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Unformatted text preview: empirical formula gives the chemical formula with the
smallest whole-number ratio of the elements of the compound. How many moles of each? The sample has 38.7 g C, 9.7 g H, and 51.6 g O. 1mol C 38.7 g C = 3.23 mol C 12.0 g C 1 mol H 9.7 g H = 9.6 mol H 1.01g H To start let’s assume a 100. g sample.
(This will make the math in the 1st step easy) 2. Since the # of atoms
are integers, divide
by the smallest value
and find the smallest
CH3O is called the
empirical formula Notice how we used
formula and the
mole to get to this
point. Empirical vs Molecular Formula A sample is composed of only 38.7% C, and 9.7% H, & ?% O,
what is the chemical formula of the compound? 1. A mass percent is just the fraction mass of an element divided by
the total mass of the molecule times 100. (Fraction*100=%)
From previous result: O + C + H = 240. + 180. + 45. = 465 g EG 240. g O *100% = 51.6% O 465 g EG 180. g C *100% = 38.7% C 465 g EG 45.g H *100% = 9.7% H 465 g EG Conservation of mass check: O + C + H =
240. + 180. + 45. = 465 g √ 1 mol O 51.6 g O = 3.23 mol O 16.0 g O Often this information is express as a mass percent. Mass ? 1 mol EG 465 g EG 62.1g EG We have determined the elemental composition by mass of a
substance using a macroscopic sample and a chemical formula. 3.23 mol O
3.23 mol 3.23 mol C
9.6 mol H
3.23 mol (We encounter a similar idea when we used the smallest neutral
fragment for the formula unit of an ionic substance).
The empirical formula results from the use of macroscopic
composition data working down toward molecular composition.
The problem is that at the macroscopic level the sample is
homogenous (a uniform mixture of elements). It is not until the
molecular level that the individual constituents can be resolved.
Some experiments are capable of finding the molecular mass, the
mass of 1 molecule (or molar mass, g/mol, of molecules). Only
with this addition information can the molecular formula be found.
The molecular formula gives the elemental composition of the
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