1. Chapter 3 Atoms: the Building Blocks of Matter

2. The parts that make up an atom are called subatomic particles. The parts that make up an atom are called subatomic particles. 1) Protons (p + ) positively charged particle 2) Neutron (n o ) neutral particle (uncharged) 3) Electrons (e - ) negatively charged particle Neutrons and Protons are located in the nucleus of an atom. Neutrons and Protons are located in the nucleus of an atom. Electrons orbit around the nucleus. Electrons orbit around the nucleus.

3. Q- How are atoms of different elements distinguished from one another? In other words, how do we distinguish a one another? In other words, how do we distinguish a helium atom from a carbon atom? helium atom from a carbon atom? A- Their number of protons, indicated by the atomic number Let’s look at helium, He. It has an atomic number of 2, which means that is has 2 protons in it’s nucleus.

4. Atomic Structure Here are the basics; you need to know these. 1H1.0076Hydrogen Atomic Number (Z): the number of protons (p + ) Atomic Mass: the number of protons (p + ) + the number of neutrons (n 0 ) ▪ measured in atomic mass units (amu) which is one twelfth ▪ measured in atomic mass units (amu) which is one twelfth the mass of a carbon-12 atom. the mass of a carbon-12 atom. ▪ the mass of electrons (1/1860 p + ) is negligible. ▪ the mass of electrons (1/1860 p + ) is negligible. Number of Neutrons: the atomic mass - the atomic number Atomic Number Atomic Symbol Atomic Mass

5. Lets practice! Find the missing information? Element Atomic # AtomicMassProtonsElectronsNeutrons Ar18 39.948 amu He22 O 15.999 amu 8

6. The Famous Gold Foil Experiment This showed us that the atom is made of mostly empty space.

7. Isotopes Atoms of the same element with different number of neutrons Because they have the same number of protons, all isotopes of an element have the same chemical properties.

8. Mass Numbers of Hydrogen Isotopes What would the masses be?

9. The Mole: A Measurement of Matter At the end of this section, you should be able to: Describe how Avogadro’s number is related to a mole of any substance Describe how Avogadro’s number is related to a mole of any substance Calculate the mass of a mole of any substance Calculate the mass of a mole of any substance

10. The Mole (aka Avagadro’s Number): 6.02 x 10 23

11. The Mole and Avogadro’s Number SI unit that measures the amount of substance SI unit that measures the amount of substance 1 mole = 6.02 x 10 23 representative particles 1 mole = 6.02 x 10 23 representative particles Representative particles are usually atoms, molecules, or formula units (ions) Representative particles are usually atoms, molecules, or formula units (ions)

12. But Why the Mole? Just as 12 = 1 dozen, or 63,360 inches = 1 mile, the mole allows us to count microscopic items (atoms, ion, molecules) on a macroscopic scale. So, 1 mole of any substance is a set number of Items, namely: 6.02 x 10 23. Chemistry = awesome

13. Examples: Substanc e Representative Particle Chemical Formula Representative Particles in 1.00 mol Atomic nitrogen AtomN 6.02 x 10 23 WaterMolecule H2OH2OH2OH2O Calcium ion Ion Ca 2+ 6.02 x 10 23

14. Solve SubstanceRepresentativeParticleFormulaUnit Representativ e Particles in 1.00 mol Nitrogen gas N2N2N2N2Molecule Calcium Fluoride CaF 2 Molecule Sucrose C 12 H 22 O 11 Molecule CarbonCMolecule

15. Answers Nitrogen gas-molecule-N 2 Nitrogen gas-molecule-N 2 Calcium fluoride-formula unit-CaF 2 Calcium fluoride-formula unit-CaF 2 Sucrose-molecule-C 12 H 22 O 11 Sucrose-molecule-C 12 H 22 O 11 Carbon-atom-C Carbon-atom-C All have 6.02 x 10 23 representative particles in 1.00 mol

16. How many atoms are in a mole? Determined from the chemical formula Determined from the chemical formula List the elements and count the atoms List the elements and count the atoms Solve for CO 2 Solve for CO 2 C - 1 carbon atom C - 1 carbon atom O - 2 oxygen atoms O - 2 oxygen atoms Add: 1 + 2 = 3 Add: 1 + 2 = 3 Answer: 3 times Avogadro’s number of atoms Answer: 3 times Avogadro’s number of atoms

17. Solve: How many atoms are in a mole of 1. Carbon monoxide – CO 1. Carbon monoxide – CO 2. Glucose – C 6 H 12 O 6 2. Glucose – C 6 H 12 O 6 3. Propane – C 3 H 8 3. Propane – C 3 H 8 4. Water – H 2 O 4. Water – H 2 O

18. How many moles of magnesium is 1.25 x 10 23 atoms of magnesium? Refer to page 174 in text Refer to page 174 in text Divide the number of atoms or molecules given in the example by 6.02 x 10 23 Divide the number of atoms or molecules given in the example by 6.02 x 10 23 Divide (1.25 x 10 23) by (6.02 x 10 23) Divide (1.25 x 10 23) by (6.02 x 10 23) Express in scientific notation Express in scientific notation Answer = 2.08 x 10 -1 mol Mg Answer = 2.08 x 10 -1 mol Mg

19. Objectives Use the molar mass to convert between mass and moles of a substance Use the molar mass to convert between mass and moles of a substance Use the mole to convert among measurements of mass, volume, and number of particles Use the mole to convert among measurements of mass, volume, and number of particles

20. Molar mass Mass (in grams) of one mole of a substance Mass (in grams) of one mole of a substance Broad term (can be substituted) for gram atomic mass, gram formula mass, and gram molecular mass Broad term (can be substituted) for gram atomic mass, gram formula mass, and gram molecular mass Can be unclear: What is the molar mass of oxygen? Can be unclear: What is the molar mass of oxygen? O or O 2 ? - element O or molecular compound O 2 ? O or O 2 ? - element O or molecular compound O 2 ?

21. Calculating the Molar Mass of Compounds (Molecular and Ionic) 1. List the elements 1. List the elements 2. Count the atoms 2. Count the atoms 3. Multiply the number of atoms of the element by the atomic mass of the element (atomic mass is on the periodic table) 3. Multiply the number of atoms of the element by the atomic mass of the element (atomic mass is on the periodic table) 4. Add the masses of each element 4. Add the masses of each element 5. Express to tenths place 5. Express to tenths place

22. What is the molar mass (gfm) of ammonium carbonate (NH 4 ) 2 CO 3 ? N 2 x 14.0 g = 28.0 g N 2 x 14.0 g = 28.0 g H 8 x 1.0 g = 8.0 g H 8 x 1.0 g = 8.0 g C 1 x 12.0 g = 12.0 g C 1 x 12.0 g = 12.0 g O 3 x 16.0 g = 48.0 g O 3 x 16.0 g = 48.0 g Add ________ Add ________ Answer 96.0 g Answer 96.0 g

23. Practice Problems 1. How many grams are in 9.45 mol 1. How many grams are in 9.45 mol of dinitrogen trioxide (N 2 O 3 ) ? of dinitrogen trioxide (N 2 O 3 ) ? a. Calculate the grams in one mole a. Calculate the grams in one mole b. Multiply the grams by the number b. Multiply the grams by the number of moles of moles 2. Find the number of moles in 92.2 g 2. Find the number of moles in 92.2 g of iron(III) oxide (Fe 2 O 3 ). of iron(III) oxide (Fe 2 O 3 ). a. Calculate the grams in one mole a. Calculate the grams in one mole b. Divide the given grams by the grams in one mole b. Divide the given grams by the grams in one mole

24. Answers 1. 718 g N 2 O 3 (one mole is 76.0g) 1. 718 g N 2 O 3 (one mole is 76.0g) 2. 0.578 mol Fe 2 O 3 (one mole is 159.6 g) 2. 0.578 mol Fe 2 O 3 (one mole is 159.6 g)

25. Volume of a Mole of Gas Varies with a change in temperature or a change in pressure Varies with a change in temperature or a change in pressure At STP, 1 mole of any gas occupies a volume of 22.4 L At STP, 1 mole of any gas occupies a volume of 22.4 L Standard temperature is 0°C Standard temperature is 0°C Standard pressure is 101.3 kPa (kilopascals), or 1 atmosphere (atm) Standard pressure is 101.3 kPa (kilopascals), or 1 atmosphere (atm) 22.4 L is known as the molar volume 22.4 L is known as the molar volume

26. 22.4 L of any gas at STP contains 6.02 x 10 23 representative particles of that gas 22.4 L of any gas at STP contains 6.02 x 10 23 representative particles of that gas One mole of a gaseous element and one mole of a gaseous compound both occupy a volume of 22.4 L at STP (Masses may differ) One mole of a gaseous element and one mole of a gaseous compound both occupy a volume of 22.4 L at STP (Masses may differ) Molar mass (g/mol) = Density (g/L) x Molar Volume (L/mol) Molar mass (g/mol) = Density (g/L) x Molar Volume (L/mol)

27. Objectives Define the terms Define the terms Calculate the percent composition of a substance from its chemical formula or experimental data Calculate the percent composition of a substance from its chemical formula or experimental data Derive the empirical formula and the molecular formula of a compound from experimental data Derive the empirical formula and the molecular formula of a compound from experimental data

28. Terms to Know Percent composition – relative amounts of each element in a compound Percent composition – relative amounts of each element in a compound Empirical formula – lowest whole- number ratio of the atoms of an element in a compound Empirical formula – lowest whole- number ratio of the atoms of an element in a compound

29. An 8.20 g piece of magnesium combines completely with 5.40 g of oxygen to form a compound. What is the percent composition of this compound? 1. Calculate the total mass 2. Divide each given by the total mass and then multiply by 100% 3. Check your answer: The percentages should total 100%

30. Answer The total mass is 8.20 g + 5.40 g = 13.60 g The total mass is 8.20 g + 5.40 g = 13.60 g Divide 8.2 g by 13.6 g and then multiply by 100% = 60.29412 = 60.3% Divide 8.2 g by 13.6 g and then multiply by 100% = 60.29412 = 60.3% Divide 5.4 g by 13.6 g and then multiply by 100% = 39.70588 = 39.7% Divide 5.4 g by 13.6 g and then multiply by 100% = 39.70588 = 39.7% Check your answer: 60.3% + 39.7% = 100% Check your answer: 60.3% + 39.7% = 100%

31. Calculate the percent composition of propane (C 3 H 8 ) 1. List the elements 1. List the elements 2. Count the atoms 2. Count the atoms 3. Multiply the number of atoms of the element by the atomic mass of the element (atomic mass is on the periodic table) 3. Multiply the number of atoms of the element by the atomic mass of the element (atomic mass is on the periodic table) 4. Express each element as a percentage of the total molar mass 4. Express each element as a percentage of the total molar mass 5. Check your answer 5. Check your answer

32. Answer Total molar mass = 44.0 g/mol Total molar mass = 44.0 g/mol 36.0 g C = 81.8% 36.0 g C = 81.8% 8.0 g H = 18.2% 8.0 g H = 18.2%

33. Calculate the mass of carbon in 52.0 g of propane (C 3 H 8 ) 1. Calculate the percent composition using the formula (See previous problem) 2. Determine 81.8% of 82.0 g Move decimal two places to the Move decimal two places to the left (.818 x 82 g) left (.818 x 82 g) 3. Answer = 67.1 g

34. 1) Find the percent composition of Aluminum Oxide (Al 3 O 2 ) Aluminum Oxide (Al 3 O 2 ) 2) How much of a 5-g piece of Iron Bromide (FeBr 3 ) is iron? Bromide (FeBr 3 ) is iron?

35. Calculating Empirical Formulas Microscopic – atoms Microscopic – atoms Macroscopic – moles of atoms Macroscopic – moles of atoms Lowest whole-number ratio may not be the same as the compound formula Lowest whole-number ratio may not be the same as the compound formula Example: The empirical formula of hydrogen peroxide (H 2 O 2 ) is HO Example: The empirical formula of hydrogen peroxide (H 2 O 2 ) is HO

36. Empirical Formulas The first step is to find the mole-to-mole ratio of the elements in the compound The first step is to find the mole-to-mole ratio of the elements in the compound If the numbers are both whole numbers, these will be the subscripts of the elements in the formula If the numbers are both whole numbers, these will be the subscripts of the elements in the formula If the whole numbers are identical, substitute the number 1 If the whole numbers are identical, substitute the number 1 Example: C 2 H 2 and C 8 H 8 have an empirical formula of CH Example: C 2 H 2 and C 8 H 8 have an empirical formula of CH If either or both numbers are not whole numbers, numbers in the ratio must be multiplied by the same number to yield whole number subscripts If either or both numbers are not whole numbers, numbers in the ratio must be multiplied by the same number to yield whole number subscripts

37. What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen? 1. Assume 100 g of the compound, so that 1. Assume 100 g of the compound, so that there are 25.9 g N and 74.1 g O there are 25.9 g N and 74.1 g O 2. Convert to mole-to-mole ratio: 2. Convert to mole-to-mole ratio: Divide each by mass of one mole Divide each by mass of one mole 25.9 g divided by 14.0 g = 1.85 mol N 25.9 g divided by 14.0 g = 1.85 mol N 74.1 g divided by 16.0 g = 4.63 mol O 74.1 g divided by 16.0 g = 4.63 mol O 3. Divide both molar quantities by the 3. Divide both molar quantities by the smaller number of moles smaller number of moles

38. 4. 1.85/1.85 = 1 mol N 4. 1.85/1.85 = 1 mol N 4.63/1.85 = 2.5 mol O 4.63/1.85 = 2.5 mol O 5. Multiply by a number that converts each to a whole number (In this case, the number is 2 because 2 x 2.5 = 5, which is the smallest whole number ) 5. Multiply by a number that converts each to a whole number (In this case, the number is 2 because 2 x 2.5 = 5, which is the smallest whole number ) 2 x 1 mol N = 2 2 x 1 mol N = 2 2 x 2.5 mol O = 5 2 x 2.5 mol O = 5 Answer: The empirical formula is N 2 O 5 Answer: The empirical formula is N 2 O 5

39. Determine the Empirical Formulas 1. H 2 O 2 1. H 2 O 2 2. CO 2 2. CO 2 3. N 2 H 4 3. N 2 H 4 4. C 6 H 12 O 6 4. C 6 H 12 O 6 5. What is the empirical formula of a compound that is 3.7% H, 44.4% C, and 51.9% N? 5. What is the empirical formula of a compound that is 3.7% H, 44.4% C, and 51.9% N?

40. Answers Compound Empirical Formula Compound Empirical Formula 1. H 2 O 2 HO 1. H 2 O 2 HO 2. CO 2 CO 2 2. CO 2 CO 2 3. N 2 H 4 NH 2 3. N 2 H 4 NH 2 4. C 6 H 12 O 6 CH 2 O 4. C 6 H 12 O 6 CH 2 O 5. HCN 5. HCN

41. Calculating Molecular Formulas The molar mass of a compound is a simple whole-number multiple of the molar mass of the empirical formula The molar mass of a compound is a simple whole-number multiple of the molar mass of the empirical formula The molecular formula may or may not be the same as the empirical formula The molecular formula may or may not be the same as the empirical formula

42. Calculate the molecular formula of the compound whose molar mass is 60.0 g and empirical formula is CH 4 N. 1. Using the empirical formula, calculate the empirical formula mass (efm) 1. Using the empirical formula, calculate the empirical formula mass (efm) (Use the same procedure used to calculate molar mass.) (Use the same procedure used to calculate molar mass.) 2. Divide the known molar mass by the efm 2. Divide the known molar mass by the efm 3. Multiply the formula subscripts by this value to get the molecular formula 3. Multiply the formula subscripts by this value to get the molecular formula

43. Answer Molar mass (efm) is 30.0 g Molar mass (efm) is 30.0 g 60.0 g divided by 30.0 g = 2 60.0 g divided by 30.0 g = 2 Answer: C 2 H 8 N 2 Answer: C 2 H 8 N 2

44. Practice Problems 1) What is the empirical formula of a compounds that is 25.9% nitrogen and 74.1% oxygen? 1) What is the empirical formula of a compounds that is 25.9% nitrogen and 74.1% oxygen? 2) Calculate the empirical formula of a compound that is 32.00% C, 42.66% O, 18.67% N, and 6.67% H. 2) Calculate the empirical formula of a compound that is 32.00% C, 42.66% O, 18.67% N, and 6.67% H. 3) Calculate the empirical formula of a compound that is 42.9% C and 57.1% O. 3) Calculate the empirical formula of a compound that is 42.9% C and 57.1% O.

45. Practice Problems 4) What is the molecular formula for each compound: 4) What is the molecular formula for each compound: a) CH 2 O, 90 g/mol b) HgCl, 472.2 g/mol c) C 3 H 5 O 2, 146 g/mol