions of neutrons or protons and this cannot happen. Answer C 8. What is the mass of 4 atom(s) of copper in grams? When asking for grams of a substance, the first thought in your brain should be g=(mol)(mm). Inspecting this formula you immediately know that you know the molar mass of the carbon atom (63.55 amu) but you don’t know the number of moles of carbon. But, you do know that you can get moles by dividing a number of particles by Avogadro’s number. Because you are only adding one additional calculation, I would do this in line with the first equation and not write the second calculation out separately. 4 SF Answer 4.221 x10-­‐22 g Copyright © Cengage Learning. All rights reserved.
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9. Iron is biologically important in the transport of oxygen by red blood cells from the lungs to the various organs of the body. In the blood of an adult human, there are approximately 2.69 × 1013 red blood cells with a total of 2.90 g of iron. On the average, how many iron atoms are present in each red blood cell? (molar mass Fe = 55.85 g/mol) This would be a really good problem to work backwards on. Think, “I want the number of iron atoms present in each red blood cell—do I know a formula that can give me this?” Yes! Fe atoms per RBC = (mol Fe per RBC)(AN). You write this formula down and then inspect it, realizing that you don’t have a number of moles of Fe per red blood cell. “Then ask, do I know a formula that gives me number of moles?” Yes! mol Fe per RBC = (g Fe per RBC)/mm. You write this formula down (I would put this all in one line of calculation) and inspect it, realizing that you don’t have g Fe per RBC—but, you do have total grams of iron and the number of red blood cells, so you can divide these to get the number of grams of irons per RBC. Now you have all of the information—plug values in and solve. 3SF Answer 1.16 x 109 atoms Fe per RBC 10. You have a sample of zinc (Zn) and a sample of aluminum (Al). You have an equal number of atoms in each sample. Which of the following statements concerning the masses of the samples is true? zinc sample. a) The mass of the aluminum sample is more than the mass of the zinc sample, but it is not twice as great. b) The mass of the zinc sample is more than twice as great as the mass of the aluminum sample. c) The mass of the zinc sample is more than the mass of the aluminum sample, but it is not twice as great. d) The mass of the aluminum sample is more than twice as great as the mass of the e) The masses of each sample are equal. If you have one atom of zinc and one atom of aluminum you would have a mass of 65.39 amu vs 26.98 amu (zinc atom more than twice the mass of an aluminum atom). This leads you to realize that it doesn’t matter what the number of atoms is, as long as the number of atoms is the same for both, the mass of zinc will always be more than twice the mass of aluminum. Answer B 11. A sample of ammonia has a mass of 45.5 g. How many molecules are in this sample? As you are reading the question your brain immediately thinks, “molecules equals (mol)(AN),” so you write this equation down. Inspecting the equation you realize Copyright © Cengage Learning. All rights reserved.
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that you do not have mol of ammonia but you have been given grams and you can determine moles from grams using mol = g/mm. This means you will also have to carry out a molar mass calculation for ammonia—but, you should be able to do this in your head because you know that mm N = 14.01 g/mol and mm of 3 H = 3.03g/mol, so mm of NH3 = 17.04 g/mol. I would put all of this into a single line of calculations. Putting this together you should show: ⎛ g ⎞
⎛
⎞
45.5g
molecules = (mol)( AN ) = ⎜
( AN ) = ⎜
(6.022x1023 molecules / mole) = 1.61x1024 molecules ⎝ mm ⎟⎠
⎝ 17.04g / mol ⎟⎠
Answer 1.61 x 1024 molecules 12. How many moles of hydrogen chloride are contained in a 52.6-­‐g sample of this gas? a) 0.693 mol b) 1.44 mol c) 89.1 mol d) 52.2 mol e) 2.89 mol Develop you skills at estimating so you can easily do multiple choice questions quickly and without a calculator. You are asked for mol of HCl and you know mol=(g/mm). You are given grams and you can make a quick estimate of molar mass in your head by knowing that the mm of Cl is about 35.5 and the mm of H is about 1, so mm of HCl is about 36.5 g/mol. You see that the number of grams in two moles would be about 73 grams, so you know that the number of moles in 52.6 g is going to be slightly less than halfway between one and two moles. Only one answer fits this requirement. You should be able to answer this particular question in about 10-­‐15 seconds without a calculator. Answer B 13. What is the molar mass of propanol (C3H7OH)? a) 59.09 g/mol b) 36.07 g/mol c) 60.09 g/mol d) 30.03 g/mol e) 149.04 g/mol Develop you skills at estimating so you can easily do multiple choice questions quickly and without a calculator. You are asked for the molar mass and you can estimate this quickly if know the molar masses of C (12.01), H (1.01), and O (16.00). 3 x 12.01 is 36.03, plus 8.08 (for 8 H’s) is 44.11, plus 16 is about 60.11 g/mol. Close enough! Answer C 14. For which compound does 0.256 mole weigh 12.8 g? a) C2H4O b) CO2 c) CH3Cl d) C2H6 e) none of these Copyright © Cengage Learning. All rights reserved.
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Develop you skills at estimating so you can easily do multiple choice questions quickly and without a calculator. You have been given that 12.8 g of a compound is .256 mol. This is just a tiny bit more than ¼ of a mole, and 12.8 g is just slightly less than 13 g. Therefore, the molar mass of the compound should be a little less than 4 times a little less than 13 equals about 50 g/mol. Going through the list, C2H4O is about 44 (24+4+16) and immediately eliminated. You should know that CO2 is 44.01 and is immediately eliminated. CH3Cl is about 35.5+12+3=about 50—a good possibility. C2H6 is about 24+6=about 30 and quickly eliminated. CH3Cl is probably close enough to the correct answer that the risk is low enough to not consider the “none of the above answer.” Answer C 15. Roundup, an herbicide manufactured by Monsanto, has the formula C3H8NO5P. How many moles of molecules are there in a 304.3-­‐g sample of Roundup? a) 0.5556 b) 2.203 c) 1.800 d) 16.91 e) none of these Develop you skills at estimating so you can easily do multiple choice questions quickly and without a calculator. You should be able to create a running sum of the approximate molar mass of the compound by adding: about 36 + about 8 is 44, + about 14 is about 58, + 80 is about 138, + about 31 is about 179 g/mol. 2 mol of compound would be about 360 g so 304 g is somewhat less than 2 mol. A reasonable answer with a low risk of the “none of the above” possibility would be 1.800 mol. Answer C 16. Calculate the molar mass of calcium perchlorate. First, you need the correct formula. The –ate form of chlorine based oxoanion is ClO3-­‐. Therefore, the per-­‐ / -­‐ate form is ClO4-­‐. As calcium is in column 2A and would have a 2+ charge, it would take 2 anions to balance it, so Ca(ClO4)2. mm Ca(ClO4 )2 = 40.08 + 2(35.45) + 8(16.00) = 238.98 g / mol Answer 238.98g/mol 17. Phosphorus has the molecular formula P4, and sulfur has the molecular formula S8. How many grams of phosphorus contain the same number of molecules as 7.88 g of sulfur? Work backwards on this one—you want g P4 and you know g = (mol P4)(mm P4). Inspecting this formula you don’t have mol P4 but have been told that you will have the same number of molecules of P4 that you have of S8, which means that you will have the same number of moles of P4 that you have of S8. This means you could Copyright © Cengage Learning. All rights reserved.
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Chapter Error! Unknown document property name.: Error! Unknown document property name.
substitute mol S8 in for mol P4. Inspecting this formula you don’t have moles of S8, but you know that mol = (g/mm), and you have g of S8 and can determine its molar mass, so substitute this in for mol S8. Plug in values and solve—3 SF ⎛ g S8 ⎞
⎛
⎞
7.88g
g P = (mol P )(mm P ) = (mol S )(mm P ) =
(mm P ) =
((4)(30.97g / mol) = 3.80g P 4
4
4
8
4
⎜⎝ mm S ⎟⎠
8
4
⎜⎝ (8)(32.07g / mol) ⎟⎠
4
(NOTE: If you follow this pattern of solving a problem like this I think it is important to keep your information straight by labeling with the formula of the P4 and S8 as I have shown. Without doing this, for listing of the calculation can quickly become confusing, especially for someone trying to grade this.) Answer 3.80 g P4 18. A given sample of a xenon fluoride compound contains molecules of a single type XeFn, where n is some whole number. Given that 9.09 × 1020 molecules of XeFn weigh 0.370 g, calculate n. This problem might appear to be difficult either working forward or backward—I will show both ways. Working forward. You have been given a number of molecules of XeFn and therefore the ability to find the number of moles: mol = (molecules)/(AN). You have also been given a number of grams that this number of moles equals, so you can find the mm. Once you find the mm, you can subtract the mass of one Xe atom from this to get the total mass of F. Then you can divide this number by the mm of F to get the number of F atoms. molecules XeFn
9.09x1020 molecules
mol XeFn =
=
= 1.51x10−3 mol
AN
6.022x1023 molecules / mol
g
.370g
mm XeFn =
=
= 245.12 g / mol
mol 1.51x10−3 mol
mass of F = mm XeFn − mmXe = 245.12 g / mol − 131.3 g / mol = 113.82g / mol
n=
gF
113.82g / mol
=
= 5.99 = 6 atoms of F → XeF6
mm F 19.00g / mol
Working backward. In order to work backward in this scenario you will need to do some reasoning. To begin with, you know that the molar mass of a compound is the sum of the molar masses of all of the components: mm XeFn = (mm Xe) + (n)(mm F). This is your starting point to work backwards—rearranging to isolate n we get: mm XeFn − mm Xe
n=
We obviously know the mm of F and Xe. However, inspecting mm F
this formula we don’t know the mm of XeFn. But, we know mm = g/mol, and we do have a number of grams (.370). We also have the ability to find the number of moles because we have number of molecules, and we could divide this by AN to get moles. Plugging the values in, solve for n. Copyright © Cengage Learning. All rights reserved.
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mm XeFn = mm Xe + (n)(mm F )
mm XeFn − mm Xe 245.12 g / mol − 131.3g / mol
=
= 5.99 = 6 → XeF6
mm F
19.00g / mol
g XeFn
.370 g
.370 g
mm XeFn =
=
=
= 245.12g / mol
mol XeFn ⎛ molecules XeFn ⎞
9.09x1020 molecules
⎜⎝
⎟⎠ 6.022x1023 molecules / mol
AN
19. Phosphoric acid can be prepared by reaction of sulfuric acid with “phosphate rock” according to the equation: Ca3(PO4)2 + 3H2SO4 → 3CaSO4 + 2H3PO4 What is the molar mass of Ca3(PO4)2? To calculate molar mass you just need to multiply the molar masses of each element by the subscript of that element and then sum these results. Remember that if there is no subscript this means a value of one atom. Remember that for polyatomic ions the subscript after the parentheses multiplies the everything inside the parenetheses. (2 P and 8 0). If you can do the multiplications in your head just write out the product in the calculation—for example, for oxygen, you should be able to calculate 8 x 16 in your head to get 128.00—just write this down in your calculation. Make sure you include units in the final answer (g/mol). For molar masses we normally limit the number of digits to the hundredths place unless something else in the calculations limits it to fewer figures.. Answer 310.18 g/mol (3)(40.08) + (2)(30.97) + 128.00 = 310.18g / mol 20. Phosphoric acid can be prepared by reaction of sulfuric acid with “phosphate rock” according to the equation: Ca3(PO4)2 + 3H2SO4 → 3CaSO4 + 2H3PO4 How many oxygen atoms are there in 3.80 ng of Ca3(PO4)2? Remember that when you are asked to find number of atoms, you are going to need to use AN. In this case, ask the question, “do I know a formula that calculates number of atoms?” Yes—atoms = (mol O)(AN). Then, inspect this formula and realize that you do not have mol of O. Ask, “can I create a formula that calculates moles O?” Yes—mol O = (8)(mol Ca3(PO4)2). You know this because you can count the number of atoms of oxygen for every formula unit of Ca3(PO4)2. If there are 8 O atoms for every 1 formula unit of Ca3(PO4)2, there will be 8 moles of oxygen for every 1 mole of Ca3(PO4)2. Then, inspect this formula and realize that you do not have mol Ca3(PO4)2. Ask, “can I create a formula that will calculate mol Ca3(PO4)2? Yes—mol Ca3(PO4)2 = g (Ca3(PO4)2)/(mm Ca3(PO4)2). Then, inspect this formula and realize you do have all of the information (that is, if you calculated the mm of Ca3(PO4)2 in the previous problem—if not, you will have to solve for mm too). Plug it in, and then use this answer to answer the previous n=
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formula, and use that answer to answer the formula before that, etc, until you answer the original formula. Make sure you convert ng to g. It would also be easy enough to put all of the calculations in one line. 3 SF Answer 5.90 x 1013 atoms O 21. What is the mass of a 6.267-­‐mol sample of sodium hydroxide? a) 40.00 g b) 250.7 g c) 150.4 g d) 6.382 g e) 0.1567 g We want g of NaOH. Remember, g NaOH= (mol NaOH)(mm NaOH). Inspecting this formula we see we have been given mol NaOH and just need to calculate the mm of NaOH. I would do this within the line of the calculation. 4SF Answer 250.7 g gNaOH = (mol NaOH )(mm NaOH ) = (6.267 mol)((22.99 + 16.00 + 1.01)g / mol) = 250.7 g 22. How many atoms of hydrogen are present in 3.18 g of ammonia? To answer this question you will need to know how many atoms of H are in one molecule of ammonia—the formula is NH3 (you should know this formula), so there are 3 moles of H for every mole of NH3. Remember that when you are asked to find number of atoms, you are going to need to use AN. In this case, ask the question, “do I know a formula that calculates number of atoms of hydrogen?” Yes—atoms = (mol H)(AN). Then, inspect this formula and realize that you do not have mol of H. Ask, “can I create a formula that calculates moles H?” Yes—mol H = (3)(mol NH3) . You know this because you can count the number of atoms of H for every molecule of NH3. If there are 3 H atoms for every 1 molecule of NH3, there will be 3 moles of H for every 1 mole of NH3. Then, inspect this formula and realize that you do not have mol NH3. Ask, “can I create a formula that will calculate mol NH3? Yes—mol NH3 = g (NH3)/(mm NH3). Then, inspect this formula and realize you do have all of the information—you can calculate the molar mass of NH3 in your head—(14.01 + 3.03=17.04 g/mol)—or, add it into the calculation. Plug it in, and then use this answer to answer the question. It would be easy enough to put the entire set of calculations in the line of the calculation. 3SF ⎛ g NH 3 ⎞
⎛
⎞
3.18g
atoms H = (mol H )( AN ) = (3)(mol NH )( AN ) = (3)
( AN ) = (3)
(6.022x1023 atoms / mol) = 3.37x1023 atoms
3
Answer 3.37 x 1023 atoms Copyright © Cengage Learning. All rights reserved.
⎜⎝ mm NH ⎟⎠
3
⎝⎜ (14.01+ 3.03)g ⎠⎟
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23. The molar mass of the compound formed by potassium and oxygen is a) 94.2 g/mol b) 71.1 g/mol c) 55.1 g/mol d) 87.1 g/mol e) 133.3 g/mol You first have to determine what the compound formed by potassium and oxygen is. As you are given a metal and a non-­‐metal it will be an ionic compound. Oxygen is in column 6A and so will have a charge of -­‐2. Potassium is in column 1A and so will have a charge of +1. You need two K+ to balance one O2-­‐-­‐-­‐so, K2O. Then just determine the molar mass in the usually way. Multiply the molar masses of each element by the subscript of that element and then sum these results. 3 SF Answer 94.2 g/mol mm K 2O = (2)(39.10) + 16.00 = 94.2 g / mol 24. What is the molar mass of cryolite (Na3AlF6)? To calculate molar mass you just need to multiply the molar masses of each element by the subscript of that element and then sum these results. Remember that if there is no subscript this means a value of one atom. Make sure you include units in the final answer (g/mol). For molar masses we normally limit the number of digits to the hundredths place unless something else in the calculations limits it to fewer figures. Answer 209.95 g/mol mm Na3 AlF6 = (3)(22.99) + 26.98 + (6)(19.00) = 209.95 g / mol 25. One molecule of a compound weighs 2.56 × 10–22 g. The molar mass of this compound is: a) 2.35 g/mol b) 649 g/mol c) 146 g/mol d) 154 g/mol e) none of these Remember that molar mass is the mass of one mole of particles (molecules in this case). If you only have 1 particle, to get one mole of particles you would multiply this times AN. Therefore, to get the mass of one mole, as you know what the mass of the one molecules is, to find the mass for a whole mole of these particles, multiply this mass time AN. 3 SF (
2.56x10−22 g 6.022x1023 molecules
)(
) = 154.16 g / mol mol
molecule
26. NaHCO3 is the active ingredient in baking soda. How many grams of oxygen are in 0.42 g of NaHCO3? Copyright © Cengage Learning. All rights reserved.
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Remember that to find number of grams, you start with g = (mol)(mm), in this case, (g O)= (mol O)(mm O). Inspecting the formula you realize that you don’t have moles of oxygen, but from inspecting the chemical formula NaHCO3, you realize that for every formula unit of compound, you have 3 atoms of oxygen. This means that for every mole of NaHCO3 , you have 3 moles of oxygen. So, even if you don’t know how many moles of oxygen you really have, you know that it is 3 times the number of moles of NaHCO3, so substitute this into the formula. Inspecting this new formula, you see that you don’t have moles of NaHCO3 either. However, you know a formula to find this, too: mol NaHCO3= (g NaHCO3/ mm NaHCO3). You’ve been given g NaHCO3 and you know how to find mm NaHCO3 (I would just include this in line with the overall calculation). It should look something like this: 2 SF Answer .24 g 27. A compound is composed of element X and hydrogen. Analysis shows the compound to be 80% X by mass, with three times as many hydrogen atoms as X atoms per molecule. Which element is element X? a) He b) C c) F d) S e) none of these An important algebra skill is being able to see when an amount can be represented by a ratio. If I know that a certain mass represents a certain fraction or percentage of a total, (for example, if I know that 5 g of a substance represents 50%, or .5 of the total amount), we can use this ratio to find the number of grams of substance that is any other percent (for example, we can then tell that 10 g equals 100%, or that 7.5 g equals 75%): If 5 g is 50% we can create a ratio and use it to help us find some
5g
x
other percent :
=
Without even calculating we can see x = 10g
50% 100%
In this problem, we are told that the mass of substance X equals 80%, and that the rest (which must be 20%) is three times as many H atoms. Lets assume that there is 1 X atom and 3 H atoms. We know that the mass of 3 H atoms is 3.03 g/mol. So, 3.03 g = 20% of the total mass. We can use the ratio of 3.03:20% to help us find the ratio of X:80%. Then, solving for X we get that the atomic mass of the element must be 12.12, which is very close to the molar mass for carbon. Answer B 3.03g
Xg
=
(3.03)(80) = ( X )(20) X = 12.12 20% 80%
28. Which compound contains the highest percent by mass of hydrogen? Copyright © Cengage Learning. All rights reserved.
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a) HCl b) H2O c) H2SO4 d) H2S e) HF Remember that to find the mass percent of an element in a compound we divide the mass of that element by the molar mass of the compound and multiply times 100—
this answers the question, what percent of the total mass, is the mass of the element? Remember that if the element in question has more than one atom in the formula, the mass of that element must be multiplied by the subscript in order to account for the mass of all of its atoms. (Note: see after the calculations to see how I would really do this problem). In this problem, we have to check the mass percent of H in all of the compounds to see which contains the greatest mass percent of H. You could do this by carrying out complete mass percent calculations for each compound—to shorten things a little, you could accomplish the mm calculation within the line of the calculation rather than as separate calculations—this would look like: mass H
1.01 g / mol
mass percent H =
x100 =
x100 = 2.77%
mm HCl
(1.01+ 35.45)g / mol
mass percent H =
mass H
(2)1.01 g / mol
x100 =
x100 = 11.2%
mm H 2O
(2.02 + 16.00)g / mol
mass percent H =
mass H
(2)1.01 g / mol
x100 =
x100 = 1.96% mm H 2 SO4
(2.02 + 32.07 + 64.00)g / mol
mass percent H =
mass H
(2)1.01 g / mol
x100 =
x100 = 5.93%
mm H 2 S
(2.02 + 32.07)g / mol
mass H
1.01 g / mol
x100 =
x100 = 5.05%
mm HF
(1.01+ 19.00)g / mol
Based on these calculations, the compound with the greatest mass percent is H2O. However, this is a multiple choice question, and you would want to streamline this process, possibly even be able to do this in your head. Here is how I would do this. In my brain I would be using mass fractions instead of mass percents. Going down the list I would think: Mass fraction of H in HCl is about 1 in 35 (about 3%) Mass fraction of H in H2O is about 2 in 18 or 1 in 9 (about 10%) Mass fraction of H in H2SO4 is about 2 in 100 or 1 in 50 (about 2%) Mass fraction of H in H2S is about 2 in 32 or 1 in 16 (about 6%) Mass fraction of H in HF is about 1 in 20 (about 5%) Water has the greatest mass percent of H—for this question, these estimates are mass percent H =
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good enough, and could take you as little as 30 seconds, without a calculator. Answer B 29. A substance contains 35.0 g nitrogen, 5.05 g hydrogen, and 60.0 g of oxygen. How many grams of hydrogen are there in a 179-­‐g sample of this substance? A basic mass fraction problem is to give you the numbers of grams of elements in one sample of a compound so you can figure out the mass fraction of one (or more) of the elements, and then use this mass fraction to determine the number of grams of that element in another sample. Here you are given the grams of N, H, and O in a compound, so you can determine the mass fraction of H. Then, multiply the mass fraction of H just determined by the mass of the other sample (179 g) to find the number of grams in it. 3SF Answer 9.03 g gH
5.05g
mass fraction H =
=
= .0505
g H + g N + g O 5.05g + 35.0g + 60.0g
g H = (mass fraction H )(sample mass) = (.0505)(179g) = 9.03 g
30. How many grams of potassium are in 21.9 g of K2CrO7? This is a different version of the basic mass fraction (mass percent) problem. In this problem, you still need to figure out a mass fraction of potassium in the compound so you can multiply this by the given sample mass to get the number of grams in the given sample. However, you are not given the masses of the elements in the compound to calculate the mass fraction, but rather, you are just given the formula. In this case then, to determine the mass fraction of potassium, you need to use the atomic weights of each element to determine the mass fraction. You know that the mass of potassium in the compound will equal (2)(39.10) g/mol, that the mass of Cr will equal 52.00 g/mol, and that the mass of oxygen will equal (7)(16.00). Using these values calculate the mass fraction of potassium, and then multiply it times the given sample mass to get the number of grams of potassium in the sample. 3SF. Answer 7.07g mass K
(2)(39.10)
mass fraction of K =
=
= .3229
mm K 2CrO7 (2)(39.10) + 52.00 + 112.00
g K = (mass fraction K )(sample mass) = (.3229)(21.9g) = 7.07g
31. Chloric acid, HClO3, contains what percent hydrogen by mass? Remember that to find the mass percent of an element in a compound we divide the mass of that element by the molar mass of the compound and multiply times 100—
this answers the question, what percent of the total mass, is the mass of the element? Remember to show the molar mass calculation for the compound. 3SF Answer 1.20% mass H
1.01 g / mol
mass percent H =
x100 =
x100 = 1.20% mm HClO3
(1.01+ 35.45 + 48.00)g / mol
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32. A mixture of KCl and KNO3 is 44.20% potassium by mass. The percentage of KCl in the mixture is closest to This is a very difficult problem and you would not be likely to see one like this on the AP exam. However, it illustrates algebra skills that it would be good for you to have. In any mixture containing two components, you could represent the amount of one component with the variable x%, and therefore, the amount of the other component with (1 – x)%. Then, realize that you have been told that the fraction of the x% that is potassium, plus the fraction of the (1-­‐x)% that is potassium, equals 44.20% of the total mass of the mixture. This means we can write the following equation. We can easily determine the mass percents to put into the equation and solve for x, which will be the percentage of the mixture that is KCl. (mass percent of K in KCl)((x)% ) + (mass percent of K in KNO3 )((1− x)%) = 44.20%
mass percent of K in KCl =
mass K
39.10g / mol
x100 =
x100 = 52.45%
mm KCl
(39.10 + 35.45)g / mol
mass percent of K in KNO3 =
mass K
39.10g / mol
x100 =
x100 = 38.67%
mm KNO3
(39.10 + 14.01+ 48.00)g / mol
(52.45)x + (38.67)(1− x) = 44.20
52.45x + 38.67 − 38.67x = 44.20
13.78x = 5.53
x = 40.13%
33. A substance, A2B, has the composition by mass of 60% A and 40% B. What is the composition of AB2 by mass? a) 40% A, 60% B b) 50% A, 50% B c) 27% A, 73% B d) 33% A, 67% B e) none of these Remember that if we are given mass percents of components of compounds, we can pretend that we have 100 g of compound (assume 100g). Normally we would then use these masses to determine numbers of moles of each component and then determine the mole ratio to determine the empirical formula. In this case, however, we have been given an actual empirical formula (A2B) and this allows us to say that if we assume 100 g, we must assign 30 g to each A and 40 g to B. This means that the ratio of the mass of A to B is 3:4. So, in the compound AB2, we would say the ratio of the mass of A:B2 is 3:8. The percent by mass of A and B would then be: 3
8
mass% A =
*100 = 27%
mass% B =
= 73% 3+ 8
3+ 8
Answer C Copyright © Cengage Learning. All rights reserved.
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34. Each molecule of fluorescein contains 20 atoms of carbon (plus other atoms). The mass percent of carbon in fluorescein is 72.29%. What is the molar mass of fluorescein? Go back to question 27 and refresh your memory about the skill of being able to use ratios to help you solve problems. In this problem, whether you realize it or not, you are told that a certain mass represents 72.29%. If you know this, then you could use this ratio to say, “as this number of grams represents 72.29%, x grams must represent 100%.” Then you would equate these ratios and solve for x. So, what is the mass that represents 72.29%? You are given that 20 atoms of carbon is this mass. Multiplying 20 by the atomic mass of carbon gives: (20)(12.01)=240.20 g of C per mol of fluorescein. Plug this into the ratios and solve. 4 SF Answer 332.3 g/mol 240.20g
x
=
(240.20)(100) = 72.29x
x = 332.27 g / mol 72.29% 100%
35. A substance contains 23.0 g sodium, 27.0 g aluminum, and 114 g fluorine. How many grams of sodium are there in a 119-­‐g sample of the substance? A basic mass fraction problem is to give you the numbers of grams of elements in one sample of a compound so you can figure out the mass fraction of one (or more) of the elements, and then use this mass fraction to determine the number of grams of that element in another sample. Here you are given the grams of Na, Al, and F in a compound, so you can determine the mass fraction of Na. Then, multiply the mass fraction of Na just determined by the mass of the other sample (119 g) to find the number of grams in it. 3SF Answer 16.7 g g Na
23.0g
mass fraction Na =
=
= .1402
g Na + g Al + g F 23.0g + 27.0g + 114g
g H = (mass fraction H )(sample mass) = (.1402)(119g) = 16.7 g
37. The mineral mimetite has the formula Pb5(AsO4)3Cl. What mass percent of oxygen does it contain? Remember that to find the mass percent of an element in a compound we divide the mass of that element by the molar mass of the compound and multiply times 100—
this answers the question, what percent of the total mass, is the mass of the element? Remember that if the element in question has more than one atom in the formula, the mass of that element must be multiplied by the subscript in order to account for the mass of all of its atoms. In this case, there is a total of 12 oxygen atoms per formula unit, so this accounts for a mass of (12)(16.00 g/mol). Remember to show the molar mass calculation for the compound. 4SF Answer 12.90% mass O
(12)(16.00)g / mol
mass percent O =
x100 =
x100 = 12.90%
mm Pb5 ( AsO4 )3 Cl
1488.21g / mol
mm Pb5 ( AsO4 )3 Cl = (5)(207.2) + (3)(74.92) + (12)(16.00) + 35.45 = 1488.21g / mol
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39. The molar mass of an insecticide, dibromoethane, is 187.9 g/mol. Its molecular formula is C2H4Br2. What percent by mass of bromine does dibromoethane contain? Remember that to find the mass percent of an element in a compound we divide the mass of that element by the molar mass of the compound and multiply times 100—
this answers the question, what percent of the total mass is the mass of the element? Remember that if the element in question has more than one atom in the formula, the mass of that element must be multiplied by the subscript in order to account for the mass of all of its atoms. In this case, there is a total of two bromine atoms per molecule, so this accounts for a mass of (2)(79.90 g/mol). Remember to show the molar mass calculation for the compound. 4SF Answer 91.9% mass Br
(2)(79.90)g / mol
mass percent Br =
x100 =
x100 = 91.9%
mm CH 2 Br2
173.83g / mol
mm CH 2 Br2 = (12.01) + (2.02) + (2)(79.90) = 173.83g / mol
41. Ammonium sulfate, (NH4)2SO4, contains what percent nitrogen by mass? Remember that to find the mass percent of an element in a compound we divide the mass of that element by the molar mass of the compound and multiply times 100—
this answers the question, what percent of the total mass is the mass of the element? Remember that if the element in question has more than one atom in the formula, the mass of that element must be multiplied by the subscript in order to account for the mass of all of its atoms. In this case, there is a total of two nitrogen atoms per molecule, so this accounts for a mass of (2)(14.01 g/mol). Remember to show the molar mass calculation for the compound. 4SF Answer 21.20% mass N
(2)(14.01)g / mol
mass percent N =
x100 =
x100 = 21.20%
mm (NH 4 )2 SO4
132.17g / mol
mm mm (NH 4 )2 SO4 = (2)(14.01) + (8)(1.01) + 32.07 + 64.00 = 132.17g / mol
43. You take an aspirin tablet (a compound consisting solely of carbon, hydrogen, and oxygen) with a mass of 1.00 g, burn it in air, and collect 2.20 g of carbon dioxide and 0.400 g water. The molar mass of aspirin is between 170 and 190 g/mol. The molecular formula of aspirin is Remember that finding the empirical formula for an organic compound requires a little more work than for an inorganic compound. The main difference is that for an inorganic compound you will either receive the mass percents (mass fractions) or grams of the elements involved, while for organic compounds you will receive the masses of the combustion products (CO2 and H2O) and have to calculate the masses of C, H, and O from the combustion product masses. Three steps to get grams of C, H, and O. Calculate grams of carbon by multiplying mass fraction of carbon in CO2 by grams of carbon dioxide (see below for calculation). Copyright © Cengage Learning. All rights reserved.
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Calculate grams of hydrogen by multiplying mass fraction of hydrogen in H2O by grams of water (see below for calculation). Calculate grams of oxygen by subtracting g C and g H from original sample mass (see below for calculation). Why is it okay to say that the number of grams of the carbon in the carbon dioxide and the number of grams of hydrogen in the water represent the number of grams of C and H in the original sample? Because the only source of carbon in the carbon dioxide and the only source of hydrogen in the water is the original sample. There is no where else but the original sample from which the carbon and hydrogen could come. Once you have grams of each component, then you can determine the moles of each, and then compare these number of moles to get the mole ratio. Remember that if a mole ratio is not immediately apparent, divide each number of moles by the smallest number of moles. This will give you a ratio that is more obvious, although you may still need to multiply each number of moles by some factor to get a ratio of whole numbers, which will be the empirical formula. (see below for calculations) In this case, all mole values needed to be divided by .222. The resulting values then needed to be multiplied by 4 to give the whole number ratio of C:H:O of 9:8:4, meaning the empirical formula is: C9H8O4 Finally, if you are given the molecular mass of the compound, you can divide the molecular mass by the empirical formula mass to determine the factor you need to multiply the empirical formula by to get the molecular formula. The molar mass of the empirical formula is: (9)(12.01) + 8.08 + 64.00 = 180.17g / mol As this value is already in the range given for the molecular formula, the empirical formula is also the molecular formula. ⎛
⎞
12.01g / mol
g C = (mass frxn C)(g CO2 ) = ⎜
(2.20g) = .600 g C
⎝ (12.01+ 32.00)g / mol ⎟⎠
⎛
⎞
2.02g / mol
g H = (mass frxn H )(g H 2O) = ⎜
(.400g) = .0448 g H
⎝ (2.02 + 16.00)g / mol ⎟⎠
g O = 1.00 g − .600g − .0448g = .355 g O
g
.600 g
mol C =
=
= .0500 mol / .0222 = 2.25 x 4 = 9
mm 12.01g / mol
g
.0448 g
mol H =
=
= .0444 mol / .0222 = 2 x 4 = 8
mm 1.01g / mol
mol O =
g
.355 g
=
= .0222 mol / .0222 = 1 x 4 = 4
mm 16.00g / mol
Answer C9H8O4 Copyright © Cengage Learning. All rights reserved.
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45. A chloride of rhenium contains 63.6% rhenium. What is the formula of this compound? a) ReCl b) ReCl3 c) ReCl5 d) ReCl7 e) Re2Cl3 Go back to question 27 (and 34) and refresh your memory about the skill of being able to use ratios to help you solve problems. In this problem we are told that a 63.6% of the mass of a compound is Re. You can look up the atomic weight of Re to find out that it is 186.2 amu. This means you can create a ratio saying the 186.2 amu represents 63.6% of the total mass of the compound. You know that the remaining fraction is 36.4%. So, you could say that there are “x” amu in 36.4%. You could then set the ratios equal to each other and solve for x. This would be the equivalent of saying: “If 186.2 amu represents 63.6% of the mass of the sample, then “x” amu represents 36.4%.” You should even be able to make a prediction that the amount of x will be about half of the amount of 186.2 amu. Then, once you know the mass of chlorine in the compound (this is what “x” gives you), you will have to divide this by the mm of chlorine to get the number of the subscript for chlorine. 186.2amu
x
=
(36.4%)(186.2amu) = 63.6%x
x = 106.57amu
63.6%
36.4%
mass chlorine 106.56amu
subscript of Cl =
=
= about 3 atoms
so ReCl3
mm chlorine
35.45amu
Answer B 47. A hydrocarbon (a compound consisting solely of carbon and hydrogen) is found to be 85.6% carbon by mass. What is the empirical formula for this compound? This problem will be easier than the typical organic empirical formula problem because you are already give mass percents rather than having to determine g of C and H from combustion products. Also, there are only C and H to consider. Because you are given the mass fraction of carbon (.856), you also already know the mass fraction of H because it is the only other element—.144. Given mass fractions you can assume 100 g of compound—therefore, in a sample of 100g, you would have 85.6 g C and 14.4 g of H. From these you can determine a mole ratio. Remember that if a mole ratio is not immediately apparent, divide each number of moles by the smallest number of moles. This will give you a ratio that is more obvious, although you may still need to multiply each number of moles by some factor to get a ratio of whole numbers, which will be the empirical formula. In this case, you can easily see that the mole ration of carbon to hydrogen is 1:2 without further calculation. Answer CH2 Copyright © Cengage Learning. All rights reserved.
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gC
85.6g
=
= 7.13 mol = 1 empirical formula : CH 2
mm C 12.01g / mol
gH
14.4g
mol H =
=
= 14.26 mol = 2
mm H 1.01g / mol
49. The empirical formula of styrene is CH; its molar mass is 104.1 g/mol. What is the molecular formula of styrene? Remember, to find the molecular formula, the subscripts of the elements in the molecular formula will be some multiple of the subscripts of the empirical formula. To find this multiplying factor, divide the molar mass of the actual compound by the molar mass of the empirical formula. Then multiply the subscripts of the empirical formula by this number to get the actual molecular formula. Answer C8H8 mm of empirical formula (CH ) = 12.01+ 1.01 = 13.02g / mol
mol C =
multiplying factor =
molar mass
104.1 g / mol
=
= about 8 empirical formula mass 13.02 g / mol
(8)(CH ) = C8 H 8
51. Vitamin C contains the elements C, H, and O. It is known to contain 40.9% C and 4.58% H by mass. The molar mass of vitamin C has been found to be about 180 g/mol. The molecular formula for vitamin C is: This problem will be easier than the typical organic empirical formula problem because you are already give mass percents rather than having to determine g of C and H from combustion products. Because you are given the mass fraction of carbon (.409) and the mass fraction of hydrogen (.0458), you can also easily find the mass fraction of oxygen. It must be 1 – mass fraction of carbon – mass fraction of hydrogen, = 1-­‐.409-­‐.0458=.545. Given mass fractions you can assume 100 g of compound—therefore, in a sample of 100g, you would have 40.9 g C, 4.58 g of H, and 54.5g of O. From these you can determine a mole ratio. Remember that if a mole ratio is not immediately apparent, divide each number of moles by the smallest number of moles. This will give you a ratio that is more obvious, although you may still need to multiply each number of moles by some factor to get a ratio of whole numbers, which will be the empirical formula. In this case, you may need this extra manipulation to the get correct mole ratio of 3:4:3. Then, remember, to find the molecular formula, the subscripts of the elements in the molecular formula will be some multiple of the subscripts of the empirical formula. To find this multiplying factor, divide the molar mass of the actual compound by the molar mass of the empirical formula. Then multiply the subscripts of the empirical formula by this number to get the actual molecular formula. Answer C6H8O6 Copyright © Cengage Learning. All rights reserved.
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mol C =
gC
40.9g
=
= 3.41/ 3.41 = 1 x 3 = 3
mm C 12.01g / mol
58
empirical formula : C3 H 4O3
gH
4.58g
=
= 4.53 / 3.41 = 1.33 x 3 = 4
mm H 1.01g / mol
gO
54.5g
mol O =
=
= 3.41/ 3.41 = 1 x 3 = 3
mm O 16.00g / mol
mm empirical formula = (3)(12.01) + 4.04 + 48 = 88.05
mol H =
multiplying factor =
mm compound
= about 2
mm empirical formula
molecular formula = (2)(C3 H 4O3 ) = C6 H 8O6
53. A 0.5242-­‐g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.9740 g of CO2 and 0.1994 g of H2O. What is the empirical formula of the compound? Remember that finding the empirical formula for an organic compound requires a little more work than for an inorganic compound. The main difference is that for an inorganic compound you will either receive the mass percents (mass fractions) or grams of the elements involved, while for organic compounds you will receive the masses of the combustion products (CO2 and H2O) and have to calculate the masses of C, H, and O from the combustion product masses. Three steps to get grams of C, H, and O. Calculate grams of carbon by multiplying mass fraction of carbon in CO2 by grams of carbon dioxide (see below for calculation). Calculate grams of hydrogen by multiplying mass fraction of hydrogen in H2O by grams of water (see below for calculation). Calculate grams of oxygen by subtracting g C and g H from original sample mass (see below for calculation). Why is it okay to say that the number of grams of the carbon in the carbon dioxide and the number of grams of hydrogen in the water represent the number of grams of C and H in the original sample? Because the only source of carbon in the carbon dioxide and the only source of hydrogen in the water is the original sample. There is no where else but the original sample from which the carbon and hydrogen could come. Once you have grams of each component, then you can determine the moles of each, and then compare these number of moles to get the mole ratio. Remember that if a mole ratio is not immediately apparent, divide each number of moles by the smallest number of moles. This will give you a ratio that is more obvious, although you may still need to multiply each number of moles by some factor to get a ratio of whole Copyright © Cengage Learning. All rights reserved.
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numbers, which will be the empirical formula. (see below for calculations) In this case, all mole values needed to be divided by .01475. The resulting values then needed to be multiplied by 2 to give the whole number ratio of C:H:O of 3:3:2, meaning the empirical formula is: C3H3O2 ⎛
⎞
12.01g / mol
g C = (mass frxn C)(g CO2 ) = ⎜
(.9740g) = .2658 g C
⎝ (12.01+ 32.00)g / mol ⎟⎠
⎛
⎞
2.02g / mol
g H = (mass frxn H )(g H 2O) = ⎜
(.1994g) = .02235 g H
⎝ (2.02 + 16.00)g / mol ⎟⎠
g O = .5242 g − .2658g − .02235g = .23605 g O
g
.2658 g
mol C =
=
= .02213 mol / .01475 = 1.5 x 2 = 3
mm 12.01g / mol
g
.02235 g
mol H =
=
= .02213 mol / .01475 = 1.5 x 2 = 3
mm 1.01g / mol
g
.23605 g
mol O =
=
= .01475mol / .01475 = 1 x 2 = 2
mm 16.00g / mol
empirical formula : C3 H 3O2
Answer C3H3O2 55. T F The molecular formula always represents the total number of atoms of each element present in a compound. True—this is basically the definition of the molecular formula—the formula “of the molecule.” 57. Balanced chemical equations imply which of the following? a) Numbers of molecules are conserved in chemical change. b) Numbers of atoms are conserved in chemical change. c) Volume is conserved in chemical change. d) A and B e) B and C Only atoms (and their associated masses) are conserved in a chemical change. The numbers of atoms are indicated by the subscripts of each atom in each chemical formula. As the relationships of each atom to the other in molecules change, molecules are not conserved. As changes of state may occur during chemical changes, volume is also not conserved. Answer B 59. What is the coefficient for water when the following equation is balanced? As(OH)3(s) + H2SO4(aq) → As2(SO4)3(aq) + H2O(l) All this question is asking you to do is balance the equation and then make a statement about what the number of water molecules in the balanced equation is. To Copyright © Cengage Learning. All rights reserved.
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start, inspection of the equation shows that there are no particles on the reactant versus the product side which are “unbalanced by one” for which we can “switch the numbers” to get coefficients. Further inspection reveals that the product formula As2(SO4)3 will at least require that we need to multiply the reactant As(OH)3 by 2 to get 2 As particles, and will require that we need to multiply the H2SO4 reactant by 3 to get 3 SO4 particles. This is a good starting point. Once we do this, and see that the As and SO4 particles are balanced. We see that we still have to balance the remaining (OH)3 (from AS(OH)3) and H2 (from H2SO4) on the reactant side with the water molecules on the product side. 2 times (OH)3 gives us 6 O atoms and 6 H atoms, and 3 times H2 gives us 6 more H atoms. 6 O atoms and 12 H atoms would be equivalent to 6 H2O molecules, so 6 is the coefficient for the product water molecules. Answer 6 (Note: you should not write out a whole new equation every time you make a modification to the equation—just add to the beginning equation as you make a modification.) Use the following to answer questions 61: Consider a specific chemical reaction represented by the equation aA + bB → cC + dD. In this equation the letters A, B, C, and D represent chemicals, and the letters a, b, c, and d represent coefficients in the balanced equation. 61. How many possible values are there for the quantity “c/d”? a) 1 b) 2 c) 3 d) 4 e) infinite For a specific chemical reaction—that is, one that contains the same specific products and reactants, the specific values of the coefficients may vary—for example, for the above general reaction we might have: 2 A + 3B → 3C + 2D, or
4 A + 6B → 6C + 4D, or
5A + 7.5B → 7.5C + 5D, etc
However, for this particular reaction, the ratio of moles of A:B:C:D must remain the same: 2:3:3:2. Therefore, the quantity c/d will also always remain the same for that particular reaction—there is only one possible value for this quantity. Answer A 63. What is the coefficient for oxygen when the following equation is balanced? NH3(g) + O2(g) → NO2(g) + H2O(g) Copyright © Cengage Learning. All rights reserved.
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All this question is asking you to do is balance the equation and then make a statement about what the number of oxygen molecules in the balanced equation is. To start, inspection of the equation shows that there are particles on the reactant versus the product side which are “unbalanced by one” for which we can “switch the numbers” to get coefficients. We notice that we have 3 H atoms on the reactant side and 2 H atoms on the product side. Switch these numbers and use them as coefficients for the respective substances. Now that you have a coefficient of 2 for the reactant, NH3, this requires a coefficient of 2 for the product NO2. Then you can inspect the number of atoms of oxygen on the product side and determine that you need 7 oxygen atoms total. This would require a fractional coefficient of 7/2 for the reactant O2 molecule. This would mean that you would need to multiply the whole equation by 2 to get whole number coefficients. You could mentally combine these last two steps into 1 step. Answer 7 (Note: you should not write out a whole new equation every time you make a modification to the equation—just add to the beginning equation as you make a modification.) 65. Determine the coefficient for O2 when the following equation is balanced in standard form (smallest whole numbers). C8H18(g) + O2(g) → CO2(g) + H2O(g) All this question is asking you to do is balance the equation and then make a statement about what the number of oxygen molecules in the balanced equation is. Recognize that this is a combustion reaction (reacting a carbon containing compound with oxygen and obtaining CO2 and H2O as products) and we have a standard procedure for balancing these. First, automatically multiply the carbon containing compound 2 two. Determine the total number of C’s this provides and add that number to the coefficient of the CO2. Then determine the number of H’s this provides and balance this with the correct coefficient for water. This then allows you to see how many oxygen atoms you need (add the oxygen atoms from the CO2’s and the H2O’s). This number will be divided by 2 to determine the coefficient of the oxygen molecules (if this number turns out to be a fraction then you would need to multiply the whole equation by the number of the denominator to get all whole number coefficients). Answer 25 Copyright © Cengage Learning. All rights reserved.
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(Note: you should not write out a whole new equation every time you make a modification to the equation—just add to the beginning equation as you make a modification.) 67. wPCl5 + xH2O → yPOCl3 + zHCl The above equation is properly balanced when: a) w = 1, x = 2, y = 2, z = 4 b) w = 2, x = 2, y = 2, z = 2 c) w = 2, x = 2, y = 2, z = 1 d) w = 1, x = 1, y = 1, z = 2 e) none of these This is just another way to create a problem that asks you to balance an equation. To avoid confusion, I would actually rewrite the equation without the coefficient variable in from them and then balance, only stating the coefficients after the equation was balanced. PCl5 + H 2O → POCl3 + HCl seems easier to deal with than
wPCl5 + xH 2O → yPOCl3 + zHCl
As I inspect this equation, my standard approaches (look for particles unbalanced by one particle, look for elements that are going to be required to be multiplied) are not going to work. However, I can see that the number of Cl atoms on the reactant side of the equation are going to have to be balanced on the product side by the sum of Cl atoms in two different compounds. At the beginning, I notice that there are 5 Cl on the reactant side and a total of 4 on the product side. I can add a coefficient of 2 to HCl and this will make the total number of Cl on the product side 5, equal to those on the reactant side. This is my starting point—lets see where this leads. Then I notice that P’s are already balanced (1 on each side); O’s are already balanced (1 on each side). And, the adding of the coefficient of 2 for HCl also balanced the number of hydrogens. Just by adding 2 in front of HCl, we have balanced the whole reaction. The coefficients for the other compounds are all 1’s. Alternatively, I could have just noticed from my inspection that P and O were already balanced and there was only one more H and one Cl on the reactant side than on the product side and just adding one more HCl would have fixed this. Answer D 69. Indium reacts with chlorine to form InCl3. In the balanced equation for this reaction, the coefficient of the indium trichloride is This is just another way to create a problem that asks you to balance an equation. The added piece in this problem is that you need to create the equation as well as balance it. Remember that when Cl is elemental in a reaction (that is, it is by itself, not part of a compound), it is diatomic—Cl2 (remember HOFBrINCl). So, to begin with, write out the equation as it is described (see below) Copyright © Cengage Learning. All rights reserved.
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Notice in this equation there is a particle that is “unbalanced” by one atom—Cl2 and Cl3. Start by “switching” these numbers as coefficients. This gives you 6 Cl atoms on each side of the equation. Then notice this unbalances In atoms—there is only one on the reactant side and there are 2 on the product side. So, just add a coefficient of 2 to the In on the reactant side. Answer 2 In + Cl2 → InCl3
In + 3 Cl2 → 2 InCl3 2 In + 3 Cl2 → 2 InCl3
(Note: you should not write out a whole new equation every time you make a modification to the equation—just add to the beginning equation as you make a modification.) 71. In the balanced equation for the reaction: xP4O6(s) + yH2O(l) → zH3PO3(aq) if x equals 2, the coefficient z equals: If the coefficient for P4O6 equals 2, this means there are 8 P atoms on the reactant side. This means there needs to be 8 P atoms on the product side as well. To accomplish this you would need a coefficient of z = 8. Answer 8 73. When the equation C4H10 + O2 → CO2 + H2O is balanced with the smallest set of integers, the sum of the coefficients is All this question is asking you to do is balance the equation and then make a statement about what the value of the sum of all of the coefficients is. Recognize that this is a combustion reaction (reacting a carbon containing compound with oxygen and obtaining CO2 and H2O as products) and we have a standard procedure for balancing these. First, automatically multiply the carbon containing compound 2 two. Determine the total number of C’s this provides and add that number to the coefficient of the CO2. Then determine the number of H’s this provides and balance this with the correct coefficient for water. This then allows you to see how many oxygen atoms you need (add the oxygen atoms from the CO2’s and the H2O’s). This number will be divided by 2 to determine the coefficient of the oxygen molecules (if this number turns out to be a fraction then you would need to multiply the whole equation by the number of the denominator to get all whole number coefficients). In this case, you will then need to sum all of the coefficients to get the final answer. Answer 25 Copyright © Cengage Learning. All rights reserved.
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75. A reaction occurs between sodium carbonate and hydrochloric acid producing sodium chloride, carbon dioxide, and water. The correct set of coefficients, respectively, for the balanced reaction is: a) 3 6 6 3 4 b) 8 6 5 10 5 c) 5 10 10 5 5 d) 1 2 2 1 1 e) none of these This is just another way to create a problem that asks you to balance an equation. The added piece in this problem is that you need to create the equation as well as balance it. You should know the chemical formulas for all of the above compounds without a second’s hesitation. Write out the equation as the description states (see below). There would be many ways to approach balancing this equation but probably the easiest is just to start by looking at the most complex compound, Na2CO3 and noticing that there are 2 Na atoms on the reactant side and only 1 on the product side, so multiply NaCl by 2. This means you need to Cl atoms on the reactant side, so multiply HCl by 2. At this point, take inventory of your atoms on both sides and realize that the equation is already balance, giving you a set of coefficients of 1, 2, 2, 1, 1. Answer D Na2CO3 + HCl → NaCl + CO2 + H 2O
Na2CO3 + HCl → 2 NaCl + CO2 + H 2O Na2CO3 + 2HCl → 2NaCl + CO2 + H 2O
77. T F When balancing a chemical equation, it is generally best to start with the least complicated molecules. False—normally if you balance by starting with least complicated molecules first, the coefficients of these will almost certainly change as you later try to balance the more complicated molecules. In general, start with more complicated molecules first and you will run into less “back and forth” changing of coefficients later in the balancing. 79. You heat 3.935 g of a mixture of Fe3O4 and FeO to form 4.162 g Fe2O3. The mass of oxygen reacted is Although the solution to this problem is easy, it is made more difficult by the way the question is worded. Remember that the law of conservation of mass says that the mass of atoms that begins a reaction must equal the mass of the products that end the reaction. Here, we have more mass at the end than when the reaction begins. The only place this mass could have come is from the atmosphere, and is most likely to come from O2 in the atmosphere. This is the oxygen the problem is talking about—
not oxygen that is coming from Fe3O4 or FeO. However, this is not explicitly stated in the problem and you can only figure out the solution if this possibility occurs to you. If this does occur to you, then it is a simple matter to find the mass of oxygen that reacts by subtracting the initial mass from the final mass. Answer .227 g Copyright © Cengage Learning. All rights reserved.
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4.162g − 3.935g = .227g 81. A chemical reaction has the equation: 2A + B → C. Which of the following figures best illustrates a stoichiometric ratio of A and B? I. II. III. IV. a) I only b) II only c) III only d) IV only e) both I and IV This question is simply asking you to interpret a visual model of a reaction described by the equation 2 particles of A react with every one particle of B to form one particle of C. Therefore, find the model that shows two particles of A for every particle of B. Only diagram I accomplishes this. (Note-­‐to interpret diagram IV, recognize that each particle of C represents the previous presence of 2 particles of A and 1 particle of B. If you substitute these values in for A and B you would end up with 12 particles of A and 12 particles of B—still not the correct stoichiometric relationship) Answer A 85. How many grams of Ca(NO3)2 can be produced by reacting excess HNO3 with 6.55 g of Ca(OH)2? As always, we begin any type of problem like this with a balanced chemical equation. We are given a description of the reactants and products, but we are not told one of the products. You know that HNO3 is nitric acid and Ca(OH)2 is calcium hydroxide. This is an acid base reaction. As with all acid base reactions, when H+ reacts with OH-­‐, water is formed, so water will be a product of this reaction too. Once you have all of the reactants and products this is easy to balance (see below). Fortunately this will not be a limiting reagent problem—we are told that there is excess HNO3 so the mass of Ca(OH)3 is all we need to work through the problem. Copyright © Cengage Learning. All rights reserved.
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We calculate g Ca(NO)3 obtained in our usual way. As is our strategy, we begin with the given value, 6.55 g Ca(OH)3. We want to get rid of g Ca(OH)3 and create mol Ca(OH)3 so we place these units in the denominator and numerator respectively of the first conversion factor, using the mm calculated for Ca(OH)3 (you can calculate the mm in the line of the calculation if you want, although I calculated it separately). Next, we want to convert mol Ca(OH)3 to mol Ca(NO)3 so we place these units in the denominator and numerator respectively of the next conversion factor along with the number of moles specified by the chemical equation. Next we want to convert mol Ca(NO)3 to g Ca(NO)3 so we place these units in the denominator and numerator respectively of the next conversion factor, along with the mm calculated for Ca(NO)3 (you can calculate the mm in the line of the calculation, although I calculated it separately). 2HNO3 + Ca(OH )2 → Ca(NO3 )2 + 2H 2O ⎛ 1 mol Ca(OH )2 ⎞ ⎛ 1mol Ca(NO3 )2 ⎞ ⎛ 164.10g Ca(NO3 )2 ⎞
g Ca(NO3 )2 : (6.55g Ca(OH )2 ) ⎜
= 14.5g
⎝ 74.10 g Ca(OH )2 ⎟⎠ ⎜⎝ 1 mol Ca(OH )2 ⎟⎠ ⎜⎝ 1 mol Ca(NO3 )2 ⎟⎠
mm g Ca(OH )2 = 40.08 + 32.00 + 2.02 = 74.10g / mol
mm g Ca(NO3 )2 = 40.08 + 28.02 + 96.00 = 164.10g / mol
89. The refining of aluminum from bauxite ore (which contains 50.% Al2O3 by mass) proceeds by the overall reaction 2Al2O3 + 3C → 4Al + 3CO2. How much bauxite ore is required to give the 5.0 × 1013 g of aluminum produced each year in the United States? (Assume 100% conversion.) We have already been given a balanced chemical equation so we do not have to worry about creating one. Also, this is not a limiting reagent problem. We would nee d to be given amounts of both reactants in order for this to be the case, but we are given the mass of neither reactant—we are basically told through the context of the problem that we want to know that mass of bauxite required to create the given amount of Al under the given circumstances—we are told, assume 100% conversion. We calculate g Al2O3 required in our usual way. As is our strategy, we begin with the given value, 5.0 x 1013 Al. We want to get rid of g Al and create mol Al so we place these units in the denominator and numerator respectively of the first conversion factor, using the mm calculated for Al (26.98 g/mol from the periodic table). Next, we want to convert mol Al to mol Al2O3 so we place these units in the denominator and numerator respectively of the next conversion factor along with the number of moles specified by the chemical equation. Next we want to convert mol Al2O3 to g Al2O3 so we place these units in the denominator and numerator respectively of the next conversion factor, along with the mm calculated for Al2O3 (you can calculate the mm in the line of the calculation). Performing the calculation would give us the g Al2O3 required to create 5.0 x 1013 g of Al. However, we were not asked for g Al2O3, we were asked for g bauxite and were told that bauxite is 50% by mass Al2O3—the other 50% is a bunch of other stuff. So, Copyright © Cengage Learning. All rights reserved.
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we need a final conversion factor to get rid of g Al2O3 and create g bauxite. We do this in the usual way, stating that 1 g Al2O3 is equivalent to 2 g bauxite (or you could say 1 g bauxite is equivalent to .5 g Al2O3). Put this last conversion factor in and solve. 2 SF. Answer 1.9x1014 g bauxite ⎛ 1 mol Al ⎞ ⎛ 2 mol Al2O3 ⎞ ⎛ ((2)(26.98) + 48.00)gAl2O3 ⎞ ⎛ 2 g bauxite ⎞
14
g bauxite : (5.0x1013 g Al) ⎜
⎟⎠ ⎜⎝ 1 g Al O ⎟⎠ = 1.9x10 g bauxite
1 mol Al2O3
⎝ 26.98 g Al ⎟⎠ ⎜⎝ 4 mol Al ⎟⎠ ⎜⎝
2 3
93. The following two reactions are important in the blast furnace production of iron metal from iron ore (Fe2O3): 2C(s) + O2 (g ) → 2CO(g )
Fe2 O3 (s) + 3CO(g ) → 2Fe(s) + 3CO2 (g )
Using these balanced reactions, how many moles of O2 are required for the production of 1.69 kg of Fe? Hopefully you will see that we need to combine these component reactions into a single reaction to be able to determine what the question is asking. Notice that, as we have discussed, there is a product of one reaction that is a reactant of another reaction—this is CO. As we have stated, such a substance is called a reaction intermediate—it is present in both reactions, but it does not appear as a beginning reactant of as a product. Importantly, also notice that as a product of the first reaction, there are 2 moles, but as reactant of the second reaction, three moles are required. Therefore will we have to make the coefficient the same in both reactions. The way we would doe this by using a strategy we have used in other circumstances where things are “unbalanced”—“switch” the numbers. Multiply the whole first reaction by 3 and the whole second reaction by 2—this will give us equal numbers of moles of CO in both reactions. Then add the two equations together to get the final equation, and follow our standard procedure for calculating the number of moles of O2 required to provide 1.69 kg of Fe. (Notice—we do not have to go all the way to g O2—just moles.) As is our strategy, we begin with the given value, 1690 g Fe. We want to get rid of g Fe and create mol Fe so we place these units in the denominator and numerator respectively of the first conversion factor, using the mm calculated for Fe (you can calculate the mm in the line of the calculation). Next, we want to convert mol Fe to mol O2 so we place these units in the denominator and numerator respectively of the next conversion factor along with the number of moles specified by the chemical equation. This is as far as we need to go. 3 SF. Answer 22.7 mol O2 3 (2C + O2 → 2CO )
2(Fe2O3 + 3CO → 2Fe + 3CO2 )
6C + 3O2 + 2 Fe2O3 → 4Fe + 6CO2
⎛ 1 mol Fe ⎞ ⎛ 3 mol O2 ⎞
mol O2 : (1690 g Fe) ⎜
= 22.7 mol O2
⎝ 55.85g Fe ⎟⎠ ⎜⎝ 4 mol Fe ⎟⎠
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97. The limiting reactant in a reaction a) has the lowest coefficient in a balanced equation b) is the reactant for which you have the fewest number of moles c) has the lowest ratio of moles available/coefficient in the balanced equation d) has the lowest ratio of coefficient in the balanced equation/moles available e) none of these (A) False—do not fall into this trap—you cannot use the reactant coefficients by themselves to say anything about limiting reagents. These only tell us about the ratio of moles in which the reactants combine to form the products—which one is limiting will be determined by the actual numbers of moles used in the reaction. (B) False—do not fall into this trap—a reactant could have a few number of actual moles but this may not be limiting if one (or more) of the other reactants requires a greater number of moles. For example—given the following balanced equation for the formation of water from H2 and O2: 2H 2 + O2 → 2H 2O If we were given 1.5 mol of H2 and 1 mol of O2, because the reaction requires 2 moles of H2 for every 1 mole of O2, there would not be enough H2—there would need to be at least 2 moles of H2. So, even though there is a smaller number of moles of O2 than H2, the 1.5 moles of H2 is the limiting reagent. (C) True—based on the above example then, ask yourself, for the limiting reagent, what was the ratio of the given moles of H2 to the coefficient of H2?—1.5:2, which is .75:1. Then ask yourself, for the other reagent, what was the ratio of the given moles of O2 to the coefficient of O2?—1:1. It should make sense, then that we might be able to apply this relationship to any chemical equation. The reactant that demonstrates the smallest ratio of actual moles (given moles) to coefficient value will be the limiting reagent. Try this rule on other limiting reagent problems and see if it works. It may be useful in helping you to predict whether or not there is a limiting reagent and what the limiting reagent might be in some problems (although there may be quicker ways to make this estimate—for example, see the explanation for question 101). (D) False—this is the opposite of the relationship expressed in answer C—this would predict the reactant that is not the limiting reactant. (E) As C is true, this is not a correct statement. Answer C 101. SO2 reacts with H2S as follows: 2H2S + SO2 → 3S + 2H2O When 7.50 g of H2S reacts with 12.75 g of SO2, which statement applies? a) 6.38 g of sulfur are formed. Copyright © Cengage Learning. All rights reserved.
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b) 10.6 g of sulfur are formed. c) 0.0216 moles of H2S remain. d) 1.13 g of H2S remain. e) SO2 is the limiting reagent. On the AP test, you cannot use a calculator on the multiple choice section so I will try to only give you multiple choice homework questions that can be answered without a calculator. To be really certain of the correct answer in this problem you would probably want to do the calculations. But, to extend your skill in completing problems without a calculator, I am going to solve the problem with and without a calculator to highlight useful non-­‐calculator skills in stoichiometric problems. Without a calculator first: After inspecting the formula you should notice that the number of moles of H2S required will be twice that the number of moles of SO2. You have been given 12.75 g of SO2, which will be about (12/60) or 1/5 mol of SO2 (mm SO2 is about 60—
32.07+32.00). So, the number of moles of H2S required would be about 2/5 of a mole. You have been given 7.5 g H2S, which ends up being about only 1/5 of a mole. (mm H2S is about 34—2.02+32.07). So, H2S is limiting, and no H2S will remain. This eliminates answers c, d, and e. The only decision left then is to estimate whether the number of grams of H2S given will provide 6.38 g of sulfur or 10.6 g. We have seen that the 7.5 g of H2S is about 1/5 of a mole. With a 3:2 ratio of S:H2S being provided by the equation, this means that the moles of S created will be about 3/10 of a mole. We know that the mm of S is 32.07, and 3/10 of this would be about 10 g, so answer b would be our choice. Lets confirm this by actually doing the math. We want g S—follow the standard procedure. As is our strategy, we begin with the given value, 7.5 g H2S. We want to get rid of g H2S and create mol H2S so we place these units in the denominator and numerator respectively of the first conversion factor, using the mm calculated for H2S (you can calculate the mm in the line of the calculation). Next, we want to convert mol H2S to mol S so we place these units in the denominator and numerator respectively of the next conversion factor along with the number of moles specified by the chemical equation. Next we want to convert mol S to g S so we place these units in the denominator and numerator respectively of the next conversion factor, along with the mm calculated for S. Performing the calculations we get 10.6 g S ⎛
⎞ ⎛ 3 mol S ⎞ ⎛ 32.07 g S ⎞
1 mol H 2 S
g S : (7.50 g H 2 S) ⎜
⎜
⎟ = 10.6g ⎝ (2.02 + 32.07)g H 2 S ⎟⎠ ⎜⎝ 2 mol H 2 S ⎟⎠ ⎝ 1 mol S ⎠
Answer B 105. A 5.95-­‐g sample of AgNO3 is reacted with BaCl2 according to the equation Copyright © Cengage Learning. All rights reserved.
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2AgNO3 (aq) + BaCl2 (aq) → 2AgCl(s) + Ba(NO3 )2 (aq) to give 4.19 g of AgCl. What is the percent yield of AgCl? Because we are looking for percent yield of AgCl, we know that we are going to need to calculate the number of grams of AgCl we are supposed to obtain, divide the number of grams we actually obtained (4.19 g AgCl) by this number, and multiply by 100. Fortunately this will not be a limiting reagent problem—while we are not told that there is excess BaCl2, we are not given any amount o BaCl2, and we cannot be expected to just come up with a random amount of this substance, so we must assume that it will be adequate to react with the given amount of AgNO3. To determine the theoretical number of grams of AgCl we only have to worry about the 5.95 g of AgNO3. We calculate g AgCl expected in our usual way. As is our strategy, we begin with the given value, 5.95 g AgNO3. We want to get rid of g AgNO3 and create mol AgNO3 so we place these units in the denominator and numerator respectively of the first conversion factor, using the mm calculated for AgNO3 (you can calculate the mm in the line of the calculation). Next, we want to convert mol AgNO3 to mol AgCl so we place these units in the denominator and numerator respectively of the next conversion factor along with the number of moles specified by the chemical equation. Next we want to convert mol AgCl to g AgCl so we place these units in the denominator and numerator respectively of the next conversion factor, along with the mm calculated for AgCl (you can calculate the mm in the line of the calculation). We find that the number of grams of AgCl we should obtain is 5.02g. Now divide the actual yield (4.19 g) by this number and multiply x 100 to get percent yield. 3SF. Answer 83.5% ⎛
⎞ ⎛ 2 mol HC2 H 3O2 ⎞ ⎛ (107.87 + 35.45)g AgCl ⎞
1 mol AgNO3
g AgCl : (5.95 g AgNO3 ) ⎜
⎜
⎟⎠ = 5.02g
1 mol AgCl
⎝ (107.87 + 14.01+ 48.00)g AgNO3 ⎟⎠ ⎜⎝ 2 mol C2 H 5OH ⎟⎠ ⎝
percent yield =
actual yield
4.19g
x100 =
x100 = 83.5%
theoretical yield
5.02g
109. The following equation describes the oxidation of ethanol to acetic acid by potassium permanganate: 3C2 H5OH + 4KMnO4 → 3HC2 H3O2 + 4MnO2 + 4KOH + H2O 5.00 g of ethanol and an excess of aqueous KMnO4 are reacted, and 5.98 g of HC2H3O2 result. What is the percent yield? Because we are looking for percent yield of HC2H3O2, we know that we are going to need to calculate the number of grams of HC2H3O2 we are supposed to obtain, divide the number of grams we actually obtained (5.98 g H2C2H3O2) by this number, and multiply by 100. Fortunately this will not be a limiting reagent problem—we are told that there is excess KMnO4, which means we already know that we base our calculation of final g of HC2H3O2 on initial g of C2H5OH. Copyright © Cengage Learning. All rights reserved.
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Calculate g HC2H3O2 in our usual way. As is our strategy, we begin with the given value, 5.00 g C2H5OH. We want to get rid of g C2H5OH and create mol C2H5OH so we place these units in the denominator and numerator respectively of the first conversion factor, using the mm calculated for C2H5OH (you can calculate the mm in the line of the calculation). Next, we want to convert mol C2H5OH to mol HC2H3O2 so we place these units in the denominator and numerator respectively of the next conversion factor along with the number of moles specified by the chemical equation. Next we want to convert mol HC2H3O2 to g HC2H3O2 so we place these units in the denominator and numerator respectively of the next conversion factor, along with the mm calculated for HC2H3O2 (you can calculate the mm in the line of the calculation). We find that the number of grams of HC2H3O2 we should obtain is 6.52g. Now divide the actual yield (5.98 g) by this number and multiply x 100 to get percent yield. 3SF. Answer 91.7% ⎛
⎞ ⎛ 3 mol HC2 H 3O2 ⎞ ⎛ ((2)(12.01) + 4.04 + 32.00)g HC2 H 3O2 ⎞
1 mol C2 H 5OH
g HC2 H 3O2 : (5.00 g C2 H 5OH ) ⎜
⎟⎠ = 6.52 g
1 mol HC2 H 3O2
⎝ ((2)(12.01) + 6.06 + 16.00)g C2 H 5OH ⎟⎠ ⎜⎝ 3 mol C2 H 5OH ⎟⎠ ⎜⎝
percent yield =
actual yield
5.98g
x100 =
x100 = 91.7%
theoretical yield
6.52g
113. Consider the following reaction: 2A + B → 3C + D 3.0 mol A and 2.0 mol B react to form 4.0 mol C. What is the percent yield of this reaction? It is not stated but you need to assume this will be a limiting reagent problem. If you inspect the chemical equation you can quickly see this will be the case. If you start with 2.0 mol B you would need two times this number of moles of A, or 4.0 mol, A to completely react. Therefore, A is the limiting reagent. However, you also need to show this. As we want to know how many moles of C we are supposed to get in order to determine the theoretical yield, use this product to test for which reactant is limiting. In this case, to test each reactant to see if it is limiting, you have already been given moles rather than grams, so you will only need to go from “moles to moles.” Given each number of moles of reactant, convert these to number of moles of product C to see which reactant is limiting. Remember that it makes things less confusing if you label each calculation with the reactant you are testing (see below). The testing calculations confirm our hypothesis that moles of A are limiting. We will only get 4.5 mol of C given 3 mol A, whereas we would get 6 mol C for 2 mol B. We can then see that the number of moles we should get from using 3.0 mol of A is 4.5 (theoretical yield), but we are given that we only obtained 4.0 mole (actual yield). To get the % yield divide the actual by the theoretical yield and multiply x 100. 2 SF Answer 80.% Copyright © Cengage Learning. All rights reserved.
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⎛ 3 mol C ⎞
mol C for mol A : (3.0 mol A) ⎜
= 4.5 mol C
⎝ 2 mol A ⎟⎠
⎛ 3 mol C ⎞
mol C for mol B : (2.0 mol A) ⎜
= 6.0 mol C ⎝ 1 mol B ⎟⎠
% yield =
actual yield
4.0 mol
=
x 100 = 89%
theoretical yield 4.5 mol
117. If 30.8 g of O2 are mixed with 30.8 g of H2 and the mixture is ignited, what mass of water is produced? As always, we first write a balanced chemical equation to represent the process described (see below) Inspecting the balanced equation, and knowing that the masses of the reactants given are equal, we should be able to see immediately that we have a far greater number of moles of H2 than O2, and that even the reaction requires more moles of H2 than it does of O2, the amount of O2 will be limiting. Normally, we would want to “test” the amounts of each reactant to see how many moles of product they would give. However, in this scenario, it is so obvious that oxygen is limiting you could probably get away with saying something like: “Inspection of given information shows that moles O2 will obviously be limiting.” However, lets go through the whole process anyway. Test each reactant for how many moles of product will result if each is the limiting reagent (see below for calculations). Because the molar masses are simple calculations, you can just include them in the line of the rest of the calculations rather than showing them separately. Remember that it makes things less confusing if you label each calculation with the reactant you are testing (see below). Also remember that we do not have to go “all the way to grams” to determine which reagent is limiting—just “go to moles.” The testing calculations confirm our hypothesis that moles of O2 are limiting—in fact, they are severely limiting. We will only get .482 mol of water given 30.8g of oxygen. Then, as we actually want grams of H2O, go ahead and calculate the number of grams of water represented by .482 moles of water. 3 SF. Answer 8.67g Copyright © Cengage Learning. All rights reserved.
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2H 2 + O2 → 2H 2O
⎛ 1 mol H 2 ⎞ ⎛ 2 mol H 2O ⎞
mol H 2O for H 2 : (30.8 g H 2 ) ⎜
= 15.25 mol H 2O
⎝ 2.02g H 2 ⎟⎠ ⎜⎝ 2 mol H 2 ⎟⎠
⎛ 1 mol O2 ⎞ ⎛ 1 mol H 2O ⎞
mol H 2O for O2 : (30.8 g O2 ) ⎜
=
.481
mol
H
O
2
⎝ 32.00g O2 ⎟⎠ ⎜⎝ 2 mol O2 ⎟⎠
⎛ 18.02g ⎞
g H 2O = (mol)(mm) = (.481mol) ⎜
= 8.67 g H 2O
⎝ mol ⎟⎠
121. T F Given the equation 3A + B → C + D, you react 1 mole of A with 3 moles of B. True or false: A is the limiting reactant because you have fewer moles of A than B. False. A is certainly the limiting reactant in this case but not because there are fewer moles of A than there are of B. You could actually have more moles of A than B and A could still be the limiting reactant. If there are 3 moles of B, this would require 9 moles of A, so as long as moles of A were less than 9 but greater than 3, there would be more moles of A than B, which falsifies the statement made. A better statement would be A is the limiting reactant because there are not enough moles of A to react with 3 moles of B—it would require 9 moles of A. Answer False 125. Vitamin B12, cyanocobalamin, is essential for human nutrition. It's concentrated in animal tissue but not in higher plants. People who abstain completely from animal products may develop anemia, so cyanocobalamin is used in vitamin supplements. It contains 4.35% cobalt by mass. Calculate the molar mass of cyanocobalamin assuming there is one cobalt per molecule. Go back to questions 27 (and 34 and 45) and refresh your memory about the skill of being able to use ratios to help you solve problems. In this problem we are told that 4.35% of the mass of a compound is cobalt, and that this represents a single cobalt atom. You can look up the atomic weight of Co to find out that it is 58.93 amu. This means you can create a ratio saying the 58.93 amu represents 4.35% of the total mass of the compound. Likewise, you could say that there are “x” amu in 100%. You could then set the ratios equal to each other and solve for x. This would be the equivalent of saying: “If 58.93 amu represents 4.35% of the mass of the sample, then “x” amu represents 100%.” You should even be able to make a prediction that the amount of x will be about 20 times of the amount of 58.93 amu. Answer 1354.71 g/mol 58.93amu
x
=
(100%)(58.93amu) = 4.35%x x = 1354.71amu 4.35%
100%
129. In order to determine the molecular formula from the empirical formula, we must know the __________. Remember, to find the molecular formula, the subscripts of the elements in the molecular formula will be some multiple of the subscripts of the empirical formula. To find this multiplying factor, divide the molar mass of the actual compound by the Copyright © Cengage Learning. All rights reserved.
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molar mass of the empirical formula. Then multiply the subscripts of the empirical formula by this number to get the actual molecular formula. Answer molar mass. 133. One of the major commercial uses of sulfuric acid is in the production of phosphoric acid and calcium sulfate. The phosphoric acid is used for fertilizer. The reaction is Ca3(PO4)2 + 3H2SO4 → 3CaSO4 + 2H3PO4. What mass of concentrated H2SO4 (98% by mass) must be used to react completely with 100.00 g of calcium phosphate? We are told that we need g concentrated H2SO4 (we will see why it is necessary to include the word “concentrated” at the end of the problem). As is our strategy, we begin with the given value, 100.00g Ca3(PO4)2. We want to get rid of g Ca3(PO4)2 and create mol Ca3(PO4)2 so we place these units in the denominator and numerator respectively of the first conversion factor, using the mm calculated for Ca3(PO4)2. Next, we want to convert mol Ca3(PO4)2 to mol H2SO4 so we place these units in the denominator and numerator respectively of the next conversion factor along with the number of moles specified by the chemical equation. Next we want to convert mol H2SO4 to g H2SO4 so we place these units in the denominator and numerator respectively of the next conversion factor, along with the mm calculated for H2SO4. Finally, we have been told that the concentrated H2SO4 is only 98% by mass. This means that for every 100g of solution, there are 98g of H2SO4. Remember, we stated at the beginning of the problem that we needed to include the word “concentrated” in the statement of what we want. The reason now becomes apparent—we don’t just want the number of grams of H2SO4, we want the number of grams of a 98% by mass solution of H2SO4. We can create a conversion factor for this by saying just this:
100 g solution
. 2 SF. Answer 97g 98 g H 2 SO4
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