Find the molecular formula for a gas with the following composition: 82.76 percent C and 17.24

| August 13, 2021 | education

Find the molecular formula for a gas with the following composition: 82.76 percent C and 17.24 percent H. The air density is 2.01.

to solve this problem, we write down the given: w (C) = 82.76 or 0.8276 w (H) = 17.24 or 0.1724 air density D (air) = 2.01.
Find: the formula of the substance CxHy
Solution:
First, let’s write down the formula by which we will make the calculation.
w = n * Ar / Mr
Where n is the number of atoms or molecules, Ar is the relative atomic mass, Mr is the relative molecular weight of the substance.
To find Mr, write the formula: D = Mr / 29
Mr = D * 29
Mr = 2.01 * 29 = 58.29
From the formula w = n * Ar / Mr, we derive n.
We get: n = w * Mr / Ar
Let’s substitute the numbers and make the calculation:
n (C) = 0.8276 * 58.29 / 12 = 4
n (H) = 0.1724 * 58.29 / 1 = 10
Thus, we get that this is the formula C4H10 – butane
Answer: C4H10



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