Molecular mass of organic compound 92.4% carbon and 7.6% hydrogen calculate its molecular formulla
Answers
plz...give molecular mass only when I will able to findout your molecular formula.
☆☆☆I am waiting plz....give mass as soon as possible☆☆☆
--------But I can calculate empirical formula and guid you how to determine molecular formula--------
Lets start,
☆C = 92.4%
molar mass of carbon = 12
=92.4/12 = 7.7
☆H = 7.6%
molar mass of H = 1
= 7.6/1 = 7.6
Now divide both no. by the least one of them,
☆7.7/7.6 = 1
☆7.6/7.6 = 1
So the Empirical formula will be = CH
☆for molecular formula☆
n = molecular mass of comp./empirical formula
***Now , we can get molecular formula by multiplying = n×empirical formula
I just hope it will help you****
Thanks!
Answer:
Assuming that is a mass amount, divide each proportion by the atomic weight of the component. Normalize to obtain the empirical formula and then use the molecular mass to determine the actual molecular formula.
93.91
12
=
7.83
relative moles of carbon
6.29
1
=
6.29
relative moles of hydrogen
Ratio of C:H is 7.83:6.29 or 1.2. Nearest whole multiple is 5, resulting in ratio of 6:5. 6-C and 5-H would have a molecular weight of
(
6
×
12
)
+
(
5
×
1
)
=
77
The actual molecular weight is 128, so the formula must be
128
77
=
1.67
×
C
6
H
5
.
C
10
H
8
Subsequent note on Whole Multiple:
A chemical compound MUST have whole numbers of atoms - we can't have a compound with a "partial" atom! The observed/calculated math just gives us ratios of the mass. We need to insure that our final mass contains WHOLE atoms. Thus, when we find our molar ratio of C:H is 7.83:6.29 or 1.2:1 we need to convert that to one with whole numbers in the numerator as well as the denominator.
In this case, with the denominator already "normalized" at 1, we only need to find a factor to make the 1.2 a whole number. The key that 0.2 part - we need to make it
1.0
.
0.2
×
?
=
1.0
?
=
1.0
0.2
=
5
Applying this to our original
1.2
1
ratio we obtain:
5
×
1.2
1
=
6
5
So, our atomic molar ratio is 6-C and 5-H, as used in the rest of the original answer derivation.