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Empirical formulae of compounds

High resolution is used to determine the exact mass of an ion species in a mixture knowledge of the exact mass of an unknown substance allows its atomic composition to be established. Target analysis exact mass determination proves the presence of a particular ion species (compound) in a mixture. Mass spectrometry is perhaps the only method that can be used to find the empirical formulae of compounds that are not completely pure. [Pg.356]

FT-ICR mass spectrometers take advantage of ion-cyclotron resonance to select and detect ions. This analyzer can be used with both ESI and MALDI interfaces. Their particular advantages are their sensitivity, extreme mass resolution, and mass accuracy. The latter allows for the determination of the empirical formulae of compounds under 1000 Da. As far as we know, this analyzer has not been applied to flavonoids. [Pg.91]

Dalton s work on relative weights, multiple proportions, and the atomic theory did not have an immediate effect on chemists of his day. Dalton s ideas did provide a framework for determining the empirical formula of compounds, but his table of relative weights was not accurate enough to give consistent results. Many scientists still debated the existence of atoms in the second half of the nineteenth century. Still, little by little, the atomic theory was adopted by chemists as a valid model for the basic structure of matter. While Dalton continued his life as a humble tutor in Manchester, other chemists used Dalton s ideas to establish the atomic theory. Foremost among these was Jons Jacob Berzelius (1779-1848) of Sweden, the foremost chemical authority of the first half of the nineteenth century. [Pg.34]

Compound X contains 69.9% carbon, 6.86% hydrogen, and 23.3% oxygen. Determine the empirical formula of compound X. [Pg.211]

OBJECTIVE To understand the meaning of empirical formulas of compounds. [Pg.227]

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. [Pg.111]

What is the empirical formula of compound 14.5.7 based on the mass spectral data ... [Pg.705]

The empirical formula of a compound is based on experiments that give the number of moles of each element in a sample of the compound. That is why we use the word "empirical," which means "based on observation and experiment" Chemists have devised a number of different experimental techniques to determine the empirical formulas of compounds. One of these is combustion analysis, which is commonly used for compounds containing principally carbon and hydrogen as their component elements. [Pg.94]

Mustroph and Bach proposed, without having any NMR results, that the reaction of diazonium salts with l-phenyl-3-methyl-4-(a-acet-ethylidene)-pyrazol-5-one gave compounds 14. Detailed analysis of 2D H and C NMR spectra revealed that the reaction gives rise the compound 15, possibly owing to nucleophilic attack of nitrogen to carbonyl group. The empirical formulae of compounds 14 and 15 are the same and, therefore, the reaction products cannot be differentiated using elemental analysis data or the w/z value in the mass spectrum. [Pg.12]

In empirical formulas of inorganic compounds, electropositive elements are listed first [3]. The stoichiometry of the element symbols is indicated at the lower right-hand side by index numbers. If necessary, the charges of ions are placed at the top right-hand side next to the element symbol (e.g., S "). In ions of complexes, the central atom is specified before the ligands are listed in alphabetical order, the complex ion is set in square brackets (e.g., Na2[Sn(OH)+]). [Pg.20]

Figure 2-39. The empirical formula of C3H6O can be expressed by seven structure diagrams and even more compound names. Figure 2-39. The empirical formula of C3H6O can be expressed by seven structure diagrams and even more compound names.
Besides structure and substructure searches, Gmclin provides a special search strategy for coordiuation compouuds which is found in no other database the ligand search system, This superior search method gives access to coordination compounds from a completely different point of view it is possible to retrieve all coordination compounds with the same ligand environment, independently of the central atom or the empirical formula of the compound. [Pg.249]

Attempts to characterize polymeric substances had been made, of course, and high molecular weights were indicated, even if they were not too accurate. Early workers tended to be more suspicious of the interpretation of the colliga-tive properties of polymeric solutions than to accept the possibility of high molecular weight compounds. Faraday had already arrived at Cs Hg as the empirical formula of rubber in 1826, and isoprene was identified as the product... [Pg.1]

Since the relative number of atoms of each element in the compound is the same as the relative number of moles of atoms of each element in the sample, we can say that the empirical formula of ethanol is... [Pg.325]

What are the empirical formulas of the three compounds shown in Figure 18-11 The molecular formulas Which are structural isomers ... [Pg.341]

To determine the empirical formula of a compound, we begin by measuring the mass of each element present in a sample. The result is usually reported as the mass... [Pg.70]

The mass percentage composition of a compound that assists in the coagulation of blood is 76.71 % C, 7.02% H, and 16.27% N. Determine the empirical formula of the compound. [Pg.72]

The empirical formula of a compound is determined from the mass percentage composition and the molar masses of the elements present. [Pg.73]

F.8 Determine the empirical formula of each of the following compounds from the data given, (a) Talc (used in talcum powder) has mass composition 19.2% Mg, 29.6% Si, 42.2%... [Pg.75]

F.10 A chemist found that 4.69 g of sulfur combined with fluorine to produce 15.81 g of a gas. (a) What is the empirical formula of the gas (b) Assuming that the empirical and molecular formulas of the compound are the same, what is its name ... [Pg.75]

F.13 Osmium forms a number of molecular compounds with carbon monoxide. One light-vellow compound was analyzed to give the following elemental composition 15.89% C, 21.18% O, and 62.93% Os. (a) What is the empirical formula of this compound (b) From the mass spectrum of the compound, the molecule was determined to have a molar mass of 907 g-mol 1. What is its molecular formula ... [Pg.75]

J.9 You are asked to identify compound X, which was extracted from a plant seized by customs inspectors. You run a number of tests and collect the following data. Compound X is a white, crystalline solid. An aqueous solution of X turns litmus red and conducts electricity poorly, even when X is present at appreciable concentrations. When you add sodium hydroxide to the solution a reaction takes place. A solution of the products of the reaction conducts electricity well. An elemental analysis of X shows that the mass percentage composition of the compound is 26.68% C and 2.239% H, with the remainder being oxygen. A mass spectrum of X yields a molar mass of 90.0 g-moF. (a) Write the empirical formula of X. (b) Write... [Pg.101]

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. [Pg.120]

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 ... [Pg.121]

We conclude that the empirical formula of the new compound is C2H6Q. [Pg.122]

Self-Test M.3A When 0.528 g of sucrose (a compound of carbon, hydrogen, and oxygen) is burned, 0.306 g of water and 0.815 g of carbon dioxide are formed. Deduce the empirical formula of sucrose. [Pg.123]

Determine the empirical formula of an organic compound containing carbon, hydrogen, and oxygen by combustion analysis (Example M.4). [Pg.123]


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See also in sourсe #XX -- [ Pg.187 , Pg.188 , Pg.189 , Pg.190 , Pg.191 , Pg.192 ]




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Compound formula

Compounds empirical formula

Empirical formula

Formulas empirical formula

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