A geologist gives you a piece of rock containing a mineral called Smithsonite, w

Question

A geologist gives you a piece of rock containing a mineral called Smithsonite, which is ZnCO_3 in its pure form. She wants to know what percentage of the rock is zinc carbonate (molar mass = 125.418 g/mol) Given the sample, you weigh out three separate sample of crushed rock, each weighing about 1 gram dissolves each in 100 mL of tilde 6 M HCl and filter off the residue. Then you transfer each filtrate solution quantitatively to a 1.00 L volumetric flask, filling to the line with 1% (v/v) HCl (see table 1 for data) Next, you remove 5.00 mL of this solution using a volumetric transfer pipet and drain it into another 1.00 L volumetric flask. You add 2.34 g of La(NO_3)_3 and fill to the line with 1% 3 (v/v) HCl. You do this for each of the three samples separately You also make a stock solution from a piece of clean, polished, zinc metal and then make a series of standard concentrations by diluting volumes of approximately 5, 4, 3, 2, and 1 mL of Zn stock solution (measured using a buret) to a final volume of 500.00 mL with 1% (v/v) HCl (see table 2 for data) after adding 1.17 g of La(NO_3)_3 to each. You analyze each of the samples, standards and some 1% (v/v) HCl by AAS recording the absorbance reading for each and plot the calibration curve below. Calculate the weight percent of ZnCO_3 mineral in the rock sample as the average of the three replicate samples .The value of s_x from the statistical analysis of the calibration curve equals 0.2064 ppm Zn. The standard deviation of the three absorbance readings is 0.001301 absorbance units. Report the uncertainty of the result using a 95% confidence level.

Explanation / Answer

Weight percentage of ZnCO3 is calculated using the molar mass and the volune of each sample in each step.

Weight percentage at first sample=(100mL soln each×percentage if HCL added×La(NO3)3 added)/molar mass

=100×0.01/125.418=0.0079

Weight percentage of second sample=(100-5)(5 mL is removed)×(0.01×2.34)/125.418

=0.0177

Weight percentage of third sample=(500mL ×0.01×1.17)/125.418=0.0466

Given that the statistical result is 0.2064ppm and the standard deviation value is 0.001301 units.

There is variation with these values because the acid and basic levels do not remain the same all the time. The levels vary and so the results show some stoichiometric changes in the concentration.

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