The Mole. Chapter 10. Dimensional Analysis. The Mole. How much mass is in one atom of carbon-12? Molar Mass of Atoms 3/1/2015

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1 The Mole Chapter 10 1 Objectives Use the mole and molar mass to make conversions among moles, mass, and number of particles Determine the percent composition of the components of a compound Calculate empirical and molecular formulas for compounds Determine the formulas for hydrates Examine relationship between volume and amount of gas (Avogadro s Law Ch. 13.2) 2 Dimensional Analysis Conversion Factors: What is 8a*5b/2a= 20 b - Real Example: 1 mile = 5280 ft Convert 5.5 miles to feet. 5.5 miles x 5280 ft = 1 mi ft 3 The Mole Mole (mol) measures the number of particles of a substance (atom, molecule, formula unit) Using mole is just a shorthand way for saying 6.02 x particles 6.02 x particles = Avogadro s number 4 How much mass is in one atom of carbon-12? Molar Mass of Atoms Definition: g of Carbon-12 is exactly 1 mol (6.02x10 23 ) of Carbon-12 atoms. Therefore: Exactly 12 amu (by definition) 5 6 1

2 Finding Molar Mass of Other Elements All other elements are determined with respect to Carbon-12 by mass spectroscopy. Finding Molar Mass of Other Elements Another element: Fe-56 has amu So it is /12 or 4.67 x the mass of C-12 If we put all the iron isotopes together, the average Fe atom is amu Each amu has a mass of 1.66 x g, so: For example: the He-4 has 4/12 or 1/3 the mass of Carbon Finding the Molar Mass of an Element. x. = 9.27 x g/atom. x. = g/mol Fe Mass of Atoms If you have two 1 g samples of different substances, what do they have in common? 9 10 Mass of Atoms 11 The atomic mass of ONE iron atom is AMU (atomic mass units). The molar mass of (a mole of) iron is grams. 12 2

3 The number of grams in a mole is called the molar mass. Why does a mole of iron weigh more than a mole of carbon? For the same reason that a dozen bowling balls weighs more than a dozen golf balls! 13 Bowling balls are heavier than golf balls. 14 Mass of Iron on Balance g of iron (iron s molar mass) on scale is: 1 mole of iron (1 mol Fe = g Fe) 6.02 x atoms of iron A conversion you already know. How many eggs are in 14 dozen eggs? 14 dozen eggs x 12 eggs = 168 eggs 1 dozen eggs How many dozens of eggs would you need to buy if you were going to feed 504 people 1 egg each? 504 eggs x 1 dozen eggs = 42 dozen eggs 12 eggs Moles and Number of Particles Examples If have 2 mole of iron how many atoms of iron do you have? 2 moles Fe x 6.02x10 23 atoms = 12.04x10 23 atoms 1mole If have x atoms of iron how many moles of iron have you? 12.04x10 23 atoms x 1 mole = 2.0 moles 6.02x10 23 atoms Mass and Moles Examples What is the mass in grams of 2.00 moles of Cu? 2.00 moles Cu x 63.5 grams = 127 grams Cu 1mole Cu How moles is 190. g of Cu? 190. g Cu x 1 mole Cu = 3.00 mol Cu 63.5 g

4 Mass Particles Examples In order to go from mass to number of particles, you have to go through moles. How many atoms are in 46.0g of Na? 46.0 g x mol Na x Mass Particles Examples What is the mass in grams of x atoms of Al? x atoms x 2.50 mol Al x Mass (grams) Multiply by molar mass from periodic table Flowchart Divide by molar mass from periodic table Moles Divide by 6.02 X Multiply by 6.02 X Atoms or Molecules 21 Mass, Moles, and Number of Particles Practice How many moles are in 25.5 g of Ag? How many grams are mol of silicon (Si)? How many atoms are in 15.0 mol of Xe? How many moles are in 7.23 x atoms of Xe? How many atoms are in 6.50 g of B? How many grams are in 5.53 x atoms of Mg? 22 Practice Practice How many moles are in 25.5 g of Ag? How many grams are mol of silicon (Si)?

5 Practice How many atoms are in 15.0 mol of Xe? Practice How many moles are in 7.23 x atoms of Xe? Practice How many atoms are in 6.50 g of B? Practice How many grams are in 5.53 x atoms of Mg? One Mole of Four Elements One-Mole Quantities of Some Elements & Compounds One mole each of helium, sulfur, copper, and mercury. How many atoms of helium are present? Of sulfur? Of copper? Of mercury?

6 We can also calculate the molar mass of compounds like carbon dioxide. The formula for carbon dioxide is CO 2. That means there is one carbon atom and two oxygen atoms in every molecule of CO We can also calculate the molar mass of compounds like carbon dioxide. Carbon weighs grams/mole Oxygen weighs grams/mole. Therefore CO 2 has a molar mass of: C + O + O = CO = g 32 Atom Ca Cl CaCl 2 Finding Molar Mass of CaCl 2 (assume 1 mole compound) # mol Atoms in 1 mol of compound Molar Mass (g/mol) Total (g) Let s find out the molar mass of glucose (C 6 H 12 O 6 ). Use the molar masses from the periodic table: Carbon 12.0 grams/mole Hydrogen 1.0 gram/mole Oxygen 16.0 grams/mole How many grams are in a mole of glucose (C 6 H 12 O 6 )? Carbon: 6 x 12.0 g/mol = 72.0 g Hydrogen: 12 x 1.0 g/mol =12.0 g Oxygen: 6 x 16.0 g/mol = 96.0 g g/mol (For consistency, round all molar masses (elements & compounds) to 0.1 g.) Moles to Mass of Compound How many grams of KCl are in 2.30 mol of KCl? First, find molar mass. K: 1 K x 39.1 g/mol = 39.1 g Cl: 1 Cl x 35.5 g/mol = 35.5 g 74.6 g/mol

7 Moles to Mass of Compound How many grams of KCl are in 2.30 mol of KCl? # of moles Molar mass 2.30 mol KCl x 74.6 g KCl = mol KCl =171.6 g KCl = 172 g KCl (proper Sig Figs) 37 Mass to Moles in a Compound How many moles of KCl are present in grams KCl? First, determine the molar mass of KCl. (Same as before, which is 74.6 g/mol) Then, divide mass by molar mass to get moles g KCl x 1 mol =3.40 mol KCl 74.6 g KCl 38 Mass to Moles in a Compound Can determine the number of moles of each of the atoms/ions that make up the compound. Multiply by ion/compound conversion factor. Mass to Moles in a Compound How many moles of Cl - ions are there 5.5 mol of CaCl 2? 5.5 mol CaCl 2 x 2 mol Cl - = 11.0 mol Cl - 1 mol CaCl 2 Conversion factor: mole of specific atom 1 mol of compound Example: 2 mol Cl - 1 mol CaCl Mass, Moles & Particles Use the flowchart from the elemental calculations and use them for compounds. Moles are still central Get to mass (g) by multiplying by molar mass Get to # molecules (covalent compounds) or formula units (ionic compounds) by multiplying by Avogadro s number. Mass to Moles in a Compound How many moles of Na + are there in g of Na 2 CO 3? First, find Molar Mass of Na 2 CO 3 : Na: 2 x 23.0 g/mol = 46.0 g C: 1 x 12.0 g/mol = 12.0 g O: 3 x 16.0 g/mol = 48.0 g Molar Mass = g/mol

8 Mass to Moles in a Compound How many moles of Na + are there in g of Na 2 CO 3? Next, find the number of moles of Na 2 CO g x 1 mol = mol Na 2 CO g Use conversion of moles of ion/mole of compound mol Na 2 CO 3 x 2 mol Na += 10.6 mol Na + Mass to Number of Particles How many Na + ions are in g of Na 2 CO 3? Note: Looking for individual # of ions. Will be a very large number. First, find # of moles of the ion or element you are interested in. In this case it was 10.6 mol Na +. 1 mol Na 2 CO Mass to Number of Particles Finally, multiply # of mol by Avogadro s Constant mol Na + x 6.02 x ions 1 mol = 6.38 x Na + ions Mass to Number of Particles How many formula units of Na 2 CO 3 are there in the 5.30 mol of Na 2 CO 3? 5.30 mol Na 2 CO 3 x 6.02 x Frm Unts 1 mol of Na 2 CO 3 = 31.9 x formula Units Determine the # of formula units, the number of moles of each ion, and the number of each ion in: a) 2.50 mol ZnCl 2 Determine the # of formula units in 2.50 mole of ZnCl 2. b) g of Fe 2 S 3 Determine the mass in grams of 2.11x10 24 formula units of Na 2 S

9 Determine the # of moles Zn 2+ ions in 2.50 mol of ZnCl 2. Determine the # of moles Cl - ions in 2.50 mol of ZnCl Determine the # of Zn 2+ ions in 2.50 mol of ZnCl 2. Determine the # of Cl - ions in 2.50 mol of ZnCl Determine the # of formula units in in 623.7g of Fe 2 S 3. Determine the # of formula units in in 623.7g of Fe 2 S

10 Determine the # of moles of Fe 3+ ions in 623.7g of Fe 2 S 3. Determine the # of moles of S 2- ions in 623.7g of Fe 2 S Determine the # Fe 3+ ions ions in 623.7g of Fe 2 S 3. Determine the # S 2- ions in 623.7g of Fe 2 S Determine the # Fe 3+ ions and S 2- ions in 623.7g of Fe 2 S 3. Determine the mass in grams of 2.11x10 24 formula units of Na 2 S. Hint: First find Molar Mass

11 Determine the mass in grams of 2.11x10 24 formula units of Na 2 S. Ch Percent Composition Percent composition is the percent, by mass, of each element in a compound. In general, it s the mass of the element/mass of the formula: Mass of element x 100 = %mass Mass of compound of element Percent Composition Example: If a 50.0 g sample of H 2 O contains 5.6 g of H and 44.4 g of O the percent composition is: 5.6 g H x 100% = 11.1% H 50.0 g H 2 O 44.5 g O x 100% = 88.9% O 50 g H 2 O Percent Composition You can calculate the percent composition by finding the mass of each element in 1 mole of a compound. Example: Water s formula is H 2 O which means there are 2 mol of hydrogen and 1 mol oxygen in one mol of water Percent Composition So, if you have 1 mol of water, you have 18.0 g of water (molar mass) The 18.0 g of water is made up of 2 mol Hydrogen (2 g) and 1 mol oxygen (16 g). H: 2 g x 100% = 11.1% 18.0 g O: 16 g x 100% = 88.9% 18 g 65 Percent Composition Example Calculate the percent composition of each element in Ca(OH) Determine the mass of each element present in 1 mol of cmpd. Ca: 1 mol x 40.1 g/mol = 40.1 g O: 2 mol x 16.0 g/mol = 32.0 g H: 2 mol x 1.0 g/mol = 2.0 g 2. Determine mass in g of one mole of compound g 66 11

12 Percent Composition Example (Continued) 3. Calculate percentage of each element Ca: 40.1 g/74.1 g * 100 = 54.1% H: 2.0 g/74.1 g * 100 = 2.7% O: 32.0 g/74.1 g * 100 = 43.2% Percent Composition Practice Determine the % Comp of each element in: 1. KBr 2. Fe 2 O 3 3. Barium nitrate Empirical Formulas An empirical formula for a compound is the formula of a substance written with the smallest integer subscripts. In other words, the simplest mole ratios. You can use the percent composition of a compound to determine is empirical formula. Empirical Formula Steps to determine formula: 1. Consider 100g of compound 2. Convert percentages of elements to grams 3. Divide each element s respective mass by its molar mass to obtain moles 4. Divide each mole value by the smallest mole value. This give the mole ratio. 5. Multiply by appropriate number to get whole number subscripts Empirical Formula Example Determining the Empirical Formula from the Masses of Elements. We have determined the mass percentage composition of calcium chloride: 36.0% Ca and 64.0% Cl. What is the empirical formula of calcium chloride? Empirical formula Calcium Chloride (36% Calcium; 64% Chlorine) Atom Mass % In grams Molar Mass Ca Cl Moles Ratio Ratio Ca : Cl Empirical Formula

13 Empirical Formula Practice Determine the empirical formula of a compound that is 36.8% nitrogen and 63.2% oxygen. More Empirical Formula Practice Determining The Empirical Formula from Percentage Composition. (General) Benzene is a widely used industrial solvent. This compound has been analyzed and found to contain 92.26% carbon and 7.74% hydrogen by mass. What is its empirical formula? Hint: Consider a 100 g sample. The empirical formula is: Molecular Formulas Compounds with different molecular formulas can have the same empirical formula, and such substances will have the same percentage composition. Remember that the molecular formula has the actual number of atoms of each element that make one molecule of that compound. Molecular Formula from Empirical Formula The molecular formula of a compound is a multiple of its empirical formula. Molecular mass = n x empirical formula mass where n = number of empirical formula units in the molecule. Eg. Compound A = C 2 H 2 Compound B = C 6 H 6 both have the empirical formula = Molecular Formula Example Determining the Molecular Formula from the Percent Composition and Molar Mass. We have already determined the mass composition and empirical formula of benzene (CH). In a separate experiment, the molar mass of benzene was determined to be What is the molecular formula of benzene Mass of empirical formula = Molar mass benzene _ = Mass of empirical formula 77 Molecular Formula Example What is the molecular formula of a compound that has an empirical formula of CH 3 and a molar mass of 30.0 g/mol. What is the mass of each empirical unit? C: 1 x 12.0 = 12.0 H: 3 x 1.0 = How many times units? 30/15 = 2 So molecular formula is C 2 H

14 Molecular Formula Practice 1. Empirical Formula is NO 2 ; molar mass is 92.0 g/mol 2. A compound contains 26.76% C, 2.21%H, 71.17% O and has a molar mass of g/mol. Determine its molecular formula. Ch Salt Hydrates Water molecules bound to a compound For example, CaCl 2 2H 2 O Each molecule of calcium chloride has two water molecules bound to it See conceptual picture next slide CaCl 2 2H 2 O Cl - Ca +2 Cl Water Water Example of Percent Composition of a Hydrate Determine the percent salt and percent water (by mass) in 1 mol of CaCl 2 2H 2 O. 1. Find the mass of 1 mole of the compound. Ca: 1 x 40.1 = 40.1 Cl: 2 x 35.5 = 71.0 = g/mol H 2 O: 2 x 18.0 = Example of Percent Composition of a Hydrate 2. Find the percent of salt g CaCl 2 = 75.5% CaCl g CaCl 2 2H 2 O 3. Find the percent of water g H 2 O = 24.5% H 2 O g CaCl 2 2H 2 O Or you can say 100% 75.5% CaCl 2 = 24.5% H 2 O. 83 Determination of Formula Example A nickel(ii) cyanide hydrate, Ni(CN) 2 XH 2 O, contains 39.4% water by mass. What is the formula of the hydrated compound? 1. Assume 100 g of the compound. If it is 39.4% water, it is or 60.6% Ni(CN) 2. So there are 60.6 g of Ni(CN)

15 Determination of Formula Example 2. Determine the number of moles of the salt and the water g Ni(CN) 2 = 0.547mol Ni(CN) g/mol Ni: 1 x 58.7 = 58.7 C: 2 x 12.0 = 24.0 N: 2 x 14.0 = g H 2 O = 2.19 mol H 2 O 18.0 g/mol Determination of Formula Example 3. Determine the mol ratio of the water to the salt mol H 2 O mol Ni(CN) 2 = 4 mol H 2 O/mol Ni(CN) 2 So the formula is Ni(CN) 2 4H 2 O Moles and Gases (Ch ) Remember Boyle s Law, Charles s Law and Combined Gas Law? Boyle: P 1 V 1 =P 2 V 2 Charles: V 1 /T 1 =V 2 /T 2 Combined: P 1 V 1 = P 2 V 2 T 1 T 2 There is another law that relates volume to moles. The Gas Laws Avogadro s Law Amedeo Avogadro ( ) studied the relationship between volume and amount of gas The Gas Laws Avogadro s Law Avogadro s Principle: Equal volumes of gas at the same P & T contain equal numbers of particles (molecules or atoms). Avogadro s Law The volume of a gas at constant P & T is directly proportional to the number of moles of gas. Volume α n; hold P & T constant V = k x n 89 Avogadro s Principle As an extension of Avogadro s Principle, 1 mol of a gas at 0 C (273 K) and 1 atm of pressure occupies a volume of 22.4 Liters (Molar Volume) 0 C (273 K) and 1 atm of pressure are standard temperature and pressure (STP) 90 15

16 Avogadro s Principle Since volume is related to moles, if you know the volume a gas occupies at STP, you know the # of moles L 1 mol Example: If you have 11.2 L of a gas at STP, how many moles are there? 11.2 L x 1 mol = mole 22.4 L Avogadro s Principle Practice Given the conditions at STP find: Volume of mol of gas # of moles of N 2 (g) in a 2.0 L flask # of atoms of Kr in 28.5 L Ch Ideal Gas Law The gas laws that we covered all have one thing in common: they relate the volume of the gas to one of the other variables. Boyle: V 1/P Charles: V T Avogadro: V n We can put them all together to get 93 Ideal Gas Law V nt P To change from a proportionality,, to an equation, we introduce, R, the proportionality constant or Universal Gas Constant. 94 Ideal Gas Law Ideal Gas Law This makes the previous proportionality relationship into V = R*nT/P Or more commonly PV = nrt the IDEAL GAS EQUATION 95 The most common value of R is L-atm Mol-K There are other values of R with different units that are listed in your book, but we ll use the one above. Note: they all are equivalent, it just depends on what unit (mostly pressure) you are using

17 Relationship to Other Gas Laws P 1 V 1 = P 2 V 2 n 1 T 1 n 2 T 2 Both equations are equal to a constant, which is R (the gas constant) 97 Ideal Gas Equation Example A deodorant can has a volume of L and a pressure of 3.80 atm at 22ºC. How many moles of gas are contained in the can? Given: P=3.80 atm; V = L; T=22+273=295 K; R = L-atm/mol-K PV=nRT Looking for n (moles) So n = PV = (3.80 atm)(0.175 L) RT (0.0821L*atm/mol*K)*295K = mol 98 Ideal Gas Equation Practice Calculate the volume that a mol sample of gas will occupy at 265 K and atm. A 47.3 L container containing 1.62 mol of He is heated until the pressure reaches 1.85 atm. What is the temperature in ⁰C? Kr gas in a 18.5 L cylinder exerts a pressure of 8.61 atm at 24.8ºC What is the mass of Kr? 99 Gas Density and Molar Mass Let M stand for molar mass (g/mol) Where n = m/m ; m = mass in grams, n is moles And we know PV=nRT So then PV= (m/m)rt Rearranging gives: M = mrt V P Finally we get: M = drt/p Mass/volume = Density 100 Gas Density & Molar Mass Example 1 What is molar mass of a gas that has a density of 1.40 g/l at STP? Given: T=273; P=1.00 atm; d=1.40 g/l; R= L-atm/mol-K Know M = DRT/P so =(1.40 g/l)( L-atm/mol-K)(273K) 1.0 atm = 31.4 g/mol Dalton s Law of Partial Pressures Dalton s Law of Partial Pressure (end of Ch. 12.1) states that the total pressure of a mixture of gases is equal to the sum of the pressures of all the gases in the mixture. In other words P total = P 1 + P 2 + P 3 P n

18 Dalton s Law of Partial Pressures John Dalton Dalton s Law of Partial Pressures partial pressure depends only on: number of moles of gas container volume temperature of the gas. It does not depend on the identity of the gas. Because gas molecules are so far apart, they don t interact with each other Dalton s Law Example 1 We have a mixture of O 2, CO 2, and N 2 in a vessel at STP. The partial pressure of CO 2 is 0.70 atm and the partial pressure of N 2 is 0.12 atm. What is the partial pressure of O 2? Answer: What is total pressure? 1 atm, so 1.00 = P(O 2 ) P(O 2 ) = = 0.18 atm 105 Partial Pressure & Mole Fraction The partial pressure of a gas component in a mixture is dependent on how much (moles) are there: P a = n A RT/V The total pressure of a gas mixture is P total = n total RT/V 106 Partial Pressure & Mole Fraction Ratio of pressure of component to total gas pressure: = / / RT & V are constants so = Partial Pressure & Mole Fraction = This means that the partial pressure of a gas is equal to its mole fraction multiplied by the total pressure: P a = n a * P total Mole Fraction n total

19 Mole Fraction Example 1 A mixture of O 2, N 2, and He gases are in an enclosed tank. The total pressure in the tank is 12.3 atm. If there are 10.6 mol of N 2, 3.3 mol O 2 and 1.2 mol He in the tank, what is the partial pressure of each gas? 109 Mole Fraction Example 1 Find total Number of moles: = 15.1 mol gas total. For N 2 : P N2 = = 8.6 atm. For O 2 : P O2 = = 2.7 atm. For He: P He = = 1.0 atm. The sum of the pressures is 12.3 atm.110 Mole Fraction Example 2 A gas mixture containing mol He, mol Ne and mol Ar is confined to a 7.00 L vessel at 25 C. a) Calculate the partial pressure of each gas in the mixture b) Calculate the total pressure of the mixture. P(He) = mol( L-atm/mol-K)(298K)/7.0 L = 1.88 atm P(Ne) = mol( L-atm/mol-K)(298K)/7.0 L = 1.10 atm Mole Fraction Example 2 Can do previous problem because the gases are at same T & V. Therefore, their mole ratios will be proportional. Greater moles means greater partial pressure! P(Ar) = mol( L-atm/mol-K)(298K)/7.0 L = 0.36 atm Tired of moles? So am I

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