Empirical Formulas and Molar Mass: Part 3-Molar Mass 1

Objectives -Be able to calculate molar mass of a compound and use it as a conversion factor -Calculate percent composition of compounds using molar mass 2

Molar Mass We can use info from the components of a compound to figure out a compound’s total molar mass- the mass in grams of 1 mole of a compound It’s the mass of one mole: grams KC 4 H 5 O 6 /mole KC 4 H 5 O 6 3

Molar Mass Calculate the molar mass of magnesium iodide, MgI 2 from its parts. How many grams are in 1 mole? Mg 1 mole (24.3g Mg/mole )= 24.3 g Mg I 2 mole (127 g I/mole) = 254 g I 278 g/mole MgI 2 (for SD use place value!) 4

Molar Mass Calculate the molar mass (grams in one mole) of ammonium sulfite, (NH 4 ) 2 SO 3 In 1 mole of the compound there are: 2 moles of N X 14.0 g N/mole= 28.0g N 8 moles of H X 1.01 g H/mole = 8.08gH 1 mole of S X 32.1 g S/mole =32.1 g S 3 moles of O X 16.0 g O/mole=48.0g O SD by place value 116.2g units always g/mole 1mole (NH 4 ) 2 SO 3 5

Molar mass can be used to convert between moles and grams For an element: 1 mole = 6.02x10 23 atoms = atomic mass For a compound: 1 mole = 6.02x10 23 formula units = molar mass Molar Mass 6

What is the mass of 1.35 moles of (NH 4 ) 2 SO 3 ? Earlier we found the molar mass: 116.2g (NH 4 ) 2 SO 3 = 1mole (NH 4 ) 2 SO 3 ? g (NH 4 ) 2 SO 3 = 1.35 moles (NH 4 ) 2 SO 3 x g (NH 4 ) 2 SO 3 1 mole (NH 4 ) 2 SO 3 = 157 g (NH 4 ) 2 SO 3 Molar Mass as a Conversion Factor 7

75.2 g (NH 4 ) 2 SO 3 is how many moles of (NH 4 ) 2 SO 3 ? ? moles (NH 4 ) 2 SO 3 = 1 mole (NH 4 ) 2 SO g (NH 4 ) 2 SO 3 X g (NH 4 ) 2 SO 3 =.647 moles (NH 4 ) 2 SO 3 Molar Mass as a Conversion Factor 8

% Composition When we have multiple elements in a compound, each one has a certain ratio (empirical formula) From this we can figure out each element’s percent composition 9

% Composition Find percent composition of Al(C 2 H 3 O 2 ) 3 (1 moles Al) (6 moles C) (9 moles H) (6 moles O) (27.0 g/mole) = (12.0 g/mole) = (1.01 g/mole) = (16.0 g/mole) = 27.0 g Al 72.0 g C 9.09 g H 96.0 g O g Al(C 2 H 3 O 2 ) 3 /204.1 g Al(C 2 H 3 O 2 ) 3 x 100 = 13.2 % 35.3 % 4.45 % 47.0 % ~100% 27.0 g Al 72.0 g C 9.09 g H 96.0 g O 10

% Composition How many grams of aluminum can be obtained from 1.50 moles of aluminum acetate? ? grams Al = 1.50 moles Al(C 2 H 3 O 2 ) 3 x = 40.5 g Al 1.50 moles Al(C 2 H 3 O 2 ) 3 x 27.0 g Al 1 mole Al = 40.4 g Al OR 11

Objectives -Be able to calculate molar mass of a compound and use it as a conversion factor -Calculate percent composition of compounds using molar mass 12