EMPIRICAL FORMULAS FROM ANALYSES

As we learned in Section 2.6, the empirical formula for a substance tells us the relative number of atoms of each element in the substance. The empirical formula H2O shows that water contains two H atoms for each O atom. This ratio also appHes on the molar level 1 mol of H2O contains 2 mol of H atoms and 1 mol of O atoms. Conversely, the ratio of the numbers of moles of all elements in a compound gives the subscripts in the compound s empirical formula. Thus, the mole concept provides a way of calculating empirical formulas. 

Mercury and chlorine, for example, combine to form a compound that is measured to be 74.0% mercury and 26.0% chlorine by mass. Thus, if we had a 100.0-g sample of the compound, it would contain 74.0 g of mercury and 26.0 g of chlorine. (Samples of any size can be used in problems of this type, but we will generally use 100.0 g to simplify the calculation of mass from percentage.) Using atomic weights to get molar masses, we can calculate the number of moles of each element in the sample  

We then divide the larger number of moles by the smaller number to obtain the Cl Hg mole ratio  

Because of experimental errors, calculated values for a mole ratio may not be whole numbers, as in the calculation here. The number 1.98 is very close to 2, however, and so we can confidently conclude that the empirical formula for the compound is HgCl2. The empirical formula is correct because its subscripts are the smallest integers that express the ratio of atoms present in the compound, ooo (Section 2.6) 

Could the empirical formula determined from chemical analysis be used to tell the difference between acetylene, C2H2, and benzene, CeHe  

How do you calculate the mole ratio of each element in any compound  

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EMPIRICAL FORMULAS FROM ANALYSES We apply the mole concept to determine chemical formulas from the masses of each element in a given quantity of a compound. 

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. 

Be able to calculate the empirical formula from percent composition data or quantities from chemical analysis. 

Example of the Calculation of the Percentage Composition and the Empirical Formula, from the. Data of Analysis... 

We can reverse the procedure just described and determine the empirical formula from the elemental analysis of a compound, as illustrated by Example 2.3. 

Empirical formulas can also be calculated from the mass of each element in a sample of a compound. Analysis of a sample of a compound shows that it contains 1.179 g Na and 0.821 g S. You could calculate the percent of each element from these numbers, but that s not necessary. The mass of each element is converted directly to moles for the empirical formula. From the table of atomic masses, we find the molar masses are 22.99 g for Na and 32.07 g for S. 

Empirical Formula from Percent Composition Methyl acetate is a solvent commonly used in some paints, inks, and adhesives. Determine the empirical formula for methyl acetate, which has the following chemical analysis ... 

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. 

The procedure developed in Example Problem 3.8 allows us to find an empirical formula from mass percentage data. As we saw in Chapter 2, though, the empirical formula does not tell us the exact composition of a molecule of the compound. In the RDX example above, we know now that the ratio of C atoms to H, N, and O atoms is 1 2, but we don t know how many of these CH2N2O2 units make up an actual molecule. To go from the empirical formula to the molecular formula, we will need an additional piece of information—the molar mass of the compound. Fortunately, modern instrumentation, such as the mass spectrometer we discussed in Section 2.2, makes it fairly simple to measure the molar mass of a substance. Once we are armed with both the elemental analysis and the molar mass, we can find the molecular formula. 

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... 

F.14 Paclitaxel, which is extracted from the Pacific yew tree Taxus brevifolia, has antitumor activity for ovarian and breast cancer. It is sold under the trade name Taxol. On analysis, its mass percentage composition is 66.11% C, 6.02% H, and 1.64% N, with the balance being oxygen. What is the empirical formula of paclitaxel ... 

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... 

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. 

Suppose that 10.0 g of an organic compound used as a component of mothballs is dissolved in 80.0 g of benzene. The freezing point of the solution is 1.20°C. (a) What is an approximate molar mass of the organic compound (b) An elemental analysis of that substance indicated that the empirical formula is C3H2C1. What is its molecular formula (c) Using the atomic molar masses from the periodic table, calculate a more accurate molar mass of the compound. 

In all cases the reaction products are mixtures of ethyl polyphosphates, and, on the basis of elementary analysis, they approximate the empirical formulas given in the above equations. In Equations 3 and 5 the product has been arbitrarily called hexaethyl tetraphosphate, which may contain 8 to 20% of the active tetraethyl pyrophosphate. In Equations 4 and 6 the products have been called technical tetraethyl pyrophosphate, which may contain up to 40% of pure tetraethyl pyrophosphate. Hexaethyl tetra-phosphate has also been made from phosphoric anhydride and diethyl ether by a process recently patented by Adler (1). 

Early reports on levan are obscured by incomplete descriptions of impure products.2 96 Greig-Smith found that Bacillus levaniformans(1) produced levan from sucrose96" in suitable nutrient solutions, but not from D-glucose, D-fructose, lactose or maltose.966 He therefore assumed that levan could only be formed from the nascent D-fructose and D-glucose resulting from the inversion of sucrose. Hydrolysis of levan yielded D-fructose only, and analysis of levan agreed with the empirical formula (C HiriOi) it was noted that levan was closely related to inulin but was not identical with it. 

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

The time-series analysis results of Merz et were expressed in first-order empirical formulas for the most part. Forecasting expressions were developed for total oxidant, carbon monoxide, nitric oxide, and hydrocarbon. Fitting correlation coefficients varied from 0.547 to 0.659. As might be expected, the best results were obtained for the primary pollutants carbon monoxide and nitric oxide, and the lowest correlation was for oxidant. This model relates one pollutant to another, but does not relate emission to air quality. For primary pollutants, the model expresses the concentrations as a function of time. 

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