Chapter 11 The Mole

Objectives 1. Describe how a mole is used in chemistry. 2. Relate a mole to common counting units. 3. Convert the number of moles to the number of particles present, and number of particles to number of moles.

Counting Particles Chemists need a way to accurately count the number of very small things, like atoms and molecules. The word “mole” just means a certain quantity. Like ‘dozen’ means 12, be it eggs or bowling balls.

The Mole The mole (mol) is the base unit used to measure the amount of a substance. It is the number of representative particles, carbon atoms, in exactly 12 grams of pure carbon-12. A mole of anything has 6.0221367 x 1023 particles. So, a mole of pencils is 6.022 X 1023 pencils!

What is the Mole? Abbreviated: mol Defined by the number of carbon atoms in 12 grams of carbon-12. A mole of anything has this many particles: 6.02 x 1023

This number is called “Avagadro’s Number”. Expanded, it looks like this: 602 000 000 000 000 000 000 000 It’s such a big number, because it counts such small things.

Examples One mole of water (H2O) One mole of sodium chloride (NaCl) These are obviously not the same mass – but there are the same number of molecules in each sample.

Converting Moles to Particles Conversion factor: 6.02 x 1023 / 1 mole # of moles x 6.02x1023 1 mole So how many atoms in 2.5 moles of Zn? How many molecules of H2O in 11.5 moles? How many molecules sucrose in 3.5 moles?

Converting Particles to Moles Conversion factor: 1 mole / 6.02 x 1023 # of particles x 1 mole 6.02x1023 atoms How many moles in each of the following? 5.75 x 1024 atoms Al 3.58 x 1023 units Zn Cl2 3.75 x 1024 molecules CO2 2.5 x 1020 atoms Fe

Checkpoint How is a mole like a dozen, or a pair? What is the relationship between Avagadro’s number and the mole? How would you convert the number of particles to number of moles? Why do we use the mole?

11.2: Mass and the Mole Do you think that one dozen eggs have the same mass as one dozen lemons? They only have the same number of items, not the same mass. Same goes for elements and compounds…

Molar Mass: mass (in grams) of one mole of any substance. (Formula mass is similar but the unit U is used instead) It is numerically equal to the average atomic mass on the periodic table. What is the molar mass of nitrogen? Potassium? Aluminum? How many particles in one mole of nitrogen? Potassium? Aluminum? What is the formula mass for calcium nitrate (Ca(NO3)2 ____________________

11.3: Moles of Compounds Remember that a chemical formula shows us the number and types of atoms in a compound. CCl2F2 has 1 Carbon atom, 2 Chlorine atoms, and 2 Fluorine atoms. To find the molar mass of a compound, we simply add the individual masses together.

CCl2F2 Mass of Carbon = 12.0 g Mass of Chlorine = 35.4 g Mass of Fluorine = 19 g Molar Mass of CCl2F2 = 12 + (2)35.4 + (2)19 = 120.8 g

What is the mass of 1 mole of H2O? What is the mass of 1 mole of sucrose (C6H12O6)?

Suppose, for a lab, you need 3. 0 moles of Manganese Suppose, for a lab, you need 3.0 moles of Manganese. Using a balance, how many grams of Manganese would you measure out? [Molar Mass of Mn] x 3.0 moles 54.9 x 3.0 moles = 165g Mn

You have 60. 04g of potassium carbonate. How many moles do you have You have 60.04g of potassium carbonate. How many moles do you have? (K2CO3)

Convert moles to mass: (# moles) x # of grams = mass 1 mole What is the mass of the following? 3.57 mol Al 3.45 mol Co 42.6 mol Si 2.45 mol Zn

Converting mass to moles: mass x 1 mole = number of moles number of grams How many moles in the following? 25.5 g Ag 125 g Zn 300.0 g S 1.0 Kg Fe

Converting mass and atoms: You can also convert from mass to number of atoms, or atoms to mass. How many atoms of gold are present in a 25.0g nugget of gold? First, convert mass to moles. Second, convert moles to number of atoms.

1. Mass to moles: 25.0g x (1 mol/196.97g) = 0.127 mol 2. Moles to atoms: 0.127 mol x (6.02x1023 atoms/1 mol) = 7.65 x 1022 atoms Ag

How many atoms in the following? 55.2 g Li 0.230 g Pb What is the mass (in grams) of the following? 1.00 x 1024 atoms Mn 1.5 x 1015 atoms N

Volume Moles, Gases and Molar Volume Avogadro proposed that at the same temperature and pressure, equal volumes of gases contain the same number of particles. It was found experimentally that: 1 mole (of any gas) = 22.4 Liters (L) or cubic decimeters (dm3) .

This value holds true only at STP, standard temperature (0C) and pressure (1 atmosphere). This volume, 22.4L is called molar volume. It is a “one size fits all” concept. One mole of any gas at STP occupies a volume of 22.4 L. It is a conversion factor

Example: A chemical reaction produces 0. 37 mol of nitrogen gas Example: A chemical reaction produces 0.37 mol of nitrogen gas. What volume will that gas occupy at STP?   0.37 mol x 22.4L = 8.3 Liters 1 mol

A container with a volume of 101L contains how many moles of argon gas at STP> How many grams of oxygen gas fills a volume of 28.7L at STP? (Follow the map, Volume, moles, to grams)

Notice that the mole is always at the center of the calculation. Mass Mole Particles 1 mole Number of grams 6.02x1023 particles 1 mol Number of grams 1 mole 1 mol 6.02x1023 particles

Checkpoint Explain what is meant by molar mass. What conversion factor is used to convert mass to moles? Moles to mass? What are the steps needed to convert mass to the number of atoms?

11.4 Empirical and Molecular Formulas Objectives: 1. Explain what is meant by the percent composition of a compound 2. Determine the empirical and molecular formulas for a compound from mass percent and actual mass data

Percent Composition – is the percent by mass of each element in a compound. The analytical chemist must identify the elements in a compound and determine their percent by mass. For example, a 100 g sample of a new compound contains 55g of element X and 45 g of element Y. The percent by mass of an element can be found by dividing the mass of the element by the mass of the compound and multiplying by 100

Percent composition for the chemical formula – You can use the chemical formula to calculate the molar mass of a compound. Because the percent composition of a compound is always the same, no matter the size of the sample, you can assume that the sample size is one mole. To find the mass of each element in a mole of water, multiply the molar mass of the element by its subscripts in the chemical formula.

Percent Composition Percent expresses the amount of a single quantity compared to an entire sample (or parts per 100 parts). Another way to say percent is to say that it is the ratio of a given quantity to the entire sample, all multiplied by 100. Let’s apply this concept to the composition of a compound.

The percent composition is a list of the mass percent of each element in a compound. Here is how to find it: 1. Determine the molar mass of the compound 2. % composition = mass of element in the compound / molar mass of compound x 100% ( To find the mass of the element, multiply the atomic mass of the element by the number of atoms of it in the compound.)

Ex. Find the % composition of potassium chromate.(K2CrO4) Molar mass = (Mass of element/molar mass ) 100   Ex. If the percent composition of water is 11% hydrogen, what is the % composition of the oxygen? (Hint: it is in %)

Empirical Formula – the formula with the smallest whole number mole ratio of the elements. The empirical formula may or may not be the same as the actual molecular formula. If the two formulas are different the molecular formula will always be a simple multiple of the empirical formula

The empirical formula of a compound is corresponds to the simplest whole-number ratio of ions in a formula unit or to the simplest ratio of atoms of each element in a molecule. It can be determined from having the number of grams given for each component of the compound, or from a percent composition of the compound

Here is a little rhyme to help you remember how to solve this type of problem: Percent to mass Mass to mole Divide by small Multiply ‘til whole! Here’s an example to follow: A compound is analyzed and found to contain 68.54 % carbon, 8.63% hydrogen, and 22.83 % oxygen. What is the empirical formula for this compound?

Percent to mass. When dealing with %, assume the sample to be 100 g Percent to mass. When dealing with %, assume the sample to be 100 g. Therefore, the composition is: Carbon 68.54 g Hydrogen 8.63 g Oxygen 22.83 g

Mass to moles: change grams to moles using the proper molar mass. 68.54g C x 1 mol C = 5.71 mol C 12.01g 8.73 g H x 1 mol H = 8.64 mol H 1.01 g 12.01 g 22.83 g O x 1 mol O = 1.43 mol O 16.0 g

Divide by small: divide each number of moles by the smallest number of moles (and hopefully you will get a whole-number ratio that will be the subscripts). Carbon 5.71 / 1.43 = 3.99 (4) Hydrogen 8.64 / 1.43 = 5.99 (6) Oxygen 1.43 / 1.43 = 1

Multiply ‘til whole: not needed since all values came out whole.   The empirical formula of the compound is C4H6O. Ex. What is the empirical formula of the compound whose percent composition is 65.2% As and 34.8 % O?

Molecular Formula The molecular formula shows exactly how many atoms of each element occurs in a molecule; sometimes it is the same as the empirical formula. Remember that molecular compounds are represented by individual molecules composed of nonmetal atoms. Here are the steps

You must know the molar or molecular mass (it will be given in the problem) If the empirical formula is NOT given, you must determine it from the information given. Determine the molar mass of the empirical formula. Divide the molecular mass given by the molar mass of the empirical formula. Multiply that number by the empirical formula to find out how many atoms of each element there are in the formula

Ex. The empirical formula for fructose, or fruit sugar is CH2O Ex. The empirical formula for fructose, or fruit sugar is CH2O. If the molar mass of fructose is 180 g/mol, find the molecular formula for the sugar.

The molar mass of the empirical formula CH2O is 12 + 2 + 16 = 30 g/mol Dividing the mass given (180 g/mol) by the molar mass of the empirical formula (30 g/mol) gives us 6. 6 (CH2O) = C6H12O6 which is the molecular formula for fructose

Ex. A compound is found by analysis to contain 5. 90% H and 94. 1% O Ex. A compound is found by analysis to contain 5.90% H and 94.1% O. Its molecular mass is determined to be 34.0 g/mol. What is its empirical formula? What is the molecular formula?