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S6 Chemistry: Instrumental Analysis (Paper 2: Intermediate)
S6 Chemistry: Instrumental Analysis (Paper 2: Intermediate)
Ten items on the Beer-Lambert law, dilution factors, isotope peak ratios and internal standards, with answers.
Direct answers Answers in order: 3000, 1.2e-3 mol dm-3, bromine, chlorine, an ethyl group, a carboxylic acid, to correct injection variation, 0.050 percent, 0.16 mol dm-3, no.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 20 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S6 Chemistry: Instrumental Analysis |
| Level | S6 |
02 Questions (10 items)
- FoundationA solution of concentration 2.0e-4 mol dm-3 in a 1.0 cm cell has absorbance 0.60. What is the molar absorption coefficient?
- FoundationA sample diluted tenfold gives absorbance 0.30 in a 1.0 cm cell and the molar absorption coefficient is 2500. What is the original concentration?
- FoundationIn a mass spectrum the M and M+2 peaks are almost equal in height. Which element is present?
- IntermediateThe M+2 peak is about one third the height of the M peak. Which element is present?
- IntermediateAn ester has its molecular ion at m/z 88 and a fragment at m/z 59. What group has been lost?
- IntermediateAn infrared spectrum has a very broad absorption from 2500 to 3300 cm-1 and a strong band at 1710 cm-1. What is it?
- ChallengeWhat is the purpose of the internal standard method in gas chromatography?
- ChallengeA 0.50 g sample is made up to 100 cm3 and atomic absorption gives 2.5 ppm of a metal. What is the percentage by mass in the sample?
- ChallengeStandards of 0.10 and 0.20 mol dm-3 give absorbances of 0.25 and 0.50. An unknown reads 0.40. What is its concentration?
- Error redoMisconception drill: does atomic absorption measure the mass of molecular ions in solution?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 3000 dm3 mol-1 cm-1 | Rearrange A equals epsilon c l, so 0.60 is divided by the product of 2.0e-4 and 1.0. |
| 2 | 1.2e-3 mol dm-3 | The diluted solution is 1.2e-4, and multiplying by the tenfold dilution gives the original value. |
| 3 | Bromine | The two bromine isotopes at 79 and 81 have similar abundance, so M and M+2 are nearly equal. |
| 4 | Chlorine | Chlorine isotopes 35 and 37 occur in about a three to one ratio, so M+2 is a third of M. |
| 5 | An ethyl group, C2H5 | 88 minus 59 gives 29, which is the mass of an ethyl group. |
| 6 | A carboxylic acid | A very broad OH absorption together with a carbonyl band is characteristic of a carboxylic acid. |
| 7 | To correct for variation in injection volume and detector response | The standard and analyte are injected together, so the ratio of peak areas cancels injection error. |
| 8 | 0.050 percent | At 2.5 ppm there is 2.5 mg per litre, so 0.1 L holds 0.25 mg, which is 0.0005 of 500 mg. |
| 9 | 0.16 mol dm-3 | Absorbance 0.40 is 1.6 times 0.25, so the concentration is 1.6 times 0.10. |
| 10 | No, it measures absorption by ground state atoms in the flame | The sample is nebulised and atomised, so the signal comes from atoms rather than molecular ions. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
Write A equals epsilon c l before substituting any number.Tutor Registration →
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