Mole, Avogadro’s Number Mass percent composition Molar Mass Mass percent composition Empirical Formula

Why is Knowledge of Composition Important? Everything in nature is either chemically or physically combined with other substances Some Applications: the amount of sodium in NaCl for diet the amount of iron in iron ore for steel production the amount of carbon in fossil fuel in terms of green house effect

Counting Pennies by Weight What if a person doesn’t have bills but pounds of pennies when he wants to buy a $10 pizza? Seinfeld: Kramer’s attempt to buy calzone with pennies https://www.youtube.com/watch?v=ywidjw9oQVw http://www.youtube.com/watch?v=kMimygVTgbU Assuming each penny weighs exactly the same (2.500 g), we can count pennies by weight the total weight of pennies…

Counting Pennies by Weight Kramer brought 2,500. g (ca. 5.5 lbs) of new pennies to the Italian restaurant. Do you think he can buy three calzones ($10.0 total)? Each new penny weighs 2.500 g.

Counting Coins by Weight What if Kramer bought a different coin? Would the mass of single dime be 2.500 g? Would there be $10.00 in 2,500. g dimes? How would this affect the weight-dollar conversion factors?

Mass of atoms Atoms of the same element have on average the same mass. Mass of atom in amu (atomic mass unit): 1 amu = 1.66054×10-27 kg Mass of a C-12 atom = 12 (exact) amu Mass of a gold atom on average = 197.0 amu Do I have to memorize these numbers?!

Counting atoms by mass Similar to how we count the number of pennies, we can count the number of atoms using their mass The number of atoms in a sample often is an astronomical number, so we use a big unit to account for the number of atoms. Just like 1 dozen = 12 items

“Chemical Dozen” – the Mole The number of particles in 1 mole: Avogadro’s Number = _______________ units 1 mole of any atomic element has 6.022 x 1023 atoms 1 mole of oxygen gas has 6.022 x 1023 O2 molecules 1 mole of NaCl solid has 6.022 x 1023 Na+ ions and 6.022 x 1023 Cl- ions

Avogadro’s Number as Conversion Factor Given number of Moles  number of Units Given number of Units  Mole # Mole ´ = #Units

1 mole Ag atoms = 6.022 x 1023 Ag atoms Example: A silver ring contains 0.041 moles of silver. How many silver atoms silver are in the ring? Information Given: 0.041 moles of Ag atoms Find: ? number of Ag atoms 1 mole Ag atoms = 6.022 x 1023 Ag atoms 2.5 x 1022 Ag atoms (2 sig. figs)

Example: How many moles of water are in 1.234 x 1020 water molecules? Information Given: 1.234 x 1020 water molecules Find: ? moles of water molecules 1 mole water = 6.022 x 1023 water molecules 2.049 x 10-4 moles water (4 sig. figs)

Mole, Atomic Mass, Mass The mass of exactly one mole of any element equals the Atomic Mass in grams. Example: Exactly 1 mole He = _______ g Exactly 1 mole Li = _______ g If given the mass of a particular number of atoms, we can determine the number of atoms in any mass of the element!

The Mass of one mole substance depends on the Element

Molar Mass The mass of one mole of atoms is called the Molar mass. Unit as g/mole or g/mol. The molar mass of an element, in grams, is numerically equal to the element’s atomic mass. Example: Molar mass of Sodium = ___________

Molar Mass as Conversion Factor Example: molar mass He = 4.003 g So 1 mole He = 4.003 gram He Given number of Moles  Mass Mass (gram) = _________________ Given Mass  Mole Mole = ___________________

Example: Calculate the number of moles of sulfur in 57.8 g of sulfur Information Given: 57.8 g S Find: ? moles S Conv. Fact.: 1 mole S = 32.06 g 1.80 moles S (3 sig. figs)

Example: Calculate the mass of 3.34 x 10-3 mole of helium gas. Information Given: 3.34 x 10-3 moles He Find: ? Grams of He Conv. Fact.: 1 mole He = 4.00 g 0.0133 g He

Practice: #atoms, moles, molar mass The largest uncut diamond (pure carbon) weighs 755 carat (1 carat = 0.2 grams exactly). How many moles of carbon atoms are in this diamond? What is mass for 1.2 x 10-3 mole gold? (optioinal) 70% of the mass of the Sun is hydrogen. The mass of the Sun is 1.99 x 1030 kg. How many moles of hydrogen atoms are in the Sun? 12.6 mol C 0.24 g mol Au 1.38 x 1033 mol H atom

Molar Mass and Avogadro’s Number Since 1 mole contains Avogadro’s number of units whilst has a mass of molar mass there is direct connection/conversion factor between Avogadro’s number and molar mass. Example: 6.022 x 1023 Al atoms = _______ g Al

Example: How many aluminum atoms are in an aluminum can with a mass of 16.2 g? Information Given: 16.2 g Al Find: ? atoms Al 1 mol Al = 26.98 g = 6.022 x 1023 Al atoms g Al mole Al atoms Al 3.62 x 1023 atoms Al

Example: What is the mass of 1.34 x 1020 silver atoms? Given: 1.34 x 1020 Ag Find: ? grams Ag 1 mol Ag = 107.87 g Ag = 6.022 x 1023 Ag atoms SM: atoms  mol  g 0.0240 g Ag

Practice: Mass vs. Molec/Atom Diamond is made of pure carbon. The largest uncut diamond weighs 755 carat (1 carat = 0.2 grams exactly). How many carbon atoms are in this diamond? What is the mass of one gold atom in grams? 7.57 x 1024 C atoms 3.2078 x 10-22 grams

Molar Mass of Compounds Molar Mass: Mass of molecules per exactly one mole of this molecule. Unit = g/mol Example: 1 mole of H2O = 2 moles H + 1 mole O Molar Mass of H2O = 2(1.008 g/mol H) + 16.000 g/mol O = 18.016 g/mol 1 mole Al(NO3)3 = 1 mole Al + 3 mole N + 9 mole O Molar Mass of Al(NO3)3 = 1(26.98) + 3(14.01) + 9(16.00) = 213.01 g/mol

Practice: Find the Molar Mass Sulfuric acid Aluminum sulfate Ammonium chlorite Hydrobromic acid Nickel(II) phosphate

Mass of Compound ↔ Moles of Compound

Example: Calculate the mass (in grams) of 1.75 mol of water Information Given: 1.75 mol H2O Find: ? g H2O C F: 1 mole H2O = 18.02 g H2O 31.5 g H2O

Example: Calculate the mole of CO2 in 454 g dry ice (solid carbon dioxide). Information Given: 454 g CO2 Find: ? mol CO2 C F: 1 mole CO2 = 44.01 g CO2 10.3 mol CO2

Mass of Compound ↔ Number of molecules

Example: Find the mass of 4.8 x 1024 NO2 molecules Information Given: 4.8 x 1024 molec NO2 Find: ? g NO2 1 mole NO246.01 g NO2 3.7 x 102 g

Example: How many water molecules in 1.00 mL DI water? Information Given: 1.00 mL H2O Find: #H2O molecules 1 mole H2O = 18.02 g H2O Density of water = ______ g/mL 3.34 x 1022 H2O molecules

Practice Online: http://www.sciencegeek.net/Chemistry/taters/Unit4GramMoleVolume.htm How many moles of water in 10.00 mL pure water if the density of water is 1.00 g/mL? Determine the mass of 2.0 × 103 mole NaCl.

Chemical Formulas as Conversion Factors 1 spider  8 legs + 1 head + 1 abdomen 1 chair  4 legs + 1 seat + 1 back 1 H2O molecule  2 H atoms + 1 O atom

Mole Relationships in Chemical Formulas Given: #moles of the compound Chemical formula of the compound #moles of a constituent element within the compound Moles of Compound Moles of Constituents 1 mol NaCl 1 mole Na, 1 mole Cl 1 mol H2O 2 mol H, 1 mole O 1 mol CaCO3 1 mol Ca, 1 mol C, 3 mol O 1 mol C6H12O6 6 mol C, 12 mol H, 6 mol O

*Example: How many oxygen atoms are in 100. g of iron(II) phosphate. Chemical formula: __________ Molar mass = 389.56 g/mol For exactly 1 mole compound, there are ___ mole of oxygen 0.2567 mol _________ 2.054 mol O 1.24 x 1024 O atoms

Mass Percent Composition Mass% represents the mass of constituent element in grams per exactly 100 g of compound: Example: mass percent of nitrogen in Al(NO3)3

Example: Find mass% Find Mass% of Nitrogen in Ammonium phosphate. Formula: Molar mass = 181.17 g/mol For exactly 1 mole compound, there are ___ mole of Nitrogen Exactly 1 mole compound weighs ______ g ____ moles nitrogen weighs 42.03 g Mass% of Nitrogen = 42.03 g/181.17 g x 100% = 23.20%

Example: Use mass% to find mass of constituent element Given the mass% of nitrogen in Ammonium phosphate (23.20%), find the mass of nitrogen in 454 g of ammonium phosphate. Meaning of mass%: A conversion factor  exact 100 g ammonium phosphate = 23.20 g N 454 g _______ x Mass Nitrogen = 105 g

Empirical Formula Chemical Formula  Mass% of each element Mass% of each element  Chemical Formula? Empirical Formula: The simplest, whole-number ratio of atoms in a molecule It shows the mole ratio in the compound It can be determined from percent composition or combining masses

Empirical Formula vs. Molecular Formula Hydrogen Peroxide (in store) Molecular Formula = H2O2 Empirical Formula = HO Benzene (in gasoline) Molecular Formula = C6H6 Empirical Formula = CH Glucose (essential energy in body) Molecular Formula = C6H12O6 Empirical Formula = CH2O

Finding an Empirical Formula mass% of each element  grams of each element skip if already grams grams  moles of each element use atomic mass of each element pseudoformula w/ moles as subscripts divide all by smallest number of moles multiply all mole ratios by number to make all whole numbers if ratio ?.5, multiply all by 2; if ratio ?.33 or ?.67, multiply all by 3, etc. skip if already whole numbers

How to Determine Empirical Formulas. A From the masses of constituents mass A (g) moles A MMA moles A moles B mass B (g) moles B MMB

Empirical Formula: Mass composition One 1.205-g sample of iron-oxygen compound contains 0.872 g iron. Find the empirical formula of this compound. Solution Map: g Fe, O mol Fe, O mol ratio empirical formula Atomic mass (g/mol): O 16.00, Fe 55.85 mass of O = 1.205 – 0.872 = 0.333 g 0.0156 mol Fe 0.0208 mol O

Example: One 1. 205-g sample of iron-oxygen compound contains 0 Example: One 1.205-g sample of iron-oxygen compound contains 0.872 g iron. Find the empirical formula of this compound. Find: 0.0156 mol Fe, 0.0208 mol O (Contd.)

Practice: A 1. 144-g sample of manganese ore contains 0. 320 g oxygen Practice: A 1.144-g sample of manganese ore contains 0.320 g oxygen. Find its empirical formula.

Example: Empirical formula from mass% Find the empirical formula of aspirin with the given mass percent composition: 60.00% C, 4.48% H, 35.53% O g C, H, O mol C, H, O mol ratio empirical formula

Example: Find the empirical formula of aspirin with the given mass percent composition. Given: 60.00% C, 4.48% H, 35.52% O 1 mol C = 12.01 g; 1 mol H = 1.008 g; 1 mol O = 16.00 g Assuming 100 (exact) grams of compound aspirin, mass of each element in grams equals percentage. Mass C = 60.00 g Mass H = 4.48 g Mass O = 35.52 g C4.996H4.44O2.221 C9H8O4

From Empirical Formula to Molecular Formula Note: Empirical formula comes from REDUCED Molecular formula The “Molar mass” based on Empirical formula is multiplied by an integer n should give the Molecular Formula

Example: Find Empirical Formula and Molecular Formula Vitamic C has the following mass percent composition: C 40.91%, H 4.58%, O 54.41%. Find the empirical formula. Ans: C3H4O3 The molar mass of vitamin C is 176.1 g/mol. Find the molecular formula of vitamin C. Ans: solve n = 2, so molecular formula C6H8O6

Same Empirical Formula, Different Molecular Formula! Name Molecular Formula Empirical Molar Mass, g glyceraldehyde C3H6O3 CH2O 90 erythrose C4H8O4 120 arabinose C5H10O5 150 glucose C6H12O6 180 Name Molecular Formula Empirical glyceraldehyde C3H6O3 CH2O erythrose C4H8O4 arabinose C5H10O5 glucose C6H12O6

Answer key: Molar Mass (all in g/mol) Sulfuric acid: 98.09 g/mol Aluminum sulfate: 342.17 g/mol Ammonium chlorite: 181.70 g/mol Hydrobromic acid: 80.91 g/mol Nickel(II) phosphate: 366.01 g/mol

Example: Find the empirical formula of aspirin with the given mass percent composition. Information Given: 60.00 g C, 4.48 g H, 35.53 g O Find: empirical formula, CxHyOz CF: 1 mol C = 12.01 g; 1 mol H = 1.008 g; 1 mol O = 16.00 g SM: g C,H,O  mol C,H,O  mol ratio  empirical formula calculate the moles of each element

Apply the Solution Map: Example: Find the empirical formula of aspirin with the given mass percent composition. Information Found: 4.996 mol C, 4.44 mol H, 2.221 mol O Find: empirical formula, CxHyOz Apply the Solution Map: write a pseudoformula C4.996H4.44O2.221

Apply the Solution Map: Example: Find the empirical formula of aspirin with the given mass percent composition. Information Given: C2.25H2O1 Find: pseudoempirical formula, C2.25H2O1 CF: Apply the Solution Map: multiply subscripts by factor to give whole number { } x 4 C9H8O4