A solution contains an unknown amount of osmium at the mu g/g level, or below. C

Question

A solution contains an unknown amount of osmium at the mu g/g level, or below. Chemically pure osmium (Os) may be separated from the solution, but the chemical yields are variable and low. The determination is done by direct isotope dilution analysis (DIDA). The isotopic abundances of ^186 Os and ^190 Os are 1.58 and 26.4%, respectively. 7.80 Times 10 9 g of isotopically pure, stable ^186 Os is added to the unknown solution and mixed thoroughly, and then a small amount of pure Os is extracted from the mixture. With a mass spectrometer, the ^190 Os/^186 Os mass ratio in the extracted sample is found to be 7.40. Calculate the amount of Os originally present in the unknown solution.

Explanation / Answer

Isotopic abundance of 186Os = 1.58%

Isotopic abundance of 190Os = 26.4%

Mass of  186Os added to the solution = 7.80x10-9 g

Let "m" g Os was originally present in the solution.

Mass of  186Os present = 1.58% of m = 0.0158*m g

Mass of  190Os present = 26.4% of m = 0.264*m g

Total mass of  186Os present = (0.0158*m + 7.80x10-9 )g

Mass of  190Os / Mass of  186Os = 7.40

0.264*m / (0.0158*m + 7.80x10-9 ) = 7.40

0.264*m = 7.40 * (0.0158*m + 7.80x10-9 )

0.264*m = 0.11692*m + 57.72x10-9

0.14708*m = 57.72x10-9

m = 3.92x10-7

Hence, mass of Os originally present in the unknown solution = 3.92x10-7 g

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