Organic Chemistry – Some Basic Principles and Techniques · Lesson 14 of 15
Measuring Halogens, Sulphur, Phosphorus, and Oxygen
“Use precipitate masses and conservation of atoms to complete an elemental composition analysis.”
• Explain Carius halogen estimation and calculate a halogen percentage. • Calculate sulphur and phosphorus percentages from weighed precipitates. • Distinguish oxygen by difference from direct oxygen estimation. • Explain the shared mass-fraction reasoning and the role of modern elemental analysers.
A weighed product carries a known fraction of the element
Many elemental methods convert the target element into a stable compound that can be isolated and weighed. The product contains other atoms supplied by the analytical reagents, so its full mass is not the mass of the original element. The formula of the product tells us the fraction that belongs to the element of interest. This is the same conservation argument used for carbon in carbon dioxide.
Halogens form silver halides in Carius analysis
In the Carius method, a known sample mass is heated with fuming nitric acid and silver nitrate in a sealed hard-glass tube. Carbon and hydrogen are oxidised, while chlorine, bromine, or iodine is converted into its silver halide. The precipitate is filtered, washed, dried, and weighed. These stages matter because adhering solution or incomplete drying would add mass that does not belong to silver halide.
One mole of AgX contains one mole of the halogen X. Therefore the halogen mass in the precipitate is its atomic mass divided by the formula mass of AgX, multiplied by the precipitate mass. Silver makes up much of the remaining mass and must not be attributed to the original sample. The method described here concerns chloride, bromide, and iodide products, not an assumption that every halogen produces a suitably insoluble silver salt.
Problem
A 0.150 g sample gives 0.120 g AgBr. Calculate its bromine percentage.
- 1.The formula mass of AgBr is . Its bromine mass fraction is .
- 2.The bromine mass is .
- 3.The bromine percentage is . The full 0.120 g is not bromine.
Sulphur is weighed as barium sulphate
Oxidation with fuming nitric acid or sodium peroxide converts the sample’s sulphur into sulphuric acid or sulphate. Adding barium ions precipitates barium sulphate, which is recovered, washed, dried, and weighed. Each formula unit of barium sulphate contains one sulphur atom. Using the approximate masses in the chapter, 233 g of BaSO₄ corresponds to 32 g of sulphur.
Problem
A 0.157 g sample gives 0.4813 g barium sulphate. Calculate sulphur percentage.
- 1.The sulphur mass is approximately.
- 2.The percentage is .
- 3.The precipitate is heavier than the original sample because barium and oxygen are supplied during analysis. This does not violate conservation of the original sulphur atoms.
Phosphorus can be weighed in two product forms
Phosphorus is oxidised to phosphoric acid. One route precipitates ammonium phosphomolybdate using ammonia and ammonium molybdate. An alternative uses magnesia mixture to precipitate magnesium ammonium phosphate, which on ignition produces magnesium pyrophosphate. The two weighed products contain different numbers of phosphorus atoms per formula unit, so they require different conversion factors.
| Weighed phosphorus product | Approximate formula mass | Phosphorus mass represented | Conversion factor |
|---|---|---|---|
| 1877 | 31 per formula mass | 31/1877 | |
| 222 | 62 per formula mass | 62/222 |
Problem
A 0.250 g sample gives 0.111 g magnesium pyrophosphate. Find its phosphorus percentage.
- 1.Each formula unit contains two P atoms, so the phosphorus fraction is , not .
- 2.The phosphorus mass is .
- 3.The percentage is . The two-phosphorus factor follows directly from the product formula.
Oxygen by difference and by direct conversion
Oxygen is often estimated by subtracting the measured percentages of every other element from 100. This works only if the composition has been fully accounted for. If nitrogen, sulphur, a halogen, or another component is present but omitted, its contribution will be incorrectly counted as oxygen. The difference also accumulates errors from the other measurements.
The direct method described in the chapter decomposes a known mass of sample in a nitrogen stream. Oxygen in the gaseous products is converted to carbon monoxide over red-hot carbon. Warm iodine pentoxide oxidises that carbon monoxide to carbon dioxide and liberates iodine. Either the carbon dioxide or iodine amount can provide the oxygen estimate after the appropriate stoichiometric accounting.
Problem
A compound known to contain only C, H, and O has 40.0% C and 6.7% H. Separately, a 0.200 g sample in the direct method yields 0.220 g carbon dioxide. Find oxygen percentage in each case.
- 1.By difference, oxygen is . This requires the stated C/H/O-only condition.
- 2.For the direct method, the sample oxygen mass is . The percentage is .
- 3.These are separate sample descriptions, not measurements of the same stated compound. The direct factor differs from carbon combustion analysis because the element being traced is different.
In the iodine route, five CO molecules give one I₂ molecule. Each CO carried one oxygen atom from the sample, so one mole of iodine corresponds to five moles of sample oxygen atoms. This relationship provides an alternative calculation when iodine is the measured product. The conversion must be quantitative for either route to be reliable.
Modern C/H/N elemental analysers can measure these elements using only about 1–3 mg of material and automatically process the combustion products. Automation improves speed and handles small samples, but it still depends on calibration, complete conversion, and correct interpretation. The underlying principle remains relating measured products to atoms in the original sample.
Never divide precipitate mass directly by sample mass and call the result the element percentage. Account for the product’s formula and the number of target atoms in it. Oxygen by difference is not valid if some other element has simply been left unmeasured.
Quiz
In Carius halogen estimation, what is weighed?
Which factor converts BaSO₄ mass to sulphur mass?
Why does the phosphorus factor for use 62?
When is oxygen by difference justified?
In the described direct oxygen method, 44 g of measured CO₂ corresponds to how much sample oxygen?
Practice Problems
- A 0.3780 g chloro compound gives 0.5740 g AgCl. Calculate chlorine percentage using Ag = 108 and Cl = 35.5.
- A 0.468 g sample gives 0.668 g barium sulphate. Calculate sulphur percentage.
- Derive both phosphorus conversion factors from the formulas of the weighed products.
- Calculate phosphorus percentage when 0.300 g sample yields 0.1877 g ammonium phosphomolybdate.
- Explain why a compound containing unmeasured nitrogen cannot have its oxygen correctly estimated from C and H percentages alone.
- Derive the sample-oxygen-to-carbon-dioxide relationship for the direct method and explain where the second oxygen in CO₂ comes from.
- Explain one advantage of a modern elemental analyser and one reason its result still requires chemical interpretation.
Key Takeaways
• A precipitate’s formula converts its total mass into the target element mass. • Halogen, sulphur, and phosphorus factors follow their specific weighed products. • Oxygen by difference requires all other constituents to be accounted for. • Direct oxygen estimation follows sample oxygen through CO to measured products.