Chapter 9Chapter 9

Problems 9.120 a, 9.18 b and Problems 9.120 a, 9.18 b and 9.38 a9.38 a

Citric acid with a molecular mass Citric acid with a molecular mass 192 amu192 amu

Mass %: C 37.50, H 4.21 and O 58.29Mass %: C 37.50, H 4.21 and O 58.29

Determine molecular formula starting from Determine molecular formula starting from 100.0 g citric acid100.0 g citric acid

How many moles of each in 100.0 g?How many moles of each in 100.0 g?

C: 100.0 g x C: 100.0 g x 37.50 g37.50 g x x 1 mole1 mole = 3.122 = 3.122100.0 g100.0 g 12.01 g12.01 g

H: 100.0 g x H: 100.0 g x 4.21 g4.21 g x x 1 mole1 mole = 4.16 = 4.168383100.0 g100.0 g 1.01 g1.01 g

Citric acid with a molecular mass Citric acid with a molecular mass 192 amu – cont’d192 amu – cont’d

O: 100.0 g x O: 100.0 g x 58.29 g58.29 g x x 1 mole1 mole = 3.643 = 3.643100.0 g100.0 g 16.00 g16.00 g

Empirical formulaEmpirical formulaC: 3.125/3.125 = 1C: 3.125/3.125 = 1H: 4.17/3.125 = 1.33H: 4.17/3.125 = 1.33O: 3.643/3.125 = 1.166O: 3.643/3.125 = 1.166What is the multiplier?What is the multiplier?66

Citric acid with a molecular mass Citric acid with a molecular mass 192 amu – cont’d192 amu – cont’d

CC66HHxxOOyy

x = 1.33 * 6 = 7.98x = 1.33 * 6 = 7.98

y = 1.166 * 6 = 6.996y = 1.166 * 6 = 6.996

CC66HH88OO77

This is the …..This is the …..

Citric acid with a molecular mass Citric acid with a molecular mass 192 amu – cont’d192 amu – cont’d

Empirical formulaEmpirical formulaEmpirical formula mass is …Empirical formula mass is …

6 * 12.01 + 8 * 1.01 + 7 * 16.00 = 6 * 12.01 + 8 * 1.01 + 7 * 16.00 = 192.14 amu192.14 amu

What is the molecular formula?What is the molecular formula?Same as empirical formulaSame as empirical formula

Citric acid CCitric acid C66HH88OO77

Going BackwardGoing Backward

Knowing the formula, can we calculate % Knowing the formula, can we calculate % mass composition?mass composition?

Does it matter if we have empirical or Does it matter if we have empirical or molecular formula?molecular formula?

Problem 9.18 b) % compositionProblem 9.18 b) % compositionof Na in NaCNof Na in NaCN

Na: 1 * 22.99 = 22.99 amuNa: 1 * 22.99 = 22.99 amu

C: 1 * 12.01 = 12.01 amuC: 1 * 12.01 = 12.01 amu

N: 1 * 14.01 = 14.01 amuN: 1 * 14.01 = 14.01 amu

Total = 49.01 amuTotal = 49.01 amu

%Na by mass =%Na by mass =

(22.99/49.01) * 100 = 46.90879%(22.99/49.01) * 100 = 46.90879%

Problem 9.38 a)Problem 9.38 a)

Mass in grams of 0.981 mole of SOMass in grams of 0.981 mole of SO22

S: 1 * 32.06 = 32.06S: 1 * 32.06 = 32.06

O: 2 * 16.00 = 32.00O: 2 * 16.00 = 32.00

Sum = 64.06Sum = 64.06

0.981 mole * (64.06 g/1mole) = 62.84286 g0.981 mole * (64.06 g/1mole) = 62.84286 g

Significant figures?Significant figures?

How many molecules of SOHow many molecules of SO22 in 0.981 mole? in 0.981 mole?