Question

vandium has 2 iostopes whose relative abundance aer v51 = 99.75% and v50 =0.25%. to determine vandiume concentration of steel, 2g of the steel was dissolved in a acid medium and 1 ug of V50 WAS ADDED TO RESULTING SOLUTION. AFTER STIRRING AN ANALYSIs by icp-ms obtained a mass spectrum with mass 50 and 51, same area size.

1.calculate % of each isotope of vandium in the unknown samole if the ratios of theareas of the peaks are the same

2. calculate a more exact answer kmowing that v50=49.947g/mol and v51=50.944 g/mol

Answer 1

The mass of V50 added = 1 g

The % of vanadium in the high-carbon steel alloys = 0.2 %

The % of V51 in the sample = 2*(0.2/100)*(99.75/100) = 0.00399 = 3990 g

The % of V50 in the sample = 2*(0.2/100)*(0.25/100) = 0.00001 g = 10 g

The total mass of V50 = 10 + 1 = 11 g

1. Therefore, the % of V51 = {3990/(3990+11)}*100 = 99.725 %

And the % of V50 = {11/(3990+11)}*100 = 0.285 %

2. The no. of millimoles of V51 = (0.00399/50.944)*1000 = 0.0783 mmol

The no. of millimoles of V50 = (0.00001/49.947)*1000 = 0.0002 mmol

The total no. of mmol of vanadium = 0.0785 mmol

Now, the more exact % of V51 = (0.0783/0.0785)*100 = 99.745 %

And the more exact % of V50 = (0.0002/0.0785)*100 = 0.255 %