Chapter 3 Topic Formula weights 3.4 The mole

Chapter 3 Topic 6 3.3 Formula weights 3.4 The mole
3.5 Calculating empirical and molecular formulas Combustion analysis

3.3 Formula Weights Formula weights tell us the mass of a molecule of that substance For example, Sulfuric acid is a highly corrosive compound found in car batteries. It has the formula H2SO4. It contains 2 hydrogen, one sulfur, and 4 oxygen atoms Thus, its formula mass is the combined atomic masses of 2 hydrogen atoms = 2(1.0) = 2.0 1 sulfur atom = 1(32.0) = 32.1 Plus 4 oxygen atoms = 4(16.0) = 98.1 Notice that in each case, the atomic masses have been rounded off to the nearest tenth. This seems to give a good balance of precision AND convenience

( ) (1.0)

Practice problems 1.0 + 14.0 + 3(16.0) = 63.0 40.1 + 2(35.5) = 110.1
Calculate the formula mass of each: HNO3 CaCl2 Cu(NO3)2 4. C6H12O6 (16.0) = 63.0 (35.5) = 110.1 [ (16.0)] = 187.6 6(12.0) + 12(1.0) +6(16.0) = 180.0

One use of formula mass is calculating Percentage composition
Percent composition is the percent by mass of each element in a chemical compound. For example, sulfur dioxide, SO2,, a byproduct of coal burning, is a serious air pollutant and component of acid rain. It is made of molecules with one sulfur and two oxygen atoms. Although 1/3rd (33%) of the atoms are sulfur and 2/3rds (67%) of the atoms are oxygen, the percentage by mass depends on the atomic masses of the individual elements in a compound. One sulfur atom has an atomic mass of 32.0 u’s (atomic mass units) While two oxygen atoms have atomic masses of 2 x 16.0 u Since the mass of sulfur is equal to the mass of oxygen, in sulfur dioxide, its composition is 50% Sulfur and 50% oxygen. How much sulfur enters the atmosphere, for every 100 kilograms of SO2 that is expelled from a coal burning plant? Right, 50 kilograms of sulfur, since SO2 is 50% sulfur by mass!

A percent (literally per hundred) is a ratio which tells us what mass of a given element is present in every 100 grams of that substance. 50% sulfur means 50 grams of sulfur in every 100 grams of sulfur dioxide To calculate percents we must know the mass of the individual element and the mass of the whole compound. Here is the mathematical formula for percent by mass composition: Finding the formula weight for sulfur dioxide: SO2 = u + 2(16.0 u ) = u And then applying the formula: % by mass of sulfur = (1 S atom)(32.0 u) x 100 64.0 u = x 100 64.0 = x 100 = 50 % Sulfur

In Regents chem we call the steps “ACE”: Analyze (and plan)
Calculate (solve with a setup) Evaluate (check for reasonable-ness)

Analyze (and Plan): Percentage by mass is calculated through a simple formula
Masses of individual elements are obtained from the periodic table Setup and calculate: % by mass of nitrogen = (2 N atoms)(14.0 u) x 100 [ (16.0)] = x 100 164.0 = x 100 Evaluate: the percent of Nitrogen is less than 100 and 17% is reasonable for 28 grams of nitrogen out of 164 grams total.

Regents level: 32:2(16) (2)127 2(12):6(1) 24 : 6

40/74 x 100

Stoichiometry Stoichiometry: quantitative information available from chemical formulas and equations Formulas show numerical composition of substances and equations are recipes for reactions…if you know how to read them. © 2012 Pearson Education, Inc.

Quantities Quantities can be represented by mass (weight) ex: 3 lbs of coffee volume ½ gallon of milk number 24 eggs gross units 2 dozen (2 units of 12) Notice: Gross units are developed for convenience of grouping small items into common units © 2012 Pearson Education, Inc.

Chemical Quantities Chemical quantities can be represented by mass in grams volume in liters or milliliters number particles (atoms, molecules, etc.) gross units are moles of particles Moles are the gross unit used for convenience in chemistry. 1 mole = 6.02 x 1023 particles 1 mole of sulfur 6.02 x 1023 S atoms 32 grams This is “Avogadro’s” number 1 mole of carbon 6.02 x 1023 C atoms 12 grams How can this strange number possibly be convenient? A simple analogy may help. Press onward! © 2012 Pearson Education, Inc.

Counting Oreos Lets say you have a cookie factory where Oreos are assembled 2 cookie Halves combine with 1 cream center to make 1 cookie Cookie halves weigh 1 oz each, while creams weigh 3 oz

The recipe for an Oreo cookie is ridiculously simple.
2 Halves combine with one Cream to form the Oreo. As an equation recipe that’s 2 H + 1 C  1 H2C Since Halves weigh 1 oz, while creams weight 3 oz we could make our recipe represent weight: 2 ounces of halves + 3 ounces of creams  5 ounces of cookies Of course if we would only produce one cookie. We’ve got to think bigger!

Since cookie halves weigh less then creams we will assign them a relative mass of 1 cookie mass unit (cmu’s) . Creams are 3 times heavier than cookies so we will assign them a relative mass of 3 cmu’s A cookie made of 2 halves + 1 cream would weight of course 5 cmu’s

2 H + 1 C  1 H2C If we change our mass units to pounds, our recipe becomes more reasonable. Now 2 cmu’s of halves + 3 cmu’s of creams becomes… 2 lb of halves + 3 lb of creams = 5 lb of cookies. Since there are 16 ounces in 1 pound, how many cookies will you make? 16 cookies!

2 H + 1 C  1 H2C Still thinking too small? How about tons?
Now 2 cmu’s of halves + 3 cmu’s of creams becomes… 2 tons of halves + 3 tons of creams = 5 tons of cookies. Since there are 32,000 ounces in 1 ton, how many cookies will you make? 32,000 cookies! Now that’s more like it

This is great if you work in a cookie factory, but it is of course is an analogy
to atoms and molecules:

Counting atoms Lets say you have a laboratory where molecules are assembled 2 H atoms combine with 1 O atom to make 1 water molecule H atoms weigh 1 amu each, while O atoms weigh 16 amu

The recipe for a water molecule is ridiculously simple
The recipe for a water molecule is ridiculously simple. 2 H atoms combine with one O atom to form the water molecule. As an equation recipe that’s 2 H + 1 O 1 H2O Since H’s weigh 1 amu, while O atoms weight 16 amu we could make our recipe represent weight: 2 amu’s of H atoms + 16 amu’s of O atoms  18 amu’s of water Of course if we would only produce one water molecule. We’ve got to think bigger!

Since H atoms weigh less then O atoms we will assign them a relative mass of 1.
O atoms are 16 times heavier than H atoms so we will assign them a relative mass of 16

2 H + 1 O  1 H2O If we change our mass units to grams, our recipe becomes more reasonable. Now 2 mass units of H atoms mass units of O atoms becomes… 2 gram of Hydrogen + 16 grams of Oxygen = 18 grams of water Since there are 6.02 x 1023 amu’s in one gram. We’ve created a recipe for a quantity we call a “mole” Aka One “gram” relative mass of water

are grouped into groups called “moles”
Atoms and molecules are grouped into groups called “moles” Latin molis meaning “pile” Read on to see how we use the mole.

Counting atoms: The Mole
Here’s the deal: 1 lb = 16 ounces but… 1 gram is equal to 6.02 x 1023 atomic mass units Got it? That’s all there is too it. That’s how amu’s are related to grams. Atoms are mighty tiny so amu’s are a tiny fraction of a gram. Good. Now: A proton weighs 1 amu right? …so 6.02 x 1023 protons = 6.02 x 1023 u or 1 gram. OK? A typical Hydrogen atom also weighs 1 amu (1 proton, 1 electron, 0 neutrons…check out the periodic table and you’ll see) …so 6.02 x 1023 H atoms = 6.02 x 1023 u or 1 gram also. OK? Just like the oreos, the number stays the same; only the units change!

Think about it…..A hydrogen atom weighs 1 amu, and 6.02 x 1023 of them weighs 1 gram. Same number. Different unit. (1 amu vs. 1 gram) 1 atom vs. 1 mole (pile) of atoms Here’s another: He weighs 4 u …so 6.02 x 1023 He atoms = 4 u/atom x 6.02 x atoms…. … x 1023 u…..which is 4 grams. Or: H2 molecules = 2 u …so x 1023 H2 molecules = 2 x x 1023 = etc.….which is 2 grams. © 2012 Pearson Education, Inc.

6 x 1023 – a mole of them would weigh 32 g
Sulfur atoms weigh 32. 6 x 1023 – a mole of them would weigh 32 g Twice as much as the same number of carbon atoms. Carbon atoms weigh 12 u So 6 x 1023 – one mole of them would weigh 12 g Half as much as the same number of Sulfur atoms. Notice also that the pile (mole) of sulfur is twice as large as the mole pile of carbon. Is this to be expected? Yes, heavier atoms usually take up more space also!

See? If we know the mass of one particle, we know the mass of one mole (the gross unit: 6.02 x 1023 particles) It works the same for molecules: For water: An H2O molecule = 2 H atoms + 1 O atom = 2(1) + 16 = 18u so….6.02 x 1023 ie. 1 mole = 18 grams 18 grams = 6.02 x 1023 molecules © 2012 Pearson Education, Inc.

Get the picture? x 1023 (the gross unit) of anything is one mole If we know the mass of one particle (atom, molecule, etc.) …. One mole, 6.02 x 1023 of them, has the same mass, but the unit is now in grams. The mass of one atom is called the “atomic mass”. ….The mass of one mole is called the “gram atomic mass” The mass of one molecule is called the “molecular mass”. ….The mass of one mole is called ….the “gram molecular mass”. A different kind of mole For simplicity, a one mole quantity of anything is sometimes called a “molar mass” Mole comes from the latin Molis meaning mass, heap or pile. Think this might be useful? Press on! © 2012 Pearson Education, Inc.

Why is this useful? A simple application might suffice. Suppose we wish to make water. H2O. I know that I need 2 H atoms for every O atom. Right? H2O? H’s weigh 1u, O’s weight 16u’s If I weigh out 16 grams of Oxygen, I have 6.02 x 1023 O’s Or one mole of oxygen. I’ll need twice as many H’s: x 1023 Or two moles of H. OK? …Since H atoms weighs 1 u, one mole weighs 1 gram. but I need two moles, so I need 2 grams of H. Make sense? So, If I start with 16 grams of oxygen (1 mole O), I’ll need 2 grams of hydrogen (2 moles H) By the way, how much water will I make? 18 grams (2 g H +16 g O), ….which is 1 mole of water. To summarize: 2 H’s + 1 O = 1 H2O 2 moles H + 1 mole O = 1 mole H2O 2 (1g) (16 g) = 18 g The mole is a way to get the right number of atoms by measuring the only way that we can – using mass and weighing things out. Cool huh? © 2012 Pearson Education, Inc.

Try some mental math mole problems:
What is the atomic mass of sulfur? (look it up!) What is the molar mass of sulfur? What is the mass of one mole of S? How many atoms are in 32 g of S? How many atoms are in 64 g? How many moles is 16 g of S? How many atoms is 16 grams of S? What is the mass of 18 x 1023 S atoms? 32 32 g 32 g – yeah, it means the same thing 6.02 x 1023 atoms (they’re really small remember?) 12.04 x 1023 atoms – do you see 2 moles? 0.5 mole or ½ mole 3.01 x 1023 atoms Do you see 3 moles? 3(32) = 96 g Now we have a connection between our big “gram” world ….and the little world of atoms.

Let’s try a few more… What is molecular mass of H2O? the molar mass? What is the mass of one mole of H2O? How many molecules are in 1 mole of H2O? So, How many molecules are in 18 g of H2O? How many moles is 36 g of H2O? So, How many molecules are in 36 g? How many molecule is 9 grams of H2O What is the mass of 18 x 1023 H2O molecules? 2(1) + 16 = 18 18 g 18 g – yeah, same like before 6.02 x 1023 g 6.02 x 1023 g yeah, its still one mole 2 moles 12.04 x 1023 molecules 3.01 x 1023 molecules Notice 3 moles? = 3 x 18 g/mole = 54 g So you need to be able to calculate the formula masses for compounds easily. © 2012 Pearson Education, Inc.

Mole conversion map There’s also a connection to gas volume 1 mole = 22.4 liters But only under conditions of standard 1 atmosphere of pressure. We’ll work this one in a later chapter So we can easily convert from mass to particles by using moles As our gross unit and molar mass. © 2012 Pearson Education, Inc.

Conversions (dimensional analysis)
Since we know the relationship between a mole and its mass, particles and volume (if it’s a gas) we can start to convert from unit to unit. 1 mole = 6.02 x 1023 Particles 1 mole = gram formula mass 1 mole = 22.4 liter of gas (under standard conditions) Ex: What is the mass of 1.5 moles of H2O? Unit Plan: moles  grams 1 mole H2O = 2(1) + 16 = 18 grams 18 1.5 moles x ___gram mole = 27 grams 1 © 2012 Pearson Education, Inc.

problems What is the mass of 3.5 moles of Cu? How many moles are 16.8 L of CO2 gas ?(at STP) 3. How many molecules in 66 grams of CO2? 3.5 moles Cu x 63.5 grams Cu 1 mole Cu = 220 grams Cu 16.8 Liters CO2 x mole CO2 22.4 Liters CO2 = 0.75 moles CO2 Unit plan: Grams  moles  molecules x x 1023 molecules 1 mole CO2 66 grams CO2 x 1 mole CO2 44 grams CO2 = 9.0 x 1023 molecules © 2012 Pearson Education, Inc.

How many H atoms in 1.5 moles of H2O molecules? Known: 1 H2O = 2 H atoms or 1 mole H2O = 2 moles H and 1 mole = 6.0 x 1023 particles Unit plan: moles H2O  moles H  H atoms Setup: 2 6.0 x 1023 1.5 moles H2O x _moles H mole H2O x __________H atoms mole H 1 1 = 18 x 1023 H atoms © 2012 Pearson Education, Inc.

Molecules of gases are far apart so they take up more space Chlorine atoms are much larger than O atoms making the molar mass and volume of NaCl a bit larger than the same 6 x 1023 / 1 mole sample of water. Figure 3.10 One mole each of a solid (NaCl), a liquid (H2O), and a gas (O2). © 2012 Pearson Education, Inc.

One mole is always the same number of molecules 6.02 x 1023 Since glucose molecules have so many atoms they’re Obviously heavier, and one mole of glucose would weigh more. Both one mole samples have 6.02 x 1023 molecules though. Answer: D © 2012 Pearson Education, Inc.

These textbook problems should be easy now. Sample 3.7 Sample Exercise 3.7 Estimating Numbers of Atoms Without using a calculator, arrange these samples in order of increasing numbers of carbon atoms: 12 g 12C, 1 mol C2H2, 9  1023 molecules of CO2. Solution Analyze We are given amounts of three substances expressed in grams, moles, and number of molecules and asked to arrange the samples in order of increasing numbers of C atoms. Plan To determine the number of C atoms in each sample, we must convert 12 g 12C, 1 mol C2H2,and 9  1023 molecules CO2 to numbers of C atoms. To make these conversions, we use the definition of mole and Avogadro’s number. © 2012 Pearson Education, Inc. 43

Solve One mole is defined as the amount of matter that contains as many units of the matter as there are C atoms in exactly 12 g of 12C. Thus, 12 g 12C of Contains 1 mol of C atoms = 6.02  1023 C atoms. One mol of C2H2 contains 6  1023 C2H2 molecules. Because there are two C atoms in each molecule, this sample contains 12  1023 C atoms. Because each CO2 molecule contains one C atom, the CO2 sample contains 9  1023 C atoms. Hence, the order is 12 g 12C (6  1023 C atoms) < 9  1023 CO2 molecules (9  1023 C atoms) < 1 mol C2H2 (12  1023 C atoms). © 2012 Pearson Education, Inc.

Sample Exercise 3.7 Estimating Numbers of Atoms Continued Practice Exercise Without using a calculator, arrange these samples in order of increasing number of O atoms: 1 mol H2O, 1 mol CO2, 3  1023 molecules O3. Answer: 1 mol H2O (6  1023 O atoms) < 3  1023 molecules O3 (9  1023 O atoms) < 1 mol CO2 (12  1023 O atoms) © 2012 Pearson Education, Inc. 45

Sample 3.8 Sample Exercise 3.8 Converting Moles to Number of Atoms Calculate the number of H atoms in mol of C6H12O6. Solution Analyze We are given the amount of a substance (0.350 mol) and its chemical formula C6H12O6. The unknown is the number of H atoms in the sample. Plan Avogadro’s number provides the conversion factor between number of moles of C6H12O6 and number of molecules of C6H12O6: 1 mol C6H12O6 = 6.02  1023 molecules C6H12O6. Once we know the number of molecules of C6H12O6, we can use the chemical formula, which tells us that each molecule of C6H12O6 contains 12 H atoms. Thus, we convert moles of C6H12O6 to molecules of C6H12O6 and then determine the number of atoms of H from the number of molecules of C6H12O6: Moles C6H12O molecules C6H12O atoms H © 2012 Pearson Education, Inc. 46

Check We can do a ballpark calculation: First, 0.35(6  1023) is about 2  1023 molecules of C6H12O6, and each one of these molecules contains 12 H atoms. So 12(2  1023) gives 24  1023 = 2.4  1024 H atoms, which agrees with our result. Because we were asked for the number of H atoms, the units of our answer are correct. The given data had three significant figures, so our Answer has three significant figures. Practice Exercise How many oxygen atoms are in (a) 0.25 mol Ca(NO3)2 and (b) 1.50 mol of sodium carbonate? Answer: (a) 9.0  1023, (b) 2.71  1024 © 2012 Pearson Education, Inc.

Sample Exercise 3.9 Calculating Molar Mass What is the molar mass of glucose, C6H12O6? Solution Analyze We are given a chemical formula and asked to determine its molar mass. Plan Because the molar mass of any substance is numerically equal to its formula weight, we first determine the formula weight of glucose by adding the atomic weights of its component atoms. The formula weight will have units of amu, whereas the molar mass has units of g/mol. Solve Our first step is to determine the formula weight of glucose: 6 C atoms = 6(12.0 amu) = amu 12 H atoms = 12(1.0 amu) = amu 6 O atoms = 6(16.0 amu) = amu 180.0 amu Because glucose has a formula weight of amu, 1 mol of this substance (6.02  1023 molecules) has a mass of g. In other words, C6H12O6 has a molar mass of g/mol. Check A magnitude below 250 seems reasonable based on the earlier examples we have encountered, and grams per mole is the appropriate unit for the molar mass. Comment Glucose, also known as blood sugar, is found in nature in honey and fruits. Other sugars used as food are converted into glucose in the stomach or liver before the body uses them as energy sources. Because glucose requires no conversion, it is often given intravenously to patients who need immediate nourishment. © 2012 Pearson Education, Inc. 48

Sample 3.10 Sample Exercise 3.10 Converting Grams to Moles Calculate the number of moles of glucose (C6H12O6) in g of C6H12O6. Solution Analyze We are given the number of grams of a substance and its chemical formula and asked to calculate the number of moles. Plan The molar mass of a substance provides the factor for converting grams to moles. The molar mass of C6H12O6 is g/mol (Sample Exercise 3.9). Solve Using 1 mol C6H12O6 = g C6H12O6 to write the appropriate conversion factor, we have Check Because g is less than the molar mass, an answer less than one mole is reasonable. The unit mol is appropriate. The original data had four significant figures, so our answer has four significant figures. Practice Exercise How many moles of sodium bicarbonate (NaHCO3) are in 508 g of NaHCO3? Answer: 6.05 mol NaHCO3 © 2012 Pearson Education, Inc. 49

Sample Exercise 3.11 Converting Moles to Grams Calculate the mass, in grams, of mol of calcium nitrate. Solution Analyze We are given the number of moles and the name of a substance and asked to calculate the number of grams in the sample. Plan To convert moles to grams, we need the molar mass, which we can calculate using the chemical formula and atomic weights. Solve Because the calcium ion is Ca2+ and the nitrate ion is NO3-, calcium nitrate is Ca(NO3)2. Adding the atomic weights of the elements in the compound gives a formula weight of amu. Using 1 mol Ca(NO3)2 = g Ca(NO3)2 to write the appropriate conversion factor, we have Check The number of moles is less than 1, so the number of grams must be less than the molar mass, g. Using rounded numbers to estimate, we have 0.5  150 = 75g, which means the magnitude of our answer is reasonable. Both the units (g) and the number of significant figures (3) are correct. Practice Exercise What is the mass, in grams, of (a) 6.33 mol of NaHCO3 and (b) 3.0  10-5 mol of sulfuric acid? Answer: (a) 532 g, (b) 2.9  10-3 g © 2012 Pearson Education, Inc. 50

Sample Exercise 3.12 Calculating Numbers of Molecules and Atoms from Mass (a) How many glucose molecules are in 5.23 g of C6H12O6? (b) How many oxygen atoms are in this sample? Solution Analyze We are given the number of grams and the chemical formula and asked to calculate (a) the number of molecules and (b) the number of O atoms in the sample. (a) Plan The strategy for determining the number of molecules in a given quantity of a substance is summarized in Figure We must convert 5.23 g to moles of C6H12O6 and then convert moles to molecules of C6H12O6. The first conversion uses the molar mass of C6H12O6, g, and the second conversion uses Avogadro’s number. Solve Molecules C6H12O6. Check Because the mass we began with is less than a mole, there should be fewer than 6.02  1023 molecules in the sample, which means the magnitude of our answer is reasonable. We can make a ballpark estimate of the answer: 5/200 = 2.5  mol; 2.5  10-2  6  1023 = 15  = 1.5  1022 molecules. The units (molecules) and significant figures (three) are appropriate. FIGURE 3.12 Procedure for interconverting mass and number of formula units. The number of moles of the substance is central to the calculation. Thus, the mole concept can be thought of as the bridge between the mass of a sample in grams and the number of formula Units contained in the sample. Sample 3.12 © 2012 Pearson Education, Inc. 51

Sample Exercise 3.12 Calculating Numbers of Molecules and Atoms from Mass Continued (b) Plan To determine the number of O atoms, we use the fact that there are six O atoms in each C6H12O6 molecule. Thus, multiplying the number of molecules we calculated in (a) by the factor (6 atoms O/1 molecule C6H12O6) gives the number of O atoms. Solve Check The answer is 6 times as large as the answer to part (a), exactly what it should be. The number of significant figures (three) and the units (atoms O) are correct. Practice Exercise (a) How many nitric acid molecules are in 4.20 g of HNO3? (b) How many O atoms are in this sample? Answer: (a) 4.01  1022 molecules HNO3, (b) 1.20  1023 atoms O © 2012 Pearson Education, Inc. 52

Calculating Empirical Formulas
One can calculate the empirical formula from the percent composition. For example, we find in experiments that water is 11.1% hydrogen and 88.9% oxygen (a 1:8 ratio). How do we find water’s chemical formula: H2O (2:1), from these percents? Recognize firstly, that percentage is by mass, But the empirical formula shows ratios by number To solve, we will convert mass percents into moles and look for the # ratio to use for the subscripts in the chemical formula. © 2012 Pearson Education, Inc.

In this simple example, we can examine the composition of water: Water is 11.1% hydrogen and 88.9% oxygen by mass. A theoretical 100 grams sample would thus contain 11.1 grams of Hydrogen and 88.9 grams of oxygen. Next we can use their respective molar masses to convert from mass to moles: 11.1 g H x 1 mole = 11.0 mol H g O x 1 mole = 5.6 mol O 1.01 g g 11.1 H : 5.6 O ratio is a 2:1 ratio – that is 2 H atoms for every O atom Inserting this ratio into the formula produces H2O1 or H2O (Its not exactly 2:1 since 11.1/5.6 = 1.98 since we’re using rounded numbers. ) This whole thing should make sense since Oxygen’s mass is 16 times heavier than hydrogen, but there are two Hydrogen atoms in a water molecule – so the mass ratio is 2:16 that is 1:8. (See it? 11.1: 88.9?) © 2012 Pearson Education, Inc.

For a more complex example from the textbook: The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA. © 2012 Pearson Education, Inc.

Once again we convert the mass percents into moles and look for the lowest ratio: Assuming g of para-aminobenzoic acid, C: g x = mol C H: g x = 5.09 mol H N: g x = mol N O: g x = mol O 1 mol 12.01 g 14.01 g 1.01 g 16.00 g © 2012 Pearson Education, Inc.

To find a more complex mole ratio divide each mole quanity by the smallest mole quantity (this makes the smallest a “1”): C: =  7 H: =  7 N: = 1.000 O: =  2 5.105 mol mol 5.09 mol 1.458 mol © 2012 Pearson Education, Inc.

Sample Exercise 3.13 Calculating an Empirical Formula Ascorbic acid (vitamin C) contains 40.92% C, 4.58% H, and 54.50% O by mass. What is the empirical formula of ascorbic acid? Solution Analyze We are to determine the empirical formula of a compound from the mass percentages of its elements. Plan The strategy for determining the empirical formula involves the three steps given in Figure 3.13. FIGURE 3.13 Procedure for calculating an empirical formula from percentage composition. The key step in the calculation is step 2, determining the number of moles of each element in the compound. Sample 3.13 © 2012 Pearson Education, Inc. 61

Ascorbic acid (vitamin C) contains 40.92% C, 4.58% H, and 54.50% O by mass. What is the empirical formula of ascorbic acid? Solve 1. For simplicity, we assume we have exactly 100 g of material (although any other mass could also be used). In 100 g of ascorbic acid, we have 40.92 g C, 4.58 g H, and g O. 2. We calculate the number of moles of each element: © 2012 Pearson Education, Inc.

Sample Exercise 3.13 Calculating an Empirical Formula Continued 3. We determine the simplest whole-number ratio of moles by dividing each number of moles by the smallest number of moles: The ratio for H is too far from 1 to attribute the difference to experimental error; in fact, it is quite close to 1 . This suggests we should multiply the ratios by 3 to obtain whole numbers: 1 3 C:H:O = 3(1:1.33:1) = 3:4:3 Thus, the empirical formula is C3H4O3 Check It is reassuring that the subscripts are moderate-size whole numbers. Also, calculating the percentage composition of C3H8O gives values very close to the original percentages. You might also notice that since this involves thirds, you might also see this ratio as 3/3 to 4/3 to 3/3. See the 3:4:3 ratio? © 2012 Pearson Education, Inc. 63

Practice Exercise A g sample of methyl benzoate, a compound used in the manufacture of perfumes, contains g of carbon, g of hydrogen, and g of oxygen. What is the empirical formula of this substance? Answer: C4H4O © 2012 Pearson Education, Inc.

Calculating Molecular formulas from empirical formulas and molar mass.
The empirical formula is only the ratio formula. For the molecular (actual) formula we also need to know how much the molecule weighs. Ex: The formula for hydrogen peroxide is H2O2. That’s a 1:1 mole ratio. During analysis, we would find an empirical formula of HO. The mass of HO is = 17 If we know that hydrogen peroxide molecules weigh 34. That’s 2 times bigger than 17 So the actual formula for hydrogen peroxide must be 2 x HO = H2O2

Sample Exercise 3.14 Determining a Molecular Formula
Mesitylene, a hydrocarbon found in crude oil, has an empirical formula of C3H4 and an experimentally determined molecular weight of 121 amu. What is its molecular formula? Solution Analyze We are given an empirical formula and a molecular weight and asked to determine a molecular formula. Plan The subscripts in a compound’s molecular formula are whole-number multiples of the subscripts in its empirical formula. We find the appropriate multiple by using Equation 3.11. Solve The formula weight of the empirical formula C3H4 is 3(12.0 amu) + 4(1.0 amu) = 40.0 amu Next, we use this value in Equation 3.11: Only whole-number ratios make physical sense because molecules contain whole atoms. The 3.02 in this case could result from a small experimental error in the molecular weight. We therefore multiply each subscript in the empirical formula by 3 to give the molecular formula: C9H12. © 2012 Pearson Education, Inc. 66

Sample Exercise 3.14 Determining a Molecular Formula Continued Check We can have confidence in the result because dividing molecular weight by empirical formula weight yields nearly a whole number. Practice Exercise Ethylene glycol, used in automobile antifreeze, is 38.7% C, 9.7% H, and 51.6% O by mass. Its molar mass is 62.1 g/mol. (a) What is the empirical formula of ethylene glycol? (b) What is its molecular formula? Answer: (a) CH3O, (b) C2H6O2 © 2012 Pearson Education, Inc. 67

Combustion Analysis Compounds containing C, H, and O are routinely analyzed through combustion in a chamber like the one shown in Figure 3.14. C is determined from the mass of CO2 produced. H is determined from the mass of H2O produced. O is determined by difference after the C and H have been determined. This seems a bit complicated but just remember you’re working backward to find the C from the CO2,, the H from the H2O, and the O from what’s left over. Go slow and follow the sample problem next: © 2012 Pearson Education, Inc.

Sample 3.15 Sample Exercise 3.15 Determining an Empirical Formula by Combustion Analysis Isopropyl alcohol, sold as rubbing alcohol, is composed of C, H, and O. Combustion of g of isopropyl alcohol produces g of CO2 and g of H2O. Determine the empirical formula of isopropyl alcohol. Solution Analyze We are told that isopropyl alcohol contains C, H, and O atoms and given the quantities of CO2 and H2O produced when a given quantity of the alcohol is combusted. We must determine the empirical formula for isopropyl alcohol, a task that requires us to calculate the number of moles of C, H, and O in the sample. Plan We can use the mole concept to calculate grams of C in the CO2 and grams of H in the H2O. These masses are the masses of C and H in the alcohol before combustion. The mass of O in the compound equals the mass of the original sample minus the sum of the C and H masses. (Notice: You can’t calculate the O from the products since you’re adding extra oxygen and you want to know how much was in the molecule to start. Ask me in class if this still doesn’t make sense to you. Its confusing.) Once we have the C, H, and O masses, we can proceed as in Sample Exercise 3.13. © 2012 Pearson Education, Inc. 69

Isopropyl alcohol, sold as rubbing alcohol, is composed of C, H, and O. Combustion of g of isopropyl alcohol produces g of CO2 and g of H2O. Determine the empirical formula of isopropyl alcohol. Solve To calculate the mass of C from the measured mass of CO2,we first use the molar mass of CO2, 44.0 g/mol, to convert grams of CO2 to moles of CO2. Because each CO2 molecule has only one C atom, there is 1 mol of C atoms per mole of CO2 molecules. This fact allows us to convert moles of CO2 to moles of C. Finally, we use the molar mass of C, 12.0 g, to convert moles of C to grams of C: © 2012 Pearson Education, Inc.

Isopropyl alcohol, sold as rubbing alcohol, is composed of C, H, and O. Combustion of g of isopropyl alcohol produces g of CO2 and g of H2O. Determine the empirical formula of isopropyl alcohol. The calculation for determining H mass from H2O mass is similar, although we must remember that there are 2 mol of H atoms per 1 mol of H2O molecules: At last, The mass of the sample, g, is the sum of the masses of C, H, and O. Thus, the O mass is rest of the original mass: Mass of O = mass of sample - (mass of C + mass of H) = g - (0.153 g g) = g O © 2012 Pearson Education, Inc.

The number of moles of C, H, and O in the sample is Therefore To find the empirical formula, we must compare the relative number of moles of each element in the sample. We determine relative number of moles by dividing each of our calculated number of moles by the smallest number: The first two numbers are very close to the whole numbers 3 and 8, giving the empirical formula C3H8O © 2012 Pearson Education, Inc. 72

Sample Exercise 3.15 Determining an Empirical Formula by Combustion Analysis Continued Check The subscripts work out to be moderate-size whole numbers, as expected. Practice Exercise (a) Caproic acid, responsible for the odor of dirty socks, is composed of C, H, and O atoms. Combustion of a g sample of this compound produces g CO2 and g H2O.What is the empirical formula of caproic acid? (b) Caproic acid has a molar mass of 116 g/mol. What is its molecular formula? Answer: (a) C3H6O, (b) C6H12O2 © 2012 Pearson Education, Inc. 73

You are working with rounded mass values to begin with gets you close, but not exact whole numbers. The more precision in the masses, the closer the mole ratios will come to perfect whole numbers. As long as you recognize the whole numbers that they come close to, you can use the values as clues as to the actual ratios. Answer: D © 2012 Pearson Education, Inc.

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Chapter 3 Topic Formula weights 3.4 The mole

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