A TAKAMUL INTERNATIONAL SCHOOL CH.10 THE MOLE PREPARED BY MR. FAHAD AL-JARAH

A TAKAMUL INTERNATIONAL SCHOOL CH.10 THE MOLE PREPARED BY MR. FAHAD AL-JARAH

Chapter Outline Section 10.1 Measuring Matter Key Concepts The mole is a unit used to count particles of matter indirectly. One mole of a pure substance contains Avogadro s number of particles. Representative particles include atoms, ions, molecules, formula units, electrons, and other similar particles. One mole of carbon-12 atoms has a mass of exactly 12 g. Conversion factors written from Avogadro s relationship can be used to convert between moles and number of representative particles. Section 10.2 Mass and the Mole Key Concepts The mass in grams of 1 mole of any pure substance is called its molar mass. The molar mass of an element is numerically equal to its atomic mass. The molar mass of any substance is the mass in grams of Avogadro s number of representative particles of the substance. Molar mass is used to convert from moles to mass. The inverse of molar mass is used to convert from mass to moles. Section 10.3 Moles of Compounds Key Concepts Subscripts in a chemical formula indicate how many moles of each element are present in 1 mol of the compound. The molar mass of a compound is calculated from the molar masses of all of the elements in the compound. Conversion factors based on a compound s molar mass are used to convert between moles and mass of a compound. Section 10.4 Empirical & Molecular Formulas Key Concepts The percent by mass of an element in a compound gives the percentage of the compound s total mass due to that element. The subscripts in an empirical formula give the smallest whole-number ratio of moles of elements in the compound. The molecular formula gives the actual number of atoms of each element in a molecule or formula unit of a substance. The molecular formula is a whole-number multiple of the empirical formula

Stoichiometry Road Map Essentially there are six possible mathematical equations that come out of this road map. o o o Mass = Moles x (Molar mass/1mole) Moles = Mass x (1mole/Molar Mass) Gas Volume = Moles x (22.4 L/1 Mole) o Moles = Gas Volume x (1 mole/22.4 L) o o No. Particles = Moles x (6.02 x 10 23 particles/1 mole) Moles = No. Particles x (1 mole/6.02 x 10 23 particles)

Finding the Empirical Formula & Molecular Formula

Ch.10 The Moles Mole: (working definition) the mass of a compound that is the same number of grams as the compound s formula or molecular mass in amu. Mole: (formal definition) the amount of matter that contains as many objects (atoms, molecules, etc.) as the number of atoms in exactly 12 g of 12 C. Avogadro s constant: 1 mole = 6.02 x 10 23 atoms, molecules, etc. Avogadro s principle: recall from the unit on gases that equal volumes of gases (at the same temperature and pressure) contain equal numbers of molecules of gas (and therefore equal number of moles). Molar volume: the space occupied by 1 mole of ANY gas. At 0 C (273 K) and 1 atmosphere of pressure, the molar volume is 22.4 L. (Memorize this number.) Molar mass (M): mass (in grams) of 1 mole of a substance for atoms, this is the same # as the atomic mass on the periodic table. For compounds, add up the mass of each atom in the compound. For example: Find the molar mass of C 6 H 12 O 6 : C 6 H 12 O 6 has 6 atoms of C, 12 atoms of H, and 6 atoms of O. This means one atom of C 6 H 12 O 6 has the same mass as 6 atoms of C plus 12 atoms of H plus 6 atoms of O. This means one mole of C 6 H 12 O 6 has the same mass as 6 moles of C (6 12 g = 72 g) plus 12 moles of H (12 1 g = 12 g) plus 6 moles of O (6 16 g = 96 g). Therefore, 1 mole of C 6 H 12 O 6 has a mass of 72 g + 12 g + 96 g = 180 g. Mole Conversions 1 mol = grams (add up the mass of the formula) 1 mol = 6.02 x 10 23 atoms, molecules, etc. 1 mol = 22.4 L of gas @ 0 C and 1 atm pressure. For example 1.) 2.5 mol of NH 3 gas occupies what volume at 0 C and 1 atm? 2.5 mol NH 3 22.4 L gas 1 1 mol gas 2.5 22.4 L NH 3 = 56.0 L NH 3

2.) What is the mass of 4.1 mol NH 3 gas? The molar mass is (1 14) + (3 1) = 17 g, so 1 mol NH 3 = 17 g NH 3. 4.1 mol NH 3 17 g NH 3 1 1 mol NH 3 4.1 17 g NH 3 = 69.7 g NH 3 3.) How many molecules are there in 0.75 mol of NH 3 gas? 0.75 NH 3 (6.02 10 23 molecules) = 0.75 mol NH 3 6.02 10 23 molecules 1 1 mol 4.52 10 23 molecules NH 3 4.) What is the volume of 25.5 g of NH 3 gas at 0 C and 1 atm? 1 mol NH 3 = (1 14) + (3 1) = 17 g 25.5 g NH 3 1 mol NH 3 22.4 L 1 17 g NH 3 1 mol 25.5 NH3 22.4 L 33.6 L NH 3 17 Percent Composition & Empirical Formula Percent Composition: the percentage by mass of each element in a compound. Empirical Formula: the formula you would have for a compound if you reduced all of the subscripts to their lowest terms. E.g., the empirical formula for both C 2 H 2 and C 6 H 6 would be CH. (You may remember that we always use empirical formulas for ionic compounds.) Determining the Empirical Formula from Percent Composition Data: 1. Write the formula as if the subscripts were in grams 2. For each element in the compound, convert grams to moles. 3. Simplify the subscripts to simple, whole numbers.

Sample problem: a 10 g sample of a hydrocarbon is analyzed and found to contain 8.56 g of carbon and the 1.44 g of hydrogen. What is the empirical formula of this compound? 1. Write the formula as C 8.56 g H 1.44 g 2. Convert grams to moles: C: H: The formula for this compound is therefore C 0.713 H 1.429 3. Convert the subscripts to simple whole numbers. The easiest way to do this is to divide them all by the smallest one and see what happens. Which you can see is just CH 2 If the problem gives percentages instead of actual mass, just pretend the percentages are out of 100 g total. E.g., if you had a compound containing 25.3% nitrogen, you would use 25.3 g of nitrogen in your calculations. Rounding: don t round your fractional subscripts by more than about 5%. If you have something like NO 2.5, double everything to get N 2 O 5. (This means you need to be able to recognize decimal equivalents for simple fractions, such as,, etc.) Empirical vs. Molecular Formula If you know the molar mass of the compound, you can use it to get from the empirical formula to the molecular formula. For example, suppose the molar mass of the above hydrocarbon was known to be 42.08 g/mol. The molar mass of the empirical formula (CH 2 ) would be (1 x 12.011) + (2 x 1.008) = 14.027. Because 42.08 is 3 times as much (14.027 x 3 = 42.08), the actual formula must have 3 times as much of each atom. This means that the actual formula would be C 3 H 6. Percent Yield When we run a reaction in the laboratory we can calculate how much product we should get from the balanced equations as shown above. We never get the same amount we calculate for many reasons. We have terms to describe these differences. The Theoretical Yield is what we calculate from the balanced equation. The Actual Yield is just that. It is what we actually have to show for our work. The Percent Yield is the (Actual Theoretical) x 100. % Yield = (Actual Yield Theoretical Yield) x 100

Chapter 10 -- The Mole Worksheet 1. The scale of atomic masses is based on the mass of what element? 2. What is the unit for atomic mass? 3. What is Avogadro s number? What does it represent? What is the unit? 4. What is the definition of mole? 5. What is the unit for molar mass? 6. What is the molar mass of a gas at STP? 7. What is the empirical formula? 8. What is the molecular formula? 9. Calculate the formula mass of potassium bicarbonate. 10. Find the molar mass of barium hydroxide.

11. Calculate the number of atoms in 0.56 mol of iron. 12. What is the mass of 7.21 x 10 21 molecules of silver acetate? 13. A sample of iron oxide has a mass of 3.192 g. On analysis it was found to contain 2.232 g of iron and 0.96 g of oxygen. Find the percentage composition of this compound. 14. Find the molecular formula of a sugar containing 40% carbon, 6.7 % hydrogen, and 53.3% oxygen. The molar mass of the sugar is 180g/mol. 15. Outline or describe the procedure you should follow to determine the molecular formula of a substance if you know the molar mass and percents of the elements in the substance.

16. Outline or draw a diagram showing how to convert grams to molecules, and molecules to grams. 17. The neurotransmitter norepinephrine is 56.8 percent carbon, 6.5 percent hydrogen, 28.5 percent oxygen, and 8.3 percent nitrogen. Its molar mass is 169 g/mol. Find the molecular formula of this substance. 18. Acetylene gas is burned in welding torches. It is composed of 92.26 percent carbon and 7.74 percent hydrogen. Its molar mass is 26.02 g/mol. Calculate the empirical and molecular formula of acetylene.