Chemistry Reference
In-Depth Information
8.
[D. CH and C
2
H
2
]—This problem can be solved by combining the work we
did for a question such as #7 with the work we did for a question such as
#5, as shown here:
92.3 g
12.0 g/mole
= 7.69 moles
# of moles of carbon =
7.7 g
1.0 g/mole
# of moles of hydrogen =
= 7.7 moles
7.69 moles
7.69 moles
subscript for carbon =
= 1
7.7 moles
7.69 moles
subscript for hydrogen =
= 1
Empirical Formula = CH
Now that we know that the empirical formula for the compound is CH,
we determine the mass of the empirical formula to be 13.0 u (12.0 u + 1.01 u
= 13.0 u). The problem stated that the molecular mass was 26.0 u, which is
double the empirical formula mass, so we multiply each subscript by 2 to get
the molecular formula of 2(CH) = C
2
H
2
.
9. [HO]—Divide each subscript by 2.
10. [CH
3
]—Divide each subscript by 4.
Lesson 5-5 Review
1.
[molecular]—This is the mass of one molecule of a substance.
2.
[formula]—The fact that ionic compounds don't contain molecules makes
this additional term necessary.
3.
[molecule]—Each water molecule contains 3 atoms.
4.
[formula unit]—The formula unit indicates the smallest whole number ratio
of the ions found in the ionic compound.
5.
[294 u]—Work shown here:
Ca = 40.1 u/atom x 1 atom =
40.1 u
I = 127 u/atom x 2 atoms =
+ 254 u
Total =
294 u
6.
[24.0 u]—Work shown here:
Li = 6.94 u/atom x 1 atom =
6.94 u
O = 16.0 u/atom x 1 atom =
16.0 u
H = 1.01 u/atom x 1 atoms =
+ 1.01 u
Total =
24.0 u
