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# Chapter 10 - Chemical Quantities

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### Chapter 10 - Chemical Quantities

1. 1. Chemical Quantities Or How I Learned to Love the Mole
2. 3. The Mole How do you measure matter? If it is something large like the number of beans in a bag, you just count them. If, however, it is something exceedingly small like an atom or ion, how do you count it? Counting individual particles is not possible, but you can “count” particles if you introduce a term that represents a specific number of particles. A “dozen” represents 12 of that item, and a mole represents 6.02 x 10 23 representative particles of a substance.
3. 4. Not the type of mole we are talking about
4. 5. What is meant by “representative particles”? Representative particles are whatever you are talking about: atoms, molecules, or formula units (ions). The representative particle of most elements is the atom. Seven elements exist as diatomic molecules, and, as such, its representative particle is the molecule. The seven: H 2 , N 2 , O 2 , F 2 , Cl 2 , Br 2 , I 2 .
5. 7. How much is a mole? A mole of pennies is very large.
6. 8. <ul><li>OK, so now we know that a mole is 6.02 x10 23 representative particles. So how many atoms are in one mole of an compound, or how many moles are in 6.02 x 10 23 atoms? </li></ul><ul><li>Example: </li></ul><ul><li>How many moles are in 6.02 x 10 23 atoms of silver? </li></ul><ul><li>Solution: </li></ul><ul><li>Determine the conversion factor. </li></ul><ul><li>1 mole = 6.02 x 10 23 representative particles </li></ul>1 mol representative particles 6.02 x 10 23 particles
7. 9. 6.02 x 10 23 Ag atoms x 1 mol Ag 6.02 x 10 23 Ag atoms
8. 10. To find the number of atoms in a mole of a compound, you must determine the number of atoms in a representative formula of that compound. Example: How many oxygen atoms are in a mole of CO 2 ? 1 mole CO 2 x = 12.04 x 10 23 atoms O 2 ; or, 1.204 x 10 24 atoms O 2 . 6.02 x 10 23 molecules CO 2 1 mole CO 2 2 oxygen atoms 1 molecule CO 2 x
9. 11. When discussing the mole, using atoms leads to very large numbers. An easier way to discuss moles is to work with grams of atoms instead. The gram atomic mass ( gam ) is the atomic mass of an element expressed in grams . For carbon, the gam is 12.0 g. For atomic hydrogen, the gam is 1.0 g. What is the gam for iron and mercury? (55.85 g & 200.6 g) How many atoms are contained in the gram atomic mass of an element? The gam contains one mole of atoms (6.02 x 10 23 atoms) of that element.
10. 12. Thus, if 12.0 g of carbon is the gam of carbon, 12.0 g is 1 mol of carbon, and has 6.02 x 10 23 atoms. What is the mass of a mole of a compound? To answer this you must know the formula of the compound. The formula tells you the number of atoms of each element in a representative particle of that compound. You calculate the mass of a mole of a compound by adding together the atomic masses of the atoms making up the compound. This is called gram molecular mass (gmm) Example: What is the molecular mass of SO 3 ?
11. 13. S (x1) = 32.1 g O (x3) = 16.0 g x 3 = 48.0 g SO 3 = 80.1 g What is the gmm of H 2 O 2 ? C 2 H 6 ? PCl 2 ? C 3 H 7 OH? N 2 O 5 ? As you know, not all compounds are molecules. Sodium chloride (NaCl) is an ionic compound. Its representative particle is the formula unit. The mass of 1 mol of an ionic compound is the gram formula mass (gfm) As with gmm, you simply sum the atomic masses of the atoms in the compound.
12. 14. What is the gfm of ammonium carbonate, (figure out the formula yourself :) )? N (x2) = 14.0 g x 2 = 28.0 g H (x8) = 1.0 g x 8 = 8.0 g C (x1) = 12.0 g = 12.0 g O (x3) = 16.0 g x 3 = 48.0 g = 96.0 g Review: gam = mole of element gmm = a mole of molecules gfm = a mole of formula units (also generic for gmm) All of the above can be described by the generic term “molar mass”. (NH 4 ) 2 CO 3
13. 15. Mole-Mass Relationships The gram formula mass of an element or compound is used to convert grams of a substance into moles grams A = mol A The gfm can also be used to convert moles of a substance into grams . mol B = grams B x 1 mole A gfm A x gfm B 1 mol B
14. 16. Example: How many grams are in 7.20 mol of N 2 O 3 ? N 2 = 28.0 g O 3 = 48.0 g 1 mole N 2 O 3 = 76.0 g 7.20 mol N 2 O 3 = 547.2 g N 2 O 3 = 5.47 x 10 2 g N 2 O 3 You Try It! 1. How many grams in .720 mol Be? 2. How many moles in 2.40 g N 2 ? 6.48 g 0.086 mol x 76.0 g N 2 O 3 1 mol N 2 O 3
15. 17. The Amount of A Mole of Gas We have seen that one-mole amounts of liquids or solids have different volumes than other solids or liquids. What about the volume of gas? Moles of gases have very predictable volumes. Changing temperature or pressure of a gas can vary the volume. That is why the volume of a gas is measured at a standard temperature and pressure (STP). Standard temperature is 0°C. Standard pressure is 101.2 kPa or 1 atmosphere (atm) At STP, 1 mol of any gas occupies a volume of 22.4 L.
16. 18. 22.4 L is known as the molar volume of a gas which means it contains 6.02 x 10 23 representative particles of that gas. Example: Determine the volume, in liters, of 0.600 mol of SO 2 gas at STP. 0.600 mol SO 2 = 13.4 L SO 2 Assuming STP, how many moles are in 67.2 L SO 2 ? 67.2 L SO 2 = 3 mol SO 2 x 22.4 L SO 2 1 mol SO 2 x 1 mole SO 2 22.4 L SO 2
17. 19. Review: 1. What volume, in liters, will 0.680 mol of a certain gas occupy? 2. How many moles is 1.33 x 10 4 mL of O 2 at STP?
18. 20. Mass/Density of a Gas Would 22.4 L of one gas also have the same mass as 22.4 L of another gas at STP? Probably not. A mole of one gas have a mass equal to its gfm. Different gases usually have different gfm’s. Measuring the volume of a gas is preferred to measuring mass. Knowing the volume can also help find the density of the gas. Density is found by dividing the mass of a gas by its volume. Because volume can change with a change in temperature, density is measured at STP.
19. 21. Density (at STP) Example: What is the density of oxygen gas at STP? (in grams per liter.) = 1.43 g/L P. 198 = molar mass molar volume = g/mol 22.4 L/mol = g L D = molar mass molar volume = 32 g/mol 22.4 L/mol
20. 22. Percent Composition To keep your lawn healthy, wealthy, and wise you need to use fertilizer. You can’t just use any ol’ fertilizer. You need to use one that has the right mixture of elements or compounds depending on what you need to do. You need to know the relative amount of each nutrient. This is the same in the laboratory. When you make a new compound, you need to determine its formula by finding the relative amounts of elements in the compound. The relative amounts are expressed as the percent composition, the percent by mass of each element in a compound.
21. 23. There are as many percent values as there are elements in the compound. The percentages must add up to 100%. % mass of element X = x 100% Example: An 8.20-g piece of magnesium combines with a 5.40-g sample of oxygen completely to form a compound. What is the percent composition of this compound? grams of element X grams of compound
22. 24. 1. Find mass of compound. 13.60 g = mass of compound (8.20 g + 5.40 g) 2. Find % of each element % Mg = = 60.3% % O = = = 39.7 % mass Mg mass cmpd. = 8.20 g 13.6 g mass O mass cmpd. 5.40 g 13.6 g
23. 25. Once you determine the percent composition of a compound you can determine the number of grams of an element in a specific amount of compound. Example: Calculate the mass of carbon in 82.0 g of propane, C 3 H 8 . 1. Determine % composition of C 3 H 8 . C 3 H 8 = 44.0 g C: H: 36 g 44 g = 81.8 % 8 g 44 g = 18.2 % H
24. 26. 2. Use conversion factor based on percent by mass of carbon in ethane. a. 81.8 % C means that for every 100 g C 3 H 8 , 81.8 g will be C. 82 g C 3 H 8 = 67.1 g C Check to see if 67.1/82 = 81.8 % x 81.8 g C 100 g C 3 H 8
25. 27. Problems: Calculate the amount of hydrogen in: a. 350 g C 2 H 6 . b. 20.2 g NaHSO 4 c. 124 g Ca(C 2 H 3 O 2 ) 2 d. 378 g HCN e. 100 g H 2 O
26. 28. Empirical Formulas Once you make a new compound in the laboratory, you can determine the percent composition information. Once you know the percent composition, you can determine the empirical formula of the compound. The empirical formula gives the lowest whole number ratio of the atoms of the elements in a compound. CO 2 is an empirical formula because it is the lowest whole-number ratio. N 2 H 4 (an explosive) has an empirical formula of NH 2 .
27. 29. What is the empirical formula of a compound that is 25.9% N and 74.1% O? 1. Remember % composition means that in 100 g of that compound each % equals that many grams. 2. Change g to moles 25.9 g N x = 1.85 mol N 74.1 g O x = 4.63 mol O N 1.85 O 4.63 = mole ratio Not empirical formula because needs to be whole -numbers. 1 mol N 14.0 g N 1 mol O 16.0 g O
28. 30. 3. Divide both molar values by smallest value to give you a “1” for element with smallest value. = 1 mol N = 2.50 mol O Is N 1 O 2.5 correct? No, not a whole number. Just multiply both values by a number to make a whole number. 1 x 2 = 2 mol N; 2.50 x 2 = 5 mol O Now you have whole numbers. empirical formula: N 2 O 5 1.85 mol N 1.85 4.63 mol O 1.85
29. 31. If grams are already given to you, just convert to moles. Example: Analysis of a compound indicates it contains 2.08 g K, 1.40 g Cr, and 1.74 g O. Find its empirical formula. 2.08 g K x = .053 mol K 1.40 g Cr x = .027 mol Cr 1.74 g O x = .109 mol O 1 mol K 39.1 g K 1 mol Cr 52 g Cr 1 mol O 16.00 g O
30. 32. Then divide by smallest mole number to get mole ratio. Multiply if you need to.
31. 33. Molecular Formula Determining the empirical formula does not always tell you the actual molecular formula. An example is methanol and glucose. CH 2 O = methanol (empirical and molecular) C 6 H 12 O 6 = glucose (molecular) Same empirical; different molecular formula. Well, then, how do you know if you have empirical or molecular? You need to know the molar mass of the compound.
32. 34. The molecular formula is some multiple of the empirical formula based on their masses. (simplest formula) x = molecular formula, where “x” = whole-number multiple (simplest-formula mass) x = molecular-formula mass
33. 35. Example: The simplest formula of a compound containing phosphorus and oxygen was found to be P 2 O 5 . The molar mass of this compound is 283.889 g/mol. What is the molecular formula of this compound? (simplest-formula mass) x = molecular-formula mass = 1.999 (P 2 O 5 ) 2 = molecular formula = P 4 O 10 x = molecular-formula mass simplest-formula mass = 283.889 g/mol 141.945 g/mol