1

Quantitative Composition of Compounds

Chapter 7

Quantitative Composition of Compounds

Chapter 7

Hein and Arena

Version 1.1

2

Chapter Outline7.1 The Mole

7.2 Molar Mass of Compounds

7.3 Percent Composition of Compounds

7.4 Empirical Formula versus Molecular Formula

7.5 Calculating Empirical Formulas

7.6 Calculating the Molecular Formula from the Empirical Formula

3

The mass of a single atom is too small to measure on a balance.

mass of hydrogen atom = 1.673 x 10-24 g

The MoleThe Mole

4

This is an

infinitesimal

mass

1.673 x 10-24 g

5

1 mole = 6.022 x 1023 objects

6

LARGE6.022 x 1023

is a very

number

7

6.022 x 1023

is

number

Avogadro’s Number

8

If 10,000 people started to count Avogardro’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.

9

1 mole of any element contains

6.022 x 1023

particles of that substance.

10

The atomic mass in grams

of any element23

contains 1 mole of atoms.

11

This is the same number of particles6.022 x 1023

as there are in exactly 12 grams of

C126

12

Species

Quantity

Number of H atoms

H

1 mole

6.022 x 1023

13

Species

Quantity

Number of H2 molecules

H2

1 mole

6.022 x 1023

14

Species

Quantity

Number of Na atoms

Na

1 mole

6.022 x 1023

15

Species

Quantity

Number of Fe atoms

Fe

1 mole

6.022 x 1023

16

Species

Quantity

Number of C6H6 molecules

C6H6

1 mole

6.022 x 1023

17

1 mol of atoms = 6.022 x 1023 atoms

6.022 x 1023 molecules

6.022 x 1023 ions

1 mol of molecules =

1 mol of ions =

18

• The molar mass of an element is its atomic mass in grams.

• It contains 6.022 x 1023 atoms (Avogadro’s number) of the element.

19

ElementAtomic mass

Molar massNumber of

atoms

H 1.008 amu 1.008 g 6.022 x 1023

Mg 24.31 amu 24.31 g 6.022 x 1023

Na 22.99 amu 22.99 g 6.022 x 1023

20

Atomic mass iron = 55.85

How many moles of iron does 25.0 g of iron represent?

Conversion sequence: grams Fe → moles Fe

1 mol Fe(grams Fe)

55.85 g Fe

1 mol Fe(25.0 g Fe)

55.85 g Fe

0.448 mol Fe

Set up the calculation using a conversion factor between moles and grams.

21

Atomic mass iron = 55.85

Conversion sequence: grams Fe → atoms Fe

236.022 x 10 atoms Fe(grams Fe)

55.85 g Fe

How many iron atoms are contained in 40.0 grams of iron?

236.022 x 10 atoms Fe(25.0 g Fe)

55.85 g Fe

232.70 x 10 atoms Fe

Set up the calculation using a conversion factor between atoms and grams.

22

Molar mass Na = 22.99 g

Conversion sequence: atoms Na → grams Na

23

22.99 g Na(atoms Na)

6.022 x 10 atoms Na

What is the mass of 3.01 x 1023 atoms of sodium (Na)?

2323

22.99 g Na(3.01 x 10 atoms Na)

6.022 x 10 atoms Na

11.5 g Na

Set up the calculation using a conversion factor between grams and atoms.

23

Atomic mass tin = 118.7

What is the mass of 0.365 moles of tin?

Conversion sequence: moles Sn → grams Sn

1 molar mass Sn(moles Sn)

1 mole Sn

118.7 g Sn(0.365 moles Sn)

1 mole Sn

43.3 g Sn

Set up the calculation using a conversion factor between grams and atoms.

24

2(2.00 mol O )23

2

2

6.022 x 10 molecules O1 mol O

2

2 atoms O1 molecule O

Conversion sequence: moles O2 → molecules O → atoms O

232

2

6.022 x 10 molecules O1 mol O

How many oxygen atoms are present in 2.00 mol of oxygen molecules?

Two conversion factors are needed:

2

2 atoms O1 mol O

24= 2.41 x10 atoms O

25

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

Molar Mass of Molar Mass of CompoundsCompounds

26

2 C = 2(12.01 g) = 24.02 g

6 H = 6(1.01 g) = 6.06 g

1 O = 1(16.00 g) = 16.00 g

46.08 g

Calculate the molar mass of C2H6O.

27

1 Li = 1(6.94 g) = 6.94 g1 Cl = 1(35.45 g) = 35.45 g4 O = 4(16.00 g) = 64.00 g

106.39 g

Calculate the molar mass of LiClO4.

28

Calculate the molar mass of (NH4)3PO4 .

3 N = 3(14.01 g) = 42.03 g12 H = 12(1.01 g) = 12.12 g

1 P = 1(30.97 g) = 30.97 g4 O = 4(16.00 g) = 64.00 g

149.12 g

29

Avogadro’s Number of Particles

6 x 1023 Particles

Molar Mass

1 MOLE

30

1 MOLE Ca

Avogadro’s Number ofCa atoms

6 x 1023 Ca atoms

40.078 g Ca

31

1 MOLE H2O

Avogadro’s Number of

H2O molecules

6 x 1023 H2O

molecules

18.02 g H2O

32

H Cl HCl

6.022 x 1023 H atoms

6.022 x 1023 Cl atoms

6.022 x 1023 HCl molecules

1 mol H atoms 1 mol Cl atoms1 mol HCl molecules

1.008 g H 35.45 g Cl 36.46 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.

33

In dealing with diatomic elements (H2, O2, N2, F2, Cl2, Br2, and I2), distinguish between one mole of atoms and one mole of molecules.

34

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 H2 molecules.

2 H = 2(1.01 g) = 2.02 g

35

How many moles of benzene, C6H6, are present in 390.0 grams of benzene?

Conversion sequence: grams C6H6 → moles C6H6

6 6

6 6

78.12 grams C HUse the conversion factor:

1 mole C H

6 6

6 6

1 mole C H 78.12 g C H

6 6(390.0 g C H ) 6 6= 5.000 moles C H

The molar mass of C6H6 is 78.12 g.

36

How many grams of (NH4)3PO4 are contained in 2.52 moles of (NH4)3PO4?

Conversion sequence: moles (NH4)3PO4

→ grams (NH4)3PO4

4 3 4

4 3 4

149.12 grams (NH ) POUse the conversion factor:

1 mole (NH ) PO

4 3 4(2.52 mol (NH ) PO ) 4 3 4

4 3 4

149.12 g (NH ) PO1 mol (NH ) PO

4 3 4= 376g (NH ) PO

The molar mass of (NH4)3PO4 is 149.12 g.

37

2(56.04 g N ) 2

2

1 mol N28.02 g N

232

2

6.022 x 10 molecules N

1 mol N

56.04 g of N2 contains how many N2 molecules?

The molar mass of N2 is 28.02 g.

Conversion sequence: g N2 → moles N2 → molecules N2

Use the conversion factors

2

2

1 mol N 28.02 g N

232

2

6.022 x 10 molecules N

1 mol N

242= 1.204 x 10 molecules N

38

2

2

1 mol N28.02 g N

2(56.04 g N )23

2

2

6.022 x 10 molecules N

1 mol N

56.04 g of N2 contains how many N2 atoms?

The molar mass of N2 is 28.02 g.

Conversion sequence: g N2 → moles N2 → molecules N2

→ atoms NUse the conversion factors

2

2

1 mol N 28.02 g N

232

2

6.022 x 10 molecules N

1 mol N 2

2 atoms N 1 molecule N

24= 2.409 x 10 atoms N2

2 atoms N1 molecule N

39

Percent composition of a compound is the mass percent of each element in the compound.

H2O11.19% H by mass 88.79% O by mass

Percent CompositionPercent Compositionof Compoundsof Compounds

40

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.

Percent Composition Percent Composition From FormulaFrom Formula

41

total mass of the element x 100 = percent of the element

molar mass

42

Step 1 Calculate the molar mass of H2S.2 H = 2 x 1.01g = 2.02 g

1 S = 1 x 32.07 g = 32.07 g34.09 g

Calculate the percent composition of hydrosulfuric acid H2S.

43

Calculate the percent composition of hydrosulfuric acid H2S.

Step 2 Divide the mass of each element by the molar mass and multiply by 100.

H

5.93%

S

94.07%

32.07g SS: (100) 94.07%

34.09g

2.02 g HH: (100) = 5.93%

34.09 g

44

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.

Percent Composition Percent Composition From FormulaFrom Formula

45

Step 1 Calculate the total mass of the compound1.52 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

46

Step 2 Divide the mass of each element by the total mass of the compound formed.

3.47g O(100) = 69.5%

4.99g

1.52 g N(100) = 30.5%

4.99 g

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.

47

• 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.

Empirical Formula versus Empirical Formula versus Molecular FormulaMolecular Formula

48

• 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.

49

C2H4Molecular Formula

CH2Empirical Formula

C:H 1:2Smallest Whole Number Ratio

50

C6H6Molecular Formula

CHEmpirical Formula

C:H 1:1Smallest Whole Number Ratio

51

H2O2Molecular Formula

HOEmpirical Formula

H:O 1:1Smallest Whole Number Ratio

52

53

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

54

55

Step 1 Assume a definite starting quantity (usually 100.0 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.

CalculatingCalculatingEmpirical FormulasEmpirical Formulas

56

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.

57

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.

FeO1.5

Fe1 x 2O1.5 x 2 Fe2O3

58

• The results of calculations may differ from a whole number.

– If they differ ±0.1 round off to the next nearest whole number.

2.93– Deviations greater than 0.1 unit from a

whole number usually mean that the calculated ratios have to be multiplied by a whole number.

59

Some Common Fractions and Their Decimal Equivalents

Common FractionDecimal

Equivalent14

13

23

12

34

0.333…

0.25

0.5

0.75

0.666…

Resulting WholeNumber

1

1

2

1

3

Multiply the decimal equivalent by the number in

the denominator of the fraction to get a whole number.

60

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 = 56.58 g

C = 8.68 g

O = 34.73 g

61

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.

K: 56.58 g K1 mol K atoms

39.10 g K

1.447 mol K atoms

C: 8.68 g C1 mol C atoms

12.01 g C

0.723 mol C atoms

O: 34.73 g O1 mol O atoms

16.00 g O

2.171 mol O atoms

C has the smallest number of moles

0.723 mol C atoms

62

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.1.447 mol

K = = 2.000.723 mol

0.723 molC: = 1.00

0.723 mol

2.171 molO = = 3.00

0.723 mol

The simplest ratio of K:C:O is 2:1:3

Empirical formula K2CO3

C has the smallest number of moles

0.723 mol C atoms

63

The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance.

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 = 25.94 g

O = 74.06 g

64

Step 2 Convert the grams of each element to moles.

N: 25.94 g N1 mol N atoms

14.01 g N

1.852 mol N atoms

O: 74.06 g O1 mol O atoms

16.00 g O

4.629 mol C atoms

The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance.

65

Step 3 Divide each number of moles by the smallest value.

1.852 molN = = 1.000

1.852 mol4.629 mol

O: = 2.5001.852 mol

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.

66

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 N2O5

N: (1.000)2 = 2.000 O: (2.500)2 = 5.000

67

Calculating the Molecular Formula Calculating the Molecular Formula from the Empirical Formulafrom the Empirical Formula

68

• 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.

molar massn = =

massof empirical formula formula units number of empirical

69

What is the molecular formula of a compound which has an empirical formula of CH2 and a molar mass of 126.2 g?

The molecular formula is (CH2)9 = C9H18

Let n = the number of formula units of CH2.

Calculate the mass of each CH2 unit

1 C = 1(12.01 g) = 12.01g

2 H = 2(1.01 g) = 2.02g

14.03g126.2 g

n 9 (empirical formula units)14.03 g

70

Key Concepts

7.1 The Mole – Avogadro’s Number

7.2 Molar Mass of Compounds – Molar mass – Avogadro’s Number

7.3 Percent Composition of Compounds – Calculations!!!

7.4 Empirical Formula versus Molecular Formula – Calculations!!!!

7.5 Calculating Empirical Formulas – Calculations!!!!

7.6 Calculating the Molecular Formula from the Empirical Formula – Calculations!!!!!