Chapter 2 · Terminology, Standards, and Mass Spectrometry

Answer 1

1. Given the relationship δ_X–B = δ_X–A + δ_A–B + 0.001·δ_X–A·δ_A–B (equation (2.3)), the δ³⁴S_CDT of the sample is 12.39 + (−4.66) + 0.001(12.39)(−4.66) = 7.67‰.
2. Not printed in the 1st ed. [By the same equation, −4.22 + (−22.68) + 0.001(−4.22)(−22.68) = −26.8‰ vs VPDB (constructed).]

Answer 2

For part (a), the problem is analogous to Problem 1(a). Given the relationship δ_X–B = δ_X–A + δ_A–B + 0.001·δ_X–A·δ_A–B (equation (2.3)), the δ¹⁸O of the sample is 38.72 + (−14.57) + 0.001(38.72)(−14.57) = 23.59‰.

For part (b), we need only find the working standard in terms of VPDB instead of VSMOW. To convert, we have δ¹⁸O_PDB = 0.970017·δ¹⁸O_SMOW − 29.98. So the working gas is 0.970017 × 38.72 − 29.98 = 7.58‰ vs VPDB. Then we use equation (2.3) with the value of the working standard relative to VPDB to finally solve the problem: −14.57 + 7.58 + 0.001(−14.57)(7.58) = −7.10‰ vs VPDB.

[Worked grid for Problems 1 and 2 (constructed). Click a cell to see its formula.]

Answer 3

The ³²S/³⁴S value of the sample is 21.56. The ³⁴S/³²S value of the sample is 0.0464, and that of CDT is 0.0450. Given that the δ value is given by equation (2.1), seawater sulfate has a

[Worked grid (constructed). Carried at full precision the grid gives +30.6‰; the printed +31.1‰ comes from rounding the ratios to 0.0464 and 0.0450 before differencing, a useful illustration of rounding error in delta arithmetic.]

Answer 4

2.83‰ vs AIR.

[Worked route (constructed): R_29/28(sample)/R_29/28(standard) = (4.573/3.014)/(4.568/3.021) = 1.00342, so the sample is +3.42‰ vs the reference gas; then 3.42 + (−0.57) + 0.001(3.42)(−0.57) = +2.84‰ vs AIR.]

[Worked grid (constructed). At full precision it reads +2.85‰ vs the printed 2.83‰; check vs print.]

Answer 5

δ¹⁸O_water = α(1000 + δ¹⁸O_vapor) − 1000 = 1.0101 × (1000 − 13.42) − 1000 = −3.46‰. reconstructed [The printed answer line is lost in the extraction; the printed value −3.46‰ survives as a fragment and matches this calculation.]

Answer 6

1. α_qz–water = (1000 + 24.4)/(1000 − 11.6) = 1.0364, so 10³lnα = 10³ln(1.0364) = 35.75. Solving 35.75 = 4.11·10⁶/T² − 3.70 for T gives 326 K, or 53 °C.
2. In place of 10³lnα we have Δ = 24.4 − (−11.6) = 36.00. Solving for T gives 325 K, essentially the same as the temperature found in part (a).
3. Not printed in the 1st ed. [Constructed: at 300 °C (573 K), 10³lnα = 4.11·10⁶/573² − 3.70 = 8.81, so δ¹⁸O_water = 1024.4/e^0.00881 − 1000 = +15.4‰.]

[Worked grid (constructed). At full precision parts (a) and (b) evaluate to 323 K and 322 K (≈50 °C); the printed answers read 326 K and 325 K (53 °C). Check vs print.]

Answer 7

For two phases a and b,

If δ_b = 0 and δ_a = 10, then ε is exactly equal to 10. If, for example, δ_b = 100 and δ_a = 110, then α = 1110/1100 = 1.00909 and ε = 9.09. But δ_a − δ_b = 10.

Answer 8

1. The δD values of snow falling in the Sierra Nevada mountains are as low as −116‰ (or average −116‰, as the case may be).
2. The carbon isotope composition of the Hawaiian nodule is −2‰, higher than the value of normal mantle material. (Note that the value of the sample still ‘is’ −2‰. There is no reason to use the past tense unless all of the sample was ground up and there is no more of this mantle nodule. Otherwise, we can all assume that the nodule has the same value as when it was originally measured.)
3. There is no apparent correlation between the δ³⁴S and δ¹³C values of these samples.

Answer 9

There are two sulfur atoms in the pyrite formula (FeS₂) and only one in sphalerite. Therefore we write the reaction as follows:

In terms of thermodynamic properties such as volume or free energy, we would use 1/2 the value reported for FeS₂, because our formula is written as 1/2 of the total formula.

Answer 10

Not recoverable from the extraction. check vs print [The derivation follows from writing α_X–B = α_X–A·α_A–B in terms of the δ definition and expanding (constructed hint).]

Answer 11

H₂¹⁸O, HD¹⁷O, and D₂¹⁶O (all mass 20).

Answer 12

For the first part, the sample has an ¹⁸O/¹⁶O ratio of 0.00200520 × 1.0197 = 2044.70·10⁻⁶. Using the definition of delta, we solve the equation

where δ = 19.7 and R_std = 2005.20·10⁻⁶. Upon rearranging terms, it is found that the value is identical to that obtained by the first method.

Answer 13

The definition assumes that there is more than one isotope of a particular element. In the case of fluorine, for example, there is only one naturally occurring isotope. There is no requirement that two isotopes of an element must exist.

Answer 14

Starting with the definition of delta and knowing that D/H of VSMOW = 155.76·10⁻⁶, we determine the D/H ratio of our starting water (δD = −90‰) to be 141.73·10⁻⁶, and the D/H ratio of water with δD = +500‰ to be 233.63·10⁻⁶. For one liter, the amount of D₂O is 0.234 ml per liter of water. The same calculation for our starting water gives 0.1417 ml per liter of water. The difference is 0.092 ml. That is, adding only ~0.09 ml of D₂O to a liter of water with an initial δD of −90‰ will increase the δD value to +500‰ vs VSMOW.

[Worked grid (constructed). The volume conversion constants are in the parameter cells; the grid reads 0.093 mL vs the printed 0.091 mL, within the rounding of the printed route.]