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Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
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Chemistry Chapter 3

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  • 1. Chapter 3 Atoms: the Building Blocks of Matter
  • 2.
    • 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.
    • 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
    • helium atom from a carbon atom?
    • A- Their number of p rotons , 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. 1 H 1.0076 Hydrogen 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 the mass of a carbon-12 atom. ▪ 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? 8 15.999 amu O 2 2 He 39.948 amu 18 Ar Neutrons Electrons Protons Atomic Mass Atomic # Element
  • 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
    • 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
    • 1 mole = 6.02 x 10 23 representative particles
    • 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: 6.02 x 10 23 Ca 2+ Ion Calcium ion 6.02 x 10 23 H 2 O Molecule Water 6.02 x 10 23 N Atom Atomic nitrogen Representative Particles in 1.00 mol Chemical Formula Representative Particle Substance
  • 14. Solve Representative Particles in 1.00 mol Formula Unit Representative Particle Substance Molecule C Carbon Molecule C 12 H 22 O 11 Sucrose Molecule CaF 2 Calcium Fluoride Molecule N 2 Nitrogen gas
  • 15. Answers
    • Nitrogen gas-molecule-N 2
    • Calcium fluoride-formula unit-CaF 2
    • Sucrose-molecule-C 12 H 22 O 11
    • 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
    • List the elements and count the atoms
    • Solve for CO 2
    • C - 1 carbon atom
    • O - 2 oxygen atoms
    • Add: 1 + 2 = 3
    • Answer: 3 times Avogadro’s number of atoms
  • 17. Solve: How many atoms are in a mole of
    • 1. Carbon monoxide – CO
    • 2. Glucose – C 6 H 12 O 6
    • 3. Propane – C 3 H 8
    • 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
    • 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)
    • Express in scientific notation
    • 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 mole to convert among measurements of mass, volume, and number of particles
  • 20. Molar mass
    • 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
    • Can be unclear: What is the molar mass of oxygen?
    • O or O 2 ? - element O or molecular compound O 2 ?
  • 21. Molar Mass
    • Gram atomic mass (gam) – atomic mass of an element taken from the periodic table
    • Gram molecular mass (gmm) – mass of one mole of a molecular compound
    • Gram formula mass (gfm) – mass of one mole of an ionic compound
    • Can use molar mass instead of gam, gmm, or gfm
  • 22. Calculating the Molar Mass of Compounds (Molecular and Ionic)
    • 1. List the elements
    • 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)
    • 4. Add the masses of each element
    • 5. Express to tenths place
  • 23. What is the molar mass (gfm) of ammonium carbonate (NH 4 ) 2 CO 3 ?
    • N 2 x 14.0 g = 28.0 g
    • H 8 x 1.0 g = 8.0 g
    • C 1 x 12.0 g = 12.0 g
    • O 3 x 16.0 g = 48.0 g
    • Add ________
    • Answer 96.0 g
  • 24. Practice Problems
    • 1. How many grams are in 9.45 mol
    • of dinitrogen trioxide (N 2 O 3 ) ?
    • a. Calculate the grams in one mole
    • b. Multiply the grams by the number
    • of moles
    • 2. Find the number of moles in 92.2 g
    • of iron(III) oxide (Fe 2 O 3 ).
    • a. Calculate the grams in one mole
    • b. Divide the given grams by the grams in one mole
  • 25. Answers
    • 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)
  • 26. Volume of a Mole of Gas
    • 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
    • Standard temperature is 0 ° C
    • Standard pressure is 101.3 kPa (kilopascals), or 1 atmosphere (atm)
    • 22.4 L is known as the molar volume
  • 27.
    • 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)
    • Molar mass (g/mol) = Density (g/L) x Molar Volume (L/mol)
  • 28. Objectives
    • Define the terms
    • 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
  • 29. Terms to Know
    • 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
  • 30. 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
    • Divide each given by the total mass
    • and then multiply by 100%
    • Check your answer: The
    • percentages should total 100%
  • 31. Answer
    • 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 5.4 g by 13.6 g and then multiply by 100% = 39.70588 = 39.7%
    • Check your answer: 60.3% + 39.7% = 100%
  • 32. Calculate the percent composition of propane (C 3 H 8 )
    • 1. List the elements
    • 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)
    • 4. Express each element as a percentage of the total molar mass
    • 5. Check your answer
  • 33. Answer
    • Total molar mass = 44.0 g/mol
    • 36.0 g C = 81.8%
    • 8.0 g H = 18.2%
  • 34. Calculate the mass of carbon in 52.0 g of propane (C 3 H 8 )
    • Calculate the percent composition using the formula (See previous problem)
    • 2. Determine 81.8% of 82.0 g
    • Move decimal two places to the
    • left (.818 x 82 g)
    • 3. Answer = 67.1 g
  • 35.
    • 1) Find the percent composition of
    • Aluminum Oxide (Al 3 O 2 )
    • 2) How much of a 5-g piece of Iron
    • Bromide (FeBr 3 ) is iron?
  • 36. Calculating Empirical Formulas
    • Microscopic – atoms
    • Macroscopic – moles of atoms
    • 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
  • 37. Empirical Formulas
    • 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 whole numbers are identical, substitute the number 1
    • 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
  • 38. 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
    • there are 25.9 g N and 74.1 g O
    • 2. Convert to mole-to-mole ratio:
    • Divide each by mass of one mole
    • 25.9 g divided by 14.0 g = 1.85 mol N
    • 74.1 g divided by 16.0 g = 4.63 mol O
    • 3. Divide both molar quantities by the
    • smaller number of moles
  • 39.
    • 4. 1.85/1.85 = 1 mol N
    • 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 )
    • 2 x 1 mol N = 2
    • 2 x 2.5 mol O = 5
    • Answer: The empirical formula is N 2 O 5
  • 40. Determine the Empirical Formulas
    • 1. H 2 O 2
    • 2. CO 2
    • 3. N 2 H 4
    • 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?
  • 41. Answers
    • Compound Empirical Formula
    • 1. H 2 O 2 HO
    • 2. CO 2 CO 2
    • 3. N 2 H 4 NH 2
    • 4. C 6 H 12 O 6 CH 2 O
    • 5. HCN
  • 42. Calculating Molecular Formulas
    • 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
  • 43. 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)
    • (Use the same procedure used to calculate molar mass.)
    • 2. Divide the known molar mass by the efm
    • 3. Multiply the formula subscripts by this value to get the molecular formula
  • 44. Answer
    • Molar mass (efm) is 30.0 g
    • 60.0 g divided by 30.0 g = 2
    • Answer: C 2 H 8 N 2
  • 45. Practice Problems
    • 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.
    • 3) Calculate the empirical formula of a compound that is 42.9% C and 57.1% O.
  • 46. Practice Problems
    • 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

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