Chemistry: Gravimetric Determination of Nickel: A steel sample contain nickel as

Question

Chemistry: Gravimetric Determination of Nickel:

A steel sample contain nickel as an alloying element. The analyst has taken three samples of the steel sample and digested them with HCl, boiling the sample with the acid until the sample goes into solution. She then adjusts the pH to a pH of 8.0. An alcoholic solution of Dimethylglyoxime is added in sufficient quantity to precipitate the nickel present in the samples as a chelated species. Determine the percent nickel in each sample and determine if the samples are statistically equivalent. Determine the mass of nickel for each sample if the chelated nickel is ashed and just the nickel is left.

3. A steel sample contain nickel as an alloying element. The analyst has taken three samples of the steel sample and digested them with HCI, boiling the sample with the acid until the sample goes into solution. She then adjusts the pH to a pH of 8.0. An alcoholic solution of Dimethylglyoxime is added in sufficient quantity to precipitate the nickel present in the samples as a chelated species. Determine the percent nickel in each sample and determine if the samples are statistically equivalent. Determine the mass of nickel for each sample ifthe chelated nickel is ashed and just the nickel is left. Test Method (htt vlab.amrita.edu/?sub 2& brch-193&tsim; 348&cnt-1;) (htt Sample No. Steel Sample wt. (g) Precipitated wt. (g) 1 13.0124 .3125 2 13.0028 .3078 3 12.9875 ,2998

Explanation / Answer

Solution :-

formula of the chelate formed as precipitate of the Nickel is Ni(C4H7O2N2)2

molar mass of Ni(C4H7O2N2)2 = 288.9146 g per mol

mass percent of the Ni in the Ni(C4H7O2N2)2 is 20.3151 %

now lets find the mass of the Nickel in the each sample

mass of nickel = (mass of precipitate * % nickel / 100 %)

mass of nickel in sample 1 = (0.3125 g * 20.3151 % /100%) = 0.0635 g

mass of nickel in sample 2 = (0.3078 g * 20.3151 % / 100%) = 0.0625 g

mass of nickel in sample 3 = (0.2998 g * 20.3151 % / 100%) = 0.0609 g

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