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The Mole What is a mole? • Well, yes, but we’re not discussing biology or dermatology now. • We want the CHEMIST’S mole. Relating Mass to Numbers of Atoms • Recall that we defined the amu in terms of the carbon-12 atom’s mass. Now we want to relate this to the number of atoms that corresponds to this mass. • Chemists devised a counting unit called the MOLE. You use counting units all the time— dozen, pair—to mean a specific number. • We are going to introduce the MOLE, the SI unit for amount of substance. The Mole A mole is the amount of a substance that contains as many particles as there are atoms in exactly 12 g of carbon-12. A mole is 6.022 x 1023 units of a substance. That’s 602 000 000 000 000 000 000 000 water molecules or 602 000 000 000 000 000 000 000 atoms of copper. This number is commonly known as Avogadro’s number, named for Amedeo Avogadro, the 19th Century Italian chemist. Molar Mass • The MOLAR MASS of a PURE substance is the mass of one mole (6.022 x 1023 units) of that substance. • The mass in grams of one mole of an element is equal to the atomic weight of that element. What is the molar mass of carbon? What is the atomic weight of carbon? The molar mass is 12.01 grams. 1 mol C = 12.01 g C or 12.01 g/mol C So 12.01 g C contains 1 mole of atoms. Practice Give the molar mass of (include units!!): a) Ca 40.08 g/mol 2 x 19.00g/mol = 38.00 g/mol b) F2 c) Xe 131.29 g/mol What is the mass of 2 moles of nitrogen? 2 x 14.01 g/mol = 28.02 g/mol What is the mass of 0.5 moles of rubidium? 0.5 x 85.47 g/mol = 42.74 g/mol But what about molecules? • Well, one mole of water (H2O) has 6.022 x 1023 molecules in it, because the unit here is a ‘molecule’, not an atom. • To find the molar mass of water, we need the atomic masses of the elements that make up water: hydrogen and oxygen. 1 mol H = 1.0079 g H = 1.01 g H 1 mol O = 16.00 g O Molar Mass of Water In each mole of water, there are two moles of H 2 mol H x 1.01 g H = 2.02 g H 1 mol H and one mole of O 1 mol O x 16.00 g O = 16.00 g O 1 mol O Therefore the molar mass of H2O is the sum = 2.02 g + 16.00g = 18.02 g or 18.02 g/mol H2O Molar Masses of Diatomic Molecules • What is the molar mass of O2? What is the atomic weight of oxygen? 16.00 g The subscript tells us we have two atoms of oxygen in each molecule, so the molar mass is 2 mol O x 16.00 g O = 32.00 g O 1 mol O so 1 mol O2 = 32.00 g O2 or 32.00 g/mol O2 What is the molar mass of carbon dioxide (CO2)? Start with the molar mass of each element: 1 mol C = 12.01 g C 1 mol O = 16.00 g O From the compound’s formula, we see that we need 1 mole of carbon and 2 moles of oxygen: 1 mol C x 12.01 g C = 12.01 g C 1 mol C 2 mol O x 16.00 g O = 32.00 g O 1 mol O So the molar mass of CO2 = (12.01 g + 32.00g) = 44.01 g CO2 or 44.01 g/mol CO2 Practice What is the molar mass of 1. NaCl? 2. CaCl2? 3. Fe2O3? 4. Ca3(PO4)2? Moles to Gram Conversions What is the mass in grams of 3.50 mol of the element copper, Cu? Given: 3.50 mol Cu Unknown: Mass of Cu in grams PLAN: • We need to convert the amount of Cu in moles into mass of Cu in grams. • We multiply the amount of moles by the molar mass (g/mol) to find the amount in grams. 𝑔𝑟𝑎𝑚𝑠 𝐶𝑢 𝑚𝑜𝑙𝑒𝑠 𝐶𝑢 × = 𝑔𝑟𝑎𝑚𝑠 𝐶𝑢 𝑚𝑜𝑙𝑒𝑠 𝐶𝑢 Remember, for an element, the molar mass is just the atomic mass: here 63.55 g for copper. molar mass 63.55 𝑔 𝐶𝑢 3.50 𝑚𝑜𝑙 𝐶𝑢 × = 222 𝑔 𝐶𝑢 1 𝑚𝑜𝑙 𝐶𝑢 Moles to Grams Practice Convert moles to grams: A. 2 moles Co 1 mol Co = 58.93 g Co 58.93 𝑔 𝐶𝑜 2 𝑚𝑜𝑙 𝐶𝑜 𝑥 = 117.86 g Co 1 𝑚𝑜𝑙 𝐶𝑜 B. 5 moles H2 2.016 𝑔 𝐻2 1𝑚𝑜𝑙 𝐻2 2 𝑥 1.008 𝑔 𝐻2 1 𝑚𝑜𝑙 𝐻2 2 𝑥 1.008 𝑔 = 5 𝑚𝑜𝑙 𝐻2 × = 10.08 𝑔 𝐻2 C. 2.1 moles H2O 2 x 1.008 + 16.00 = 18.02 g H2O/1 mol H2O 2.1 𝑚𝑜𝑙 𝐻2 O × 18.02 𝑔 𝐻2 𝑂 1 𝑚𝑜𝑙 𝐻2 𝑂 = 37.84 𝑔 𝐻2 O Grams to Moles Conversions A chemist produced 11.9 grams of aluminum. How many moles of aluminum is this? Given: 11.9 g Al Unknown: amount of Al in moles PLAN: mass of Al in grams → moles of Al 𝑚𝑜𝑙𝑒𝑠 𝐴𝑙 𝑔𝑟𝑎𝑚𝑠 𝐴𝑙 × = 𝑚𝑜𝑙𝑒𝑠 𝐴𝑙 𝑔𝑟𝑎𝑚𝑠 𝐴𝑙 1 𝑚𝑜𝑙 𝐴𝑙 11.9 𝑔 𝐴𝑙 × = 0.441 𝑚𝑜𝑙 𝐴𝑙 26.98 𝑔 𝐴𝑙 This is just the inverse of the molar mass! Grams to Moles Conversions Another way to view this conversion is as DIVIDING the mass of aluminum in grams by the molar mass to get the number of moles of Al: 11.9 𝑔 𝐴𝑙 = 0.441 𝑚𝑜𝑙 𝐴𝑙 26.98 𝑔 𝐴𝑙 1 𝑚𝑜𝑙 𝐴𝑙 Mathematically, it’s the same! Grams to Moles Practice Find the number of moles for these masses: A. 54.2 g N2 14.01 x 2 = 28.02 g N2/1 mol N2 54.2 𝑔 𝑁2 × 1 𝑚𝑜𝑙 𝑁2 28.02 𝑔 𝑁2 = 1.93 𝑚𝑜𝑙 𝑁2 B. 202.2 g MgO 24.30 g + 16.00 = 40.30 g MgO/1 mol MgO 202.2 𝑔 𝑀𝑔𝑂 × 1 𝑚𝑜𝑙 𝑀𝑔𝑂 40.3 𝑔 𝑀𝑔𝑂 = 5.02 𝑚𝑜𝑙 𝑀𝑔𝑂 C. 519.3 g FeCl3 55.85 + 3(35.45) = 162.2 g FeCl3/1 mol FeCl3 519.3 𝑔 𝐹𝑒𝐶𝑙3 × 1 𝑚𝑜𝑙 𝐹𝑒𝐶𝑙3 162.2 𝑔 𝐹𝑒𝐶𝑙3 = 3.20 𝑚𝑜𝑙 𝐹𝑒𝐶𝑙3 What?!? I have to enter Avogadro’s number EVERY time?!?!! Want to avoid that?? From the home screen 1) Enter Avogadro’s number into your calculator. 2) Press STO→ 3) Press ALPHA and then whatever key corresponds to the letter you want 4) ENTER To use this feature, enter the number by which you’ll be multiplying/dividing Avogadro’s number, then press ALPHA [your letter] to multiply or ÷ ALPHA [your letter] Using registers on your calculator 6.022 ×1023 𝑎𝑡𝑜𝑚𝑠 1 𝑚𝑜𝑙𝑒 For instance, 4.32 𝑚𝑜𝑙𝑒𝑠 × would be entered like this: 4.32 ALPHA Q ENTER [if I put Avogadro’s number in register Q] which equals 2.60 x 1024 [2.6E24 on your calculator] Moles to Number of Atoms Here’s where Avogadro’s number comes in handy: How many moles of silver (Ag) are in 3.01 x 1023 atoms of silver? Given: 3.01 x 1023 atoms Unknown: amount of Ag in moles 𝑚𝑜𝑙𝑒𝑠 𝐴𝑔 𝑎𝑡𝑜𝑚𝑠 𝐴𝑔 × = 𝑚𝑜𝑙𝑒𝑠 𝐴𝑔 𝐴𝑣𝑜𝑔𝑎𝑑𝑟𝑜′𝑠 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝐴𝑔 3.01𝑥1023 1 𝑚𝑜𝑙𝑒 𝐴𝑔 𝑎𝑡𝑜𝑚𝑠 𝐴𝑔 × = 0.500 𝑚𝑜𝑙𝑒𝑠 𝐴𝑔 23 6.022𝑥10 𝑎𝑡𝑜𝑚𝑠 𝐴𝑔 Moles to Number of Atoms Practice Determine how many atoms/molecules or moles in the following samples: A. 1.32 moles Si 6.022 𝑥 1023 𝑎𝑡𝑜𝑚𝑠 1.32 𝑚𝑜𝑙 𝑆𝑖 × = 7.95 × 1025 𝑎𝑡𝑜𝑚𝑠 1 𝑚𝑜𝑙 B. 7.29 mol CO2 6.022 𝑥 1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 7.29 𝑚𝑜𝑙 𝐶𝑂2 × = 4.39 × 1024 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠𝐶𝑂2 1 𝑚𝑜𝑙 C. 3.33 x 1028 formula units AuCl 3.33 × 1028 form. units AuCl × = 5.53 × 104 moles 1 𝑚𝑜𝑙 6.022 ×1023 𝑓𝑜𝑟𝑚.𝑢𝑛𝑖𝑡𝑠 Conversions Mass of element or molecule in grams 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 = × 1 𝑚𝑜𝑙 1 𝑚𝑜𝑙 × = 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 Amount of element or molecule in moles = 1 𝑚𝑜𝑙 × 6.022 𝑥 1023 𝑢𝑛𝑖𝑡𝑠 6.022 𝑥 1023 𝑢𝑛𝑖𝑡𝑠 × = 1 𝑚𝑜𝑙 Number of atoms of element or molecule How to Calculate Percent Mass • Let’s use carbon dioxide as an example. First, we need to know the molar mass of CO2: 1 mol C x12.01g/mol C+ 2 mol O x16.00g/mol O= 44.01 g/mol CO2 Second, now find the percentage of each element in the compound. Carbon: Mass of carbon % carbon = x 100% Mass of CO2 12.01 % carbon = x 100% 27.3% 44.01 Oxygen: Mass of oxygen % oxygen = x 100% Mass of CO2 32.00 % oxygen = x 100% 72.7% 44.01 NOTE: The total percent should equal 100% Practice 1. PbCl2 2. Ba(NO3)2 3. Ammonia Determining Chemical Formulae • When a new substance is discovered or synthesized, it is analyzed quantitatively to reveal its percentage compositions. • From these data, the empirical formula is then determined. An empirical formula consists of the symbols for the elements combined in a compound, with subscripts showing the smallest whole-number mole ratio of the different atoms in the compound. Empirical Formulae • For an IONIC COMPOUND, the formula unit is usually the compound’s empirical formula. • For a MOLECULAR COMPOUND, unfortunately, the empirical formula doesn’t necessarily show the actual numbers of atoms present in each molecule. For instance, the gas diborane has an empirical formula of BH3 but the molecular formula is B2H6. We will only focus on empirical formulas. Calculation of Empirical Formulae Starting with a compound’s percentage composition, you convert this to a mass composition, then a molar composition. STEP 1: Using diborane again as an example, the percentage composition is 78.1% B and 21.9% H. Using a 100.0g sample as a basis, this would give 78.1 g of B and 21.9 g of H. STEP 2: Convert these masses into moles: 1 𝑚𝑜𝑙 𝐵 78.1𝑔 𝐵 × = 7.22 𝑚𝑜𝑙 𝐵 10.81 𝑔 𝐵 1 𝑚𝑜𝑙 𝐻 21.9𝑔 𝐻 × = 21.7 𝑚𝑜𝑙 𝐻 1.01 𝑔 𝐻 Calculation of Empirical Formulae STEP 3: These values give a mole ratio of 7.22 𝑚𝑜𝑙 𝐵 ∶ 21.7 𝑚𝑜𝑙 𝐻 but this is not a ratio of smallest whole numbers. To find such a ratio, divide each number of moles by the smaller/smallest number in the existing ratios: 7.22 𝑚𝑜𝑙 𝐵 21.7 𝑚𝑜𝑙 𝐻 : = 1 𝑚𝑜𝑙 𝐵: 3.01 𝑚𝑜𝑙 𝐻 7.22 7.22 The molar ratios may not be exact whole numbers due to rounding or experimental error. You get to ignore these differences and assume they are exact whole numbers! Here “3.01” is assumed to be exactly 3! Calculation of Empirical Formulae STEP 4: Convert the mole ratio into the empirical formula 1 𝑚𝑜𝑙 𝐵: 3 𝑚𝑜𝑙 𝐻 Which means that in this compound, for every 1 mole of boron, there are three moles of hydrogen, so the empirical formula is 𝐵𝐻3 Calculation of Empirical Formulae Example 1: Quantitative analysis shows that a compound contains 32.38% sodium, 22.65% sulfur, and 44.99% oxygen. Find the empirical formula of this compound. Plan: Convert the percentage composition into mass composition (based on a 100 g sample), then convert into moles, and then find the smallest whole-number mole ratio of atoms. STEP 1: Convert % to mass: For a 100 g sample, we would have: 32.38 g Na, 22.65 g S, and 44.99 g O. For a 100 g sample, we would have: 32.38 g Na, 22.65 g S, and 44.99 g O. STEP 2: Convert these masses into moles: 1 𝑚𝑜𝑙 𝑁𝑎 32.38 𝑔 𝑁𝑎 × = 1.408 𝑚𝑜𝑙 𝑁𝑎 22.99 𝑔 𝑁𝑎 1 𝑚𝑜𝑙 𝑆 22.65 𝑔 𝑆 × = 0.7063 𝑚𝑜𝑙 𝑆 32.07 𝑔 𝑆 1 𝑚𝑜𝑙 𝑂 44.99 𝑔 𝑂 × = 2.812 𝑚𝑜𝑙 𝑂 16.00 𝑔 𝑂 STEP 3: Find the smallest whole-number ratio of atoms by dividing all by the smallest number of moles: 1.408 𝑚𝑜𝑙 𝑁𝑎 0.7063 𝑚𝑜𝑙 𝑆 2.812 𝑚𝑜𝑙 𝑂 : : 0.7063 0.7063 0.7063 1.408 𝑚𝑜𝑙 𝑁𝑎 0.7063 𝑚𝑜𝑙 𝑆 2.812 𝑚𝑜𝑙 𝑂 : : 0.7063 0.7063 0.7063 This reduces to 1.993 mol Na : 1 mol S : 3.981 mol O Rounding each number to the nearest whole number gives us 2 mol Na : 1 mol S : 4 mol O STEP 4: Write the formula. So the empirical formula of the compound is therefore Na2SO4 Calculation of Empirical Formulae Example 2: Analysis of a 10.150 g sample of a compound known to contain only phosphorus and oxygen gives a phosphorus content of 4.433 g. What is the empirical formula of this compound? Plan: Use the masses of the sample and the phosphorus content to find the mass of the oxygen content. Then convert these masses to moles, and then find the smallest whole-number mole ratio of atoms. Givens: 10.150 g sample 4.433 g phosphorus Subtracting gives an oxygen mass of 5.717 g Now convert mass to moles: 1 𝑚𝑜𝑙 𝑃 4.433 𝑔 𝑃 × = 0.1431 𝑚𝑜𝑙 𝑃 30.97 𝑔 𝑃 1 𝑚𝑜𝑙 𝑂 5.717 𝑔 𝑂 × = 0.3573 𝑚𝑜𝑙 𝑂 16.00 𝑔 𝑂 Find the smallest whole-number ratio of atoms: 0.1431 𝑚𝑜𝑙 𝑃 0.3573 𝑚𝑜𝑙 𝑂 : 0.1431 0.1431 1 𝑚𝑜𝑙 𝑃: 2.497 𝑚𝑜𝑙 𝑂 1 𝑚𝑜𝑙 𝑃: 2.497 𝑚𝑜𝑙 𝑂 Hmm, not as neat as before. We need the smallest WHOLE-number ratio here. We can round this to 1 𝑚𝑜𝑙 𝑃: 2. 5 𝑚𝑜𝑙 𝑂 but certainly not up to 3 or down to 2. Now what? Aha, if we multiply both numbers by 2, we get 2 𝑚𝑜𝑙 𝑃: 5 𝑚𝑜𝑙 𝑂 which is a whole-number ratio. The empirical formula is therefore 𝑃2 𝑂5 Consider the following … • What if you get a whole number ratio of 2.33 𝑚𝑜𝑙 𝐶: 2 𝑚𝑜𝑙 𝐻 ? You should recognize ‘.33’ as the decimal equivalent of one-third (1/3). Multiply through by 3 to get 7 𝑚𝑜𝑙 𝐶: 6 𝑚𝑜𝑙 𝐻 • What if you get a whole number ratio of 1.75 𝑚𝑜𝑙 𝐾: 3 𝑚𝑜𝑙 𝑆 ? You should recognize ‘.75’ as the decimal equivalent of three-fourths (3/4). Multiply through by 4 to get 7 𝑚𝑜𝑙 𝐾: 12 𝑚𝑜𝑙 𝑆 Molecular Formulae • We have just practiced finding empirical formulas that contain the smallest possible whole-number ratios of atoms. • The molecular formula is the actual formula of a molecular compound. The empirical formula may or may not be a correct molecular formula. • We used diborane earlier as an example of this (empirical formula of BH3 but the molecular formula is B2H6). Ethene, C2H4, and cyclopropane, C3H6, have the same atomic ratio (1C:2H) yet they are very different substances. Relationship between Empirical and Molecular Formulas 𝑥 𝑒𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 = 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 The number represented by x is a whole number indicating the factor by which subscripts in the empirical formula are multiplied to get the molecular formula. So we can find the factor x by comparing the formula masses (same as molar mass for the compound). 𝑥 𝑒𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 = 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎r 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑥= 𝑒𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 Calculation of Molecular Formula Example 3: In the last problem, the empirical formula of a compound of phosphorus and oxygen was found to be 𝑃2 𝑂5 . Experimentation shows that the molar mass of this compound is 283.89 g/mol. What is the compound’s molecular formula? Plan: Using the empirical formula, find the empirical molar mass and compare it to the molecular formula mass to find the multiplying factor. Empirical molar mass (based on 𝑃2 𝑂5 ): P has a molar mass of 30.97 g O has a molar mass of 16.00 g 30.97g/mol x 2 + 16.00 g/mol x 5 = 141.94 g/mol 𝑥 𝑒𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 = 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑥= 𝑒𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 283.89 𝑔/𝑚𝑜𝑙 𝑥= = 2.0001 141.94 𝑔/𝑚𝑜𝑙 Therefore the compound’s molecular formula is 2 × 𝑃2 𝑂5 = 𝑃4 𝑂10 Example 4: If 4.04 g of N combine with 11.46 g of O to produce a compound with the formula mass of 108.0 amu, what is the molecular formula of the compound? 1 𝑚𝑜𝑙 𝑁 4.04 𝑔 𝑁 × = 0.2884 𝑚𝑜𝑙 𝑁 14.01 𝑔 𝑁 1 𝑚𝑜𝑙 𝑂 11.46 𝑔 𝑂 × = 0.71625 𝑚𝑜𝑙 𝑂 16.00 𝑔 𝑂 0.2884 𝑚𝑜𝑙 𝑁 0.71625 𝑚𝑜𝑙 𝑂 : = 0.2884 0.2884 1 mol N: 2.4835 mol O The empirical formula is thus N2O5. The empirical formula is thus N2O5. The empirical molar mass would be: Nitrogen 14.01 g/mol Oxygen 16.00 g/mol 2 x 14.01 g/mol + 5 x 16.00 = 108.02 g/mol The empirical formula mass is 108.02 amu. 𝑥= 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑚𝑎𝑠𝑠 𝑒𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑚𝑎𝑠𝑠 = 108.0 𝑎𝑚𝑢 108.02 𝑎𝑚𝑢 = 0.9998 Therefore, the molecular formula is the same as the empirical formula, N2O5.