Empirical formula from percentage composition

EXAMPLE F.2 Determining the empirical formula from mass percentage composition... 

If 1.00 g of the unknown contains 0.817 g carbon, the mass percent of carbon is 81.7 percent, leaving the remaining 18.3 percent as hydrogen. Therefore, we need to use the procedures for determining an empirical formula from a percentage composition. The problem will progress as follows (remember with percents, assume a 100 g sample) ... 

Finding a Compound s Empirical Formula from Percentage Composition Part A... 

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

Empirical Formula.—The calculation of the empirical formula from the percentage composition is as follows ... 

Determine a compound s empirical formula from its percentage composition. 

Let s say that you want to find an empirical formula from the percentage composition. First, convert the mass percentage of each element to grams. Second, convert from grams to moles using the molar mass of each element as a conversion factor. (Keep in mind that a formula for a compound can be read as a number of atoms or as a number of moles.) Third, as shown in Sample Problem C, compare these amounts in moles to find the simplest whole-number ratio among the elements in the compound. 

Sample Problem G Determining an Empirical Formula from Percentage Composition... 

FIGURE 3.13 Procedure for calculating an empirical formula from percentage composition. The key step in the calculation is step 2, determining the number of moles of each element in the compound. 

A Figure 3.13 Procedure for calculating an empirical formula from percentage composition. 

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. 

Determining the empirical formula from masses of elements and percentage composition Given the masses of elements in a known mass of compound, or given its percentage composition, obtain the empirical formula. (EXAMPLES 3.10,3.11)... 

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. 

To determine a compound s empirical formula from its percentage composition, begin by converting percentage composition to a mass composition. Assume that you have a 100.0 g sample of the compound. Then calculate the amoimt of each element in the sample. For example, the percentage composition of dihorane is 78.1% B and 21.9% H. Therefore, 100.0 g of diborane contains 78.1 g of B and 21.9 g of H. 

The easiest way to calculate an empirical formula from percentage composition is to consider a 100.00 g sample of the compound. In this case, a 100.00 g sample would contain 62.04 g of carbon, 10.41 g of hydrogen, and 27.55 g of oxygen. Convert each mass to moles so that you can compare the mole ratio of the three elements. 

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... 

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

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... 

Empirical formulae may also be calculated from percentage composition by mass in a similar way. 

In the problem above, we determined the percentage data from the chemical formula. We can determine the empirical formula if we know the percent compositions of the various elements. The empirical formula tells us what elements are present in the compound and the simplest whole-number ratio of elements. The data may be in terms of percentage, or mass or even moles. However, the procedure is still the same—convert each element to moles, divide each by the smallest, and then use an appropriate multiplier if necessary. We can then determine the empirical formula mass. If we know the actual molecular mass, dividing the molecular formula mass by the empirical formula mass, gives an integer (rounded if needed) that we can multiply each of the subscripts in the empirical formula. This gives the molecular (actual) formula, which tells what elements are in the compound and the actual number of each. 

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. 

Calculate the empirical formula of a compound from its mass percentage composition, Self-Test F.3. 

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. 

Answer The first step is to calculate the empirical formula. To begin, we need to determine the number of moles of each element. However, the percentage composition does not tell us a mass. To get around this, the commonly used technique is to assume that you have a 100.0-gram sample of the substance. Because 100.0 grams is equal to 100%, you can simply take the percentages and say that the masses are equal to that same amount in grams. From there, the problem proceeds just like the previous example ... 

In the previous section, you learned how to calculate the percentage composition of a compound from its chemical formula. Now you will do the reverse. You will use the percentage composition of a compound, along with the concept of the mole, to calculate the empirical formula of the compound. Since the percentage composition can often be determined by experiment, chemists use this calculation when they want to identify a compound. 

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

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