Chapter 9 – Chem 160

Chemical Calculations: The Mole Concept and Chemical

Formulas

The Law of Definite Proportions

Compounds are pure substances and

They are a chemical combination They can be broken down They have a definite, constant

elemental composition

Problem 9.5 – which two of three experiments produced the same compound?

ExperimentX grams Q gramsCompound mass

1 3.37 8.90 12.272 0.561 1.711 2.2723 26.9 71.0 97.9

Problem 9.5 cont’d

ExperimentX grams Q gramsCompound mass X/Q X/Q (s.f.)

1 3.37 8.90 12.27 0.378652 0.3792 0.561 1.711 2.272 0.327878 0.3283 26.9 71.0 97.9 0.378873 0.379

Calculation of Formula Masses

Definition

Calculation

Example 9.3 a) C7H6O2

a) benzoic acid formula mass

7 * C 7 x 12.01 amu = 84.07 amu6 * H 6 x 1.01 amu = 6.06 amu2 * O 2 x 16.00 amu = 32.00 amu

Sum 122.13 amu

Significant Figures andAtomic Mass

Uncertainty in data vs. uncertainty in formula mass

We’ll use atomic masses rounded to hundredths

Consider formula mass calculations as a pure addition

Percent composition Definition

Calculation - % composition of Au(NO3)3

Au N 3*O NO3 3*(NO3)196.97 14.01 48.00 62.01 186.03Formula Mass 383.00

% Au 51.43%

The Mole

Avogadro’s number # of molecules in 1.20 moles of CO 1.20 moles CO x 6.022x1023 CO

molecules 1 mole CO

Cancel “moles CO” Answer 7.2264 x 1023 – what units? Significant figures?

Mass of a Mole

Molar mass of an element is …

Molar mass of a compound is …

Problem 9.37 (d)

Mass of 1.357 moles of Na3PO4

3 x Na = 3 x 22.99 = 68.97 g P = 30.97 g 4 x O = 4 x 16.00 = 64.00g 1 mole 68.97+30.97+64.00=163.94 g 1.357 mole x 163.94 g = 222.466 g Note significant figures

Significant Figures and Avogadro’s Number

The mole is the amount of substance …

Avogadro’s number should never be the limiting factor in s.f. considerations

A.M.U. and Gram Units

6.022 x 1023 amu = 1.000 g Proof 6.022x1023 atoms N x 1 mole N x 14.01

amu 1 mole N 14.01 g N 1 atom N

6.022 x 1023 amu/g

A.M.U. and Gram Units (cont’d)

What is the mass, in grams, of a molecule whose mass on the amu scale is 104.00 amu? (Example 9.8)

104.00 amu x 1.000 g 6.022 x 1023 amu

1.7270009 x 10-22 g 1.727 x 10-22 g accounting for s.f.

Counting Particles by Weighing

Atomic ratio to mass ratio Table 9.2 Cl (35.45) / Na (22.99)

Extension to molecules

Counting Particles by Weighing – Problem 9.58 b)

Grams of Cu that will contain twice as many atoms as 20.00 g of Zn

20.00 g Zn x 6.022 x 1023 atoms x 63.55 g Cu 65.41 g Zn 6.02 x 1023 atoms

= 19.43128 g Cu contains ….. as many atoms as 20.00 g Zn Answer is 19.43 x 2 = 38.86 g Cu

Mole and Chemical Formulas

Microscopic level interpretation

Macro level

Mole and Chemical Formulas – Problem 9.60

6 mole to mole conversion factors from the formula K2SO4

2 moles of K / 1 mole of K2SO4

1 mole of S / 1 mole of K2SO4

4 moles of O / 1 mole of K2SO4

2 moles of K / 1 mole of S 2 moles of K / 4 moles of O 1 mole of S / 4 moles of O

Mole andChemical Calculations

Particles of A

Moles of A Moles of BParticles of

B

Grams of A Grams of B

Avogadro’s number Formula subscript Avogadro’s number

Mol

ar

mas

s

Mol

ar

mas

s

Mole & Chemical Calculations Problem 9.70 d)

Mass of 989 molecules of H2O

{(2x1.01)+16.00} g x 989 molecules 6.022 x 1023 molecules

= 2.95945 x 10-20 g

Purity of Samples

Definition Problem 9.86 a) calculate the mass

in grams of Cu2S present in a 25.4 g sample of 88.7% pure Cu2S

25.4 g of sample Cu2S x 88.7 g Cu2S 100 g of sample Cu2S

= 22.5298 g

Empirical andMolecular Formulas

Empirical formula – smallest whole number ratio of atoms

Molecular formula – actual number of atoms in a formula unit

Empirical andMolecular Formulas – cont’d

Problem 9.96 a) write empirical formula for P4H10

P2H5

9.96 d) C5H12? No change

Determination ofEmpirical Formulas

Elemental composition data

Empirical formula + molecular mass

Empirical andMolecular Formulas – cont’d

Problem 9.98 b) determine the empirical formula if 40.27% K, 26.78% Cr and 32.96% O

Mass (g)

Molar mass

# of moles

K 40.27 39.10 1.030Cr 26.78 52.00 0.5150O 32.96 16.00 2.060

Empirical andMolecular Formulas – cont’d

Mass (g)

Molar mass

# of moles

Divide by Cr

K 40.27 39.10 1.030 2.000Cr 26.78 52.00 0.5150 1.000O 32.96 16.00 2.060 4.000

Empirical formula for 9.98 b) is K2CrO4

Determination ofMolecular Formulas

Molecular mass information needed

molecular formula = (empirical formula)x; where x is a whole number

x= molecular formula (experimental)empirical formula(calculat’d from atomic masses)

For example (CH2)5 = C5H10

Determination ofMolecular Formulas – 9.116 b)

P2O3, empirical formula; molecular mass is 220 amu

P: 2 x 30.97 = 61.94 amu O: 3 x 16.00 = 48.00 amu Empirical formula = 109.94 amu Molecular/Empirical = 2.00 Molecular formula is P4O6

Determination ofMolecular Formulas – 9.120 b)

Citric acid, molecular mass 192 amu; 37.50% C, 4.21% H and 58.29% O

MoleCitric acid molar mass

Mass % of comp.

Atomic mass of comp.

Moles of comp.

units Mole g/mole g/100 g mole/g moles

C 1 192 0.3750 12.01 5.995004

H 1 192 0.0421 1.01 8.003168

O 1 192 0.5829 16.00 6.9948

divide by