AP Chemistry A Review of Analytical Chemistry

AP Chemistry Ch 1 (Prentice Hall)

What Temperature Do You Read? A measurement always has some amount of uncertainty To indicate the uncertainty of a single measurement scientists use a system called significant figures What temperature do you read?

Are you certain there is a 30? Are you certain about 32? Are you UNcertain about the decimal point?

Uncertainty in Measurement Consider: 32 degrees 32.5 degrees 32.3 degrees 32. 7 degrees It is important to know which digits of the reported number are uncertain The last number may be different from person to person and is a visual estimate Therefore, the third digit is an uncertain number

Reporting Measurements Unless stated otherwise, uncertainty is in the last digit 1 For example, 32.5 0.1, means the temperature ranges from 32.4 to 32.6 degrees The numbers recorded in a measurement (all the certain numbers plus the first uncertain number) are called significant figures

Uncertainty comes from limitations of the techniques used for comparison (the limitation is due to the instrument) Let s weigh the same item using three different scales Let s write down these number & we sill use them for our practice calculations

Cu = Cu = Cu =

Rules For Counting Significant Figures

1. Nonzero integers are always significant: For example: 1234 has four sig figs

2. Dealing With Zeros. There are three classes: Leading zeros never significant, for example: 0.0025 has 2 sig figs Captive zeros are significant, for example: 1.008 has 4 sig figs Trailing zeros are significant if the number has a decimal point, for example: 100 has 1 sig fig 100. Has 3 sig figs

3. Exact Numbers are numbers known with certainty and were not obtained by measuring devices: (They have an unlimited number of significant figures) They arise from definitions (the periodic table) Amu ( g/1 mole) 22.4 L/1 mole They arise from counting numbers Number of moles Number of coefficients

4. Exponents do not count as sig. figs. For example: 6.02 x 10 23 = 3 sig. figs 6.02214199 x 10 23 = 9 sig. figs

Rules for Rounding Off If the number is less than 5, the preceding digit stays the same, for example: 1.33 rounds to 1.3 If the number is equal to or greater than 5, the preceding digit is increased by 1, for example: 1.36 rounds to 1.4 (worked problems in the text show the correct number of sig figs in each step)

Calculations Using Significant Figures Calculators do not know about sig. Figs For example, try this on your calculator: 8.315 = 0.027902685 298 Answers to calculations must be rounded to the proper number of significant figures (answers must reflect the measuring device)

Multiplication and Division Using Sig Figs The answer has the same number of sig figs as the number in the problem with the least number of sig figs For example: 4.56 x 1.4 = 6.384 Final answer 6.4 Now try this rule using our three measuring devices, convert to moles:

Adding and Subtracting Using Sig Figs Go with the least number of decimals For example: 12.11 18.0 least number of decimals + 1.013 31.123 Final answer 31.1 Now let s add our measured masses together

Accuracy vs. Precision Accuracy refers to how close a measurement is to some accepted value, we ll use % error % error = exp value accepted value x 100% accepted value Precision refers to the repeatability of a measurement Using sig figs Using Absolute Deviation

Remember Scientific Notation: is a technique used to express very large or very small numbers is based on the power of 10 To use numbers written in scientific notation (calculator): Use the EE or Exp button, this means x 10 Never hit x 10 or you ll be off by 10 times!

Let's practice scientific notation...

You will need to memorize the following common SI Prefixes & conversions: nm m 1,000,000,000 nm = 1 m or 1.0 x 10 9 nm = 1 m Let's do a few... 1. 500 nm to m (color of blue) 2. 3.4 m to nm (FM radiowave)

Required way to show your work You have two jobs in this class: 1. To be able to perform the conversions 2. To be able to prove that you know why the answer is correct 3. In short, SHOW YOUR WORK!!! 4. Note the 2006 AP Exam Scoring Guidelines

Note the grading rubric & how units are given, even for multiplying

Note how this answer earned all points Note how this answer lost points due to significant figures

AP Chem Ch 4-1 (Prentice Hall)

1,000,000,000 nm = 1 m This conversion comes in handy when dealing with wavelengths with atomic spectra chemistry: Frequency (v) = c = speed of light (3 x 10 8 m/s) wavelength What s the frequency of a wavelength of light of 500 nm? The frequency of a red light is 4.74 x 10 14 calculate its wavelength. sec

AP Chemistry Solution Chemistry Ch 15.1-15.4 (Prentice Hall)

Solution: is a homogeneous mixture, where the components are uniformly intermingled. Solvent: the substance present in the largest amount (usually the component doing the dissolving). Solute: the substance present in the smaller amount (usually the component being dissolved).

Solutions can be categorized by their physical state: Solid solutions Gas solutions Liquid solutions

Solid solutions: contains two or more metals and are called alloys The more abundant element = solvent The less abundant element = solute For example steel = carbon + iron What is the solvent? What is the solute?

Gaseous solutions: contain two or more gases that do not react with each other For example: Air: 78% N 2 21% O 2 1% other gases

Liquid solutions: have a liquid solvent and the solute can consist of either a gas, liquid, or solid Aqueous solutions are special solutions that have water as the liquid solvent

You will need to memorize the following common SI Prefixes & conversions: ml L 1000 ml = 1 L Let's practice a few, use the "picket fence method" 1. 459 L to ml 2. 0.0032 ml to L

You will need to memorize the following common SI Prefixes & conversions: ml L 1000 ml = 1 L For example, this conversion comes in handy when dealing with solution chemistry: Molarity (M) = moles of solute = mol Liters of solution L 8.320g NiCl 2 added to 250 ml of water, what s the concentration? How many grams of KCl are needed to prepare 0.750 L of a 1.50 M

1000 g = 1 kg Molality (m) = moles of solute = mol Kg of solvent kg What is the molality of a solution containing 75.2 g of AgClO 4 dissolved in 885 g of benzene? What is the molality of a solid solution of 0.125 g Cr & 81.3 g of Fe?

Mass Percent: is one way of describing a solution s composition, sometimes called weight percent, (%w/w) which is the mass of a solute in a given mass of solution. Mass Percent = mass of solute x 100% mass of solution or Mass Percent = g of solute x 100% g of solute + g of solvent Does this look familiar? part/whole x 100%

Milk is a solution it is composed of a dissolved sugar called lactose. Cow s milk typically contains 4.5% by mass of lactose. Calculate the mass of lactose present in 175 g of milk. Use the formula below Mass Percent = mass of solute x 100% mass of solution

Dissolved sugar called lactose = solute Mass % = 4.5% Solution mass = 175 g of milk 4.5% = mass of solute (sugar) x 100% 175 g 7.9 g = mass of solute (sugar lactose)

Switch To Clickers A sample of brass contains 68 g copper and 7 g zinc. What is the mass percent of this solid solution, (alloy)? (remember to first identify the solute and solvent)

A sample of brass contains 68 g copper and 7 g zinc. % Cu = 68 g x 100% 68 g + 7 g = 68 g x 100% 75 g % Cu = 90.67%

10-1 Calculating AMU

Next, let s determine the mass of each atom. As we learned, there are different types of atoms known as elements. Each element has a different mass and it is called it s amu atomic mass units

Another name for amu is molar mass Molar mass is obtained by summing the masses of the component atoms.

For example: NH 3 has the following amu: N = 1 atom x 14.01 amu = 14.01 amu H = 3 atoms x 1.01 amu = 3.03 amu total amu = 17.04 amu or 17.04 molar mass

Percent Composition Ch 10-3

If we wanted to know the percent boys in the class currently, how could we do this? How about the percent girls in the class?

Correct, count the number of boys/girls (part) Count total number in the class (whole) Then divide the part over the whole and multiply by 100 % Percent composition = Part x 100% whole Does this formula make sense?

Let s apply this formula % Composition = part x 100% whole by using the formula of the compound

What is the % N to % H of Windex, which is ammonia, (NH 3 )

First let s determine the number of elements in the formula NH 3 : N:1 atom H:3 atoms

Now do part x 100% whole N = 1 atom x 14.01 amu = 14.01 x 100% 17.03 = 82.21 % N H = 3 atoms x 1.01 amu = 3.03 x 100% 17.03 = 17.79 % H

The percentages may not always total to 100% due to rounding, for example if you go to 1 decimal spot 82.2% N + 17.7% H 99.9 % total But if you go to 2 decimal spots you get closer to 100% 82.21% N + 17.79% H 100.00 % total Overall your numbers should add up close to 100%

Switch To Clickers You and your lab partner find the %Cl in this formula (CCl 2 F 2 ).

You should have calculated 58.64 % Cl

Calculating % Composition By Given Masses

http://videos.howstuffworks.com/sciencechannel/29291-100-greatest-discoveriesatomic-weight-video.htm Dalton s discovery of relative weights

According to the Law of Constant Composition, any sample of a pure compound always consists of the same elements combined in the same proportion by mass.

Percent composition can be determined of each element in a compound by its mass: % Composition = part mass x 100% whole mass This formula can be applied by using the formula of the compound or by experimental mass analysis of the compound

A sample of butane (C 4 H 10 )--lighter fluid-- contains 288 g carbon and 60 g hydrogen. Find %C and %H in butane

First find total mass of sample ( whole ) 288 g C + 60 g O = 348 g

Next find the percent of each element in the formula Use the following equation: Percent composition = Part x 100% whole 288 g C + 60 g O = 348 g Part Part whole Can you figure it out?

288 g C (C part) x 100 % = 82.8 % C 348 g (whole) 60 g O (O part) x 100% = 17.2 % O 348 g (whole) + --------------- 100.0 % Percents should add up close to 100% (It may be a little over or under, that s ok )

Switch To Clickers Now you and your lab partner try this one What is the percentage composition of a carbon and oxygen compound, that contains 40.8 g of carbon and 54.4 g of oxygen. The total mass of the compound is 95.2 g. % C =? And % O =?

You should have calculated 42.86 % C 57.14 % O

Ch 10-1 The Mole

Molar Mass The mass of one mole of atoms of any element is the molar mass which is numerically equal to the atomic mass unit (amu), but in grams Molar mass = g = 1 amu 1 mole Therefore CO 2 has an amu = 44.01 or 44.01 g 1 mol

Using The Mole Map

Converting from grams to moles

EXAMPLE: A student weighs out 88 grams of solid CO 2 (dry ice), how many moles does the student have?

First, find the amu of CO 2 : C = 1 atom x 12.01 amu = 12.01 amu O = 2 atoms x 16.00 amu = 32.00 amu CO 2 total amu = 44.01 amu CO 2 = 44g 1 mol

Then convert from grams to moles Use your mole map

Using The Mole Map

(What s Given) x 1 mole (molar mass) 88 g CO 2 x 1 mole = 2 moles CO 2 44 g

Now you try to find the moles of various compounds ALWAYS SHOW YOUR WORK!!!

Converting To Particles

A particle can be defined as: an atom, a molecule, or formula unit (ionic compounds)

Next, convert from mol to molecules: Use your mole map

(moles calculated) x (Avogadro s #) 1 mole 2 mole CO 2 x 6.02 x 10 23 molecules 1 mole = 1.204 x 10 24 molecules CO 2

Empirical Formulas and Molecular Formulas Ch 10-3

Empirical Formulas are the simplest (lowest) whole number ratio of atoms in a molecule or ionic compound For example: C 6 H 6 = CH H 2 O 2 = HO C 6 H 12 O 6 = CH 2 O

Empirical Formulas can be determined from % composition, here is the process: 1. % is the same as grams 2. Convert from grams to moles 3. Next divide by the smallest # of moles 4. this gives the empirical formula

Empirical Formula of Eugenol a Component of Clove Oil? is 73.14% C, 7.37% H and 19.49 g O Remember, % is the same as grams (g)

Empirical Formula of Eugenol, continued Next, convert to the central unit, the mole 73.14 g C x 1 mole = 6.09 mol C 12.01 g 7.37 g H x 1 mole H = 7.31 moles H 1.0079 g 19.49 g O x 1 mole O = 1.22 moles O 15.9994 g

Empirical Formula of Eugenol, continued Finally divide by the smallest # of moles 6.09 mol C 7.31 moles H 1.22 moles O 1.22 mol 1.22 mol 1.22 moles C: 4.99 H: 5.99 O: 1.00 Or C: 5 atoms H: 6 atoms O: 1.00 atoms Therefore C 5 H 6 O is Eugenol s empirical formula

Eugenol Count the number of carbon atoms, hydrogen atoms and oxygen atoms, does this fit the empirical formula that we just derived? No, because we did not find the molecular formula

Molecular Formulas Molecular formulas are also known as the true formula of a molecule. To derive this use amu: Molecular Formula = True amu empirical amu

True Formulas The molar mass of Eugenol is 164.2 g/mol, what s the molecular formula of Eugenol? Use: True amu empirical amu 164.2 g/mol = 164.2 g/mol = 2 C 5 H 6 O 82 g/mol Therefore 2(C 5 H 6 O) = C 10 H 12 O 2

Hydrated Compounds Ch 7-3 & Ch 14-3

Hydrated Compounds If ionic compounds are prepared water solutions and then isolated as solids, the crystals often have molecules of water trapped in the lattice, for example CuSO 4

CuSO 4 5 H 2 O Should you add the water when figuring the molar mass? Yes, therefore CuSO 4 5 H 2 O has an amu of 249.7 g/mol Notice the color difference of the dehydrated crystals

Hydrates A compound that is hydrated is called a hydrate, they form solids that includes water in their crystal structure Water can be driven from a hydrate to leave an anhydrous compound

Naming Hydrates To name hydrates: 1. Name the compound 2. Plus the word hydrate use prefixes to indicate how many waters are associated with the compound 3. Example: Copper (II) Sulfate pentahydrate 4. To write their formulas Write: the name of the compound number of H 2 O CuSO 4 5 H 2 O

Units of Hydration A student heats hydrated crystals of CuSO 4, how many moles of water are associated with the crystals? Step 1: Find the mass of the crystals: 1.023 g of CuSO 4 x H 2 O Step 2: Subtract the dehydrated crystal mass from the initial crystal mass = mass of water 1.023 g of CuSO 4 x H 2 O 0.654 g of CuSO 4 = 0.369 g water

Units of Hydration Continued Step 3: Determine the number of moles 0.369 g H 2 O x 1 mol 18.02 g = 0.0205 mol H 2 O 0.654 g CuSO 4 x 1 mol/159.6 g = 0.00410 mol CuSO 4 Step 4: Determine the molar ratio (see above) 0.0205 mol H 2 O 0.00410 mol CuSO 4 0.00410 mol CuSO 4 0.00410 mol CuSO 4 1 CuSO 4 5 H 2 O

Ch 11-1 Mole-Mole Relationships

Using Mole Ratios Ammonia (NH 3 ) is used in huge quantities as a fertilizer. It is manufactured by combining nitrogen and hydrogen according to the following equation: N 2 + 3 H 2 2 NH 3 How many moles of NH 3 can be made from 1.30 mol H 2?

N 2 + 3 H 2 2 NH 3 1.30mol? mol Write what s given first: 1.30 mol H 2 Next look at the balanced equation to convert from moles of one thing to moles of another: 1.30 mol H 2 x 2 mol NH 3 = 0.867 mol NH 3 3 mol H 2

http://spaceflight.nasa.gov/gallery/video/shuttle/sts-107/html/fd11.html What if you had the task of figuring out how much LiOH was needed on board a space shuttle flight? Solid LiOH is used to take out CO 2 from the shuttle s environment.

SAMPLE PROBLEM Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide from the living environment. STEP 1 Balance the equation for the reaction: LiOH + CO 2 Li 2 CO 3 + H 2 O

STEP 1 Balance the equation for the reaction: 2 LiOH + CO 2 Li 2 CO 3 + H 2 O

STEP 1 Balance the equation for the reaction: 2 LiOH + CO 2 Li 2 CO 3 + H 2 O You need 41.8 mol LiOH in order to maintain the atmosphere of the space shuttle for the astronauts How many moles of CO 2 will be taken out of the space shuttle?

Next Step Using this mole ratio, we can calculate the moles of CO 2 needed to react with the given moles of LiOH: 2 LiOH + CO 2 Li 2 CO 3 + H 2 O 41.8 mol LiOH X 1 MOL CO 2 = 20.9 mol CO 2 2 mol LiOH

What if we start off with grams?

11-2 Mass-Mass Calculations

Steps for Calculating the Masses of Reactants & Products in Chemical Reactions STEP 1 Balance the equation for the reaction. STEP 2 Convert the masses of reactants or products to moles STEP 3 Use the balanced equation to set up the appropriate mole ratio(s). STEP 4 Use the mole ratio(s) to calculate the number of moles of the desired reactant or product. STEP 5 Convert from moles back to mass.

SAMPLE PROBLEM Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide from the living environment. The products are solid lithium carbonate and liquid water. What mass of gaseous carbon dioxide can 1000 g of lithium hydroxide absorb?

What mass of gaseous carbon dioxide can 1000 g of lithium hydroxide absorb? 2 LiOH + CO 2 Li 2 CO 3 + H 2 O 1000 g? g

STEP 2 Convert the masses of reactants or products to moles: 1.00 x 10 3 g LiOH x 1 mol LiOH = 41.8 mol LiOH 23.95 g LiOH STEP 3 The appropriate mole ratio is: 2 LiOH + CO 2 Li 2 CO 3 + H 2 O 1 mol CO 2 2 mol LiOH

Step 4 Using this mole ratio, we calculate the moles of CO 2 needed to react with the given mass of LiOH: 41.8 mol LiOH X 1 MOL CO 2 = 20.9 mol CO 2 2 mol LiOH

Step 5 We calculate the mass of CO 2 by using its molar mass (44.01 g): 20.9 mol CO 2 x 44.01 g = 920. g CO 2 1 mol CO 2 Thus 1.00x10 3 g of LiOH(s) can absorb 920. g of CO 2 (g).

Now You Try! Phosphorus is placed in a flask of chlorine gas, heat and light is given off forming phosphorus trichloride: Step 1: Write and balance the Equation: P 4 + Cl 2 PCl 3 QuickTime and a Cinepak decompressor are needed to see this picture.

Step 1: Write and balance the Equation: The Balanced Equation P 4 + 6 Cl 2 4 PCl 3

Now You Try! How many grams of Cl 2 will react with 1.24 g of P 4? P 4 + 6 Cl 2 4 PCl 3 1.24 g? g

P 4 + 6 Cl 2 4 PCl 3 1.24 g? g g P 4 mol P 4 mol: mol ratio g Cl 2 1.24 g P 4 x 1 mol P 4 = 0.0100 mol P 4 123.88 g 0.0100 mol P 4 x 6 mol Cl 2 = 0.0601 mol Cl 2 1 mol P 4 0.0601 mol Cl 2 x 70.90 g Cl 2 = 4.26 g Cl 2 1 mol Cl 2

Limiting Reactants Ch 11-3

Stoichiometry: is the process of using a chemical equation (mole ratio) to calculate the masses of reactants and products in a reaction There are three parts to a reaction when using stoichiometry

The reactant that runs out first limits the amount of product that can be formed and is called the limiting reactant (or limiting reagent).

The reactant that is left over when the reactions stops is called the excess of reactant (or reagent).

A stoichiometric quantity is when reactants are mixed in exactly correct amounts so that all reactants run out at the same time. The balanced equation is the correct stoichiometric quantity nothing is left over nothing runs out first

Let s Do A Little Activity You will receive a small kit of molecular models to build. We will investigate a simple reaction. But we will build by using different mole relationships to determine the Limiting Reactant and Excess Reactant. Let s begin

1. 3 H 2 + 1 O 2 -->? H 2 O * What is the limiting and excess reactant??

1. 3 H 2 + 2 O 2 -->? H 2 O * What is the limiting and excess reactant??

1. 2 H 2 + 1 O 2 -->? H 2 O * What is the limiting and excess reactant??

Now Balance The Equation (This is the correct Stoichiometric Quantity) H 2 + O 2 --> H 2 O Are There Any Limiting or Excess Reactants? 2 H 2 + O 2 --> 2 H 2 O

Using this concept we can complete a special mass to mass calculation to figure out how much of our reactants and products we need to get the greatest amount of our product.

What is the correct ratio of Zn to HCl to produce the maximum amount of H 2? (moles of HCl will be held constant ) 7.00 g 3.27 g 1.31 g 7.00 g 3.27 g 1.31 g

0.107 mol Zn 0.100 mol HCl 0.050 mol Zn 0.100 mol HCl 0.020 mol Zn 0.100 mol HCl Excess Zinc Reactants match Excess HCl perfectly; all is consumed

Steps To Determine The Limiting 1. Compare the moles of reactant 1 to reactant 2 2. This will determine how many moles needed for reactant 2 3. Then determine how many moles you actually have for reactant 2 4. Compare the moles of needed to actual 5. This will help you to determine what is limiting and what is excess

Steps for Solving Stoichiometry Problems Involving Limiting Reactants STEP 1 Write and balance the equation for the reaction. STEP 2 Convert known masses of reactants to moles. STEP 3 Using the numbers of moles of reactants and the appropriate mole ratios, determine which reactant is limiting. STEP 4 Using the amount of the limiting reactant and the appropriate mole ratios, compute the number of moles of the desired product. STEP 5 Convert from moles of product to grams of product, using the molar mass (if this is required by the problem).

Calculate the mass of ammonia produced 25,000 g of nitrogen gas and 5000 g of hydrogen gas are mixed and reacted to form ammonia. Remember, the limiting reactant determines the amount of product formed N 2 (g) + 3 H 2 (g) 2 NH 3 (g) 25,000 g 5000 g?

N 2 (g) + 3 H 2 (g) 2 NH 3 (g) 25,000 g 5000 g What is the Limiting & Excess Reactant?

N 2 (g) + 3 H 2 (g) 2 NH 3 (g) 25,000 g 5000 g N 2 25,000 g N 2 x 1 mol N 2 28.02 g N 2 = 892 mol N 2 Mol of N 2 available H 2 5000 g H 2 x 1 mol H 2 2.016 g H 2 = 2480 mol H 2 Mol of H 2 available 892 mol N 2 x 3 mol H 2 1 mol N 2 = 2680 mol H 2 needed to react We have 2480 mol H 2 but NEED 2680 mol H 2. Therefore H 2 will run out first. N 2 is excess

Now determine the maximum amount of product produced if hydrogen is the limiting reactant Remember 2480 mol H 2 Mol of H 2 available N 2 (g) + 3 H 2 (g) 2 NH 3 (g)

Calculate the mass of ammonia produced if hydrogen is the limiting reactant 25 g of nitrogen gas and 5 g of hydrogen gas are mixed and reacted to form ammonia. Remember, the limiting reactant determines the amount of product formed N 2 (g) + 3 H 2 (g) 2 NH 3 (g) 25 g 5 g?

N 2 (g) + 3 H 2 (g) 2 NH 3 (g) 25 g 5 g (limiting)? 5 g H 2 x x 1 mol H 2 = 2.48 mol H 2 2.016 g H 2 2.48 mol H 2 x 2 mol NH 3 = 1.65 mol NH 3 3 mol H 2 1.65 mol NH 3 x 17.03 g NH 3 = 28.1 g NH 3 1 mol NH 3 Therefore: N 2 (g) + 3 H 2 (g) 2 NH 3 25 g 5 g 28.1 g NH 3

Percent Yield Chapter 11-3

Theoretical Yield The amount of product formed is controlled by the limiting reactant products stop forming when on reactant runs out. The amount of product calculated in this way is called the theoretical yield. This is the amount of product predicted from the amount of reactants used.

Actual Yield However, the amount of product predicted (the theoretical yield) is seldom obtained. One reason for this is the presence of side reactions (other reactions that consume one or more of the reactants or products). The actual yield of product, is the amount of product actually obtained.

Percent Yield The comparison of the product actually obtained and theoretically obtained is called the percent yield: Percent Yield = Actual Yield x 100% Theoretical Yield

Chapter 1 Summary