8 | 1 CHAPTER 8 CHEMICAL COMPOSITION

8 | 2 Atomic Masses Balanced equations tell us the relative numbers of molecules of reactants and products. C + O 2  CO 2 1 atom of C reacts with 1 molecule of O 2 to make 1 molecule of CO 2

8 | 3 Atomic Masses (cont.) If you want to know how many O 2 molecules you will need, or how many CO 2 molecules you can make, you will need to know how many C atoms are in the sample of carbon you are starting with. You can either count the atoms (!) or weigh the carbon and calculate the number of atoms – if you knew the mass of one atom of carbon.

8 | 4 Atomic Masses (cont.) Atomic masses allow us to convert weights into numbers of atoms. Unit is the amu (atomic mass unit) –1 amu = 1.66 x g

8 | 5 Atomic Masses (cont.)

8 | 6 Atomic Masses (cont.) If our sample of carbon weighs 3.00 x amu, we will have 2.50 x atoms of carbon. Since our equation tells us that 1 C atom reacts with 1 O 2 molecule, if we have 2.50 x C atoms, we will need 2.50 x molecules of O 2

8 | 7 Example #1 Determine the mass of 1 Al atom 1 atom of Al = amu Use the relationship as a conversion factor Calculate the mass (in amu) of 75 atoms of Al.

8 | 8 Moles The mass of 1.0 mole of an element is equal to the atomic mass in grams. A mole is the number of particles equal to the number of carbon atoms in 12 g of C-12. One mole = x units. This number is called Avogadro’s number. 1 mole of C-12 atoms weighs 12.0 g and has 6.02 x atoms. 1 atom of Carbon-12 weighs 12.0 amu.

8 | 9 Example #2 Compute the number of moles and the number of atoms in 10.0 g of Al.

8 | 10 Example #2 (cont.) Use the periodic table to determine the mass of 1 mole of Al. 1 mole Al = g Use this as a conversion factor for grams-to-moles.

8 | 11 Example #2 (cont.) Use Avogadro’s Number to determine the number of atoms in 1 mole. 1 mole Al = 6.02 x atoms Use this as a conversion factor for moles-to-atoms.

8 | 12 Compute the number of moles in and the mass of 2.23 x atoms of Al. Example #3

8 | 13 Example #3 (cont.) Use Avogadro’s Number to determine the number of atoms in 1 mole. 1 mole Al = 6.02 x atoms Use this as a conversion factor for atoms-to-moles.

8 | 14 Example #3 (cont.) Use the periodic table to determine the mass of 1 mole of Al. 1 mole Al = g Use this as a conversion factor for moles-to-grams.

8 | 15 Molar Mass Molar mass: the mass in grams of one mole of a compound The relative weights of molecules can be calculated from atomic masses water = H 2 O = 2(1.008 amu) amu = amu 1 mole of H 2 O will weigh g, therefore the molar mass of H 2 O is g 1 mole of H 2 O will contain g of oxygen and 2.02 g of hydrogen

8 | 16 Percent Composition Percentage by mass of each element in a compound Can be determined from the formula of the compound or by experimental mass analysis of the compound The percentages may not always total to 100% due to rounding.

8 | 17 Determine the percent composition from the formula C 2 H 5 OH. Example #4

8 | 18 Example #4 (cont.) Determine the mass of each element in 1 mole of the compound. 2 moles C = 2(12.01 g) = g 6 moles H = 6(1.008 g) = g 1 mol O = 1(16.00 g) = g Determine the molar mass of the compound by adding the masses of the elements. 1 mole C 2 H 5 OH = g

8 | 19 Divide the mass of each element by the molar mass of the compound and multiply by 100% Example #4 (cont.)

8 | 20 Empirical Formulas Empirical formula: the simplest, whole-number ratio of atoms in a molecule –Can be determined from percent composition or by combining masses Molecular formula: a multiple of the empirical formula % A mass A (g) moles A 100g MM A % B mass B (g) moles B 100g MM B moles A moles B

8 | 21 Determine the empirical formula of benzopyrene, C 20 H 12 Example #5

8 | 22 Example #5 (cont.) Find the greatest common factor (GCF) of the subscripts. factors of 20 = (10 x 2), (5 x 4) factors of 12 = (6 x 2), (4 x 3) GCF = 4 Divide each subscript by the GCF to get the empirical formula. C 20 H 12 = (C 5 H 3 ) 4 Empirical Formula = C 5 H 3

8 | 23 Determine the empirical formula of acetic anhydride if its percent composition is 47% carbon, 47% oxygen, and 6.0% hydrogen. Example #6

8 | 24 Example #6 (cont.) Convert the percentages to grams by assuming you have 100 g of the compound. –Step can be skipped if given masses

8 | 25 Convert the grams to moles Example #6 (cont.)

8 | 26 Divide each by the smallest number of moles. For this example, 2.9 is the smallest. Example #6 (cont.)

8 | 27 If any of the ratios is not a whole number, multiply all the ratios by a factor to make it a whole number. –If ratio is ?.5, then multiply by 2; if ?.33 or ?.67 then multiply by 3; if ?.25 or ?.75, then multiply by 4 Multiply all the Ratios by 3 Because C is 1.3 Example #6 (cont.)

8 | 28 Use the ratios as the subscripts in the empirical formula. C4H6O3C4H6O3 Example #6 (cont.)

8 | 29 Molecular Formulas The molecular formula is a multiple of the empirical formula. To determine the molecular formula you need to know the empirical formula and the molar mass of the compound.

8 | 30 Determine the molecular formula of benzopyrene if it has a molar mass of 252 g and an empirical formula of C 5 H 3 Example #7

8 | 31 Example #7 (cont.) Determine the empirical formula -May need to calculate it as previous C 5 H 3 Determine the molar mass of the empirical formula 5 C = g, 3 H = g C 5 H 3 = g

8 | 32 Divide the given molar mass of the compound by the molar mass of the empirical formula Round to the nearest whole number Example #7 (cont.)

8 | 33 Multiply the empirical formula by the calculated factor to give the molecular formula (C 5 H 3 ) 4 = C 20 H 12 Example #7 (cont.)