Problem 1 solution: Use the periodic table to obtain the molar mass of lithium,carbon and oxygen.
( molar mass is the number of grams equal to the atomic mass of that element)
molar mass of Li = 6.8 g , C = 12.0 g and O = 16.0 g
in the formula we have 2 atoms of Li , 1 atom of C and 3 atoms of O.S0,
molar mass of Li in Li2CO3 is =2 x 6.9 g = 13.8 g ofi , molar mass of C in Li2CO3 = 1 x 12.0 = 12.0 g of , molar mass of O in Li2CO3 = 3 x 16.0=48.0 g of O
Obtain molar mass of Li2CO3: add the masses of 2 moles of Li,1 mole of C and 3 mole of O.
13.8 + 12.0 + 48.0 = 78. 8 g/mol
Problem 2 solution :
a) SO3 : using periodic table molar mass of sulfur and oxygen are 32.0 g and 16.0 g respectively.
we have one atom of sulfur as subscript of sulfur in the formula is 1 and we have 3 atoms of oxygen as subscript of oxygen in the formula is 3.So,
molar mass of sulfur in SO3 = 1 x 32.0g = 32.0 g of S and molar mass of O in SO3 = 3 x 16.0 g = 48.0 g of S
Obtain molar mass of SO3 by adding masses of one mole of S and 3 moles of O
32.0 + 48.0 = 80.0 g of SO3
b) NaF : Molar mass of Na is 23 g and F is 19 g from periodic table
Molar mass of Na in NaF = 1x 23g = 23.0 g of Na , molar mass of F in NaF = 1x 19 g = 19g of F
Obtaining molar mass of NaF :
23.0 + 19 . 0 = 42 g of NaF per mole
c) Mg3(PO4)2 : Molar mass of Mg is 23.0 g , P is 31.0 g and O is 16.0 g using peridoic table.
Molar mass of Mg in Mg3(PO4)2 is = 24.0 x 3 = 72.0 g of Mg , molar mass of P in Mg3(PO4)2 is = 31.0 g x 2 = 62 .0 g and molar mass of O in Mg3(PO4)2 is = 16 x 8 = 128.0 g of O
obtaining molar mass of Mg3(PO4)2:
72.0g + 62.0 g + 128.0 g = 262 g of Mg3(PO4)2 per mole
d) H20 : Add together the atomic masses of all of the atoms of hydrogen and oxygen in a molecule of water: 2 x 1.008 g (hydrogen) + 1 x 16.00 g (oxygen) = 18.02 g
2 x 1.008 g (hydrogen) + 1 x 16.00 g (oxygen) = 18.02 g