Empirical formula from combustion analysis

EMPIRICAL FORMULAS FROM ANALYSIS (SECTION 3.5) The empirical formula of any substance can be determined from its percent composition by calculating the relative number of moles of each atom in 100 g of the substance. If the substance is molecular in nature, its molecular formula can be determined from the empirical formula if the molecular weight is also known. Combustion analysis is a special technique for determining the empirical formulas of compounds containing only carbon, hydrogen, and/or oxygen. 

C03-0043. Explain in words the reasoning used to deduce an empirical formula from combustion analysis of a compound containing C, H, and O. 

These are very simple examples, but provide a plan of attack for the problems at the end of the chapter. It is highly unlikely that an analyst can identify a complete unknown by its IR spectrum alone (especially without the help of a spectral library database and computerized search). For most molecules, not only the molecular weight, but also the elemental composition (empirical formula) from combustion analysis and other classical analysis methods, the mass spectrum, proton and C NMR spectra, possibly heteroatom NMR spectra (P, Si, and F), the UV spectrum, and other pieces of information may be required for identification. From this data and calculations such as the unsaturation index, likely possible structures can be worked out. 

In Chapter 3, graphic elements highlighting the correct approach to problem solving have been added to Sample Exercises on calculating an empirical formula from mass percent of the elements present, combustion analysis, and calculating a theoretical yield. 

Obtaining an Empirical Formula from Combustion Analysis... 

Obtaining an Empirical Formula from Combustion Analysis (3.10) Examples 3.20, 3.21 For Practice 3.20, 3.21 Exercises 95-98... 

To convert the mass percentage composition obtained from a combustion analysis into an empirical formula, we must convert the mass percentages of each type of atom into the relative numbers of atoms. The simplest procedure is to imagine that we have a sample of mass 100 g exactly. That way, the mass percentage... 

In a combustion analysis, the amounts of C, H, and O atoms in a sample of a compound, and thus its empirical formula, are determined from the masses of carbon dioxide and water produced when the compound bums in excess oxygen. 

From the combustion analysis we can determine the empirical formula. Note that the mass of oxygen is determined by difference. 

After we receive the results of a combustion analysis from the laboratory, we need to convert the mass percentage composition to an empirical formula. For this step, we need to determine the relative number of moles of each type of atom. The simplest procedure is to imagine that we have a sample of mass 100 g exactly. That way, the mass percentage composition tells us the mass in grams of each element. Then we can use the molar mass of each element to convert these masses into moles and go on to find the relative numbers of moles of each type of atom. Let s do that for vitamin C, which was once identified in this way, and suppose that the laboratory has reported that the sample you supplied is 40.9% carbon, 4.58% hydrogen, and 54.5% oxygen. 

PROBLEM 3.25 Menthol, a flavoring agent obtained from peppermint oil, contains carbon, hydrogen, and oxygen. On combustion analysis, 1.00 g of menthol yields 1.161 g of H20 and 2.818 g of C02. What is the empirical formula of menthol ... 

Coniine, a toxic substance isolated from poison hemlock, contains only carbon, hydrogen, and nitrogen. Combustion analysis of a 5.024 mg sample yields 13.90 mg of C02 and 6.048 mg of H20. What is the empirical formula of coniine ... 

A theoretical model of the combustion of biomass is illustrated by the complete oxidation of giant brown kelp. Note that kelp, for which complete analytical data were available, is used here simply to illustrate the utility of the model, which is applicable to all biomass species. Based on the empirical formula derived from the elemental analysis of dry kelp at an assumed molecular weight of 100, the combustion stoichiometry is... 

Empirical Formula Determined from Elemental Analysis by Combustion... 

The empirical formula of a compound can be determined if the percent composition of the compound is known (Section 3.7). But where do the percent composition data come from Various methods are used, and many depend on reactions that decompose the unknown but pure compound into known products. Assuming the reaction products can be isolated in pure form, the masses and the number of moles of each can be determined. Then, the moles of each product can be related to the number of moles of each element in the original compound. One method that works well for compounds that burn in oxygen is analysis by combustion. Each element (except oxygen) in the compound combines with oxygen to produce the appropriate oxide. 

Combustion Analysis of Organic Compounds Still another type of compositional data is obtained through combustion analysis, a method used to measure the amounts of carbon and hydrogen in a combustible organic compound. The unknown compound is burned in pure O2 in an apparatus that consists of a combustion chamber and chambers containing compounds that absorb either H2O or CO2 (Figure 3.5). All the H in the unknown is converted to H2O, which is absorbed in the first chamber, and all the C is converted to CO2, which is absorbed in the second. By weighing the contents of the chambers before and after combustion, we find the masses of CO2 and H2O and use them to calculate the masses of C and H in the compound, from which we find the empirical formula. 

From the masses of elements in an unknown compound, the relative amounts (in moles) can be found and the empirical formula determined. If the molar mass is known, the molecular formula can also be determined. Methods such as combustion analysis provide data on the masses of elements in a compound, which can be used to obtain the formula. Because atoms can bond in different arrangements, more than one compound may have the same molecular formula (constitutional isomers). 

The empirical formula of a compound is the simplest whole number ratio of the atoms it contains (Chapter 1). For some alkanes, methane and propane for instance, the empirical formula is the same as the actual molecular formula. However, for others this is not true. The empirical formula of ethane, whose molecular formula is C2Hg, is CHj. In practical terms the empirical formula is the formula that can be derived from the percentage composition data obtained from combustion analysis. In order to use this to establish the actual formula, data on the relative molecular mass (M ) of the compound is required. 

Think back to what we have just been learning about combustion analysis and the information that such analysis gave us — the empirical formula of a compound. We can now combine the results from the two techniques combustion analysis gives us an empirical formula of C3H6O, and from mass spectrometry we know that the compound has a molecular mass of 58.0417. We can see straight away that the empirical formula must also be the molecular formula, i.e. C3H6O. 

Since the carbon to hydrogen ratio of 1 2.67 is too far from a whole number to be attributed to experimental error, the empirical formula must be the lowest multiple to give a whole number ratio, and 2 x (1 2.67) gives 2 5.34, still not a whole number, but 3 x (1 2.67) gives 3 8.01, which is a whole number ratio within the experimental error. It is difficult to be precise about the experimental error that is acceptable in determinations of this kind, but the percentage of C, H, or N determined by combustion analysis is usually quoted to 0.1%. 

Suppose that we want to determine the formula of an unknown hydrocarbon. With combustion analysis we can establish the mass percent composition, and from this, we can determine the empirical formula. The method of Example 6-7 gives us a molar mass, in g moP, which is numerically equal to the molecular mass, in u. This is all the information we need to establish the true molecular formula of the hydrocarbon (see Exercise 96). 

Empirical and molecular formulas can be derived from a process called combustion analysis. You can learn about this process at the textbook s Web site. 

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