1 Quantitative Composition of Compounds Chapter 7 Hein and Arena Eugene Passer Chemistry Department Bronx Community College © John Wiley and Sons, Inc Version th Edition

2 Chapter Outline 7.1 The MoleThe Mole 7.2 Molar Mass of CompoundsMolar Mass of Compounds 7.3 Percent Composition of CompoundsPercent Composition ofCompounds 7.4 Empirical Formula versus Molecular FormulaEmpirical Formula versusMolecular Formula 7.5 Calculating Empirical FormulasCalculating EmpiricalFormulas 7.6 Calculating the Molecular Formula from the Empirical FormulaCalculating the MolecularFormula from the EmpiricalFormula

3 7.1 The Mole

4 The mass of a single atom is too small to measure on a balance. mass of hydrogen atom = x g

5 This is an infinitesimal mass x g

6 Chemists have chosen a unit for counting atoms. That unit is the Chemists require a unit for counting which can express large numbers of atoms using simple numbers. MOLE

7 1 mole = x objects

8 LARGE x is a very number

x is number Avogadro’s Number

10 If 10,000 people started to count Avogadro’s number and counted at the rate of 100 numbers per minute each minute of the day, it would take over 1 trillion years to count the total number.

11 1 mole of any element contains x particles of that substance.

12 The atomic mass in grams of any element 23 contains 1 mole of atoms.

13 This is the same number of particles x as there are in exactly 12 grams of

14ExamplesExamples

15 Species Quantity Number of H atoms H 1 mole x 10 23

16 Species Quantity Number of H 2 molecules H2H2 1 mole x 10 23

17 Species Quantity Number of Na atoms Na 1 mole x 10 23

18 Species Quantity Number of Fe atoms Fe 1 mole x 10 23

19 Species Quantity Number of C 6 H 6 molecules C6H6C6H6 1 mole x 10 23

20 1 mol of atoms =6.022 x atoms x molecules x ions 1 mol of molecules = 1 mol of ions =

21 The molar mass of an element is its atomic mass in grams. It contains x atoms (Avogadro’s number) of the element.

22 Element Atomic mass Molar mass Number of atoms H1.008 amu1.008 g6.022 x Mg24.31 amu24.31 g6.022 x Na22.99 amu22.99 g6.022 x 10 23

23ProblemsProblems

24 Atomic mass iron = How many moles of iron does 25.0 g of iron represent? Conversion sequence: grams Fe → moles Fe Set up the calculation using a conversion factor between moles and grams.

25 Atomic mass iron = Conversion sequence: grams Fe → atoms Fe How many iron atoms are contained in 25.0 grams of iron? Set up the calculation using a conversion factor between atoms and grams.

26 Molar mass Na = g Conversion sequence: atoms Na → grams Na What is the mass of 3.01 x atoms of sodium (Na)? Set up the calculation using a conversion factor between grams and atoms.

27 Atomic mass tin = What is the mass of moles of tin? Conversion sequence: moles Sn → grams Sn Set up the calculation using a conversion factor between grams and atoms.

28 Conversion sequence: moles O 2 → molecules O → atoms O How many oxygen atoms are present in 2.00 mol of oxygen molecules? Two conversion factors are needed:

Molar Mass of Compounds

30 The molar mass of a compound can be determined by adding the molar masses of all of the atoms in its formula.

31 2 C = 2(12.01 g) = g 6 H = 6(1.01 g) = 6.06 g 1 O = 1(16.00 g) = g g Calculate the molar mass of C 2 H 6 O.

32 1 Li = 1(6.94 g) = 6.94 g 1 Cl = 1(35.45 g) = g 4 O = 4(16.00 g) = g g Calculate the molar mass of LiClO 4.

33 Calculate the molar mass of (NH 4 ) 3 PO 4. 3 N = 3(14.01 g) = g 12 H = 12(1.01 g) = g 1 P = 1(30.97 g) = g 4 O = 4(16.00 g) = g g

34 Avogadro’s Number of Particles 6 x Particles Molar Mass 1 MOLE

35 1 MOLE Ca Avogadro’s Number of Ca atoms 6 x Ca atoms g Ca

36 1 MOLE H 2 O Avogadro’s Number of H 2 O molecules 6 x H 2 O molecules g H 2 O

37 HClHCl x H atoms x Cl atoms x HCl molecules 1 mol H atoms1 mol Cl atoms 1 mol HCl molecules g H35.45 g Cl36.46 g HCl 1 molar mass H atoms 1 molar mass Cl atoms 1 molar mass HCl molecules These relationships are present when hydrogen combines with chlorine.

38 In dealing with diatomic elements (H 2, O 2, N 2, F 2, Cl 2, Br 2, and I 2 ), distinguish between one mole of atoms and one mole of molecules.

39 Calculate the molar mass of 1 mole of H atoms. 1 H = 1(1.01 g) = 1.01 g Calculate the molar mass of 1 mole of H 2 molecules. 2 H = 2(1.01 g) = 2.02 g

40ProblemsProblems

41 How many moles of benzene, C 6 H 6, are present in grams of benzene? Conversion sequence: grams C 6 H 6 → moles C 6 H 6 The molar mass of C 6 H 6 is g.

42 How many grams of (NH 4 ) 3 PO 4 are contained in 2.52 moles of (NH 4 ) 3 PO 4 ? Conversion sequence: moles (NH 4 ) 3 PO 4 → grams (NH 4 ) 3 PO 4 The molar mass of (NH 4 ) 3 PO 4 is g.

g of N 2 contains how many N 2 molecules? The molar mass of N 2 is g. Conversion sequence: g N 2 → moles N 2 → molecules N 2 Use the conversion factors

g of N 2 contains how many nitrogen atoms? The molar mass of N 2 is g. Conversion sequence: g N 2 → moles N 2 → molecules N 2 → atoms N Use the conversion factor

Percent Composition of Compounds

46 Percent composition of a compound is the mass percent of each element in the compound. H2OH2O 11.19% H by mass88.79% O by mass

47 Percent Composition From Formula

48 If the formula of a compound is known, a two-step process is needed to calculate the percent composition. Step 1 Calculate the molar mass of the formula. Step 2 Divide the total mass of each element in the formula by the molar mass and multiply by 100.

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50 Step 1 Calculate the molar mass of H 2 S. 2 H = 2 x 1.01g = 2.02 g 1 S = 1 x g = g g Calculate the percent composition of hydrosulfuric acid H 2 S.

51 Calculate the percent composition of hydrosulfuric acid H 2 S. Step 2 Divide the mass of each element by the molar mass and multiply by 100. H 5.93% S 94.07%

52 Percent Composition From Experimental Data

53 Percent composition can be calculated from experimental data without knowing the composition of the compound. Step 1 Calculate the mass of the compound formed. Step 2 Divide the mass of each element by the total mass of the compound and multiply by 100.

54 Step 1 Calculate the total mass of the compound 1.52 g N 3.47 g O 4.99 g A compound containing nitrogen and oxygen is found to contain 1.52 g of nitrogen and 3.47 g of oxygen. Determine its percent composition. = total mass of product

55 Step 2 Divide the mass of each element by the total mass of the compound formed. N 30.5% O 69.5% A compound containing nitrogen and oxygen is found to contain 1.52 g of nitrogen and 3.47 g of oxygen. Determine its percent composition.

Empirical Formula versus Molecular Formula

57 The empirical formula or simplest formula gives the smallest whole- number ratio of the atoms present in a compound. The empirical formula gives the relative number of atoms of each element present in the compound.

58 The molecular formula is the true formula of a compound. The molecular formula represents the total number of atoms of each element present in one molecule of a compound.

59ExamplesExamples

60 C2H4C2H4 Molecular Formula CH 2 Empirical Formula C:H 1:2 Smallest Whole Number Ratio

61 C6H6C6H6 Molecular Formula CH Empirical Formula C:H 1:1 Smallest Whole Number Ratio

62 H2O2H2O2 Molecular Formula HOEmpirical Formula H:O 1:1 Smallest Whole Number Ratio

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64 Two compounds can have identical empirical formulas and different molecular formulas.

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Calculating Empirical Formulas

67 Step 1 Assume a definite starting quantity (usually g) of the compound, if the actual amount is not given, and express the mass of each element in grams. Step 2 Convert the grams of each element into moles of each element using each element’s molar mass.

68 Step 3 Divide the moles of atoms of each element by the moles of atoms of the element that had the smallest value. – If the numbers obtained are whole numbers, use them as subscripts and write the empirical formula. – If the numbers obtained are not whole numbers, go on to step 4.

69 Step 4 Multiply the values obtained in step 3 by the smallest numbers that will convert them to whole numbers Use these whole numbers as the subscripts in the empirical formula. FeO 1.5 Fe 1 x 2 O 1.5 x 2 Fe 2 O 3

70 The results of calculations may differ from a whole number. –If they differ ±0.1, round off to the next nearest whole number. 2.9 33 –Deviations greater than 0.1 unit from a whole number usually mean that the calculated ratios have to be multiplied by a whole number.

71 Some Common Fractions and Their Decimal Equivalents Common Fraction Decimal Equivalent 0.333… … Resulting Whole Number Multiply the decimal equivalent by the number in the denominator of the fraction to get a whole number.

72ProblemsProblems

73 The analysis of a salt shows that it contains 56.58% potassium (K); 8.68% carbon (C); and 34.73% oxygen (O). Calculate the empirical formula for this substance. Step 1 Express each element in grams. Assume 100 grams of compound. K = g C = 8.68 g O = g

74 The analysis of a salt shows that it contains 56.58% potassium (K); 8.68% carbon (C); and 34.73% oxygen (O). Calculate the empirical formula for this substance. Step 2 Convert the grams of each element to moles. C has the smallest number of moles

75 The analysis of a salt shows that it contains 56.58% potassium (K); 8.68% carbon (C); and 34.73% oxygen (O). Calculate the empirical formula for this substance. Step 3 Divide each number of moles by the smallest value. The simplest ratio of K:C:O is 2:1:3 Empirical formula K 2 CO 3 C has the smallest number of moles

76 The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance. Step 1 Express each element in grams. Assume 100 grams of compound. N = g O = g

77 Step 2 Convert the grams of each element to moles. The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance.

78 Step 3 Divide each number of moles by the smallest value. The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance. This is not a ratio of whole numbers.

79 Step 4 Multiply each of the values by 2. The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance. Empirical formula N 2 O 5 N: (1.000)2 = O: (2.500)2 = 5.000

Calculating the Molecular Formula from the Empirical Formula

81 The molecular formula can be calculated from the empirical formula if the molar mass is known. The molecular formula will be equal to the empirical formula or some multiple, n, of it. To determine the molecular formula evaluate n. n is the number of units of the empirical formula contained in the molecular formula.

82 What is the molecular formula of a compound which has an empirical formula of CH 2 and a molar mass of g? The molecular formula is (CH 2 ) 9 = C 9 H 18 Let n = the number of formula units of CH 2. Calculate the mass of each CH 2 unit 1 C = 1(12.01 g) = 12.01g 2 H = 2(1.01 g) = 2.02g 14.03g

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