1. Chemical Calculations: The Mole Concept and Chemical Formulas
Chapter 9 – Chem 160
Chemical Calculations: The Mole Concept and Chemical Formulas

2. 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

3. Problem 9.5 – which two of three experiments produced the same compound?
X & Q react to produce two different compounds, depending on conditions. Which two of the three experiments produced the same compound?

4. Problem 9.5 cont’d
Experiments 1 & 3

5. Calculation of Formula Masses
Definition
Calculation

6. Example 9.3 a) C7H6O2

7. Significant Figures and Atomic 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

8. Percent composition Definition Calculation - % composition of Au(NO3)3
% composition by weight

9. The Mole Avogadro’s number # of molecules in 1.20 moles of CO
1.20 moles CO x 6.022x1023 CO molecules mole CO
Cancel “moles CO”
Answer x 1023 – what units?
Significant figures?

10. Mass of a Mole Molar mass of an element is …
Molar mass of a compound is …

11. Problem 9.37 (d) Mass of 1.357 moles of Na3PO4
3 x Na = 3 x = g
P = g
4 x O = 4 x = 64.00g
1 mole = g
1.357 mole x g = g
Note significant figures

12. 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

13. A.M.U. and Gram Units 6.022 x 1023 amu = 1.000 g Proof
6.022x1023 atoms N x 1 mole N x amu mole N g N 1 atom N
6.022 x 1023 amu/g
Note that we can change “N” with another atom and corresponding atomic mass; the cancellations will still occur and we will get the same answer.

14. A.M.U. and Gram Units (cont’d)
What is the mass, in grams, of a molecule whose mass on the amu scale is amu? (Example 9.8)
amu x g x 1023 amu
x g
1.727 x g accounting for s.f.

15. Counting Particles by Weighing
Atomic ratio to mass ratio
Table 9.2
Cl (35.45) / Na (22.99)
Extension to molecules

16. Counting Particles by Weighing – Problem 9.58 b)
Grams of Cu that will contain twice as many atoms as g of Zn
20.00 g Zn x x 1023 atoms x g Cu g Zn x 1023 atoms
= g Cu contains …..
as many atoms as g Zn
Answer is x 2 = g Cu

17. Mole and Chemical Formulas
Microscopic level interpretation
Macro level

18. 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

19. Mole and Chemical Calculations
Two general types of problems that this pathway will help in solving.
Calculations for which information (moles, grams, particles) is given about a particular substance, and information (moles, grams, particles) is needed concerning the same substance.
Calculation for which information (moles, grams, particles) is given about a particular substance, and information is needed concerning a component of that substance.
For the first type of problem, only the left side of the figure (A boxes) is needed. For the second type of problems, both sides of the diagram are used.

20. Mole & Chemical Calculations Problem 9.70 d)
Mass of 989 molecules of H2O
{(2x1.01)+16.00} g x 989 molecules x 1023 molecules
= x g
Text answer is 2.96; must use “3” significant figures for I took it to be an exact number.
Can you go from the mass of an element to the mass of a compound and vice versa?

21. 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 g of sample Cu2S
= g

22. Empirical and Molecular Formulas
Empirical formula – smallest whole number ratio of atoms
Molecular formula – actual number of atoms in a formula unit
Table 9.4

23. Empirical and Molecular Formulas – cont’d
Problem 9.96 a) write empirical formula for P4H10
P2H5
9.96 d) C5H12?
No change
Table 9.4

24. Determination of Empirical Formulas
Elemental composition data
Empirical formula + molecular mass

25. Empirical and Molecular Formulas – cont’d
Problem 9.98 b) determine the empirical formula if 40.27% K, 26.78% Cr and 32.96% O
Table 9.4

26. Empirical and Molecular Formulas – cont’d
We would use the same approach to solve this type of problem if weights rather than weight percentages are given.
Empirical formula for 9.98 b) is K2CrO4

27. Determination of Molecular 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

28. Determination of Molecular Formulas – 9.116 b)
P2O3, empirical formula; molecular mass is 220 amu
P: 2 x = amu
O: 3 x = amu
Empirical formula = amu
Molecular/Empirical = 2.00
Molecular formula is P4O6

29. Determination of Molecular Formulas – 9.120 b)
Citric acid, molecular mass 192 amu; 37.50% C, 4.21% H and 58.29% O
Mole
Citric acid molar mass
Mass % of comp.
Atomic mass of comp.
Moles of comp.
units
g/mole
g/100 g
mole/g
moles C 1
192
0.3750
12.01 H
0.0421
1.01 O
0.5829
16.00
6.9948
divide by