1. THE MOLE “n”  The mole term is similar to the “dozen” term.  Just as a dozen represents “12”; the mole represents 6.022 x 10 23. A very large amount.  This is due to atoms & molecules being very small. ÖThe mole is also referred to as Avogadro’s number, N A Ö1 mole = N A = n = 6.022 x 10 23 particles ÖParticles could be atoms, molecules, ions, electrons, even eggs.

2. THE MOLE & ATOMIC MASS  The atomic mass (weight) is measured in reference to the mole. MOLAR MASS  The atomic mass is also known as the MOLAR MASS.  The atomic mass = mass of an element DIVIDED by 1 mole of that element.  Atomic mass = mass/mole or m/n.  Units for atomic mass is grams per mole or g/mol. For example: H = 1.008 amu = 1.008 g/mol = 6.022 x 10 23 atoms = 1 molar mass

3. CALCULATIONS INVOLING THE MOLE. Fill in the blank. ¬A mole of oxygen atoms contain _______ atoms and weighs ______ grams. ­A mole of oxygen contains _____ molecules and ____ atoms and weighs _______ grams. A mole of fluorine atoms contain _________ atoms and weighs _____grams. ­A mole of fluorine contains __________ molecules and ________ atoms and weighs ______ grams. 6.022 x 10 23 15.999 6.022 x 10 23 2(6.022 x 10 23 ) 31.998 or 2(15.999) 6.022 x 10 23 19.00 6.022 x 10 23 12.044 x 10 23 38.00

4. CALCULATIONS INVOLING THE MOLE. Dimensional Analysis Approach. ¬How many moles of Mg are contained in 15.0 grams of Mg? Answer: 15.0 g (1 mole/24.3g) = 0.617 moles. Remember if “g” is on top originally then inside the parenthesis it must go on the bottom. ­How many atoms of Mg are contained in 15.0 grams of Mg? Answer: 15.0 g (1 mole/24.3g) = 0.617 moles. then 0.617 mol (6.022 x 10 23 atoms/ 1 mole) = 3.72 x 10 23 atoms.

5. MOLAR MASS FOR COMPOUNDS The molar mass of any substance is the mass in grams for one mole of that substance. H 2 SO 4 contains 2-H atoms, 1-S atoms & 4-O atoms. Each of those atoms contribute mass to the whole compound. 2H + 1S + 4O = H 2 SO 4 so 2(1.008 g/mol) + 1(32.06 g/mol) + 4(15.99 g/mol) = 98.04 g/mol ¬Calculate the molar mass of the following compounds. 1. K 3 PO 4 2. Ca(OH) 2 3. (NH 4 ) 2 SO 3 4. SrCl 2. 6H 2 O answer on next page.

6. MOLAR MASS FOR COMPOUNDS Answer to molar mass question. 1. 3K + P + 4O = 212 g/mol 2. Ca + 2O + 2H = 74 g/mol 3. 2N + 8H + S + 3O = 116 g/mol 4. Sr + 2Cl + 12H + 6O = 267 g/mol Word Problems: ¬How many grams does one molar mass of BaCO 3 weigh? Answer: : 197 g/mol

7. CALCULATIONS INVOLING THE MOLE for COMPOUNDS. Dimensional Analysis Approach. ¬How many grams of NH 4 NO 3 are contained in 3.15 moles of NH 4 NO 3 ? Answer: 2N + 4H + 3O = 80 g/mol 3.15 mol (80 g/mol) = 252 g Remember if “mol” is on top originally then inside the parenthesis it must go on the bottom. ­How many atoms are in 6.34 g of (NH 4 ) 3 PO 4 Answer: 3N + 12H + P + 4O = 149 /mol then 6.34 g (1 mole/149 g) = 0.0426 moles. then 0.0426 mol (6.022 x 10 23 formula units/ 1 mole) (20 atoms/1 formula unit) = 5.12 x 10 23 atoms.

8. What mass of sodium will contain the same number of atoms as 100.0 g of potassium? 100.0g K (1 mole K / 39.1 g K) (6.022x10 23 K-atoms/1 mole K) = 1.540 x 10 24 K-atoms. Now since atoms of K = atoms of Na 1.540x10 24 atoms Na (1 mol/ 6.02 x10 23 atoms) (22.98 g/ 1 mol) = 58.77 g of Na Challenge MOLE problems.

9. A solution of sulfuric acid contained 65% H 2 SO 4 by mass and had a density of 1.56 g/mL. How many moles of acid are present in 1.00 L of the solution? 1.00L (1000 mL/1L) = 1000 mL of solution dV = m 1000 mL (1.56 g/mL) =1560 g of solution but only 65% of the solution is H 2 SO 4 therefore: 65 % = (x / 1560 g) 100 so x= mass of H 2 SO 4 = 1014 g 1014 g H 2 SO 4 (1 mole / 98.04 g) = 10.3 moles

10. PRACTICE PROBLEM #10 1. What is the molar mass of Fe(NO 2 ) 3.6H 2 O 2. How many moles are contained in 750.0 g of Fe(NO 2 ) 3 ? 3. How many grams of iron are contained in 0.097 mol of Fe 2 (CO 3 ) 3 ? 4. How many atoms are contained in 25.0 g of nitrogen (N 2 )? 5. How many grams of potassium nitrite will contain 6.78 x 10 24 potassium atoms? 6. If 25.0 g of calcium chloride reacted with 25.0 g of sodium carbonate, how many moles of each will be reacting? 301.9 g/mol 3.868 mol 11 g 1.07 x 10 24 N-atoms 957 g CaCl 2 = 0.225 mol & Na 2 CO 3 = 0.236 mol

11. GROUP STUDY PROBLEM #10 ______1. What is the molar mass of Ba(NO 3 ) 2. 4H 2 O a) 185.0b) 261.0c) 333.3 d) 1044 ______2. How many moles are contained in 54.78g of Cu(NO 3 ) 2 ? a) 0.2921molb) 3.424molc) 187.5mold) 0.399 ______3. Which contains the larger number of molecules? a) 125.0 g HCl b) 15.0 g C 6 H 12 O 6 c) 170.0 g of I 2 ______4. How many grams of silver are contained in 0.754 mol of AgNO 3 ? a) 0.488 gb) 81.3 gc) 128gd) 170g ______5. 7.0 g of nitrogen (N 2 ) contains a) 7.0 atoms of N c) 3.0 x 10 23 atoms of N b) 0.25 mole of N d) 7.0 atomic masses of N ______6. How many atoms are in 56.9 g of potassium nitrite? ______7. If 16.054 g of calcium chloride reacted with 25.004 g of sodium carbonate, how many moles of each will be reacting?