# IIIIIIIV The Mole I. Molar Conversions What is the Mole? A counting number (like a dozen or a pair) Avogadro’s number 6.02  10 23 1 mole = 6.02  10.

## Presentation on theme: "IIIIIIIV The Mole I. Molar Conversions What is the Mole? A counting number (like a dozen or a pair) Avogadro’s number 6.02  10 23 1 mole = 6.02  10."— Presentation transcript:

IIIIIIIV The Mole I. Molar Conversions

What is the Mole? A counting number (like a dozen or a pair) Avogadro’s number 6.02  10 23 1 mole = 6.02  10 23 things. (things can be atoms, molecules, formula units) A large amount!!!!

1 mole of hockey pucks would equal the mass of the moon! A. What is the Mole? 1 mole of pennies would cover the Earth 1/4 mile deep! 1 mole of basketballs would fill a bag the size of the earth!

Example n How many molecules are in 2.50 moles of C 12 H 22 O 11 ? 2.50 mol 6.02  10 23 molecules 1 mol = 1.51  10 24 molecules C 12 H 22 O 11

Mole- Volume n At the same temperature:  Equal volumes of gases contain equal numbers of moles n STP  Standard temperature and pressure  1 mole of any gas at STP has a volume of : 22.4 L

Example n How many moles of neon are in 44.8L of neon at STP? 44.8 L 1 mol 22.4 L = 2 mol

Molar Mass How much do I weigh??

Molar Mass n Mass of 1 mole (6.02 X 10 23 atoms or molecules) of an element or compound. n Atomic mass tells the...  atomic mass units per atom (amu)  grams per mole (g/mol) n Round to 2 decimal places when calculating

Molar Mass Examples This is easy! Just look at the periodic table! Phosphorus Aluminum Helium Silver 30.97 g/mol 26.98 g/mol 4.00 g/mol 107.87 g/mol

Molar Mass Examples WATER: H2OH2O  Ca(OH) 2 Ca  40.08 + O  2(16) + H  2(1.01) = 74.10 g/mol H → O → = 2.02 g/mol 2 x 1.01 g/mol 1 x 16.0 g/mol = 16.0 g/mol 18.02 g/mol

Molar Mass n sodium bicarbonate n sucrose  NaHCO 3  84.01 g/mol  C 12 H 22 O 11  342.34 g/mol

Example n How many moles of carbon are in 26 g of carbon? 26 g C 1 mol C 12.01 g C = 2.2 mol C

Examples n Find the mass of 2.1  10 24 molecules of NaHCO 3. 2.1  10 24 molecules 1 mol 6.02  10 23 molecules = 290 g NaHCO 3 84.01 g 1 mol

A. Percentage Composition n the percentage by mass of each element in a compound

 100 = A. Percentage Composition %Cu = 127.10 g Cu 159.17 g Cu 2 S  100 = %S = 32.07 g S 159.17 g Cu 2 S 79.852% Cu 20.15% S n Find the % composition of Cu 2 S.

%Fe = 28 g 36 g  100 = 78% Fe %O = 8.0 g 36 g  100 = 22% O n Find the percentage composition of a sample that is 28 g Fe and 8.0 g O. A. Percentage Composition

n How many grams of copper are in a 38.0-gram sample of Cu 2 S? (38.0 g Cu 2 S)(0.79852) = 30.3 g Cu Cu 2 S is 79.852% Cu A. Percentage Composition

IIIIIIIV The Mole Empirical/Molecular Formulas

B. Empirical Formula C2H6C2H6 CH 3 reduce subscripts n Smallest whole number ratio of atoms in a compound

Empirical Formula 1. Find mass (or %) of each element. (may be given) 2. Find moles of each element. (g to moles) 3. Divide moles by the smallest # to find subscripts. 4. When necessary, multiply subscripts by 2, 3, or 4 to get whole #’s.

Empirical Formula n Find the empirical formula for a sample of 25.9% N and 74.1% O. 25.9 g 1 mol 14.01 g = 1.85 mol N 74.1 g 1 mol 16.00 g = 4.63 mol O 1.85 mol = 1 N = 2.5 O

Empirical Formula N 1 O 2.5 Need to make the subscripts whole numbers  multiply by 2 N2O5N2O5

Examples n An organic compound is found to contain 92.25% carbon and 7.75% hydrogen. Determine the empirical formula.  Empirical Formula: CH n What is the empirical formula of a compound that contains 53.73% Fe and 46.27% S?  Empirical Formula: Fe 2 S 3

Molecular Formula n “True Formula” - the actual number of atoms in a compound CH 3 C2H6C2H6 empirical formula molecular formula ?

C. Molecular Formula 1. Find the empirical formula. 2. Find the empirical formula mass. 3. Divide the molecular mass by the empirical mass. 4. Multiply each subscript by the answer from step 3.

Molecular Formula n The empirical formula for ethylene is CH 2. Find the molecular formula if the molecular mass is 28.1 g/mol? 28.1 g/mol 14.03 g/mol = 2.00 empirical mass = 14.03 g/mol (CH 2 ) 2  C 2 H 4

Download ppt "IIIIIIIV The Mole I. Molar Conversions What is the Mole? A counting number (like a dozen or a pair) Avogadro’s number 6.02  10 23 1 mole = 6.02  10."

Similar presentations