Empirical formulas 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. 

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 Section F, we saw that one technique used in modern chemical laboratories to determine the empirical formulas of organic compounds is combustion analysis. We are now in a position to understand the basis of the technique, because it makes use of the concept of limiting reactant. 

EXAMPLE M.4 Determining an empirical formula by combustion analysis... 

A combustion analysis was carried out on 1.621 g of a newly synthesized compound, which was known to contain only C, H, and O. The masses of water and carbon dioxide produced were 1.902 g and 3.095 g, respectively. What is the empirical formula of the compound ... 

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. 

Determine the empirical formula of an organic compound containing carbon, hydrogen, and oxygen by combustion analysis (Example M.4). 

M.12 A compound produced as a by-product in an industrial synthesis of polymers was found to contain carbon, hydrogen, and iodine. A combustion analysis of 1.70 g of the compound produced 1.32 g of C02 and 0.631 g of H20. The mass percentage of iodine in the compound was determined by-converting the iodine in a 0.850-g sample of the compound into 2.31 g of lead(II) iodide. What is the empirical formula of the compound Could the compound also contain oxygen Explain your answer. 

In a combustion analysis, 3.21 g of a hydrocarbon formed 4.48 g of water and 9.72 g of carbon dioxide. Deduce its empirical formula and state whether it is likely to be an alkane, an alkene, or an alkyne. Explain your reasoning. 

Compounds that do not decompose cleanly into their elements must be analyzed by other means. Combustion analysis is particularly useful for determining the empirical formulas of carbon-containing compounds. In combustion analysis, an accurately known mass of a compound is burned in a stream of oxygen gas. The conditions are carefully controlled so that all of the carbon in the sample is converted to carbon dioxide, and all of the hydrogen is converted to water. Certain other elements present in the sample are also converted to their oxides. 

This flowchart summarizes the method for using combustion analysis to determine the empirical formula of a compound that contains no elements other than C, H, and O. 

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

C03-0089. Combustion analysis of 0.60 g of an unknown organic compound that contained only C, H, and O gave 1.466 g of carbon dioxide and 0.60 g of water in a combustion analysis. Mass spectral analysis showed that the compound had a molar mass around 220 g/mol. Determine the empirical formula and molecular formula. 

C03-0150. A sample of a component of petroleum was subjected to combustion analysis. An empty vial of mass 2.7534 g was filled with the sample, after which vial plus sample had a mass of 2.8954 g. The sample was burned in a combustion train whose CO2 trap had a mass of 54.4375 g and whose H2 O trap had a mass of 47.8845 g. At the end of the analysis, the CO2 trap had a new mass of 54.9140 g and the H2O trap had a new mass of 47.9961 g. Determine the empirical formula of this component of petroleum. 

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. 

In Section F, we remarked that one technique used in modern chemical laboratories or the agencies that carry out contract work on behalf of other chemists is combustion analysis. This technique—which has been used since the earliest days of chemistry—is used to establish the empirical formulas of organic compounds and, in combination with mass spectrometry, their molecular formulas. It is used both for newly synthesized compounds and to identify active compounds in natural products. We are now in a position to understand the basis of the technique, for it makes use of the concept of limiting reactant. 

M.9 In a combustion analysis of a 0.152-g sample of the artificial sweetener aspartame, it was found that 0.318 g of carbon dioxide, 0.084 g of water, and 0.0145 g of nitrogen were produced. What is the empirical formula of aspartame The molar mass of aspartame is 294 g-mol. What is its molecular formula ... 

One of the most common methods used to determine percent composition and empirical formulas, particularly for compounds containing carbon and hydrogen, is combustion analysis. In this method, a compound of unknown composition is burned with oxygen to produce the volatile combustion products C02 and H20, which are separated and weighed by an automated instrument called a gas chromatograph. Methane (CH4), for instance, burns according to the balanced equation... 

Caproic acid, the substance responsible for the aroma of dirty gym socks and running shoes, contains carbon, hydrogen, and oxygen. On combustion analysis, a 0.450 g sample of caproic acid gives 0.418 g of H20 and 1.023 g of C02. What is the empirical formula of caproic acid If the molecular mass of caproic acid is 116.2 amu, what is the molecular formula ... 

A hydrocarbon of unknown formula CxHy was submit-ted to combustion analysis with the following results. What is the empirical formula of the hydrocarbon ... 

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 0.539 g sample of a compound that contained only carbon and hydrogen was subjected to combustion analysis. The combustion produced 1.64 g of carbon dioxide and 0.807 g of water. Calculate the percentage composition and the empirical formula of the sample. 

Data which establishes the compound s molecular formula is required. Traditionally an accurate combustion analysis (within 0.3 - 0.5%) has been used to determine the empirical formula of a compound, and also to justify that the compound is of high chemical purity. However, combustion analysis data will be identical for structural isomers of any type (geometrical isomers, diastereoisomers, enantiomers etc.) and other spectroscopic or chromatographic methods will therefore be required in order to determine levels of isomeric impurities. 

Nitrogen by the Kjeldahl Method.— The absolute method, while the more accurate and often used when the results of analysis are for the purpose of determining the empirical formula of a compound, is sometimes replaced by the Kjeldahl or wet combustion process. In outline the method is as follows The organic compound is decomposed and oxidized by heating for some time in about 30 cc. of pure con-... 

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... 

Problem 2.7 Calculate the percentage composition and then the empirical formula for each of the following compounds (a) Combustion of a 3.02-mg sample of a compound gave 8.86 mg of carbon dioxide and 5.43 mg of water, (b) Combustion of an 8.23-mg sample of a compound gave 9.62 mg of carbon dioxide and 3.94 mg of water. Analysis of a 5.32-mg sample of the same compound by the Carius method gave 13.49 mg of silver chloride. 

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. 

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