1. An unknown sample of ore with a mass of 11.3629 g is allowed to react with ex

Question

1. An unknown sample of ore with a mass of 11.3629 g is allowed to react with excess dimethylglyoxime, HC4H7O2N2, in order to analyze for nickel, according to the following precipitation reaction:

Ni2+(aq) + 2HC4H7O2N2(aq) Ni(C4H7O2N2)2(s) + 2H+(aq)

After the precipitated Ni(C4H7O2N2)2 is carefully rinsed and dried, the mass is determined to be 0.05169 mg.
What was the original mass (mg) of Nickel in the ore sample?

2. An unknown sample of ore with a mass of 506.1 g is allowed to react with excess magnesium and ammonia, NH3, in order to analyze for phosphate, PO43-, according to the following precipitation reaction:

2PO43-(aq) + 2Mg2+(aq) + 2NH3(aq) + 2H2O(l) Mg2(P2O7)(s) + 2NH3(aq) + 2OH-(aq) + H2O(l)

After the precipitated Mg2(P2O7) is carefully rinsed and dried, the mass is determined to be 0.09870 mg.
Assuming all phosphorous originally present in the ore was in the form of phosphate, what was the original mass (g) of phosphorous in the ore sample?

3. An unknown organic compound was produced by students in the O-chem lab. According to their records, this compound should only contain carbon and hydrogen atoms. A combustion analysis is performed on the sample.

When 0.1727 g of the sample is burned in the presence of excess oxygen, the magnesium perchlorate cartridge increases 0.24889 g and the ascarite cartridge increases 0.53094 g.

What is the percent C (w/w) in the sample?

4. What is the percent H (w/w) in the sample?

5. What is the empirical formula for the unknown compound?

Explanation / Answer

1) Ni2+(aq) + 2HC4H7O2N2(aq) Ni(C4H7O2N2)2(s) + 2H+(aq)

Molar mass of Ni(C4H7O2N2)2 = 288.69 g/mole

moles of Ni(C4H7O2N2)2 recovered = mass/molar mass = (0.05169*10-3)/288.69 = 1.791*10-7

Now, each mole of Ni(C4H7O2N2)2 contains 1 mole of Ni

Thus, moles of Ni in 1.791*10-7 moles of Ni(C4H7O2N2)2 = 1.791*10-7

Now, molar mass of Ni = 58.69 g/mole

Thus, mass of Ni in the ore sample = moles*molar mass = (1.791*10-7)*58.69 = 1.051*10-5 g

Hence % of Ni in the ore = (mass of Ni/mass of ore)*100 = 9.25*10-5 %

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