Answer:
%C = 85.71 wt%; %H = 14.29 wt%; Empirical Formula => CH₂; Molecular Formula => C₃H₆
Explanation:
%Composition
Wt C = 6 g
Wt H = 1 g
TTL Wt = 6g + 1g = 7g
%C per 100wt = (6/7)100% = 85.71 wt%
%H per 100wt = (1/7)100% = 14.29 wt % or, %H = 100% - %C = 100% - 85.71% = 14.29 wt% H
What you should know when working empirical formula and molecular formula problems.
Empirical Formula=> smallest whole number ratio of elements in a compound
Molecular Formula => actual whole number ratio of elements in a compound
Empirical Formula Weight x Whole Number Multiple = Molecular Weight
From elemental %composition values given (or, determined as above), the empirical formula type problem follows a very repeatable pattern. This is ...
% => grams => moles => ratio => reduce ratio => empirical ratio
for determination of molecular formula one uses the empirical weight - molecular weight relationship above to determine the whole number multiple for the molecular ratios.
Caution => In some 'textbook' empirical formula problems, the empirical ratio may contain a fraction in the amount of 0.25, 0.50 or 0.75. If such an issue arises, multiply all empirical ratio numbers containing 0.25 and/or 0.75 by '4' to get the empirical ratio and multiply all empirical ration numbers containing 0.50 by '2' to get the final empirical ratio.
This problem:
Empirical Formula:
Using the % per 100wt values in part 'a' ...
% => grams => moles
%C => 85.71% => 85.71 g* / 100 g Cpd => (85.71 / 12) = 7.14 mol C
%H => 14.29% => 14.29 g / 100 g Cpd => (14.29 / 1) = 14.29 mol H
=> Set up mole Ratio and Reduce to Empirical Ratio:
mole ratio C:H => 7.14 : 14.29
To reduce mole values to the smallest whole number ratio, divide all mole values by the smaller mole value of the set.
=> 7.14/7.14 : 14.29/7.14 => Empirical Ration=> 1 : 2
∴ Empirical Formula => CH₂
Molecular Formula:
(Empirical Formula Wt)·N = Molecular Wt => N = Molecular Wt / Empirical Wt
N = 42 / 14 = 3 => multiply subscripts of empirical formula by '3'.
Therefore, the molecular formula is C₃H₆
