Deducing Organic Structures | Edexcel International A Level Chemistry Revision Notes 2017

Combined Techniques

You will be expected to deduce the empirical, molecular and structural formulae of compounds from data such as:
- Combustion analysis
- Elemental percentage composition - also known as calculating empirical formulae
- Characteristic reactions of functional groups - this can include test-tube reactions as well as interconversions
- Infrared (IR) spectra
- Mass spectra (MS)
- Nuclear magnetic resonance (NMR) spectra
The normal progression for the analysis of a compound is:
1. Find the empirical formula
2. Determine the molecular formula
3. Identify the functional groups / structural components present
4. Deduce the overall structure

Sometimes combustion analysis is performed on an unknown compound to determine the elemental percentage composition
The elemental percentage composition is then used to calculate the empirical formula
In combustion analysis, a known mass of the compound is burned in an excess of dry oxygen
The mass of carbon dioxide and water are then used to determine the percentage content of carbon, hydrogen and oxygen in the compound
- This can also be applied to include sulfur and nitrogen, although you are not expected to do this, at this level
To convert combustion analysis data to elemental percentage composition:
1. Calculate the mass of carbon in the sample from the combustion analysis results
2. Calculate the percentage of carbon in the sample
3. Calculate the mass of hydrogen in the sample from the combustion analysis results
4. Calculate the percentage of hydrogen in the sample
5. Use the percentage of carbon and hydrogen to deduce the percentage of oxygen in the sample

Combustion analysis was performed on 2.90 g of an unknown carbohydrate, A.

6.60 g of carbon dioxide and 2.70 g of water were produced.

Calculate the carbohydrates percentage composition.

Answer

Step 1: Calculate the mass of carbon in the sample

- Mass of carbon = $\frac{mass of carbon}{molecular mass of carbon dioxide} \times mass of carbon dioxide produced$
- Mass of carbon = $= \frac{12.0}{44.0} \times 6.60 = 1.80 g$

Step 2: Calculate the percentage of carbon in the sample

- Percentage carbon = $\frac{mass of carbon in sample}{mass of sample} \times 100$
- Percentage carbon $= \frac{1.80}{2.90} \times 100 = 62.1 %$

Step 3: Calculate the mass of hydrogen in the sample

- Mass of hydrogen = $\frac{TOTAL mass of hydrogen}{molecular mass of water} \times mass of water produced$
- Mass of hydrogen = $= \frac{2.0}{18.0} \times 2.70 = 0.30 g$

Step 4: Calculate the percentage of hydrogen in the sample

- Percentage hydrogen = $\frac{mass of hydrogen in sample}{mass of sample} \times 100$
- Percentage hydrogen $= \frac{0.30}{2.90} \times 100 = 10.3 %$

Step 5: Calculate the percentage of oxygen in the sample

- Percentage oxygen = 100 - percentage of carbon - percentage of hydrogen
- Percentage oxygen = 100 - 62.1 - 10.3 = 27.6%
- Therefore, the percentage composition of the carbohydrate is 62.1% carbon, 10.3% hydrogen and 27.6% oxygen

Don't forget to use 2.0 g for the mass of hydrogen as there are two hydrogen atoms in a water molecule

Calculate the empirical formula for the unknown carbohydrate, A

Answer

- The percentage composition of the unknown carbohydrate, A, is 62.1% carbon, 10.3% hydrogen and 27.6% oxygen

edexcel-7-6-1-worked-example-1

These will include:
- Infrared spectroscopy: to identify functional groups and certain bond types
- Mass spectrometry: to identify molecular formula and fragments
- Carbon-13 (¹³C) nuclear magnetic resonance: to identify compound structure
- Proton (¹H) nuclear magnetic resonance: to identify compound structure