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

Practice How many atoms are present in 23.46 grams of Mn2(SO4)7? How many kilograms are in 3.76 x 1024 particles of Fe(OH)3?

How many atoms are present in 4.56 moles of Fe? Practice How many moles are present in 3.45 x 1023 atoms of Ar? How many atoms are present in 4.56 moles of Fe? Convert 6.84 moles of Ca to particles.

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 11-3-99, 3rd, 5th, 7th

Molar Volume Pressure: amount of force per unit area STP- is a certain set of conditions Standard Pressure is 101.3 kPa or 1atm Standard Temperature is 0C or 273 K Begin 1st 11-3-99, 3rd, 5th, 7th

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

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?

Classwork 3) How many liters are in 1.5 x 10-10 moles of O2 at STP? 4) How many particles are present in 23.4 L of a gas at STP?

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

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

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

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 1024 particles of SO3?

Practice Determine the number of particles in 34.5 g of H2SO4. How many grams are present in 2.34 x 1025 particles of Br2?

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 0.54 moles of K2S? 3). How many grams of H2SO4 are present in 9.524 x 1025 atoms of H2SO4? 14

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

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)

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

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

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

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

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

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

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

Magnesium = 2.46 moles / 2.46 moles = 1 mole Magnesium Oxygen = 2.50 moles / 2.46 moles = 1.02 moles Oxygen Mg1O1 MgO is the empirical formula

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

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

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

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

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

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

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

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

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?

Find the Empirical Formula of the following: Additional Examples Find the Empirical Formula of the following: a. 71.65% Cl, 24.27% C, 4.07% H b. 43.64% P, 56.36% O

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