The Mole the mole is the SI unit representing atomic amounts. ie. 1 dozen donuts = 12 donuts 1 pair = 2 1 year = 365 days 1 ream = 500 sheets 1 mole = 6.022 x 1023 particles

6.022 x 1023 is referred to as Avogadro’s Number. Molar Mass The mass of one mole of any substance is called the molar mass. (same value as the atomic mass) ie: Molar mass of He = Molar mass of Hg = Molar mass of Li = 4 g/mol 200.59 g/mol 6.94 g/mol

Mole Conversions x molar mass = = (1/molar mass) x x Avogadro’s # = Amount in moles Mass in grams = (1/molar mass) x x Avogadro’s # = Amount in moles Number of atoms = (1/Avogadro’s #) x

If you are given moles, you……………MOLE-TIPLY!

Examples: What is the mass in grams of 3.50 mol of the element copper? 3.50 mol Cu x (63.5 g Cu / mol Cu) = 222 g Cu A chemist produced 11.9 g of aluminum. How many moles of aluminum have been produced? 11.9 g Al x (1 mol Al/ 27.0 g Al) = 0.441 mol Al How many moles of silver are there in 3.01 x 1023 atoms? 3.01 x 1023 atoms Ag x (1 mole Ag/ 6.02 x 1023 atoms) = 0.500 mol Ag

What do I want to know? Answer: cost per party ($/party) $14.78 pizza 4 slices 15 persons X X X pizza 12 slices person party = $73.90 per party

Answer: kilograms of Ca(NO3)2? What do I want to know? Answer: kilograms of Ca(NO3)2? 164 g Ca(NO3)2 1 mol Ca(NO3)2 1 molecule Ca(NO3)2 X X 1 mol Ca(NO3)2 6.022 x 1023 molecules Ca(NO3)2 6 atoms Oxygen Did you Get it? 5.00 x 1025 atoms Oxygen 1 kg X sample 1000 g = 2.27 kg Ca(NO3)2 sample

How about How about a quiz? a quiz? Remember The Fun We Had With Mole Conversions?!?!? MOLES x molar mass x Avogadro's # / molar mass / Avogadro's # MOLECULES MASS How about a quiz? How about a quiz?

Percent Composition to find the percentage by mass of each element in a compound: Mass of element in sample of compound Total mass of compound x 100 Ex: Find the percentage composition of copper (I) sulfide. copper (I) sulfide  Cu2S % Cu = 2(63.55 amu) / 159.2 amu = 79.84 % % S = 32.07 amu / 159.2 amu = 20.16 % Ex: Calculate the percent composition of sodium nitrate. sodium nitrate  NaNO3 % Na = 23 amu / 85 amu = 27.1 % % N = 14 amu / 85 amu = 16.5 % % O = 3(16 amu) / 85 amu = 56.4%

Calculate the percent composition of silver sulfate. silver sulfate  Ag2SO4 % Ag = 69.2 % % S = 10.3 % % O = 20.5 %

Percent Error a calculation used to see how accurate your measurements are. Try this: A student measures the mass of a sample as 9.67 g. Calculate the percent error, given that the correct mass is 9.82 g. Answer: 1.5%

Percent Error Try this: A handbook gives the density of calcium as 1.54 g/cm3. What is the percent error of a density calculation of 1.25 g/cm3 based on lab measurements? Answer: 19% What is the pecent error of a length measurement of 0.229 cm if the correct value is 0.225 cm? Answer: -2%

x 100 Total mass of compound H20: 5 x 18 = 90 Cu: 1 x 63.5 S: 1 x 32.0 Mass of molecule in sample of compound Total mass of compound x 100 H20: 5 x 18 = 90 Cu: 1 x 63.5 S: 1 x 32.0 O: 4 x 16.0 = 63.5 = 32.0 = 64.0 159.5 Mass of molecule in sample of compound = 90 Total mass of compound = 159.5 + 90 = 249.5 90/249.5 = 0.3607 x 100 = 36.07%

DETERMINATION OF EMPIRICAL & CHEMICAL FORMULAS empirical formula: a formula that shows the smallest whole-number ratio of the different atoms in the compound. ex: C6H12O6  CH2O To find the empirical formula use % compositions. Ex: Find the empirical formula of a substance containing 78.1 % B and 21.9 % H. Step1: Convert percentages to grams. Assume 100g  thus, 78.1 g B and 21.9 g H

Easy right????? Step 2: Find the moles of each. B: 78.1 g / 10.81 g/mol = 7.22 mol B H: 21.9 g / 1.01 g/mol = 21.7 mol H Step 3: Divide by the lowest mole value. B: 7.22 mol / 7.22 mol = 1 H: 21.7 mol / 7.22 mol = 3.01 Step 4: Show the ratio in a formula (empirical formula) empirical formula  BH3 Easy right?????

Percent to mass Mass to mole Divide by small Multiply 'til whole!

Ex: Find the empirical formula of a compound that is 32. 38g Na, 22 Ex: Find the empirical formula of a compound that is 32.38g Na, 22.65g S, and 44.99g O. Step 1: Already have masses! Step 2: Convert to moles. Na: 32.38g / 23g/mol = 1.41 moles S: 22.65g / 32.1g/mol = 0.71 moles O: 44.99g / 16.0g/mol = 2.81 moles Step 3: Divide by the lowest mole value. Na: 1.41 moles / 0.71 moles = 1.99  2 S: 0.71 moles / 0.71 moles = 1 O: 2.81 moles / 0.71 moles = 3.96  4 Step 4: Use step 3 to make the empirical formula. Na2SO4

A compound with a mass of 10. 15g contains P and O. There is 4 A compound with a mass of 10.15g contains P and O. There is 4.43g of P, what is the empirical formula of this compound? Step 1: Find the grams of each. P: 4.43g O: 10.15g – 4.43 = 5.72g Step 2: Find the moles of each. P: 4.43g / 30.97 g/mol = 0.14 mol O: 5.72g / 16 g/mol = 0.36 mol Step 3: Find the mole ratio. P: 0.14 mol / 0.14 mol = 1 O: 0.36 mol / 0.14 mol = 2.5  Be careful here!! Need to double anything that ends in a .5 P: 2 O: 5 Step 4: Find the empirical formula. P2O5

Calculating Molecular Formulas an empirical formula may or may not be the correct molecular formula. to determine the molecular formula of a compound, you must know the compound’s formula mass. Ex: The empirical formula of a compound was found to be P2O5. Experimentation shows that the molar mass of this compound is 283.89 g/mol. What is the compound’s molecular formula? Need to know the empirical formula’s molar mass: P2O5: 2(30.97g/mol) + 5(16g/mol) = 141.94 g/mol 2. Compare the molecule’s molar mass to the empirical formula’s molar mass: (283.89 g/mol) / (141.94 g/mol) = 2 Find the molecular formula: (P2O5)2 P4O10

Ex: A sample of a compound with a formula mass of 34 Ex: A sample of a compound with a formula mass of 34.00 amu is found to consist 0.44g H and 6.92g O. Find its molecular formula. Find the empirical formula: H: 0.44g H /1.01 g/mol = 0.436 mol O: 6.92g O / 16 g/mol = 0.433 mol Divide by smallest mole amount: H: 0.436 mol / 0.433 mol = 1.007 O: 0.433 mol / 0.433 mol = 1 Empirical Formula  HO (molar mass = 16.99 g/mol) Compare e.f. to m.f. using molar masses: 34.00 g/mol / 16.99 g/mol = 2 Write molecular formula: (HO)2  H2O2

Ex: If 4.04 g N combine with 11.46 g O to produce a compound with a formula mass of 108.0 amu, what is the molecular formula of this substance? Find the empirical formula. N: 4.04g / 14.01 g/mol = 0.288 mol O: 11.46g / 15.99 g/mol = 0.717 mol Ratio of N:O N: 0.288 mol / 0.288 mol = 1 O: 0.717 mol / 0.288 mol = 2.49  2.5!! Empirical Formula N2O5 (formula mass = 107.97 amu) Compare the empirical formula mass to the molecular formula mass. 108.0 amu / 107.97 amu = 1.00 Molecular Formula  N2O5

An unknown compound is found to be 91. 8% silicon and 8. 2% hydrogen An unknown compound is found to be 91.8% silicon and 8.2% hydrogen. The molar mass of the compound is 122 g/mol. Find the molecular formula of the compound. Find the empirical formula. Si: 91.8 g / 28.09 g/mol = 3.27 mol H: 8.2 g / 1.01 g/mol = 8.1 mol Ratio of moles. Si: 3.27 mol/ 3.27 mol = 1 H: 8.1 mol/ 3.27 mol = 2.48  2.5 Si2H5 (molar mass = 61.23 g/mol) Find the molecular formula. 122 g/mol / 61.23 g/mol = 2 Si4H10