1. Chapter 7 Chemical Formula Relationships

2. Number of Atoms in a Formula
In writing the formula of a substance, subscript numbers are used to indicate the number of atoms or groups of atoms of each element in the formula unit.

3. Number of Atoms in a Formula
How many atoms of each element are in a formula unit of ammonium carbonate?
The formula of the ammonium ion is NH4+
The formula of carbonate is CO32–
The formula of the ammonium carbonate is (NH4)2CO3.
Each element in the parentheses is multiplied by two.
Total number of atoms of each element:
2 nitrogen atoms, 8 hydrogen atoms,
1 carbon atom, 3 oxygen atoms

4. Atomic Mass Atomic Mass The average mass of atoms of an element
By definition the mass of a carbon-12 atom is 12 u.

5. Molecular & Formula Mass
Molecular Mass
The sum of the atomic masses of each atom in the molecule.
Formula mass
The sum of atomic masses in the formula unit

6. Molecular & Formula Mass
What is the formula mass of calcium phosphate?
Calcium ion: Ca2+ Phosphate ion: PO43–
Calcium phosphate: Ca3(PO4)2
Ca 3 × u = u
P 2 × u = u
O 8 × u = u
Ca3(PO4) u

7. Defenition of Mole
The mole is the amount of substance that contains as many elementary entities as there are atoms in exactly 12 grams of carbon-12.
In 12 g of carbon-12 there are x atoms
1 mole of any substance = x units of that substance.
When the mole is used the elementary entities must be specified: atoms, molecules, ions…

8. Avogadro’s Number NA The number of elementary units in one mole
 1023 units/mol
The Avogadro constant is a conversion factor between units and mole.

9. Conversion of Mole to Molecules
How many carbon dioxide molecules are in 2.0 moles of carbon dioxide?
2.0 mol CO2 × =
1.2 × 1024 molecules CO2

10. Molar Mass
Molar mass of a substance is the mass in grams of one mole of the substance.
Units: g/mol
Molar mass of an element is the mass of the element per mole of its atoms.

11. Atomic mass unit and gram
The mass of one atom of carbon-12 is exactly 12 atomic mass units.
The mass of one mole of carbon-12 atoms (6.022 x 1023 atoms of carbon-12) is exactly 12 grams
(6.022 x 1023 atoms) x (12 u/atom) = 12 grams
6.022 x 1023 u = 1 gram

12. Molar Mass
The mass of one atom of carbon-12 is exactly 12 atomic mass units.
The mass of one mole of carbon-12 atoms is exactly 12 grams.
This leads to the conclusion:
The molar mass of any substance in grams per mole is numerically equal to the atomic, molecular or formula mass of that substance in atomic mass units.

13. Molar Mass
Calculate the mass of one NH3 molecule and the mass of one mole of NH3 molecules.
14.01 u + 3x1.008 u = u
The mass of one ammonia molecule is u.
To change from molecular mass to molar mass, change the units from u to g/mol: g/mol.
One mole of ammonia molecules has a mass of
17.03 g.

14. Molar Mass

15. Mass, # of Moles, # of Units
Molar mass, MM, links mass in grams with the number of moles.
Avogadro’s number, NA, links the number of moles with the number of particles.

16. Mass, # of Moles, # of Units
How many molecules are in 454 g of water?
g  moles  molecules
454g x x
= 1.52 x 1025 molecules H2O

17. Mass↔Moles↔ Units How many hydrogen atoms are in 1.0 kg of ammonia?
1.0 kg NH3 × × ×
× =
1.1 × 1026 atoms H

18. Percentage Composition
The percentage composition of a compound is the
percentage by mass of each element in the compound.

19. Percentage Composition
Determine the percentage composition of calcium fluoride Ca F2.
Solution:
In one mole of Ca F2
1x(40.08 g Ca) + 2x(19.00 g F)
= g CaF2
(Solution continued on the next slide)

20. Percentage Composition
 100 = % Ca
 100 = 48.67% F
Check: 51.33% % = %

21. The simplest ratio of atoms of the elements in a compound.
Empirical Formula
Empirical Formula
The simplest ratio of atoms of the elements in a compound.
The empirical formula of C2H4 is CH2.
Likewise, the empirical formula of C3H6 is CH2.
All compounds with the general formula CnH2n
have the same empirical formula and
therefore the same percentage composition.

22. Empirical Formula
Write the empirical formulas of benzene, C6H6, and octane, C8H18.
Look for the simplest whole-number ratio of elements:
For C6H6, the 6/6 ratio can be reduced to 1/1: CH.
For C8H18, the 8/18 ratio can be divided by 2 on top and bottom to be reduced to 4/9: C4H9.

23. Find Empirical Formula
To find empirical formula, you need to find the ratio of atoms of the elements
ratio of atoms = ratio of moles of atoms

24. How to Find an Empirical Formula
Find the masses of different elements in a sample of the compound.
Convert the masses into moles of atoms.
Determine the ratio of moles of atoms.
Express the moles ratio as the smallest possible ratio of integers.
Write the empirical formula, using the number in the integer ratio as the subscript in the formula.

25. Find Empirical Formula
What is the empirical formula of a compound that has 85.6% carbon, 14.4 % hydrogen?
Solution:
It is usually helpful to organize the calculations in a table with the following headings:
Mole Formula Empirical
Element Grams Moles Ratio Ratio Formula

26. Find Empirical Formula
Mole Formula Empirical
Element Grams Moles Ratio Ratio Formula C
H CH2

27. Find Empirical Formula
What is the empirical formula of a compound that analyzes as 20.0% carbon, 2.2% hydrogen, and 77.80% chlorine?
Mole Formula Empirical
Element Grams Moles Ratio Ratio Formula C H
Cl C3H4Cl4

28. Molecular Formula The molecular formula of a compound can be
found by determination of the number of
empirical formula units in the molecule.

29. How to Find the Molecular Formula
Determine the empirical formula of the compound.
Calculate the molar mass of the empirical formula unit.
Divide the molar mass of the compound by the molar mass of the empirical formula unit to get n, the number of empirical formula units per molecule.

30. Molecular Formula
What is the molecular formula of a compound with the empirical formula C2H5and a molar mass of g/mol?
The molar mass of the empirical formula unit is
2(12.01 g/mol C) + 5(1.008 g/mol H) =29.06 g/mol
The number of empirical formula units per molecule is
(C2H5)2 = C4H10

31. Find Molecular Formula
An unknown compound is found to be 40.0% of carbon, 6.71% of hydrogen and the remainder is oxygen. The molar mass of the compound is g/mol. Find the empirical and molecular formulas of the compound.

32. First Find Empirical Formula
Mole Formula Empirical
Element Grams Moles Ratio Ratio Formula C H
O CH2O

33. Calculate Molar Mass/Empirical Formula Mass
The molar mass of the empirical formula unit is
(12.01 g/mol C) + 1(1.008 g/mol H) + (16.00 g/mol O) = g/mol
The number of empirical formula units per molecule is
(CH2O) 6 = C6H12O6

34. Molecular Formula
Another way to find molecular formula is to consider one mole of the compound ( g of compound). The numbers of moles are also numbers of atoms in the molecule.
The masses of carbon, hydrogen and oxygen are :
40.0% x g = g of C
6.71% x g = g of H
53.3% x g = g of O

35. Molecular Formula C 72.06 6 H 12.09 12 O 96.03 6 C6H12O6
Element Grams Moles Molecular Formula C H
O C6H12O6

36. Homework
7, 15, 21, 23, 43, 57, 61, 63, 65, 68, 74.