Natural samples of copper contain two isotopes. 63Cu has a mass of 62.930 amu and 65Cu has a mass of 64.928 amu. The percent abundance of 63Cu is 69.09% and 65Cu is 30.91%. Calculate the average atomic mass of copper

Reliable trace metal analysis of environmental samples is dependent upon the availability of high accuracy, matrix reference standards. Here, we present Cd, Cu, Ni, Pb and Zn isotope dilution determination for an estuary water certified reference material (LGC 6016). This work highlights the need for high-accuracy techniques in the development of trace element CRMs rather than conventional inter-laboratory trials. Certification of the estuary water LGC6016 was initially determined from a consensus mean from 14 laboratories but this was found to be unsatisfactory due to the large discrepancies in the reported concentrations. The material was re-analysed using isotope dilution ICP-MS techniques. Pb and Cd were determined using a conventional quadrupole ICP-MS (Elan 5000). Cu, Zn and Ni were determined using a magnetic sector ICP-MS (Finnigan Element), which allowed significant polyatomic interferences to be overcome. Using the magnetic sector instrument, precise mass calibration to within 0.02 amu permitted identification of the interferences. Most interferences derived from the sample matrix. For example, the high Na content causes interferences on 63Cu, due to the formation of 40Ar23Na and 23Na216O1H, which in a conventional quadrupole instrument would relate to an erroneous increase in signal intensity by up to 20%. For each analyte a combined uncertainty calculation was performed following the Eurachem/GTAC and ISO guideline. For each element a combined uncertainty of 2–3% was found, which represents a 10-fold improvement compared to certification by inter-laboratory comparison. Analysis of the combined uncertainty budget indicates that the majority of systematic uncertainty derives from the instrumental isotope ratio measurements.

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.