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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. The mole is also referred to as Avogadro’s number, N A 1 mole = N A = n = 6.022 x 10 23 particles

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THE MOLE “n”

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  1. The mole term is similar to the “dozen” term. Just as a dozen represents “12”; the mole represents 6.022 x 1023. A very large amount. This is due to atoms & molecules being very small. The mole is also referred to as Avogadro’s number, NA 1 mole = NA= n = 6.022 x 1023 particles Particles could be atoms, molecules, ions, electrons, even eggs. THE MOLE“n”

  2. The atomic mass (weight) is measured in reference to the mole. 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 1023 atoms = 1 molar mass THE MOLE & ATOMIC MASS

  3. 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. CALCULATIONS INVOLING THE MOLE. 6.022 x 1023 6.022 x 1023 15.999 19.00 6.022 x 1023 6.022 x 1023 2(6.022 x 1023 ) 12.044 x 1023 31.998 or 2(15.999) 38.00

  4. 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 1023 atoms/ 1 mole) = 3.72 x 1023 atoms. CALCULATIONS INVOLING THE MOLE.

  5. The molar mass of any substance is the mass in grams for one mole of that substance. H2SO4 contains 2-H atoms, 1-S atoms & 4-O atoms. Each of those atoms contribute mass to the whole compound. 2H + 1S + 4O = H2SO4 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. K3PO4 2. Ca(OH)2 3. (NH4)2SO3 4. SrCl2 . 6H2O answer on next page. MOLAR MASS FOR COMPOUNDS

  6. 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 BaCO3 weigh? MOLAR MASS FOR COMPOUNDS Answer: : 197 g/mol

  7. Dimensional Analysis Approach. How many grams of NH4NO3 are contained in 3.15 moles of NH4NO3? 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 (NH4)3PO4 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 1023 formula units/ 1 mole) (20 atoms/1 formula unit) = 5.12 x 1023 atoms. CALCULATIONS INVOLING THE MOLE for COMPOUNDS.

  8. Challenge MOLE problems. 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) (6.022x1023 K-atoms/1 mole K) = 1.540 x 10 24 K-atoms. Now since atoms of K = atoms of Na 1.540x1024 atoms Na (1 mol/ 6.02 x1023 atoms) (22.98 g/ 1 mol)=58.77 g of Na

  9. Challenge MOLE problems. A solution of sulfuric acid contained 65% H2SO4 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 H2SO4 therefore: 65 % = (x / 1560 g) 100 so x= mass of H2SO4 = 1014 g 1014 g H2SO4 (1 mole / 98.04 g) = 10.3 moles

  10. PRACTICE PROBLEM #10 301.9 g/mol 1. What is the molar mass of Fe(NO2)3.6H2O 2. How many moles are contained in 750.0 g of Fe(NO2)3? 3. How many grams of iron are contained in 0.097 mol of Fe2(CO3)3? 4. How many atoms are contained in 25.0 g of nitrogen (N2)? 5. How many grams of potassium nitrite will contain 6.78 x 1024 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? 3.868 mol 11 g 1.07 x 1024 N-atoms 957 g CaCl2 = 0.225 mol & Na2CO3 = 0.236 mol

  11. GROUP STUDY PROBLEM #10 ______1. What is the molar mass of Ba(NO3)2.4H2O a) 185.0 b) 261.0 c) 333.3 d) 1044 ______2. How many moles are contained in 54.78g of Cu(NO3)2? a) 0.2921mol b) 3.424mol c) 187.5mol d) 0.399 ______3. Which contains the larger number of molecules? a) 125.0 g HCl b) 15.0 g C6H12O6 c) 170.0 g of I2 ______4. How many grams of silver are contained in 0.754 mol of AgNO3? a) 0.488 g b) 81.3 g c) 128g d) 170g ______5. 7.0 g of nitrogen (N2) contains a) 7.0 atoms of N c) 3.0 x 1023 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?

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