Formula combustion analysis

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. 

Example 3-15 demonstrates how to use combustion analysis to determine the formula of a compound containing only C and H, and Example shows how combustion analysis is conducted when the compound contains 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. 

C03-0132. Police officers confiscate a packet of white powder that they believe contains heroin. Purification by a forensic chemist yields a 38.70-mg sample for combustion analysis. This sample gives 97.46 mg CO2 and 20.81 mg H2 O. A second sample is analyzed for its nitrogen content, which is 3.8%. Show by calculations whether these data are consistent with the formula for heroin, C21 H22 NO5. ... 

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. 

The assumption is made at present that elemental combustion analysis for carbon, hydrogen, and fluorine provides a good approximation to the extent of incorporation of fluoroalkyl residues, i.e. alcohols and ethers. We have ruled out trifluoromethylcarbonyl groups since no evidence is seen for their presence in either the infrared spectra or the 19F-NMR spectra. Thus, our values for percent modification reflect the best fit of the combustion data to an idealized stoichiometry for the product in Equation 1, where (m+n+o) = 100, and the percent modification (% mod.) is given by the expression [100 x (m+o)/(m+n+o)], equivalent to the number of fluoroalkyl residues per one hundred methylenes. An appropriately normalized formula was used to fit the data for polypropylene (sample 10). 

This structure derives from the formula, (MeSiHNH)g>3g(MeSiHNMe)Q g (MeSlN)g>57 established by both NMR and combustion analysis. 

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 ... 

An organic compound was found by combustion analysis to contain 38.7% C, 9.7% H, and the remainder oxygen. In order to determine its molecular formula, a 1.00 g sample was added to 10.00 g of water. The freezing point of the solution was found to be —2.94°C. What is the compound s molecular formula ... 

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. 

An 874 mg sample of cortisol was subjected to carbon-hydrogen combustion analysis. 2.23 g of carbon dioxide and 0.652 g of water were produced. The molar mass of cortisol was found to be 362 g/mol using a mass spectrometer. If cortisol contains carbon, hydrogen, and oxygen, determine its molecular formula. 

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