What is the empirical formula for a compound containing only carbon and hydrogen if it is known to contain 84.21% carbon? If the molar mass is 114 g/mol, what is the molecular formula of this compound?
Answers
Answer:
Since the compound contains carbon & hydrogen only, it’s a hydrocarbon. As the Carbon is 84.21%, the hydrogen must be 100–84.21=15.79%.
To get the total number of each atom in the compound, divide the % by the molar mass of each atom.
Carbon = 84.21 divided by 12.01= 7.012 & Hydrogen = 15.79 divided by 1 = 15.79
By approximation, C=7 & H=16, and this gives the simplest ratio of each atom in the compound, ie, it’s Empirical Formula, EF=C7H16
The molar mass of the compound is 114g/mol. Using this information, we can work out the molecular formula as follows:
Molecular Formula, MF= (C7H16) times X=114.
ie, (12.01 X 7 + 16 X 1) times X = 114.
(84.07+16) times X = 114.
Therefore X= 114 divided by 100.07 = 1.1392, which is approximately equal to 1. ie, the MF is C7H16 & the name is heptane which is an Alkane hydrocarbon.
Explanation:
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Answer:
C8H18
Explanation:
Carbon = 84.21% so Hydrogen = 15.79%
--Assuming it is a 100 gram sample--
C: 84.21g C x 1 mol C/12.01g C = 7.012 mol C
H: 15.79g H x 1 mol H/1.01 g H = 15.63 mol H
--Then divide the mole values by the lowest value--
C: 7.012 mol/7.012 mol= 1
H: 15.63 mol/7.012 mol = 2.23
--Then multiply both to get 2.23 around a whole number--
C: 1 x 4 = 4
H: 2.23 x 4 = 8.9 = 9
Empirical Formula = C4H9
--Determine the molar mass of the Empirical Formula--
(12.01 g/mol x 4) + (1.01 g/mol x 9) = 57.13 g/mol
--Then divide the mass by the molar mass--
114/57.13 = 1.995 = 2
--Multiply the Empirical Formula's subscripts by the value--
C4H9 --> C8H18 (the molecular formula)
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