1. Warm-Up
Which best represents the electron configuration for an atom of Mg+2?
A)1s22s22p63s2
B) 1s21p62s22p63s2
C) 1s22s22p6
D) 1s22s22p63s23p64s24d6

2. Practice
How many atoms are present in grams of Mn2(SO4)7? How many kilograms are in 3.76 x 1024 particles of Fe(OH)3?

3. How many atoms are present in 4.56 moles of Fe?
Practice
How many moles are present in x atoms of Ar?
How many atoms are present in 4.56 moles of Fe?
Convert 6.84 moles of Ca to particles.

4. Molar Volume
Molar Volume is often used to describe a gas
Gases can change volume based on temperature and pressure
STP-standard temperature and pressure
Begin 1st , 3rd, 5th, 7th

5. Molar Volume
Pressure: amount of force per unit area
STP- is a certain set of conditions
Standard Pressure is kPa or 1atm
Standard Temperature is 0C or 273 K
Begin 1st , 3rd, 5th, 7th

6. Molar Volume Continued
The volume of 1 mole at STP is 22.4 L
Therefore,
1 mole (at STP) = 22.4 L

7. 1) How many moles are in 45 L H2O at STP?
Examples
1) How many moles are in 45 L H2O at STP?
2) How many liters are in 12.3 moles of H2SO4 at STP?

8. Classwork
3) How many liters are in 1.5 x moles of O2 at STP?
4) How many particles are present in 23.4 L of a gas at STP?

9. Example
What volume is occupied by 90 g CO2 at STP?
1) Determine conversion factors
2) Use factor label

10. Gas Density and Molar Mass
Recall,
1 mole = 22.4 L
1 mole = gfm

11. 1 mole CO2 22. 4 L CO2 90 g CO2 44 g CO2 1 mole CO2 = 45.8 L CO2
1 mole CO2 = 44 g CO2
1 mole CO2 (at STP) = 22.4 L CO2
1 mole CO2
22. 4 L CO2
90 g CO2
44 g CO2
1 mole CO2
= 45.8 L CO2

12. How many grams are 3.4 x 1024 particles of SO3?
Review Questions
How many particles are in 54.0 g NH3?
How many grams are 3.4 x particles of SO3?

13. Practice
Determine the number of particles in g of H2SO4.
How many grams are present in x 1025 particles of Br2?

14. 1) How many atoms are in 19.5 moles of Ca?
Practice
1) How many atoms are in 19.5 moles of Ca?
2) How many grams are present in moles of K2S?
3). How many grams of H2SO4 are present in x 1025 atoms of H2SO4? 14

15. Empirical Formula
The empirical formula {also called the simplest formula} gives the lowest whole number ratio of atoms of the elements in a compound

16. Empirical Formula
In an empirical formula, the number of atoms do not reduce any further
The subscript used must be expressed as a whole number (no decimal places)

17. Steps to Solve
1) First determine the number of moles of each substance present
Convert the grams of each element to moles (using the gfm or molar mass)
2) Then divide each number of moles present by the smallest number of moles
3) (If Necessary) multiply the number of atoms by a number that will yield whole numbers

18. Example #1
What is the empirical formula of a compound that is 60.0 % Magnesium and 40.0% oxygen?

19. Assume that 100 g of the compound are present
Solution
Assume that 100 g of the compound are present
60.0% Magnesium= 60.0 g Magnesium
40.0 % Oxygen = 40.0 g Oxygen

20. = 2.46 moles Magnesium Solution Continued 60.0 g Magnesium
1 mole Magnesium
24.3 g Magnesium
= 2.46 moles Magnesium

21. = 2.50 moles Oxygen Solution Continued 40.0 g Oxygen 1 mole Oxygen

22. Solution Continued
3) Now divide the number of moles for each element by the smallest number of moles.
2.46 moles Magnesium
2.50 moles Oxygen

23. Solution Continued
Magnesium has the smallest number of moles {2.46 moles}
So divide the number of moles of Magnesium and Oxygen by 2.46

24. Magnesium = 2.46 moles / 2.46 moles = 1 mole Magnesium
Oxygen = moles / 2.46 moles = 1.02 moles Oxygen
Mg1O1
MgO is the empirical formula

25. Example #2
What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen?

26. Assume that you have 100 g of the compound
Solution
Assume that you have 100 g of the compound
25.9% Nitrogen = 25.9 g Nitrogen
74.1 % Oxygen = 74.1 g Oxygen

27. Convert the grams of each element to moles (using the gfm)
Solution Continued
Convert the grams of each element to moles (using the gfm)
25.9 g Nitrogen
1 mole Nitrogen
14 g Nitrogen
= 1.85 Moles Nitrogen

28. 74.1 g Oxygen 1 mole Oxygen 16 g Oxygen = 4.63 Moles Oxygen
Solution Continued
Oxygen
74.1 g Oxygen
1 mole Oxygen
16 g Oxygen
= 4.63 Moles Oxygen

29. 1.85 moles Nitrogen 4.63 moles Oxygen Solution Continued
3. Now divide the number of moles for each element by the smallest number of moles.
1.85 moles Nitrogen
4.63 moles Oxygen

30. Nitrogen has the smallest number of moles
Solution Continued
Nitrogen has the smallest number of moles
So divide the number of moles of Nitrogen and Oxygen by 1.85

31. Solution Continued
Nitrogen = 1.85 moles / 1.85 moles = 1 mole Nitrogen
Oxygen = 4.63 moles / 1.85 moles = 2.50 moles Oxygen
N1O2.5

32. Atoms can never appear in fractions or decimals
Solution Continued
Atoms can never appear in fractions or decimals
Thus, you must multiply the number of atoms by a number that will yield whole numbers
In N1O2.5, multiply the atoms by “2”
Thus N2O5is your final answer

33. Example # 3
An 8.20 g piece of magnesium combines completely with 5.40 g of oxygen to form a compound. What is the empirical formula?

34. Find the Empirical Formula of the following:
Additional Examples
Find the Empirical Formula of the following:
a % Cl, 24.27% C, 4.07% H
b % P, 56.36% O

35. Practice
Find the empirical formula of a compound that contains:27.4% Na, 1.19% H, 14.3% C, and
57.1% O