Determining Empirical and Molecular Formulas Quiz

PDF page 154: The elemental makeup of a compound defines its chemical identity, and chemical formulas are the most succinct way of representing this elemental makeup. When a compound’s formula is unknown, measuring the mass of each of its constituent elements is often the first step in the process of determining the formula experimentally. The results of these measurements permit the calculation of the compound’s percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows: mass H % H = × 100% mass compound mass C % C = × 100% mass compound If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C: 2.5 g H % H =

PDF page 154: The elemental makeup of a compound defines its chemical identity, and chemical formulas are the most succinct way of representing this elemental makeup. When a compound’s formula is unknown, measuring the mass of each of its constituent elements is often the first step in the process of determining the formula experimentally. The results of these measurements permit the calculation of the compound’s percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows: mass H % H = × 100% mass compound mass C % C = × 100% mass compound If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C: 2.5 g H % H =

PDF page 155: The analysis results indicate that the compound is 61.0% C, 15.4% H, and 23.7% N by mass. Check Your Learning A 24.81-g sample of a gaseous compound containing only carbon, oxygen, and chlorine is determined to contain 3.01 g C, 4.00 g O, and 17.81 g Cl. What is this compound’s percent composition? Answer: 12.1% C, 16.1% O, 71.8% Cl

PDF page 155: The analysis results indicate that the compound is 61.0% C, 15.4% H, and 23.7% N by mass. Check Your Learning A 24.81-g sample of a gaseous compound containing only carbon, oxygen, and chlorine is determined to contain 3.01 g C, 4.00 g O, and 17.81 g Cl. What is this compound’s percent composition? Answer: 12.1% C, 16.1% O, 71.8% Cl

PDF page 155: Determining Percent Composition from Molecular or Empirical Formulas Percent composition is also useful for evaluating the relative abundance of a given element in different compounds of known formulas. As one example, consider the common nitrogen-containing fertilizers ammonia (NH3), ammonium nitrate (NH4NO3), and urea (CH4N2O). The element nitrogen is the active ingredient for agricultural purposes, so the mass percentage of nitrogen in the compound is a practical and economic concern for consumers choosing among these fertilizers. For these sorts of applications, the percent composition of a compound is easily derived from its formula mass and the atomic masses of its constituent elements. A molecule of NH3 contains one N atom weighing 14.01 amu and three H atoms weighing a total of (3 × 1.008 amu) = 3.024 amu. The formula mass of ammonia is therefore (14.01 amu + 3.024 amu) = 17.03 amu, and its percent composition is: 14.01 amu N % N =

PDF page 156: % C = 60.00% C 8 × 1.008 g/mol 8.064 g/mol 8 mol H × molar mass H % H =

PDF page 156: % O = 35.52% Note that these percentages sum to equal 100.00% when appropriately rounded. Check Your Learning To three significant digits, what is the mass percentage of iron in the compound Fe2O3? Answer: 69.9% Fe

PDF page 156: As previously mentioned, the most common approach to determining a compound’s chemical formula is to first measure the masses of its constituent elements. However, keep in mind that chemical formulas represent the relative numbers, not masses, of atoms in the substance. Therefore, any experimentally derived data involving mass must be used to derive the corresponding numbers of atoms in the compound. This is accomplished using molar masses to convert the mass of each element to a number of moles. These molar amounts are used to compute whole-number ratios that can be used to derive the empirical formula of the substance. Consider a sample of compound determined to contain 1.71 g C and 0.287 g H. The corresponding numbers of atoms (in moles) are: 1 mol C 1.17 g C × = 0.142 mol C 12.01 g C 1 mol H 0.287 g H × = 0.284 mol H 1.008 g H

PDF page 156: As previously mentioned, the most common approach to determining a compound’s chemical formula is to first measure the masses of its constituent elements. However, keep in mind that chemical formulas represent the relative numbers, not masses, of atoms in the substance. Therefore, any experimentally derived data involving mass must be used to derive the corresponding numbers of atoms in the compound. This is accomplished using molar masses to convert the mass of each element to a number of moles. These molar amounts are used to compute whole-number ratios that can be used to derive the empirical formula of the substance. Consider a sample of compound determined to contain 1.71 g C and 0.287 g H. The corresponding numbers of atoms (in moles) are: 1 mol C 1.17 g C × = 0.142 mol C 12.01 g C 1 mol H 0.287 g H × = 0.284 mol H 1.008 g H

PDF page 156: Thus, this compound may be represented by the formula C0.142H0.248. Per convention, formulas contain wholenumber subscripts, which can be achieved by dividing each subscript by the smaller subscript: C 0.142 H 0.284 or CH 2 0.142 0.142 (Recall that subscripts of “1” are not written but rather assumed if no other number is present.) The empirical formula for this compound is thus CH2. This may or not be the compound’s molecular formula as well; however, additional information is needed to make that determination (as discussed later in this section). Consider as another example a sample of compound determined to contain 5.31 g Cl and 8.40 g O. Following the same approach yields a tentative empirical formula of: C1 O

PDF page 156: 0.150 0.525 3.5 0.150 0.150 In this case, dividing by the smallest subscript still leaves us with a decimal subscript in the empirical formula. To convert this into a whole number, multiply each of the subscripts by two, retaining the same atom ratio and yielding Cl2O7 as the final empirical formula. In summary, empirical formulas are derived from experimentally measured element masses by:

PDF page 158: 2⎝Fe O ⎠ = Fe O 2 3 1 1.5 The empirical formula is Fe2O3. Check Your Learning What is the empirical formula of a compound if a sample contains 0.130 g of nitrogen and 0.370 g of oxygen? Answer: N2O5

PDF page 158: 2⎝Fe O ⎠ = Fe O 2 3 1 1.5 The empirical formula is Fe2O3. Check Your Learning What is the empirical formula of a compound if a sample contains 0.130 g of nitrogen and 0.370 g of oxygen? Answer: N2O5