Unit 3: Holey Moley How does the mole concept illustrate constant composition and conservation of mass? How can we predict the relative amounts of substances in a chemical reaction?

Drill 1: 11/21 (A Day) 11/27 (B Day) Prepare for dimensional analysis quiz Outcome: I can convert between units of grams, liters, and number of particles using the mole.

CW 1: The Mole

CW 1: The Mole How many trunks are found in one dozen elephants? 1 elephant = 1 trunks, 12 elephants = 12 trunks How many legs are found in one dozen elephants? 1 elephant = 4 legs… So 12 elephants have 4 X 12 legs, 48 legs How many carbon atoms are found in one dozen methane (CH4) molecules? 1 methane = 1 C, 12 methanes = 12 C atoms How many hydrogen atoms are found in one dozen methane molecules? 1 methane = 4 H atoms … So 12 methanes have 12 × 4 hydrogens, 48 hydrogens

CW 1: The Mole How many trunks are found in one mole of elephants? 1 mole = 6.02x1023 elephants. Each have 1 trunk… 6.02x1023 trunks How many legs are found in one mole of elephants? 1 elephant = 4 legs. 4 legs × (6.02X1023) = 24.08x1023 legs How many carbon atoms are found in one mole of methane molecules? 1 methane = 1 C atom. 1 atom × (6.02x1023) = 6.02x1023 C atoms How many hydrogen atoms are found in one mole of methane molecules? 1 methane = 4 H atoms. 4 atoms × (6.02x1023)= 24.08x1023 H atoms

CW 1: The Mole How is “a mole” similar to “a dozen”? A dozen is always 12, like a mole is always 6.02x1023.  A mole is equal to 6.02x1023 items, which is a very large number. Why would chemists want to use moles as the unit to count atoms in? Atoms are very small, so we need a lot of them before they are measurable.

Particles = atoms, molecules, etc. Liters At standard temperature and pressure (STP) X g = 1 mol

CW 1: The Mole Find the molar mass of the following compounds. Sulfur Dioxide Lead (II) Nitrate Phosphoric Acid Ammonium Sulfate SO2 = 64.1 g/mol Pb(NO3)2 = 331.2 g/mol H3PO4 = 97.9 g/mol (NH4)2SO4 = 132.1 g/mol

CW 1: The Mole Explain in words or pictures how to convert from 1.56x1030 particles of sodium chloride to grams of sodium chloride. 1.56x1030 particles NaCl Avogadro’s Number moles NaCl Molar Mass grams NaCl

Particles = atoms, molecules, etc. Liters At standard temperature and pressure (STP) X g = 1 mol

CW 1: The Mole Explain in words or pictures how to convert from 1.56x1030 particles of sodium chloride to grams of sodium chloride. Solve. The answer is given so you can check your work. particles moles grams 1.56×10 30 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 1 × 1 𝑚𝑜𝑙 6.02×10 23 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 × 58.4 𝑔 1 𝑚𝑜𝑙 𝑁𝑎𝐶𝑙 =151335548.2 𝑔

CW 1: The Mole Question 14 Perform the following conversions. How many oxygen molecules are in 3.36 L of oxygen gas at STP? (Answer: 9.03x1021 molecules) Find the mass in grams of 2.00x1023 molecules of F2. (Answer:12.6 g) 3.36 𝐿 𝑔𝑎𝑠 1 × 1 𝑚𝑜𝑙 22.4 𝐿 × 6.02×10 23 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 1 𝑚𝑜𝑙 = 9.03×10 21 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 2.00×10 23 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝐹 2 1 × 1 𝑚𝑜𝑙 6.02×10 23 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 × 38 𝑔 1 𝑚𝑜𝑙 𝐹 2 =12.6 𝑔

CW 1: The Mole Question 14 Perform the following conversions. Determine the volume in liters occupied by 14 g of nitrogen gas at STP. (Answer: 11 L) Find the mass, in grams, of 1.00x1023 molecules of N2. (Answer: 4.65 g) 14 𝑔 𝑁 2 𝑔𝑎𝑠 1 × 1 𝑚𝑜𝑙 𝑁 2 28 𝑔 × 22.4 𝐿 1 𝑚𝑜𝑙 =11 𝐿 1.00×10 23 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑁 2 1 × 1 𝑚𝑜𝑙 6.02×10 23 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 × 28 𝑔 1 𝑚𝑜𝑙 𝑁 2 =4.65 𝑔

Summary 1: 11/21 (A Day) 11/27 (B Day) HW 1: Mole Calculations Practice Complete CW 1 8 Formal Report due on Edmodo @ 11:45 on 12.12 Outcome: I can convert between units of grams, liters, and number of particles using the mole.

Drill 2: 11/28 (A Day) 11/29 (B Day) Use the mole road map to set up (DO NOT SOLVE) the following: How many grams does 3 L of oxygen gas weigh? What is the mass of 2.5 moles of calcium chloride? Outcome: I can calculate the empirical formula from percent composition.

CW 2: Finding Empirical Formula Find the % composition of oxygen in H2O. Molar Mass of Compound: H2O Molar Mass of Element: O 𝐻:1.01×2=2.02 𝑔/𝑚𝑜𝑙 O:16.00×1=16.00 𝑔/𝑚𝑜𝑙 O:16.00×1=16.00 𝑔/𝑚𝑜𝑙 16.00+2.02=18.02 𝑔/𝑚𝑜𝑙 % 𝐶𝑜𝑚𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑂= 16.00 18.02 ×100%=88.79%

CW 2: Finding Empirical Formula Verify that the % composition given for ethyne in Table 1 is correct. Molar mass of C2H2:  2(12.01) + 2(1.01) = 26.04   Molar mass of C in C2H2:  2(12.01) = 24.02 % Composition C: Molar Mass of H in C2H2: 2(1.01) = 2.02 % Composition H: 24.02 26.04 ×100%=92.24 % 2.02 26.04 ×100%=7.76 %

CW 2: Finding Empirical Formula Fill in the missing info in Table 1. C6H6 92.26 7.74 85.63 14.37 C4H8 14.37 C8H16 85.63 14.37

CW 2: Finding Empirical Formula Can you determine the % composition by mass of H for 2-butene without using the equation given in model 1? If so, how? The total will always be 100 %. Subtract the % C from 100, what is left is % H.

CW 2: Finding Empirical Formula Compare the molecular formulas. Can you determine the molecular formula of a compound solely from its percent composition? Why or why not? No, cyclobutane, 2-butene, and 1-octene all have the same % composition, but they all have different molecular formulas.

CW 2: Finding Empirical Formula Agree or disagree: compounds with the same percent composition have the same molecular formula. Provide an example to support your position. Disagree! The formula is different (ethyne and benzene), but they do have the same ratio of elements, 1C : 1 H ratio

CW 2: Finding Empirical Formula Molecular Formula Simplest whole number ratio of elements CH3 Some multiple of the empirical formula C2H6; C3H9; etc.

CW 2: Finding Empirical Formula What feature related to the composition of a compound can be determined solely by percent composition? The simplest ratio of elements: empirical formula Determine the empirical formula of each of the molecules in Table 1. Name Molecular Formula Empirical Formula Ethyne C2H2 CH Benzene C6H6 Cyclobutane C4H8 CH2 2-butene 1-octene C8H16

CW 2: Finding Empirical Formula A molecule containing only nitrogen and oxygen contains (by mass) 36.8% N. How many grams of N would be found in a 100 g sample of the compound? How many grams of O would be found in the same sample? 36.8% of 100 g = 36.8 g N 100% - 36.8 = 63.2% O 63.2% of 100 g = 63.2 g O

CW 2: Finding Empirical Formula A molecule containing only nitrogen and oxygen contains (by mass) 36.8% N. How many moles of N would be found in a 100 g sample of the compound? How many moles of O would be found in the same sample? 36.8 𝑔 𝑁 1 × 1 𝑚𝑜𝑙 14.007𝑔 𝑁 =2.63 𝑚𝑜𝑙 𝑁 63.2 𝑔 𝑂 1 × 1 𝑚𝑜𝑙 15.999 𝑔 𝑂 =3.95 𝑚𝑜𝑙 𝑂

CW 2: Finding Empirical Formula A molecule containing only nitrogen and oxygen contains (by mass) 36.8% N. What is the ratio of the number of moles of O to the number of moles of N? What is the empirical formula of the compound? Need to make the 1.5 a whole # 2.63 𝑚𝑜𝑙 𝑁÷2.63 𝑚𝑜𝑙=1 ×2 =2 3.95 𝑚𝑜𝑙 𝑂÷2.63 𝑚𝑜𝑙=1.5 ×2 =3 𝑁 2 𝑂 3

CW 2: Finding Empirical Formula The molecule 2-hexene has the molecular formula C6H12. Refer to Table 1 and determine the percent composition of H in this molecule. Determine the percent composition of carbon in acetic acid, HC2H3O2.  Molecular Formula Empirical Formula % H C6H12 CH2 14.37 Molar Mass: 60.06 g/mol C: 12.01×2 = 24.02 g/mol (24.02/60.06)×100% = 39.99%

CW 2: Finding Empirical Formula A compound used as a dry-cleaning fluid was analyzed and found to contain 18.00% C, 2.27% H, and 79.73% Cl. Determine the empirical formula of the fluid. C2H3Cl3 18.00 𝑔 𝐶 1 × 1 𝑚𝑜𝑙 𝐶 12.011 𝑔 =1.499 𝑚𝑜𝑙÷1.499=1×2=2 2.27 𝑔 𝐻 1 × 1 𝑚𝑜𝑙 𝐻 1.008 𝑔 =2.252 𝑚𝑜𝑙÷1.499=1.5×2=3 79.73 𝑔 𝐶𝑙 1 × 1 𝑚𝑜𝑙 𝐶𝑙 35.453 𝑔 =2.249 𝑚𝑜𝑙÷1.499=1.5×2=3

CW 2: Finding Empirical Formula A compound was analyzed and found to contain 13.5 g Ca, 10.8 g O, and 0.675 g H. 13.5 𝑔 𝐶𝑎 1 × 1 𝑚𝑜𝑙 𝐶𝑎 40.08 𝑔 =0.337 𝑚𝑜𝑙÷0.337=1 CaO2H2 or Ca(OH)2 10.8 𝑔 𝑂 1 × 1 𝑚𝑜𝑙 𝑂 15.999 𝑔 =0.675 𝑚𝑜𝑙÷0.337=2 0.675 𝑔 𝐻 1 × 1 𝑚𝑜𝑙 𝐻 1.008 𝑔 =0.670 𝑚𝑜𝑙÷0.337=2

CW 2: Finding Empirical Formula The empirical formula of a compound is NO2. Its molecular mass is 92 g/mol. What is its molecular formula? Determine the molar mass of NO2. Divide the molecular mass by the molar mass. You should get a whole number. Multiply the subscripts in NO2 by the whole number you got in the last step. 𝑁:1×14.007=14.007 𝑔 14.007+31.998=46.005 𝑔/𝑚𝑜𝑙 O:2×15.999=31.998 𝑔 92 𝑔/𝑚𝑜𝑙 46.005 𝑔/𝑚𝑜𝑙 =2 N2O4

CW 2: Finding Empirical Formula The empirical formula of a compound is CH2. Its molecular mass is 70 g/mol. What is its molecular formula? 2. Divide molecular mass by molar mass 1. Find molar mass C:1×12.011=12.011 𝑔 70 𝑔/𝑚𝑜𝑙 14.027 𝑔/𝑚𝑜𝑙 =5 H:2×1.008=2.016 𝑔 12.011+2.016=14.027 𝑔/𝑚𝑜𝑙 3. Multiply each subscript by the number found above. C5H10

CW 3: Finding Empirical Formula A compound is found to be 40.0% carbon, 6.7% hydrogen and 53.5% oxygen. Its molecular mass is 60 g/mol. What is the empirical formula? What is the molecular formula? 40.0 𝑔 𝐶 1 × 1 𝑚𝑜𝑙 𝐶𝑎 12.011 𝑔 =3.33 𝑚𝑜𝑙÷3.33=1 6.75 𝑔 𝐻 1 × 1 𝑚𝑜𝑙 𝐻 1.008 𝑔 =6.69 𝑚𝑜𝑙÷3.33=2 CH2O 53.5 𝑔 𝑂 1 × 1 𝑚𝑜𝑙 𝑂 15.999 𝑔 =3.34 𝑚𝑜𝑙÷3.33=1 Molar Mass CH2O: 30.026 g/mol 60 𝑔/𝑚𝑜𝑙 30.026 𝑔/𝑚𝑜𝑙 =2 C2H4O2

Summary 2: 11/28 (A Day) 11/29 (B Day) Complete CW 1 to 3 if not done 8 Formal Report due on Edmodo @ 11:45 on 12.12 HW 2: The Strange Case of Mole Airlines Outcome: I can calculate the empirical formula from percent composition.