Answers to selected problems, Chapter 3

Review questions
3. In 1.00 mol of O₂ there are (by definition) 1.00 mol of oxygen molecules. Since each molecule contains 2 atoms of oxygen, there are 2.00 mol of the atoms in 1.00 mol of O₂.

8. The empirical formula is the simplest relation between the numbers of atoms of the various elements in a compound (molecule). It is called empirical because that is the way the relation comes out of typical chemical tests. For example, you would get a 1:1 relationship between two elements, not a 2:2 relationship (which is really the same relationship). For (a) H₂O₂ (hydrogen peroxide), the empirical formula would be HO.  For (b) C₈H₁₆, it would be CH₂. For (c) C₁₀H₈, it would be C₅H₅. For (d) C₆H₁₆O, it would be the same thing because O can't have a subscript less than 1. 

19. The two sets of units are the same because both are just divided by 1000 in the second one, and those two factors cancel out because they are in the numerator and the denominator. In symbols, g/L = [g/1000]/[L/1000] = mg/mL. If you don't see how the 1000s cancel out, write the expression out in longhand.

Problems
21. To get the mass of each of the substances given in this problem, just multiply the atomic masses of each element by its coefficient and add. I will do just the first one as a demonstration. I will also use the integer versions of the atomic weights for simplicity. (a) The molecular weight of C₆H₅Br is 6(Wt. C) + 5(Wt. H) + 1(Wt. Br) = (6x12) + (5x1) + (1x80) = 157 u. To the proper number of significant figures, the answer is 157.00 u. Don't forget that in formulas like (b) Ca(HCO₃)₂, the entire group of atoms in the parenthesis must be multiplied by two (which gives 2H, 2C, and 6O).

25. As with No. 21, I will do just part (a) as an example. (a) To get the mass in grams of 1.12 mol CaH₂, just calculate its molecular or formula weight (which can be understood as grams per mole) and multiply by 1.12. MW = 40 + 2(1) = 42. 1.12x42 = 47.04 g. To get the official answer of 42.10 g, use the exact atomic weights and the correct number of significant figures.

31. The problem without detailed calculations that we did in class. The trick is to read the instructions precisely and see that it is asking for the greatest number of atoms, not molecules or formula units. (a) 1.0 mol of N₂ contains 1 mol of N₂ molecules, which is 2 mol of N atoms. (b) 50.0 g of Na is about 2 mol of Na atoms because Na's atomic weight is about 23. (c) 17.0 mL of H₂O is about 1 mol of H₂O because its density is 1 g/mL, which makes 17 g of it, and its molecular weight is 16 + 2 = 18. But because there are three atoms in each molecule of H₂O, we have about 3 mol of atoms here. (d) 1.2x10²⁴ atoms of Mg is about 2 moles because 1 mole of anything is 6.0x10²³ units (atoms, molecules, etc., whatever the substance in question is) = 0.6x10²⁴ units. I strongly suggest that any of you who can't handle these exponents in your head write them out and practice with them. You may also see Appendix A.1. To review the exponential math, (1.2x10²⁴)/(6.0x10²³) = (1.2x10²⁴)/(0.6x10²⁴) = 2. You can also do it so as to get 0.2x10¹ = 2.

37. (a) To get the mass percents (or the more-fundamental mass fractions), just multiply the atomic weights of each of the elements by their coefficients in the formula for beryl, add, and divide the result for each elements by the total. The term for Be is 3xA_Be = 3x9 = 27 u. For Al it is 2xA_Al = 2x27 = 54 u. For Si it is 6xA_Si  = 6x28 = 168 u. For O it is 18xA_O = 18x16 = 288 u. The total wt. is 27 + 54 + 168 + 288 = 537 u (actually 537.54 u). Be's fraction is 27/537 = 0.0503 = 5.03%. (b) 1.00 kg of beryl would contain 0.0503x1000 g = 50.3 g of Be. (Notice that "beryl" begins with Be? Neat, eh?) 

43. This problem, determining the molecular formula of resorcinol from its percent composition, is the reverse of problem 37. Here you divide the percentages by the atomic weights of the elements (to get the molar proportions of the elements) and convert those numbers to small whole numbers. For C, we have (65.44 mass %)/(12 g/mol) = 5.45 mol% (relative). For H, 5.49%/1 = 5.49 (same units). For O, 29.06%/16 = 1.82. You should be able to see without using a calculator that the mole proportions of C, H, and O (essentially their coefficients in the formula) are 3, 3, 1 because 3x1.8 = 5.4, which is very close to the other two numbers. So the empirical formula is C₃H₃O. But does this molecular weight add up to the 110 u given in the problem? 3x12 + 3x1 + 1x16 = 55 u, which is one-half the 110 u given. So the molecular formula must be double the empirical formula, or C₆H₆O₂. You can check for yourself this this molecular weight is 110.

53. (a) To rank the three organic compounds methanol (CH₃OH), ethanol (CH₃CH₂OH), and MTBE (CH₃OC(CH₃)₃) by increasing mass fraction oxygen,  you just need to note that all three have one atom of oxygen plus various numbers of carbon and hydrogen atoms. Thus, the greater the mass of the other two elements in the formula, the lower will be the fraction of oxygen. What extra atoms does each compound have? Methanol is the simplest, with an extra CH₃ and H. Ethanol has that plus a CH₂ unit. MTBE has three extra carbons plus eight extra hydrogens. Thus methanol has the least extra mass (beyond the one oxygen) and MTBE the greatest extra mass. So MTBE has the lowest fractional mass of oxygen, ethanol has an intermediate percentage, and methanol has the greatest.
(b) EPA wants fuels for cars to contain oxygen in order to help transform some of the CO (bad stuff) it emits to CO₂ (good stuff, unless you are worried about the greenhouse effect). If the original fuel contains no oxygen and methanol is added to it until it makes up 10.5% of the total mass, you can find the fraction of the total that is oxygen by multiplying its fraction in methanol by its methane's fraction in the total fuel (0.105). If the result meets or exceeds 0.027, the mixed fuel is satisfactory. The fraction oxygen of methanol is just the mass ratio O/CH3OH, or 16/(12 + 4 + 16) = 16/32 = 0.5. Thus the mass fraction O in the mixed fuel is (0.5 O in methane)(0.105 methane in mixed fuel) = 0.0525 = 5.25%. The mixed fuel meets the standard of 2.7% oxygen. [Tip: don't forget to learn the atomic weights of the common elements, as we discussed in class. It will speed your basic calculations.]
(c) Getting the mass percent of MTBE needed to allow gasoline to meet the requirement of 2.7% oxygen is the reverse of the previous problem, but with the same equation. Remember my early tip about setting up the equation first and solving for the variable of interest before inserting any numbers? The equation here is (mass fraction O in mixed fuel) = (mass fraction O in the oxygenated component MTBE)(mass fraction of oxygenated component in the mixed fuel). Solving for the second term on the right gives (mass fraction of oxygenated component in the mixed fuel) = (mass fraction O in mixed fuel)/(mass fraction O in the oxygenated component MTBE) = (0.027)/(mass fraction O in the oxygenated component MTBE). To get the denominator, just find the mass fraction of O in CH₃OC(CH₃)₃, which is 16/[(5x12) + (1x16) + (12x1) = 16/88 = 0.182. The full equation then becomes (mass fraction of oxygenated component in the mixed fuel) = 0.027/0.182 = 0.148 = 14.8% of MBTE in the mixed fuel. To two significant figures, as required by 0.027, the answer is 15% 

55. Balancing equations.
(a) Cl₂O₅ + H₂O ® HClO₃. (1) Following the suggestions on page 102 of the text, look for an element that is present in only one compound on each side of the equation, and balance it first. That could be Cl or H here, either of which requires a coefficient of 2 for the HClO₃. That gives Cl₂O₅ + H₂O ® 2 HClO₃. (2) Since that takes care of two of the three elements, check whether the third, O, is balanced. It is, because 5 + 1 on the left = 2x3 on the right.
(b) V₂O₅ + H₂ ® V₂O₃ + H₂O. (1) V appears only once on each side and is balanced. H also appears only once and is balanced. O appears twice on the right and is not balanced. Work on the O because we should balance pure elements (H₂ here) last--second rule on page 102. (2) If we balance O by doubling H₂O on the right, we can then balance H by doubling H₂ on the left:
V₂O₅ + H₂ ® V₂O₃ + 2 H₂O
V₂O₅ + 2 H₂ ® V₂O₃ + 2 H₂O
On each side we now have 2 V, 5 O, and 4 H.
(c) Al + O₂ ® Al₂O₃. (1) Easy to balance Al, so work on O first. Lowest number that balances is 6 O's, or:
Al + 3 O₂ ® 2 Al₂O₃
(2) That requires 4 Al on the left to balance it:
4 Al + 3 O₂ ® 2 Al₂O₃
The equation is now balanced, with 4 Al and 6 O on each side.

59. This is a two-step problem. First you must determine the formula for the black residue, and second you use it to balance the equation for the reaction.
(a) Determining the formula for the black residue. As in problem 43, divide the abundances of Fe and O by their atomic weights and adjust the results to small whole numbers. Fe: 72.3%/56 = 1.29. O: 27.7%/16 = 1.73. 1.73/1.29 = 1.34/1. This is equivalent to 4/3 and so to Fe₃O₄. Thus the reaction is
Fe₂O₃ + H₂ ® Fe₃O₄ + H₂O.
(b) To balance this reaction, start with Fe because it is in only one compound on each side. The "common denominator" of 2 and 3 is 6:
3 Fe₂O₃ + H₂ ® 2 Fe₃O₄ + H₂O
Note that both H and O are balanced, so all is done.

61. Complete combustion of octane: 2 C₈H₁₈ + 25 O₂ ® 16 CO₂ + 18 H₂O.
(a) To find how many moles of CO₂ are produced when 1.8x10₄ mol of C₈H₁₈ are burned, you need to know only the molar ratios of these two compounds in the balanced equation, which is 16 of CO₂ for 2 of octane (just their coefficients). You now set up the simple scaling equation (16 mol CO₂)/(2 mol octane) = (x mol CO₂)/(1.8x10⁴ mol octane), which says that the molar proportions of the two substances are preserved in large and small reactions, and solve for x:
x = (16/2)(1.8x10⁴ ) = 1.4x10⁵ mol CO₂.
(b) Done in the same general way.

67. Use kerosene's density to calculate its mass and then its number of moles. Balance the equation. Use proportions to calculate the moles of CO₂, and then molecular weight to change to grams.

75. The proportions of elemental mercury and elemental oxygen that react together cannot possibly be understood until the equation for the reaction is balanced. 
Hg(l) + O₂(g) ® HgO(s)
To balance note that Hg is already balanced. To balance O₂, just put a 0.5 as coefficient of the O₂. (In spite of that you probably have been taught, it is OK to use fractional coefficients; it's just not the way the final equation should look.)
Hg(l) + 0.5 O₂(g) ® HgO(s)
To get rid of the 0.5, multiply all the terms by 2:
2 Hg(l) + O₂(g) ® 2 HgO(s)
This equation means that two moles of liquid mercury react with one mole of gaseous oxygen to yield 2 moles of mercury(II) oxide.
Now we can address the question of what happens when 0.2 mol Hg react with 4 g O₂. First we must express the amount of oxygen in moles: (4 g/32 g per mole) = 0.125 moles of O₂. So we have 0.2 mol of Hg and 0.125 mol of O₂. Since 0.2 mol of Hg requires 0.1 mol of O₂, the 0.125 mol of O₂ that we have is more than enough. This makes Hg the limiting reagent. So the 0.2 mol of Hg will react with 0.1 mol of O₂ to give 0.2 mol of HgO and leave 0.025 mol of O₂ unreacted. That corresponds to 0.8 g of O₂. Thus, (d) is the correct answer. (a) is wrong because it gives the same amount of Hg as before the reaction, which cannot be. (b) is wrong because it leaves half the Hg unreacted, and we know that it all reacts because it is the limiting reagent. (c) is wrong because it says that all the O₂ reacts, but we know that 0.025 mol is left over.

81. To solve this problem with theoretical vs. actual yields, you first calculate the theoretical yield of ammonium bicarbonate (i.e., 100% of the expected value). Then you multiply it by 0.747 to get the actual yield.
First check to see if the equation given in the problem is balanced (the skeptic's approach). It is, because there is one N on each side, one C, 3 O, and 5 H. The problem states that water is present in excess, so we don't have to determine whether it could be limiting. Next, we see whether NH₃ or CO₂ limits, or whether they are present in the exact proportions to react fully. In the reaction NH₃ + CO₂ + H₂O ® NH₄HCO₃,  equimolar amounts of the three reactants are needed. How many moles of NH₃ and CO₂ does the student start with? For NH₃ it is (14.8 g)/(17 g per mol) = 0.87 mol. For CO₂ it is (41.3 g)/(44 g per mol) = 0.94 mol. Thus CO₂ is present in slight excess. The result should be 0.87 mol of NH₄HCO₃, which is (0.87)(79 g per mol) = 68 g. But she gets only 0.747 x 68 g = 51 g of ammonium bicarbonate. The exact answer is 51.3 g.

87. First, a 0.215-molar solution of a substance contains 0.215 mol of that substance per liter of solution, or 0.000215 mol per mL (dividing amount and volume by the same 1000). (a) If you need 0.0876 mol of NaBr and you have 0.000215 mol per mL, you just divide the total by the concentration to get the volume needed: (0.0876 mol)/(0.000215 mol per mL) = 403 mL. Note that we didn't have to know that it was NaBr because no weights were involved--the answer will be the same for ANY substance.
(b) Given the mass of CO(NH₂)₂ needed, convert to moles and then divide by moles per mL to find how many mL are needed. (I suggest you verify the form of this calculation by checking out its units first, in the way we did in class. It's a simple but powerful aid.) The MW of the substance is 12 + 16 + 32 = 60. 32.1 g is then a bit over 0.5 mol, or 32.1/60 = 0.535 mol. At 0.000215 mol per mL, (0.535 mol)/(0.000215 mol per mL) = 2490 mL of solution would be needed. 
(c) Same procedure as (b). 715 mg methanol = 0.715 g methanol (by dividing by 1000 mg per g), which equals (0.715 g)/(32 g methanol per mol) = 0.0223 mol. It will take (0.0223 mol)/(0.000215 mol per mL) = 104 mL of solution to get 715 mg of methanol.

99. The statement that an excess of BaCl₂ is added means that it will not be the limiting reactant. Thus the same number of moles of BaSO₄ are formed as moles of Na₂SO₄ that were present (because of the equal coefficients in the equation). How many moles were present? (0.635 L)(0.314 mol per L) = 0.2000 moles of BaSO₄. Convert to grams by multiplying by its molecular weight: (0.2 mol)(233 g per mol) = 46.6 g of BaSO₄. (Exact answer = 46.5 g)

Additional problem
115. The sample of impure Ag₂O contains Ag₂O plus something else--it doesn't matter what. The problem states that the Ag₂O part of the sample decomposes to Ag and O₂ gas. The unstated assumption is that all the Ag₂O decomposes. The decomposition yields 0.183 g of oxygen. From that, you can calculate the amount of Ag₂O that must have been present. You then compare with the total original weight of the impure sample to determine what fraction of it was Ag₂O. That's not so bad, is it?
First get the mass fraction of O in Ag₂O, which is O/(2 Ag + O) = 16/(216 + 16) = 0.06897. Thus the 0.183 g of O₂ generated during decomposition corresponds to (0.183 g O₂)/(0.06897 Ag₂O per O) = 2.65 g Ag₂O, which is 2.65/2.95 = 0.900 = 90.0% of the total impure sample. Thus the sample was 90.0% Ag₂O. (The exact answer is 89.8% Ag₂O.)

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