Percentage composition data

The approximate molar mass, calculated from the gas density data, is 89 g/mol. The empirical formula, calculated from the percentage composition data, is C2H3O with the empirical formula unit mass of 43.0. The exact molar mass must be (2)(43) = 86.0 g/mol since this is the only multiple of 43.0 (whole-number multiple) reasonably close to the approximate molecular formula of 89 g/mol. The molecule must be the equivalent of 2 empirical formulas CqHgO. 

In section 6.2, you learned how to calculate the empirical formula of a compound based on percentage composition data obtained by experiment. In section 6.3, you will learn how chemists use the empirical formula of a compound and its molar mass to determine the molecular formula of a compound. 

You can also use the mole concept to calculate the empirical formula of a compound using the percentage composition data for that compound — the percentage by weight of each element in the compound. (The empirical formula indicates the different types of elements in a molecule and the lowest whole-number ratio of each kind of atom in the molecule. See Chapter 7 for details.)... 

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. 

What three types of information are used to find an empirical formula from percentage composition data ... 

Rather than giving students straight percentage composition data for determining the empirical formula of a compound (see Question 7), sometimes chemistry... 

Instrumental Analysis. It is difficult to distiaguish between the various acryhcs and modacryhcs. Elemental analysis may be the most effective method of identification. Specific compositional data can be gained by determining the percentages of C, N, O, H, S, Br, Cl, Na, and K. In addition the levels of many comonomers can be estabhshed usiag ir and uv spectroscopy. Also, manufacturers like to be able to identify their own products to certify, for example, that a defective fiber is not a competitor s. To facihtate this some manufacturers iatroduce a trace of an unusual element as a built-ia label. 

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... 

The data on the flow-rate of each individual component, on the total stream flow-rate, and the percentage composition, can be shown on the flow-sheet in various ways. The simplest method, suitable for simple processes with few equipment pieces, is to tabulate the data in blocks alongside the process stream lines, as shown in Figure 4.1. Only a limited amount of information can be shown in this way, and it is difficult to make neat alterations or to add additional data. 

Note Derivation assumes data available on the percentage composition by chemical elements of each gas and the symbols and symbol weights of the elements contained. 

Based on the method of internal normalization of peak areas, the percentage composition (less the diethyl ether solvent shown as the initial large component in Figures 1 and 2) of odor concentrates was estimated. The average composition of samples of odor isolates from individual preparations as well as from several different preparations of the same type—i.e., concurrent or noncurrent—was determined on the basis of all preparations which could be compared on a fair analytical basis (same gas chromatographic detector, column, and conditions). These results were then combined (traps plus distillate) to provide the rough estimations shown in Table IV. These data represent the best estimation of the composition of the total volatile odor concentrates from each processing method studied. 

Such data, in addition to the observed boiling and melting points, furnish a better criterion of purity than the analysis for silicon or bromine, since a small quantity of tetra-bromide, if present, will not alter appreciably the percentage composition. 

The compositional data for the aromatic hydrocarbons present in the upgraded anthracene oil are presented in Table XIII. Comparison of the data in Tables XII and XIII reveals significant increases and decreases in the amounts of hydroaromatic and aromatic hydrocarbons, respectively, under all reaction conditions. This is especially evident by the presence and absence of the hydroaromatic -10(H) series in the products and feeds, respectively, and by the large increase in the weight percentage of the -8(H) compounds in the products compared to the feeds. The presence of significant amounts of dihydro-, tetrahydro-, and octahydrophenanthrenes in the products is indicated by GC/MS, by the increase in the weight percents at C-14 in the -16(H) and -14(H) series, and by the presence of C- H g, -10(H). 

All composition data provided as percentages [%] in this chapter are based on gaseous compositions unless indicated otherwise and can be considered as molar percentages. 

In the previous Practice Problems, you used mass data to calculate percentage composition. This skill is useful for interpreting experimental data when the chemical formula is unknown. Often, however, the percentage composition is calculated from a known chemical formula. This is useful when you are interested in extracting a certain element from a compound. For example, many metals, such as iron and mercury, exist in mineral form. Mercury is most often found in nature as mercury(II) sulfide, HgS. Knowing the percentage composition of HgS helps a metallurgist predict the mass of mercury that can be extracted from a sample of HgS. 

The percentage composition calculated from the empirical formula closely matches the given data. The formula is reasonable. 

If we calculated the percent compositions of C2H2 and CeHg (Figure 7.3), we would find that both have the same percentages of carbon and the same percentages of hydrogen (compare Problem 7.100 at the end of the chapter). Both have the same empirical formula—CH. This result means that we cannot tell these two compounds apart from percent composition data alone. However, if we also have a molar mass, we can use that information with the percent composition data to determine not only the empirical formula but also the molecular formula. 

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

It is apparent from the compositional data on the raw MSW in Table 5.1 that the combustible materials make up the bulk of the MSW on a weight percentage basis, about 85 wt % in the 1990s. The amount of the individual MSW components recovered since 1960, presumably for sale of marketable components, has increased to about one-fifth of the total amount generated. The largest components by weight in the recovered material include paper and paperboard and noncombustible metals and glass. Much of this material is recycled. 

Data for percentage composition allow you to calculate the simplest ratio among the atoms found in a compound. The empirical formula shows this simplest ratio. For example, ammonium nitrite, shown in Figure 9, has the actual formula NH4NO2 and is made up of ammonium ions, NHj, and nitrite ions, NO2, in a 1 1 ratio. 

Solubility data are expressed as a solubility-concentration curve or as phase diagrams. The latter are preferable since a three-component phase diagram completely describes the effect of varying all three components of the system - namely, the soluhilisate, the solubiliser and the solvent. The axes of the phase diagram form an equilateral triangle (see Fig. 6.37), each side of which is divided into 100 parts to correspond to percentage composition. 

M.ll Calculate the mass percentage composition doing so eases the comparison of data from multiple analyses. 

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