Dimensional Analysis (a.k.a. Factor Labeling) Many problems in chemistry involve using relationships to convert one unit of measurement to another. Conversion factors are relationships between two units. For our purposes conversion factors are exact. Hence conversion factors have unlimited significant figures. Conversion factors generated from equivalence statements. e.g., 1 inch = 2.54 cm can give or

Dimensional Analysis (a.k.a. Factor Labeling) Arrange conversion factors so the starting unit cancels. Arrange conversion factor so the starting unit is on the bottom of the conversion factor. May string conversion factors. So we do not need to know every relationship, as long as we can find something else the starting and desired units are related to :

( ) ______ How many cm are in 1.32 meters? equality: 1 m = 100 cm (or 0.01 m = 1 cm) applicable conversion factors: ______ 1 m 100 cm ______ 1 m 100 cm or ( ) ______ 1 m 100 cm 132 cm X cm = 1.32 m = We use the idea of unit cancellation to decide upon which one of the two conversion factors we choose.

Again, the units must cancel. How many meters is 8.72 cm? equality: 1 m = 100 cm applicable conversion factors: ______ 1 m 100 cm ______ 1 m 100 cm or ( ) ______ 1 m 100 cm 0.0872 m X m = 8.72 cm = Again, the units must cancel.

Again, the units must cancel. How many feet is 39.37 inches? equality: 1 ft = 12 in applicable conversion factors: ______ 1 ft 12 in ______ 1 ft 12 in or ( ) ____ 1 ft 12 in 3.28 ft X ft = 39.37 in = Again, the units must cancel.

( ) ( ) ____ ______ How many kilometers is 15,000 decimeters? 10 dm 1 km 1.5 km X km = 15,000 dm =

How many seconds is 4.38 days? ____ ( ) ( ) _____ ( ) ____ 24 h 1 d 1 h 60 min 1 min 60 s X s = 4.38 d 378,432 s = If we are accounting for significant figures, we would change this to… 3.78 x 105 s

Counting Atoms Chemistry is a quantitative science - we need a "counting unit." The MOLE 1 mole is a sample of a substance that contains as many particles as there are in 12.0 g of C-12. A mole equates a gram to an amu. How many particles are in a mole?

Amedeo Avogadro (1776 – 1856) Particles in a Mole A more precise value of Avagadro’s number = 6.022045 E23. How could you determine the number of sugar crystals in a bowl of sugar? Is it possible to get a reasonably close estimate? quadrillions thousands trillions billions millions 1 mole = 602213673600000000000000 or 6.02 x 1023 There is Avogadro's number of particles in a mole of any substance.

How Big is a Mole? One mole of marbles would cover the entire Earth (oceans included) for a depth of two miles. One mole of $1 bills stacked one on top of another would reach from the Sun to Pluto and back 7.5 million times. It would take light 9500 years to travel from the bottom to the top of a stack of 1 mole of $1 bills.

Visualizing a Chemical Reaction 2 Na + Cl2 NaCl 2 ___ mole Na 10 10 ___ mole Cl2 5 5 ___ mole NaCl 10 10 ?

Molecular Weight and Molar Mass Molecular weight (or Formula mass) is the sum of atomic weights of all atoms in the molecule or formula unit. example: NaCl has a molecular weight of 58.5 amu. this is composed of a single formula unit of NaCl Molar mass = molecular weight in grams. Units = g/mol example: NaCl has a molar mass of 58.5 g/mol this is composed of a 6.02 x1023 formula units of NaCl; it is also composed of 6.02 x1023 sodium atoms and 6.02 x1023 chlorine atoms

Mole Diagram 1 mole = 6.022 x 1023 particles (atoms or molecules) 1 mole = molar mass (g) Mass Mole Particles

Percentage Composition (by mass...not atoms) Mg magnesium 24.305 12 Cl chlorine 35.453 17 % Mg = x 100 24 g 95 g % = x 100 part whole 25.52% Mg Mg2+ Cl1- 74.48% Cl MgCl2 It is not 33% Mg and 66% Cl 1 Mg @ 24.305 amu = 24.305 amu 2 Cl @ 35.453 amu = 70.906 amu 95.211 amu

Empirical and Molecular Formulas A pure compound always consists of the same elements combined in the same proportions by weight. Therefore, we can express molecular composition as PERCENT BY WEIGHT. The simplest formula is called the Empirical Formula Ethanol, C2H6O 52.13% C 13.15% H 34.72% O

Empirical Formula Quantitative analysis shows that a compound contains 32.38% sodium, 22.65% sulfur, and 44.99% oxygen. Find the empirical formula of this compound. sodium sulfate 32.38% Na 22.65% S 44.99% O 32.38 g Na 22.65 g S 44.99 g O = 1.408 mol Na / 0.708 mol = 2 Na Na2SO4 Na2SO4 = 0.708 mol S = 1 S = 2.812 mol O = 4 O Step 1) %  g Step 2) g  mol Step 3) mol mol

Empirical Formula A sample weighing 250.0 g is analyzed and found to contain the following: 27.38% sodium 1.19% hydrogen 14.29% carbon 57.14% oxygen 27.38 g Na 1.19 g H 14.29 g C 57.14 g O Assume sample is 100 g. Determine the empirical formula of this compound. Step 1) convert %  gram Step 2) gram  moles Step 3) mol / mol NaHCO3 / 1.19 mol = 1 Na / 1.19 mol = 1 H / 1.19 mol = 1 C / 1.19 mol = 3 O

Empirical & Molecular Formula (contains only hydrogen + carbon) (~17% hydrogen) A 175 g hydrocarbon sample is analyzed and found to contain ~83% carbon. The molar mass of the sample is determined to be 58 g/mol. Determine the empirical and molecular formula for this sample. Determine the empirical formula of this compound. Step 1) convert %  gram Step 2) gram  moles Step 3) mol / mol 2 C @ 12 g = 24 g 5 H @ 1 g = 5 g 29 g Assume sample is 100 g. Then, 83 g carbon and 17 g hydrogen. MMempirical = 29 g/mol / 6.917 mol = 1 C / 6.917 mol = 2.5 H (2.4577 H) CH2.5 C2H5 MMmolecular = 58 g/mol 58/29 = 2 Therefore 2(C2H5) = C4H10 butane

Find the molar mass and percentage composition of zinc acetate Zn2+ CH3COO1- acetate = CH3COO1- Zn(CH3COO)2 1 Zn @ 65.4 g/mol = 65.4 g / 183.4 g x 100% = 35.6 % Zn / 183.4 g x 100% = 26.2 % C / 183.4 g x 100% = 3.3 % H / 183.4 g x 100% = 34.9 % O 4 C @ 12 g/mol = 48 g 6 H @ 1 g/mol = 6 g 4 O @ 16 g/mol = 64 g Zn(CH3COO)2 183.4 g

A compound is found to be 45.5% Y and 54.5% Cl. Its molar mass (molecular mass) is 590 g. Assume a 100 g sample size a) Find its empirical formula 45.5 g Y 1 mol Y = 0.5118 mol Y / 0.5118 mol = 1 Y 88.9 g Y YCl3 54.5 g Cl 1 mol Cl = 1.535 mol Cl = 3 Cl 35.5 g Cl 1 Y @ 88.9 g/mol = 88.9g b) Find its molecular formula 3 Cl @ 35.5 g/mol = 106.5 g 590 / 195.4 = 3 YCl3 195.4 g 3 (YCl3) Y3Cl9

Molar Mass vs. Atomic Mass 6.02x1023 Molar Mass vs. Atomic Mass H2 = _____ 2 g H2 = _______ 2 amu H2O = _____ 18 g H2O = ________ 18 amu MgSO4 = _____ 120 g MgSO4 = ________ 120 amu (NH4)3PO4 = _____ 149 g (NH4)3PO4 = ________ 149 amu Percentage Composition (by mass) Empirical Formula %  g g  mol mol % = x 100 % part whole Empirical vs. Molecular Formula (lowest ratio)