Questions & explanations
1. A mass spectrum shows a molecular ion peak at m/z 44. What is the molar mass? If the compound contains only C and H, and the empirical formula is CH2, find the molecular formula.
The M+ peak at m/z 44 means the molar mass is 44 g/mol. The empirical formula CH2 has a mass of 14.03 g/mol (12.01+2×1.008). Divide 44 by 14.03 ≈ 3.14? Wait, recalc: 44/14.03 = 3.14, not whole. Actually, CH2 mass is 14.03, but 44/14.03 = 3.14, so maybe it's not CH2? Let's check: For hydrocarbon, possible formulas: C3H8 (propane, mass 44.1) – that gives empirical CH2.67? Actually propane is C3H8, empirical CH2.67, not CH2. So the empirical formula given might be wrong. Instead, let's use correct: If empirical is CH2, then molecular formula must be (CH2)n, n integer. For n=3, mass 42.08; n=4, 56.11. 44 is not a multiple. So the compound is likely C2H4O? No. For simplicity, assume it's C3H8, empirical C3H8? Actually, the question is to practice: given empirical CH2 and molar mass 44, the molecular formula would be C3H8? But C3H8 mass 44.1, empirical is C3H8? No, empirical of C3H8 is C3H8 (since 3:8 already simplest). So to make it work, let's change example: empirical CH2 and molar mass 56 gives C4H8. For m/z 44, it could be CO2? But compound contains only C and H. So, I'll adjust answe
2. Give an example of the law of reciprocal proportions using hydrogen, oxygen, and sulfur.
Hydrogen and oxygen form water with H:O = 1:8 by mass. Hydrogen and sulfur form hydrogen sulfide (H₂S) with H:S ≈ 1:16 by mass (since H:S mass ratio in H₂S is about 1:16). Oxygen and sulfur form sulfur dioxide (SO₂) with O:S = 1:1 by mass (since atomic masses: O=16, S=32, so SO₂ has 32:32 = 1:1). The ratio of oxygen to sulfur that combine with 1 g H is 8 g O and 16 g S, giving ratio 8:16 = 1:2. In SO₂, O:S ratio is 1:1, which is double the 1:2? Actually, reciprocal would expect the ratio of combining masses of O and S with H to be the same as the ratio in which O and S combine. Here, mass of O that combines with 1 g H is 8 g, mass of S that combines with 1 g H is 16 g, ratio O:S = 1:2. In SO₂, the mass ratio O:S is 1:1 (or 2:2?). Actually, check: In SO₂, 32 g S combines with 32 g O, ratio O:S = 1:1. But our ratio from H is 1:2, which is a simple multiple (2 times) of 1:1? Actually, 1:2 is half of 1:1? Clearer: The law says the ratio should be the same or a simple multiple. Here, the combining ratio of O:S from H is 8:16 = 1:2. In SO₂, O:S is 32:32 = 1:1. 1:1 is a simple multiple (2 t
3. What is the R/S designation for the chiral carbon in lactic acid (CH₃CH(OH)COOH)?
Lactic acid has one chiral carbon (the one with OH, H, CH₃, COOH). Assign priorities: OH (8) highest, then COOH (6), then CH₃ (6? Actually, COOH carbon is attached to O,O,O? Wait: COOH: the carbon is bonded to O (8), O (8), and OH? Let's do properly: For the chiral carbon, groups: -OH (O, priority 1), -COOH (the C is bonded to O, O, and OH? Actually, carboxylic acid group: the carbon is doubly bonded to one O and singly to another O (from OH). So the carbon has three oxygens? No, the carbon of COOH is bonded to O (double bond), O (single bond from OH), and the rest of the chain. But for priority, we consider atoms directly attached. The COOH group as a whole: the first atom is C, which has O, O, and something. The CH₃ group first atom is C with H, H, H. So C in COOH has higher priority than C in CH₃ because O > H. So order: 1. OH, 2. COOH, 3. CH₃, 4. H. Now orient so H is away. Looking, if the sequence OH→COOH→CH₃ is clockwise, then R; if counterclockwise, S. The naturally occurring L-lactic acid is S. The mirror image would be R.
4. You have a sample with an unknown amount of strontium. You add 1.00 mg of a spike containing 84Sr (enriched isotope). After mixing, the mass spectrum shows a 84Sr/88Sr ratio of 0.200. The natural ratio of 84Sr/88Sr is 0.007. How much strontium was in the sample?
Let x be the mg of natural Sr in sample. Natural Sr has abundance of 84Sr = 0.007 × (moles 88Sr?) Actually, we need the isotopic abundances. Assume natural Sr has 84Sr/88Sr = 0.007, meaning for every 1 mg of natural Sr, the mass of 84Sr is negligible? Better to define: Let R_sample be the ratio after spiking: R = (84Sr from spike + 84Sr from sample) / (88Sr from sample + 88Sr from spike). But spike only contains 84Sr, so 88Sr from spike=0. Then R = (mass_spike * (1) / (mass_sample * (1/0.007)? This gets messy. Simplify: Assume spike is pure 84Sr, sample has natural Sr with known isotopic abundances. The equation: R_meas = (M_spike + f84_nat * M_sample) / (f88_nat * M_sample), where f84_nat = 0.007, f88_nat = 0.826 (for natural Sr). Given M_spike=1.00 mg, R_meas=0.200. Solve: 0.200 = (1.00 + 0.007*M_sample) / (0.826*M_sample). => 0.1652 M_sample = 1.00 + 0.007 M_sample => (0.1652-0.007)M_sample = 1.00 => M_sample = 1.00/0.1582 = 6.32 mg. So the sample contained 6.32 mg Sr.
5. What is the structure type if a borane has formula B₅H₉?
B₅H₉ is a nido borane. According to Wade's rules, for 5 boron atoms, a closo structure would have 6 electron pairs, a nido has 7 pairs. B₅H₉ has 5×3 + 9 = 24 valence electrons = 12 pairs. Wait, that's too many. Actually, B₅H₉ is known as nido-pentaborane(9). The number of pairs matches n+2 = 7? Let me recalc: For B₅H₉, each BH gives 3, so 5×3=15, plus 9 H gives 9, total 24 electrons, 12 pairs. But n=5, so n+1=6, n+2=7, n+3=8. 12 is not matching. I need to be accurate: For neutral boranes, Wade's uses skeletal electrons: each BH contributes 2 skeletal electrons (since 1 goes to terminal B-H), actually the rule is complex. I'll adjust answer: B₅H₉ is a nido cluster. Its structure is a square pyramid missing one vertex. This matches the nido classification for five boron atoms.
6. How does the Hantzsch-Widman name differ from the common name for pyridine?
Pyridine is a six-membered ring with one nitrogen and three double bonds. Its common name is 'pyridine', but the Hantzsch-Widman name is 'pyridine' also? Actually, pyridine is already a systematic name in Hantzsch-Widman? Wait, pyridine is the common name, and the Hantzsch-Widman name for a six-membered unsaturated ring with one nitrogen is 'azine'. However, pyridine is still widely used. So the difference is that the systematic name would be 'azine' for such a ring, but pyridine is a specific compound. So the common name 'pyridine' refers to a specific structure, while the Hantzsch-Widman name 'azine' could refer to any six-membered unsaturated nitrogen heterocycle. So systematic names are more general and follow rules, while common names are unique to each compound.
7. In a Latimer diagram for nitrogen in acid: NO3- (+0.80) -> NO2- (+0.87) -> NO (+0.75) -> N2O (+1.59) -> N2 (-1.87) -> NH4+. Which oxidation state is most stable?
Stability is indicated by a low tendency to change. Look for species where potentials to left and right are both unfavorable. For N2, reduction to NH4+ has -1.87 V (very negative), and oxidation to N2O has +1.59 V (large positive). Since both directions have large magnitudes, N2 is kinetically inert but thermodynamically stable in between? Actually species that are low in free energy are stable. The Latimer diagram gives stepwise potentials; N2 is in the middle. Often the most stable form is the one that appears in the middle of the diagram with no large driving force to either side. Here, N2 has a very negative potential to NH4+ (reduction hard) and positive to N2O (oxidation also requires energy). So N2 is very stable thermodynamically.
8. How did Rutherford's nuclear model replace Thomson's plum pudding model?
Rutherford's nuclear model replaced Thomson's plum pudding model because the experimental evidence did not match Thomson's predictions. Thomson's model had positive charge spread throughout the atom, so alpha particles should pass through with only small deflections. But Rutherford observed large angle scattering and backscattering, which required a concentrated positive charge. His model placed all positive charge in a tiny nucleus, leaving most of the atom empty. This explained why some alpha particles bounced back. The nuclear model also accounted for the atom's mass being concentrated in the nucleus, which Thomson's model could not. So Rutherford's model was accepted as correct.
9. Compare the systematic name of D-glucose and L-glucose.
D-glucose and L-glucose are enantiomers, meaning they are mirror images. D-glucose has the hydroxyl on the fifth carbon (the last chiral center) pointing to the right in the Fischer projection, while L-glucose has it pointing to the left. Their systematic names are based on the same parent chain, but the D or L prefix shows the configuration. For example, D-glucose is (2R,3S,4R,5R)-2,3,4,5,6-pentahydroxyhexanal? Actually, IUPAC names use R,S designations. But in common usage, D-glucose and L-glucose are distinguished by the D/L system. They have opposite optical rotations. D-glucose is naturally occurring and used in metabolism, while L-glucose is rare and not used by the body.
10. What suffix indicates a saturated heterocycle in the Hantzsch-Widman system?
In the Hantzsch-Widman system, saturated heterocycles have specific suffixes based on ring size. For a three-membered ring, the suffix is '-irane'; four-membered is '-etane'; five-membered is '-olane'; six-membered is '-ane'; seven-membered is '-epane'; eight-membered is '-ocane'; nine-membered is '-onane'; ten-membered is '-ecane'. These suffixes are used after the heteroatom prefixes. For example, a saturated six-membered ring with one oxygen is 'oxane'. If the ring is unsaturated, suffixes change, like '-ene' for one double bond or '-ine' for aromatic rings. So the suffix '-ane' generally means a saturated six-membered ring, but for other sizes the suffix is different.
11. What is the equivalence point in a titration?
The equivalence point is the moment during a titration when the moles of added base exactly equal the moles of acid originally present, or vice versa. At this point, the reaction between the acid and base is complete. For a strong acid-strong base titration, the pH at the equivalence point is 7 because both the cation and anion do not affect water's pH. But for a weak acid-strong base titration, the pH is greater than 7 because the conjugate base of the weak acid hydrolyzes water to form hydroxide ions. The equivalence point is identified by a sharp change in pH on the titration curve. It is different from the endpoint, which is when the indicator changes color.
12. How does the activity coefficient change with composition for a non-ideal solution?
In a non-ideal solution, the activity coefficient varies with the mole fraction of each component. At very dilute concentrations, the activity coefficient of a solute often approaches 1 (ideal behavior). As concentration increases, interactions become significant, and the coefficient either increases or decreases. For example, in a solution of acetone and chloroform, the activity coefficient of acetone is less than 1 at intermediate concentrations due to strong attraction. The change is not linear; it depends on the specific intermolecular forces. Plots of activity coefficient versus composition show how the solution deviates from ideality throughout the range.