1. THE MOLE CHAPTER 10 Chemistry Class Mrs. Gonsalves

2. Measuring Matter The mole is the SI Base unit for measuring the amount of a substance One MOLE OF ANYTHING has 6.02 X 10 23 REPRESENTATIVE PARTICLES A representative particle is an atom, molecule, formula unit, electron or ion. This number is known as Avogadro’s number named after the Italian physicist. 1 mole is the amount of substance that contains as many particles (atoms or molecules) as there are in 12.0 g of C-12.

3. Why Moles? A). Chemical reactions might not be measurable in grams, molecules or atoms, which is why they are measured in moles. B). Moles provide a consistent method for converting between grams and atoms or molecules.

4. How Big is a Mole? ■One mole of marbles would cover the entire Earth (oceans included) for a depth of three miles. ■One mole of $100 bills stacked one on top of another would reach from the Sun to Pluto and back 7.5 million times.

5. Particles to Moles One mole = 6.02 x 10 23 representative particles Formula: moles = representative particles x 1 mole/6.02 x 10 23 molecules Example: How many moles of Magnesium is in 1.25 x 10 23 atoms of Mg? Moles of Mg = 1.25 x 10 23 atoms x 1 mole/6.02 x 10 23 atoms Moles = 0.208 mol of Mg

6. Moles to particles Formula: Representative particles = moles * 6.02 x 10 23 particles /1 mole Example: How many molecules are in 3.50 mol of sucrose? 3.50 mol Sucrose * 6.02 x 10 23 molecules/1mole = 2.11 x 10 24 molecules of sucrose

7. Mass of a Mole Atomic mass = is the atomic weight of the element AMU in grams is the mass of a mole of that element known as its’ Molar Mass Example: One mole of Carbon has a molar mass of 12g One mole of H 2 O is 2--H atoms = 2 g 1-- O atom = 16g Molar Mass = 18g

8. Mole –Mass relationship 1. Convert the mass of a substance to the moles of a substance Formula: mass(g) = number of moles x mass(g)/1 mole Example: What is the mass of 3 moles of NaCl if the molar mass is 58.5g/mol? mass of NaCl = 3 mol x 58.5g/mol = 176 g Example: How many moles are in a 10g sample of Na 2 SO 4 ? Find the molar mass of Na 2 SO 4 : Na – 2 mol x 23g/mol = 46g S – 1 mol x 32g/mol = 32g O – 4 mol x 16g/mol = 64g = 142g # of moles = 10g x 1mol/142g = 0.07 mol of Na 2 SO 4

9. Finding individual ions in compound ■# of atoms of compound x # of mol for individual ion /1 formula unit ■Example: ■ What is the the Na + ion, S ion and O ion of Na 2 SO 4 if there are 1.62 x 10 23 molecules of Na 2 SO 4 ? ■Na ion: 1.62 x 10 23 molecules of Na 2 SO 4 x 2 mol Na ion/1 mol Na 2 SO 4 = 3.24 x 10 23 molecules ■S ion: 1.62 x 10 23 molecules of Na 2 SO 4 x 1 mol Na ion/1 mol Na 2 SO 4 = 1.62 x 10 23 molecules ■O ion: 1.62 x 10 23 molecules of Na 2 SO 4 x 4 mol Na ion/1 mol Na 2 SO 4 = 6.48 x 10 23 molecules

10. Molar Mass of Compounds ■Look up the mass of each element in the chemical formula ■Then multiply by how many moles of each element ■Add the totals together ■Example: Molar mass for Ca(OH0 2 ■ 1 mol Ca x 40.08g Ca/1mol Ca = 40.08 g ■2 mol of O x 16.00 g O /1 mol O = 32.00g ■2 mol of H x 1.008 g H/1 mol H = 2.016 g ■Add the masses together: Molar Mass of Ca(OH) 2 = 74.10g/mol

11. Percent Composition ■Percent by Mass: –% of mass (element) = mass of element / mass of compound x 100 –Will always add up to 100 Percent by chemical formula: % by mass = mass of element in 1 mol of compound /molar mass of compound x 100 Example: Calculate the percent composition of propane (C 3 H 8 ) Molar mass of propane: C – 3 x 12 = 36g H – 8 x 1 = 8g = 44g/mol % of C = 36g/44g x 100 = 81.8% % of H = 8g/44g x 100 = 18%

12. Calculating Grams from percent ■Example: Calculate how much in grams of H and C are present in a 82g sample of propane. ■Use the prior calculations to determine the gram sample—81.8% of C in C 3 H 8 so in a 100 g sample of C 3 H 8 would have 81.8g of C. 82g of C 3 H 8 x 81.8g of C/100g of C 3 H 8 = 67.1g of C 82g of C 3 H 8 – 67.1g of C = 14.9g of H

13. Empirical Formula ■Empirical Formula Smallest whole-number ratio of atoms in a compound (not the what the molecule actually looks like) Example: What is the empirical formula a compound that contains 25.9% N and 74.1% O? N --- 25.9g N x 1mol/14g N = 1.85mol of N O --- 74.1g O x 1mol/16g O = 4.63mol of O Divide each by the smallest number of moles 1.85mol of N /1.85mol = 1 mol N 4.63mol of O / 1.85mol = 2.5mol of O Still not in whole numbers so: N – 1mol of N x 2 = 2mol of N O – 2.5mol of O x 2 = 5 mol of O Therefore the empirical formula for this compound is N 2 O 5

14. Molecular Formula ■Molecular Formula shows the molecule as it actually exists. Can be either the same as the empirical formula or the whole number multiple of its empirical formula ■ ■Example: A combustion analysis gives the following mass %: ■H= 9.15% C = 54.53% O = 36.32%. ■Determine the molecular formula knowing that: ■molecular mass = 132.16empirical formula = C 2 H 4 O

15. Solution to Molecular Formula ■Assume a 100g sample, which will convert the given percentages to gram amounts (9.15 gram H)/(1 gram/mole) = 9.15 moles (54.53 gram C)/(12 gram/mole) = 4.54 moles (36.32 gram O)/(16 gram/mole) = 2.27 moles ■Determination of the simplest moles ratio: ■Divide each of the moles figures by the lowest of the three (in this case 2.27). 2.27 moles O /2.27 =1 9.15 moles H/2.27 = 4 4.54 moles C/2.27 = 2 2*12.0 + 4*1.0 + 1*16.0 = 44

16. Solution to Molecular Formula ■Calculating the common factor: ■The common factor defines the ratio (molecular mass)/(empirical formula): ■132.16/ 44 = 3 ■Determination of the molecular formula : ■The solution: Multiply C 2 H 4 O by the common factor – ■-> C (2* 3 ) H (4* 3 ) O (1* 3 ) = C 6 H 12 O 3

17. Hydrates ■Is a compound that has a specific number of water molecules bound in its atoms. ■It’s chemical formula is written as the formula of the compound followed by a dot then the amount of H 2 O ■Example: ■CaCl 2 2H 2 O the name is Calcium chloride dihydrate ■Analyzing a hydrate: Heat the hydrate and remove the water (anhydrous compound)

18. Calculating a Formula for Hydrates ■Determining the formula of a hydrate you will need to know the number of moles of water associated with 1 mol of the hydrate. ■Heat the substance and calculate the difference of original sample with the heated massed sample. This lets you know the mass of water. ■Convert the masses of water and anhydrous hydrate to moles ■Then calculate the ratio x = mol of water/ mol of anhydrous hydrate ■This will let you know how many water molecules are in the hydrate.