MOLE AND STOIKIOMETRI We measure mass in grams. g We measure volume in liters. L We count pieces in MOLES. mol

MOLE CONCEPT A mole of Carbon atoms has a mass of 12 grams A Carbon-12 atom has a mass of 12amu C = 12amu A mole of Carbon atoms has a mass of 12 grams C

Gram Atomic Mass Equals the mass of 1 mole of an element Written on the Periodic Table 12.01 grams of C has the same number of pieces as 1.008 grams of H and 55.85 grams of iron. We can write this as 12.01 g C = 1 mole C We can count things by weighing them.

Atomic Mass Mole of atoms in grams 27.0g A mole of aluminum = The Mole Concept Atomic Mass in grams Mole of atoms 27.0g A mole of aluminum = A mole of gold = A mole of silver = A mole of boron = 197g 108g 10.8g

If a Helium atom is 4 times heavier than a Hydrogen atom, then a dozen Helium atoms is 4 times heavier than a dozen Hydrogen atoms. He H

And a mole of Helium atoms is 4 times heavier than a mole of Hydrogen atoms.

The number of atoms in a mole is called, So, the number of atoms in a mole is always the same The number of atoms in a mole is called, Avogadro's Number Avogadro's Number = 6.022 x 1023

Mole – Particle Conversions The mole concept allows us to count atoms, molecules, formula units and ions. 1 mole of Al = 6.022 x 1023 atoms of Al So how many atoms in 3 mol of Al?

Examples How much would 2.34 moles of carbon weigh? How many moles of magnesium is 24.31 g of Mg? How many atoms of lithium is 1.00 g of Li?

What about compounds? In 1 mole of H2O molecules there are 2 moles of H atoms and 1 mole of O atoms To find the mass of one mole of a compound determine the moles of the elements they have Find out how much they would weigh add them up

Types of Molar Mass Gram Molecular Mass (GMM)- the total number of atoms in a molecular compound Cl2 = 2 x 35.4 g = 70.8 g CH4= Gram Formula Mass (GFM)- the total number of atoms in an ionic compound NaCl = (1 x 23.0g) + (1 x 35.4g) = 58.4g Ca(OH)2= (1 x 12.0g) + (4 x 1.0g) = 16g (1 x 40.0g) + (2 x (16.0g+1.0g))= 74g

Molar Mass Molar mass is the generic term for the mass of one mole of any substance (in grams) The same as: 1) gram molecular mass 2) gram formula mass 3) gram atomic mass

= 78.1 g = 92.0 g = 164.1 g = 180.0 g = 149.0 g Examples Na2S N2O4 Calculate the molar mass of the following and tell what type it is: Na2S N2O4 Ca(NO3)2 C6H12O6 (NH4)3PO4 = 78.1 g = 92.0 g = 164.1 g = 180.0 g = 149.0 g

Gases Many of the chemicals we deal with are gases. They are difficult to weigh. Need to know how many moles of gas we have. Two things effect the volume of a gas Temperature and pressure We need to compare them at the same temperature and pressure.

Standard Temperature and Pressure 0ºC and 1 atm pressure 0oC = 32oF= 273 K 1 atm= 101.3 kPa = 760 mmHg (torr) At STP 1 mole of gas occupies 22.4 L Called the molar volume 1 mole = 22.4 L of any gas at STP

Examples What is the volume of 4.59 mole of CO2 gas at STP? How many moles is 5.67 L of O2 at STP? 4.59 mol CO2 22.4 L CO2 =103 L CO2 1 mol CO2 5.67 L O2 1 mol O2 =0.253 mol O2 22.4 L O2

Stoichiometry “Stoichiometry” is Greek for “Measuring Elements” It starts with a balanced equation 2H2 + O2 ® 2H2O 2 moles of hydrogen reacts with 1 mole of oxygen Forming 2 moles of water.

Stoichiometry 3 g of reactants can’t make only 2 g of products The coefficients tell us how many moles of each substance. Not Grams! For example: 2H2 + O2 ® 2H2O 2 g of H2 + 1 g of O2 = 2 g of H2O 3 g of reactants can’t make only 2 g of products

Mass of a Product 2 moles H2 2.02 g H2 = 4.04 g H2 1 mole H2 1 mole O2 The Law of Conservation of Mass applies Convert the moles to grams and the equation does work. 2H2 + O2 ® 2H2O 2 moles H2 2.02 g H2 = 4.04 g H2 1 mole H2 1 mole O2 32.00 g O2 = 32.00 g O2 1 mole O2 36.04 g H2+O2

Mass of a Product 2H2 + O2 ® 2H2O 2H2 + O2 ® 2H2O 18.02 g H2O 2 moles H2O 18.02 g H2O 36.04 g H2O = 1 mole H2O 2H2 + O2 ® 2H2O 36.04 g H2 + O2 = 36.04 g H2O

2Na + Cl2  2NaCl 2Na (s) + Cl2(g) 2NaCl (s) How many grams of sodium are needed to react with .071g of chlorine gas? 2Na (s) + Cl2(g) 2NaCl (s) Mole Ratio 2 1 2 Mass Ratio 23 x 2= 46 35.5 x 2 x 1= 71 58.5 x 2= 117 1 mol Cl2 2 mol Na 0.071 g Cl2 46 g Na 71 g Cl2 1 mol Cl2 2 mol Na = 0.046 g Na

Chemical Yield Actual Yield x 100 Theoretical Yield 3. Percent yield – a percentage/ratio between the actual yield and the theoretical yield. % Yield = % yield tells us how “efficient” a reaction is. % yield can not be bigger than 100 %. Actual Yield x 100 Theoretical Yield

Example 11.8 g NaCl x 100 13.6 g NaCl Answer = 86.7 % According to your calculations the theoretical yield for the production of NaCl is 13.6 grams. In the laboratory your actual yield is 11.8 grams of NaCl. What is the percent yield? 11.8 g NaCl x 100 13.6 g NaCl Answer = 86.7 %

Calculating Percent Composition of a Compound Like all percent problems: Part whole Find the mass of each component, then divide by the total mass. x 100 %

Calculating Percent Composition of a Compound Find the mass percent for the elements in Sodium hydrogen carbonate: NaHCO3 -molar mass is 84 g/mol -mass of: Na= 23g H= 1g C= 12g O= 3 x 16= 48g %C = (12g/84g) x 100= 14.3% %Na = (23g/84g) x 100= 27.4% %H = (1g/84g) x 100= 1.2% %O = (48g/84g) x 100= 57.1%

The Empirical Formula The lowest whole number ratio of elements in a compound. The molecular formula = the actual ratio of elements in a compound. The two can be the same. CH2 is an empirical formula C2H4 is a molecular formula C3H6 is a molecular formula H2O is both empirical & molecular

Calculating Empirical Just find the lowest whole number ratio C6H12O6 CH4N It is not just the ratio of atoms, it is also the ratio of moles of atoms. In 1 mole of CO2 there is 1 mole of carbon and 2 moles of oxygen. In one molecule of CO2 there is 1 atom of C and 2 atoms of O.

Calculating Empirical We can get a ratio from the percent composition. Assume you have a 100 g. The percentages become grams. Convert grams to moles. Find lowest whole number ratio by dividing by the smallest mole value

Example Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N. Assume 100 g 38.67 g C x 1mol C = 3.220 mole C 12.01 g C 16.22 g H x 1mol H = 16.09 mole H 1.01 g H 45.11 g N x 1mol N = 3.219 mole N 14.01 g N

Example The ratio is 3.220 mol C = 1 mol C 3.219 mol N 1 mol N The ratio is 16.09 mol H = 5 mol H 3.219 mol N 1 mol N = C1H5N1 = CH5N Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula?