| Question: What is the mass of 3.01 x 1023 oxegen atoms? |
| Method: Use the atomic mass of oxygen
from the periodic table. (O: 16.0 g/mol) |
| Solution:
3.01 x 1023 O atom 1 mole 16.0 g
------------------ x ---------------- x ------ = 8.00 g
6.02 x 1023 atom mole
|
| Question: How many flourine atoms are in a sample of 57.0 g of flourine? |
| Method: Use the atomic mass of flourine
from the periodic table. (F: 19.0 g/mol) |
| Solution:
57.0 g mole 6.02 x 1023 atom
------ x ------ x ---------------- = 1.81 x 1024 atom
19.0 g mole
|
| Question: How grams of hydrogen are in a sample of 42.0 g of ammonia? |
| Method: Use the molecular mass of
ammonia (NH3) caluculated from the atomic masses
of nitrogen and hydrogen from the periodic table (H: 1.01 g/mol; N: 14.0 g/mol) |
| Solution:
42.0 g mole NH3 3 mole H 1.01 g
------ x -------- x -------- x ------ = 7.49 g
17.0 g mole NH3 mole H
|
| Question: How grams of oxygen are in a sample of 97.0 g of lead (II) nitrate? |
| Method: Use the formula weight of
lead (II) nitrate [Pb(NO3)2] caluculated
from the atomic masses of lead, nitrogen and oxygen
from the periodic table (Pb: 207.19 g/mol; N: 14.01 g/mol; O: 16.00 g/mol) |
| Solution:
97.0 g mole Pb(NO3)2 6 mole 0 16.00 g
------ x ------------- x ------------- x ------- = 28.1 g
331.21 g mole Pb(NO3)2 mole O
|