The molecular mass of a compound is 92 g/mol. Analysis of the compound shows that it consists of 0.608 g of N and 1.388 g of O.
1 answer:
The empirical and molecular formula would be and
respectively.
The compound contains N and O.
N O
0.608/14 = 0.0434 1.388/16 = 0.0867
Divide by the smallest.
N = 1 O = 2
Thus, the empirical formula would be
To get the molecular formula:
Empirical formula mass = 14 + (16x2) = 46
n = molar mass/empirical formula mass
= 92/46 = 2
Thus, the molecular formula would be
More on empirical formulas can be found here: brainly.com/question/14044066
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Answer:
2.5 % butane, 42.2 % pentane and 55.3 % hexane
Explanation:
Hello,
In this case, the mass balance for each substance is given by:
Whereas accounts for the fractions at the outlet distillate and
for the fractions at the outlet bottoms. Moreover, with the 90 % recovery of butane, we can write:
So we can compute the product of the molar fraction of butane at the distillate by total distillate flow by assuming a 100-mol feed:
The total distillate flow:
And the total bottoms flow:
Next, by using the mass balance of butane, we compute the molar fraction of butane at the bottoms:
Then, the molar fraction of pentane and hexane:
Therefore, the molar composition of the bottom product is 2.5 % butane, 42.2 % pentane and 55.3 % hexane.
NOTE: notice the result is independent of the value of the assumed feed, it means that no matter the basis, the compositions will be the same for the same recovery of butane at the feed, only the flows will change.
Regards.