Unit 6 Chemical Analysis. Chapter 8

Transcription

1 Unit 6 Chemical Analysis Chapter 8

2 Objectives 39 Perform calculations using the mole to calculate the molar mass 40 Perform calculations using the mole to convert between grams, number of particles, volume, and moles 41 Perform advanced calculations using empirical formula, molecular formula, and percent composition

3 The Mole Society prefers to work with simple numbers. When a very large or very small numbers are involved, society converts them to easier numbers. For example: 2 shoes = 1 pair 12 eggs = 1 dozen 10 years = 1 decade 100 years = 1 century

4 The Mole Science operates under the same idea. As that atoms are so small, their average atomic mass is very hard to work with in lab. A scientist by the name of Amedeo Avogadro proposed an idea that a given volume of gas is proportional to the number of atoms or molecules regardless of the chemical identity of the gas assuming a constant temperature and pressure were held.

5 The Mole Avogadro s idea inspired scientists to look for the exact number of atoms or molecules in a given volume. In the mid-1800s, scientists first determined the number. Over time, the number was modified slightly to be more accurate and is now: x 10 23

6 The Mole The number was called Avogadro s constant. Because the number was so large, the number was set equal to 1 mole and thus we can say: 1 mole = 6.02 x particles Often Avogadro s constant is written as 6.02 x instead of writing out all of the known numbers.

7 The Mole One of the more useful aspects of the mole is what it can do for masses. Since Avogrado s number is a proportionality factor molar mass to actual mass, it allows for conversions from amu s to grams. For instance, It is known that a proton has a mass of 1 atomic mass unit. If there were 6.02 x protons, it would have a mass of 1 gram.

8 Calculating molar mass The molar mass of a chemical looks at how much mass it would have if there was 6.02 x units of that chemical. For instance, Assume, we have 1 atom of sodium. Sodium has a mass of amu s If there were 6.02 x sodium atoms (or 1 mole) it s mass would be grams. We would say that sodium has a molar mass of grams/mole.

9 Calculating molar mass Practice The molar mass of a compound or molecule works the same way. Assume, we had carbon monoxide. If we had one molecule, there would be 1 carbon atom and 1 oxygen atom. To detemine the mass of the molecule, we would add the mass of each atom (12.01 amu amu = amu s) Now if we had 6.02 x molecules of carbon monoxide, we would have a mass of grams. Thus the molar mass of carbon monoxide is grams/mole.

10 Molar Masses Calculate the molar masses of the following: 1. Mg 2. H 2 O 3. CaCl 2 4. Mg(NO 3 ) 2

11 Molar Masses Calculate the molar masses of the following: Mg grams/mole H 2 O (1 Mg x 24.31g) grams/mole (2 H x 1.01 g) + (1 O x g) CaCl grams/ mole (1 Ca x g) + (2 Cl x g) Mg(NO 3 ) grams/mole Return (1 Mg x 24.31g) + (2 N x 14.01g) + (6 O x 16g)

12 Using the mole In science, it is rare to find exactly one mole of a substance or to have the exact molar mass. Therefore, it is necessary to be able to convert from molar mass to moles or to particles or to volume depending on what you are looking for. In order to do this, we use dimensional analysis.

13 Dimensional Analysis In Unit 1, we learned a technique for converting known as dimensional analysis. It had the basic setup shown below: given amount x unit wanted unit you have = Conversion Factor The conversion factor to remember is: 1 mole = 6.02 x particles = Molar Mass (g) = 22.4 L *Particles can stand for atoms, molecules, or formula units.

14 Using the mole To see how dimensional analysis works, assume we have 15 grams of water, and we would want to know how many moles this is. We know the given is 15 grams of water We can calculate the water has a molar mass of grams/mole (our conversion factor) Since we want moles, that unit will go on top. We have grams so that unit will go on the bottom. At this point, we can calculate. 15 g water 1 mole water x = g water 0.83 moles water

15 Using the mole As long as you remember the conversion factor and the setup for dimensional analysis, you should be able to convert. Click the link to the left for additional practice. Practice

16 Mole Conversions Convert 2.5 moles to atoms: x = Convert 24 grams CO 2 to moles: x = Convert 6.75 x 1022 molecules to moles: x = Convert 48 liters to grams: x =

17 Mole Conversions Convert 2.5 moles to atoms: 2.5 moles x 6.02 x 1023 atoms 1 mole = 1.5 x 1024 atoms Convert 24 grams CO 2 to moles: 24 g CO 2 x 1 mole CO g CO 2 = 0.55 moles CO 2 Return Convert 6.75 x molecules to moles: 6.75 x 1022 molecules x 1 mole 6.02 x 1023 molecules Convert 48 liters to grams of hydrogen gas: 48 liters x 2.02 g H liters = 4.3 g H 2 = moles

18 Chemical Analysis A large part of chemistry is analyzing unknown materials. There are several analytical instruments that help with this process that will be talked about later in the year. We are going to take a look at how to use the data these instruments provide.

19 Percent Composition In Unit 4, we discussed the relative size of atoms. When a compound is created, such as KF, the size can help us visualize what percent is potassium and percent is fluorine. In this case, it is clear that even though we have one of each atom, potassium makes up a larger percentage of this compound. K + F -

20 Percent Composition A more exact percentage can be calculated using the molar mass of the compound. KF would have a molar mass of g/mol. If we divide the mass of each element by the whole, we will get the percentage of each: g K = x 100 = 67.30% K KF g mol gf = x 100 = 32.70% F KF g mol

21 Percent Composition-Review 1. Calculate the molar mass of the compound. 2. Calculate the total mass of each element in the compound. 3. Divide the total mass of each element by the molar mass of the compound. 4. Multiply that answer by 100 Practice

22 Percent Composition Calculate the % composition of CaCl 2 : Calculate the % composition of CaSO 4 :

23 Percent Composition Return Calculate the % composition of CaCl 2 : g Ca g mol CaCl 2 = x 100 = 36.11% Ca 2 x g Cl g mol CaCl 2 = x 100 = 63.89% Cl Calculate the % composition of CaSO 4 : g Ca g mol CaCl g S g mol CaCl 2 4 x g O g mol CaCl 2 = x 100 = 29.44% Ca = x 100 = 23.55% S = x 100 = 47.01% O

24 Empirical Formula When doing a chemical analysis, the percent composition is often what is provided. The empirical formula is the lowest whole number ratio of the elements that make up the compound.

25 Empirical Formula Assume we are given the following analysis for a chemical: The chemical is 13.20% magnesium. The chemical is 86.80% bromine. We want to know the empirical formula. Once we have the % s, assume you have exactly 100. grams for the analysis. By doing this, we can easily convert the % s into grams. If I have 13.20% of 100. grams, I have grams.

26 Empirical Formulas Since we assumed we had 100. grams, We can say that we have 13.20g of Mg and 86.80g of Br. Our goal is to find a ratio of Mg to Br. Since this is looking for a number of each element, we can use the mole to determine this ratio g Mg x 1 mole Mg g Mg g Br x 1 mole Mg g Br = moles Mg = moles Br

27 Empirical Formulas Now we know how many moles of each we have: moles Mg moles Br We can now look at the ratio between the two. Since we want a whole number ratio, we need to modify our numbers. To do this, divide each by the smaller number / = 1Mg 1.086/ = 2 Br Therefore, our empirical formula would be MgBr 2.

28 Empirical Formulas-Review Practice 1. Convert the % s to grams. Assume 100 grams 2. Convert grams into moles for each element. 3. Divide each part by the smallest number from step #2. Round to nearest whole number if close i.e.: 3.99 should be written as 4 4. The answers from step #3 are the subscripts for each element.

29 Determine the empirical formulas 11.2% H and 88.8% O 36.48% Na, 25.44% S, and 38.08% O

30 Determine the empirical formulas 11.2% H and 88.8% O g H and 88.8 g O g H x 1 mole H 1.01 g H 88.8 g O 1 mole O x g O / 5.55 = 2 H 5.55 / 5.55 = 1 O = 11.1 moles H = 5.55 moles O H 2 O

31 Determine the empirical formulas 36.48% Na, 25.44% S, and 38.08% O g Na, g S and g O g Na x 1 mole Na g Na g S 1 mole S x g S g O 1 mole O x g O / = 2 Na / = 1 S / = 3 O = moles Na = moles S = moles O Return Na 2 SO 3

32 Molecular Formulas Calculating the empirical formula will always work if your compound is ionic. However, there are a few more steps to determining covalent or organic formulas from the percent composition. Here is the problem: If there was 14.4% H and 85.6% C, the empirical formula would be CH 2. Unfortunately, C 2 H 4, C 3 H 6, C 4 H 8,. will all give a percent composition of 14.4% H and 85.6 % C and they are all known molecules.

33 Molecular Formulas One other piece of information that is obtained while doing a chemical analysis is the typically the mass of the unknown compound. If we know the mass of the compound and the empirical formula, we will be able to determine the correct molecular formula.

34 Molecular Formulas For the example from before, we knew that the empirical formula was CH 2. Assume, our analysis also provided the mass of the unknown compound to be grams. If we divide the mass of the unknown by the molar mass of the empirical formula we will determine how many CH 2 s make up this molecule g g CH 2 = 3 Therefore, the compound is 3(CH 2 ). Since we do not write formulas in that way, multiple the 3 by each subscript to get C 3 H 6.

35 Molecular Formula-Review 1. Determine the empirical formula. 2. Divide the mass of the unknown by the molar mass of the empirical formula. 3. Multiple the subscripts on the empirical by the answer to step #2. Practice

36 Determine the molecular formula Empirical formula CH 3 Mass of unknown compound: grams Empirical Formula SiH 3 Cl Mass of unknown compound: grams

37 Determine the molecular formula Empirical formula CH 3 Mass of unknown compound: grams g = g CH 3 C 4 H 12 Empirical Formula SiH 3 Cl Mass of unknown compound: grams g = g SiH 3 Cl Si 3 H 9 Cl 3 Return

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