The Mole Concept Exam 1 and Problem Solutions
The Mole Concept Exam 1 and Problem Solutions
1. If atomic mass of Mg atom is 24 g, find mass of 1 Mg atom.
Solution:
We can solve this problem in to ways;
1st way:
6,02x1023 amu is 1 g
24 amu is ? g
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
?=4x10-23 g
2nd way;
1 mol Mg (6,02x1023Mg atoms) is 24 g
6,02x1023 Mg atoms 24 g
1 Mg atom ? g
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
?=4x10-23 g
2. Find mass of 1 molecule C2H6. (C=12, H=1)
Solution:
C2H6=2.12+6.1=30
6,02x1023 C2H6 molecule is 30 g
1 C2H6 molecule is ? g
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
1 C2H6 molecule is =?=5.10-23 g
3. Find mole of 6,9 g Na. (Na=23)
Solution:
23 is the atomic mass of Na, in other words 1 mole Na is 23 g.
23 g Na is 1 mol
6,9 g Na is ? mol
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
?=0,3 mol
4. Find mass of 0,2 mol P4 . (P=31)
Solution:
Molecule mass of P4 =4.31=124 g
1 mol P4 is 124 g
0,2 mol P4 is ? mol
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
?= 24,8 g
5. Find mole of 4,48 liters O2 under normal conditions.
Solution:
Under normal conditions, 1 mol gas is 22,4 liters. We use following formula to find moles of gas under normal conditions;
n=V/22,4
n=4,48/22,4=0,2 mol
6. Find mass of Fe in the compound including 4,8x1023O atoms ;Fe3O4 .
Solution:
We first find mole of O in the compound;
nO=(4,48x1023)/(6,02x1023)=0,8mol
Now we find mole of compound that contains 0,8mol O;
nFe3O4=(moles of O)/(moles of O in compound)=0,8/4=0,2mol
Mole of Fe in compound is;
There are 3 mol Fe in 1 mol Compound
there are ? mol Fe in 0,2mol compound
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
?=0,6 mol Fe in compound
mass of 0,6 mol Fe is;
1 mol Fe is 56 g
0,6 mol Fe is ? g
¯¯¯¯¯¯¯¯¯¯¯¯¯
?=33,6g There are 33,6 g Fe in compound