Percent Composition and Empirical Formula

If the formula of a compound is known, it is a fairly straightforward task to determine the percent composition of each element in the compound. For example, suppose you want to calculate the percentage hydrogen and oxygen in water, H20. First calculate the molecular mass of water  

As a good check, add the percentages together. They should equal to 100% or be very close. 

Determine the mass percent of each of the elements in C6H12Og Formula mass (FM) = 180.158 amu 

In the problems above, the percentage data was calculated from the chemical formula, but the empirical formula can be determined if the percent compositions of the various elements are known. 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. But the procedure is still the same convert each to moles, divide each by the smallest number, then use an appropriate multiplier if needed. The empirical formula mass can then be calculated. If the actual molecular mass is known, dividing the molecular mass by the empirical formula mass gives an integer (rounded if needed) that is used to multiply each of the subscripts in the empirical formula. This gives the molecular (actual) formula, which tells which elements are in the compound and the actual number of each. 

For example, a sample of a gas was analyzed and found to contain 2.34 g of nitrogen and 5-34 g of oxygen. The molar mass of the gas was determined to be about 90 g/mol. What are the empirical and molecular formulas of this gas  

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Worked Examples 3.15 and 3.16 show further conversions between percent composition and empirical formulas. 

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... 

Would it surprise you to learn that two or more substances with distinctly different properties can have the same percent composition and the same empirical formula How is this possible Remember that the subscripts in an empirical formula indicate the simplest whole-number ratio of moles of the elements in the compound. But the simplest ratio does not always indicate the actual number of moles in the compound. To identify a new compound, a chemist must go one step further and determine the molecular formula, which specifies the actual number of atoms of each element in one molecule or formula unit of the substance. Figure 11-11 shows an important use of the gas, acetylene. It has the same percent composition and empirical formula, CH, as benzene which is a liquid. Yet chemically and structurally acetylene and benzene are very different. 

A 0.1888 g sample of a hydrocarbon produces 0.6260 g CO2 and 0.1602 g H2O in combustion analysis. Its molecular mass is found to be 106 u. For this hydrocarbon, determine its (a) mass percent composition (b) empirical formula (c) molecular formula. 

Making particle numbers manageable with Avogadro s number Converting between masses, mole counts, and volumes Dissecting compounds with percent composition Moving from percent composition to empirical and molecular formulas... 

In the previous section the problems presented a chemical formula and asked for the percent composition. There is a method for going from percent composition to chemical formula however, you will obtain only the empirical formula from this. The three steps in determining the empirical formula of a compound from the percent composition are as follows ... 

To determine the molecular formula from percent composition and molecular mass data or from the empirical formula and molecular mass data... 

Even if we cannot see how to solve this problem completely at first glance, we can tell immediately that the empirical formula can be calculated from the percent composition and that the number of moles can be calculated from its pressure-volume-temperature data. 

Formulas describe the composition of compounds. Empirical formulas give the mole ratio of the various elements. However, sometimes different compounds have the same ratio of moles of atoms of the same elements. For example, acetylene, C2H2, and benzene, CeHe, each have 1 1 ratios of moles of carbon atoms to moles of hydrogen atoms. That is, each has an empirical formula CH. Such compounds have the same percent compositions. However, they do not have the same number of atoms in each molecule. The molecular formula is a formula that gives all the information that the empirical formula gives (the mole ratios of the various elements) plus the information of how many atoms are in each molecule. In order to deduce molecular formulas from experimental data, the percent composition and the molar mass are usually determined. The molar mass may be determined experimentally in several ways, one of which will be described in Chap. 12. 

If we know the composition of a compound in terms of the masses (or mass percentages) of the elements present, we can calculate the empirical formula but not the molecular formula. For reasons that will become clear as we consider Example 6.15, to obtain the molecular formula we must know the molar mass. Next we will consider compounds where both the percent composition and the molar mass are known. 

The sum of the percentages is 5.926% + 94.06% = 99.99%. The small discrepancy from 100 percent is due to the way we rounded off the molar masses of the elements. If we had used the empirical formula HO for the calculation, we would have obtained the same percentages. This is so because both the molecular formula and empirical formula tell us the percent composition by mass of the compound. 

Calculate the percent composition and determine the molecu- 42. lar formula and the empirical formula for the nitrogen-oxygen compound that results when 12.04 g of nitrogen are reacted with enough oxygen to produce 39.54 g of product. The molar mass of the product is 92.02 g. 

Calculate the percent composition and determine the molecular formula and the empirical formula for the carbon-hydrogen-oxygen compound that results when 30.21 g of carbon, 40.24 g of oxygen, and 5.08 g of hydrogen are reacted to produce a product with a molar mass of 180.18 g. 

If the mass percent composition and molar mass of a compound are known, its empirical and molecular formulas can be determined. 

Explain why, if the percent composition and the molar mass of a compound are both known, the molecular formula can be determined much more readily by using the molar mass rather than a 100 g sample in the first step of the five-step procedure in Example 3-5. In this instance, how would you then obtain the empirical formula ... 

Vitamin C is essential for the prevention of scurvy. Combustion of a 0.2000 g sample of this carbon-hydrogen-oxygen compound yields 0.2998 g CO2 and 0.0819 g H2O. What are the percent composition and the empirical formula of vitamin C ... 

Suppose that we want to determine the formula of an unknown hydrocarbon. With combustion analysis we can establish the mass percent composition, and from this, we can determine the empirical formula. The method of Example 6-7 gives us a molar mass, in g moP, which is numerically equal to the molecular mass, in u. This is all the information we need to establish the true molecular formula of the hydrocarbon (see Exercise 96). 

Empirical and molecular formulas Percent composition Working mole problems... 

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