1. Chapter 3 A whole lotta stuff

2. Parts of an atom Nucleus: Almost all of the mass, almost none of the volume. Protons: Positive charge. Mass of 1 amu. Atomic number tells the number of protons Number of protons identifies the element. Neutrons: Neutral charge. Mass of 1 amu. Atomic mass – atomic number = Number of neitrons Atoms of the same element with different numbers of neutrons are called isotopes The ratio of protons to neutrons determines the nuclear stability (radioactivity) of an isotope Electrons: Negative charge. Nearly no mass compared to protons and neutrons. In a neutral atom, there are the same number of electrons as protons Atomic number tells the number of electrons in a neutral atom Electrons determine the properties of elements.

3. Periodic table Information The periodic table gives information on an element: Atomic number is the number of protons and the number of neutrons in a neutral atom Atomic mass – atomic number = neutrons Atomic notation gives the same information Atomic notation gives the same information

4. Mass Spectrometry Mass spectrometry is used to separate isotopes and figure out abundance. The most abundant mass in a sample is set at 100%, so to determine actual abundance. Neon-20 = 100/110.5 = 90.5% Neon-21 = 0.3/110.5 = 0.3% Neon-22 = 10.2/110.5 = 9.2%

5. Average Atomic Mass To determine the average atomic mass that is recorded on the periodic table: Use a weighted average of the isotopes From Neon on the previous slide Neon-20 = 100/110.5 = 90.5% Neon-21 = 0.3/110.5 = 0.3% Neon-22 = 10.2/110.5 = 9.2% So: So: (.905 x 20) + (.003 x 21) + (.092 x 22) = 20.19 (Close!)

6. Moles! A mole is a measurement of how many atoms are contained in a certain mass of a substance. Avogadro’s number, we’ll use 6.022 x 10 23, is the number of particles in a mole of anything! 1 mole of carbon = 6.02 x 10 23 carbon atoms (element) 1 mole of water = 6.02 x 10 23 water molecules (covalent compound) 1 mole of salt (NaCl) = 6.02 x 10 23 formula units of salt (ionic compound) Mass of 1 mole of any element = the atomic mass on the periodic table (in grams) 1 mole of carbon has a mass of 12.02 grams 1 mole of water has a mass of 18.02 grams (2 H = 2.02 and O = 16.00) Volume of gas of 1 mole of anything at STP = 22.4 L STP is standard temperature and pressure: 0°C (or 273 K) and 1 atm (atmosphere) pressure You will use these conversion factors over and over and over and over and over.

7. Percent Composition To figure out percent composition of elements in a compound: Determine the mass of each element in the compound Determine the total mass of the compound Divide the element mass by the total mass (and multiply by 100) to determine the percent composition. Example: Glucose C 6 H 12 O 6 C: 6 atoms x 12.01 mass = 72.06 H: 12 atoms x 1.01 mass = 12.12 O: 6 atoms x 16.00 mass = 96.00 Total Mass = 180.2 (Sig figs!)

8. % Composition Example Glucose C 6 H 12 O 6 Element masses C: 6 atoms x 12.01 mass = 72.06 H: 12 atoms x 1.01 mass = 12.12 O: 6 atoms x 16.00 mass = 96.00 Total Mass = 180.2 (Sig figs!) Percent compositions: C = 72.06/180.2 = 39.99% H = 12.12/180.2 = 6.726% O = 96.00/180.2 = 53.27% Total of percentages should equal 100% (or very close).

9. Formulas Empirical formulas are the simplest whole number ratios of atoms in a compound. The empirical formula is an ionic compound’s formula Molecular Formulas are the actual numbers of atoms in a covalent compound. Glucose’s molecular formula is C 6 H 12 O 6 Glucose’s empirical formula is C 1 H 2 O 1 Structural formulas show how the atoms are arranged in a covalent compound. Glucose’s structural formula is

10. Determining Formulas You may be given pectent compositions (or actual compositions) and be asked to determine the empirical and/or molecular formula. Empirical Formula Determination: 1.Change the mass of each element to moles using molar mass. 2.Divide the moles of each substance by the smallest number of moles to get ratios. 3.Play with the numbers to get a whole number ratio

11. Empirical Formula Determination Example You are given a sample containing 43.64% phosphorous and 56.36% oxygen. What is the empirical formula for this compound? First, if you’re given percentages, just treat them as masses. 1.Change the mass of each element to moles using molar mass. Phosphorous: 43.64 g x 1 mole = 1.409 mol P Phosphorous: 43.64 g x 1 mole = 1.409 mol P 30.97 g 30.97 g Oxygen: 56.36 g x 1 mole = 3.523 mol O Oxygen: 56.36 g x 1 mole = 3.523 mol O 16.00 g 16.00 g 2.Divide the moles of each substance by the smallest number of moles to get ratios. P = 1.409/1.409 = 1 P = 1.409/1.409 = 1 O = 3.523/1.409 = 2.5 O = 3.523/1.409 = 2.5 3.Play with the numbers to get a whole number ratio If I double the ratios, I get 2 P to 5 O If I double the ratios, I get 2 P to 5 O My empirical formula is P 2 O 5 My empirical formula is P 2 O 5

12. Molecular Formula Determination Sometimes, along with determining empirical formula, you will be asked for the molecular formula. You are given a sample containing 43.64% phosphorous and 56.36% oxygen, with a molar mass of 283.88 g/mol. What are the empirical and molecular formulas for this compound? 1.Determine the empirical formula 2.Determine the the molar mass for your empirical formula. 3.Divide the empirical formula’s molar mass by the actual molar mass to get how many times you have to multiply the empirical formula by.

13. Molecular Formula Determination Example You are given a sample containing 43.64% phosphorous and 56.36% oxygen, with a molar mass of 283.88 g/mol. What are the empirical and molecular formulas for this compound? 1.Determine the empirical formula We did this and determined it to be P 2 O 5 We did this and determined it to be P 2 O 5 2.Determine the the molar mass for your empirical formula. P = (2 x 30.97) P = (2 x 30.97) O = (5 x 16.00) O = (5 x 16.00) Total = 141.9 g/mol Total = 141.9 g/mol 3.Divide the empirical formula’s molar mass by the actual molar mass to get how many times you have to multiply the empirical formula by. 141.9 g/mol = 2 141.9 g/mol = 2 283.88 g/mol So you double your empirical formula and get P 4 O 10 So you double your empirical formula and get P 4 O 10