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