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  1. Problems
  2. /A 0.175 g sample of a compound holds 16 % nitrogen by mass…
HardChemistryHigh School (11–12)Organic Chemistry: Some Basic Principles and Techniques › Purification and Analysis of Organic CompoundsNumericalA-LevelIBNEET

A 0.175 g sample of a compound holds 16 % nitrogen by mass…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A 0.175 g0.175\ \text{g}0.175 g sample of a compound holds 16 %16\ \%16 % nitrogen by mass. All the nitrogen turns into cyanide ions in the extract. Find the amount of cyanide, and the largest mass of Prussian blue, Fe4[Fe(CN)6]3\text{Fe}_4[\text{Fe(CN)}_6]_3Fe4​[Fe(CN)6​]3​, with a molar mass of 860 g/mol860\ \text{g/mol}860 g/mol, that it could make. Eighteen cyanide ions go into one unit of Prussian blue.

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mmol
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Need a reminder? The key ideaShowHide

Carbon is shown by carbon dioxide turning lime water milky, and hydrogen by water turning white copper sulphate blue. Fusion with sodium turns N, S and the halogens into CN, S and X ions. Nitrogen: Prussian blue. Sulphur: black PbS or violet with nitroprusside. Nitrogen and sulphur together: blood red [Fe(SCN)]2+[\text{Fe(SCN)}]^{2+}[Fe(SCN)]2+. Halogens: boil with nitric acid first, then silver nitrate. Phosphorus: a yellow colour with ammonium molybdate.

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Step 1. Nitrogen. 0.175×16100=0.028 g0.175 \times \dfrac{16}{100} = 0.028\ \text{g}0.175×10016​=0.028 g.

Step 2. Cyanide. 0.02814=0.00200 mol\dfrac{0.028}{14} = 0.00200\ \text{mol}140.028​=0.00200 mol, which is 2.00 mmol2.00\ \text{mmol}2.00 mmol.

Step 3. Units of Prussian blue. 2.0018=0.111 mmol\dfrac{2.00}{18} = 0.111\ \text{mmol}182.00​=0.111 mmol.

Step 4. Mass. 0.111×860=95.6 mg0.111 \times 860 = 95.6\ \text{mg}0.111×860=95.6 mg.

Answer: 2.00 mmol2.00\ \text{mmol}2.00 mmol of cyanide, and at most 95.6 mg95.6\ \text{mg}95.6 mg of Prussian blue.

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  1. Qualitative analysisPurification and Analysis of Organic Compounds

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