Chemistry » Chemical Reactions and Stoichiometry » Quantitative Chemical Analysis

# Combustion Analysis

## Gravimetric Analysis Continued: Combustion Analysis

The elemental composition of hydrocarbons and related compounds may be determined via a gravimetric method known as combustion analysis. In a combustion analysis, a weighed sample of the compound is heated to a high temperature under a stream of oxygen gas, resulting in its complete combustion to yield gaseous products of known identities.

The complete combustion of hydrocarbons, for example, will yield carbon dioxide and water as the only products. The gaseous combustion products are swept through separate, pre-weighed collection devices containing compounds that selectively absorb each product (see figure below). The mass increase of each device corresponds to the mass of the absorbed product and may be used in an appropriate stoichiometric calculation to derive the mass of the relevant element. This schematic diagram illustrates the basic components of a combustion analysis device for determining the carbon and hydrogen content of a sample. Image credit: OpenStax, Chemistry

## Combustion Analysis Example

Polyethylene is a hydrocarbon polymer used to produce food-storage bags and many other flexible plastic items. A combustion analysis of a 0.00126-g sample of polyethylene yields 0.00394 g of CO2 and 0.00161 g of H2O. What is the empirical formula of polyethylene?

### Solution

The primary assumption in this exercise is that all the carbon in the sample combusted is converted to carbon dioxide, and all the hydrogen in the sample is converted to water:

$$\mathrm{C_xH_y}(s) + \mathrm{excess \; O_2}(g) \longrightarrow x\mathrm{CO_2}(g) + \cfrac{y}{2}\mathrm{H_2O}(g)$$

Note that a balanced equation is not necessary for the task at hand. To derive the empirical formula of the compound, only the subscripts x and y are needed.

First, calculate the molar amounts of carbon and hydrogen in the sample, using the provided masses of the carbon dioxide and water, respectively. With these molar amounts, the empirical formula for the compound may be written as described in the previous chapter of this text. An outline of this approach is given in the following flow chart:

$$\text{mol C} = 0.00394\mathrm{\; g \; CO_2} × \cfrac{\mathrm{1 \; mol \; CO_2}}{\mathrm{44.01 \; g/mol}} × \cfrac{\mathrm{1 \; mol \; C}}{\mathrm{1 \; mol \; CO_2}} = 8.95 × 10^{-5}\text{ mol C}$$

$$\text{mol H} = 0.00161\mathrm{\; g \; H_2O} × \cfrac{\mathrm{1 \; mol \; H_2O}}{\mathrm{18.02 \; g/mol}} × \cfrac{\mathrm{2 \; mol \; H}}{\mathrm{1 \; mol \; H_2O}} = 1.79 × 10^{-4}\text{ mol H}$$

The empirical formula for the compound is then derived by identifying the smallest whole-number multiples for these molar amounts. The H-to-C molar ratio is

$$\cfrac{\text{mol H}}{\text{mol C}} = \cfrac{1.79 × 10^{-4}\text{ mol H}}{8.95 × 10^{-5}\text{ mol C}} = \cfrac{\text{2 mol H}}{\text{1 mol C}}$$

and the empirical formula for polyethylene is CH2.