Unit #3 CDA Review Material

MgCl2 Mg – 1 (24.31) = 24.31 Cl – 2 (35.45) = 70.90 or 95.21 g/mol Molar Mass MgCl2 Mg – 1 (24.31) = 24.31 Cl – 2 (35.45) = 70.90 or 95.21 g/mol 1 mole MgCl2 = 95.21 g MgCl2

Percent Composition The percent BY MASS of each element in a compound divide the element’s total mass (part) by the molar mass (whole) then multiple by 100 to get the percent. Example – Find the % composition of MgCl2 (Using the info from molar mass on the previous slide Molar mass = 95.21 g/mol) Mg - 24.31 / 95.21 x 100 = 25.53% Mg Cl - 70.90 / 95.21 x 100 = 74.47% Cl (PART) (WHOLE)

Percent of an element in a compound Taking the % OF an amount Change the percent to decimal (move 2x to left) Multiply that decimal by the given amount What is the mass of chlorine in 27.8g of MgCl2 (Use the % of Cl you just found on the previous slide) 74.47% Cl .7447 x 27.8g = 20.7g Cl

~ Empirical (CANNOT be reduced) ~ Molecular (CAN be reduced) Identifying Empirical & Molecular Formulas ~ Empirical (CANNOT be reduced) ~ Molecular (CAN be reduced) C2H4 NO3 S9Cl12 C3Cl9 N4S9 CH2 - empirical molecular empirical S3Cl4 - empirical molecular CCl3 - empirical molecular empirical

Calculating Empirical Formula (% comp  empirical formula) Change %  g for each element Divide by molar mass of that element Identify the smallest number of moles and divide all by that number Round each to the nearest whole # (sometimes you have to multiply to get a whole number - special) The resulting whole #are the subscripts for that element in the empirical formula

Calculating Empirical Formula 63.5% Silver 8.2% Nitrogen 28.3% Oxygen 63.5 g Ag 8.2 g N 28.3 g O 107.87 14.01 16.00 .589 mole Ag .59 mole N 1.77 mole O .589 .589 .589 1 1 3 AgNO3

Calculating Molecular Formula Find the empirical formula Calculate the molar mass of your empirical formula Identify the molar mass of your molecular (GIVEN in the problem every time!) Divide the molecular mass / empirical mass Round to the nearest whole # Multiply the whole # by the subscripts in the Empirical formula

formula: NO3 molar mass: 62.01g Practice If a compound has an empirical formula of NO3 and a molecular mass of 186g – what is the molecular formula? formula: NO3 molar mass: 62.01g Molecular mass (given) 186g = 3 empirical mass 62.01 3 x NO3 = N3O9

Mole Conversions

Mole Conversions 1 mole = molar mass (grams) 1 mole = 6.02 x 1023 particles (atoms, molecules, fu) 1 mole = 22.4 Liters

Using molar mass How many grams are in 15.7 mole MgCl2 How many moles are in 0.75 grams of silver? What is the mass of 30.7 mole water?

Using Avagadro’s Number Remember there are 6.02 x 1023 particles (atoms, molecule, f.u) in 1 mole How many atoms are in 1 mole? How many atoms are in 2.10 moles of Copper? How many moles are in 4.21 x 1026 atoms of aluminum?

Using molar volume How many liters are in 1 mole of any gas? How many liters are in 12.95 L of oxygen gas? How many moles are in 0.758 moles of nitrogen gas?