Download Percent Composition and empirical Formula

Percent Composition and empirical Formula
Whenever a new compound is made, it must be analyzed for its elemental
composition.
The percent composition - the mass percent of each element present in a compound.
Knowing the percent composition it is possible to calculate its
empirical formula- this tells the simplest ratio of atoms in a molecule.
[Example NO2 is empirical formula for N2O4]
STRATEGY ( if you are given only the percent composition of the compound)
1.
Calculate the mass of each element in a 100 g sample
2.
Find the # of moles of each element
3.
Find the simplest mole ratio of the elements by dividing all by the smaller
number
4.
If necessary, multiply the subscripts by small integers until whole numbers
are found
Mass percent
Mass per 100 g
moles
mole ratio
subscripts in formula
EXAMPLE: The percent composition of a solid is known to be 68.4 % Ba, 10.3 % P,
and 21.3% O. What is the empirical formula of the compound?
SOLUTION
step 1. Assume a 100 g sample; this gives 68.4 g Ba, 10.3 g P and 21.3 g O
step 2.
68.4 g Ba
x 1 mol Ba = 0.498 mol Ba
137.33 g Ba
10.3g P x 1 mol P = 0.333 mol P
30.97 g P
21.3 g O x 1 mol O = 1.33 mol O
16.00 g O
step 3. Knowing the relative mole ratio divide all by the smallest number
0.498 = 1.5 mol Ba
0.333
step 4.
1.33 = 4 mol O
0.333
0.333 = 1 mol P
0.333
This gives a mole ratio (as formula subscripts) of Ba1.5 PO4
We need whole number subscripts-- start multiplying the subscripts
Ba1.5 PO4 x 2 = Ba3P2O8
The empirical formula is Ba3P2O8
[the actual ionic formula is Ba3 (PO4)2 ]
1
Molecular Formula- gives the actual numbers of atoms in a molecule
may be a multiple of the empirical formula
where: multiple = molecular mass
empirical mass
EXAMPLE: The empirical mass of a compound is C3H4O3. If the molecular mass is
176 amu what is the molecular formula?
Ans: What is the multiple?
The empirical formula mass of C3H4O3
multiple = 176 amu = 2
88.0 amu
= 88.0 amu
therefore the molecular formula is = C(2x3) H(2x4) O(2x3) = C6H8O6
PROBLEM
The formula for sodium hydrogen carbonate is NaHCO3 . What is its percent mass
composition?
Ans: 27. 37 %Na, 1.20 % H, 14.29 %C and 57.13 %O
Determining Empirical Formulas from Elemental Composition
To determine the amount of C and H in a sample, a compound of unknown composition is
burned in pure oxygen to produce combustion products CO2 and H2O
for example:
CH4 (g) + 2 O2 (g)
CO2 (g)
+ 2 H2O (g)
The amount of CO2 and H2O produced is measured and used to calculate the empirical
formula of the compound
--------------------------
STRATEGY to find x, y and z in CxHyXz where element X may be O, N ,S or
halogen
1. From grams of H2O and CO2 produced, perform a gram to mole conversion to find molar
amounts of C in CO2 and H in H2O
2. Carry out mole to mass conversion to find the grams of C and H in the original sample
3. Subtract the sum of the masses of C and H from the original sample and determine the
gramsof element unaccounted for [the identity of element X will be given to you].
4. Determine the number of moles of the remaining element X .
5. Find the mole ratio of the elements by dividing the larger # s by the smallest #
2
EXAMPLE: When 4.50 g of ethyl butyrate, a compound containing C, H, and O,
undergoes combustion, 10.24 g of CO2 and 4.19 g H2O are produced.
(a) Determine the empirical formula of this compound
(b) If the molecular mass of ethyl butyrate is 116 amu, what is its molecular formula?
Solution to (a)
step 1. Find molar amounts of C and H
10.24 g CO2 x 1mol CO2 x 1mol C =
44.0 g CO2 1 mol CO2
0.233 mol C
4.19 g H2O x 1 mol H2O x 2 mol H = 0.466 mol H
18.0 g H2O 1 mol H2O
step 2. mol
mass
0.233 mol C x 12.0 g C = 2.79 g C 0.466 mol H x 1.01 g H = 0.471 g H
1 mol C
1 mol H
step 3. Subtract masses of C and H from mass of sample combusted to calculate
mass of oxygen
4.50 g sample - 2.79 g C - 0.471 g H = 1.24 g O
mass mole
1.24 g O x 1 mol O = 0.0775 mol O
16.0 g O
4.
The mole ratio is 0.233 mol C, 0.466 mol H and 0.0775 mol O
Divide all three by 0.0775 mol O (lowest number)
0.233 mol C = 3 mol C
0.466 mol H
= 6 mol H
0.0775 mol O 1 mol O
0.0775 mol O
1 mol O
5. This gives a mol ratio of C:H: O of 3:6:1 These are all whole integers .
Therefore the empirical formula is C3H6O for ethyl butyrate
Solution to (b) The empirical formula mass of C3H6O is 58.0 amu
multiple = molecular mass = 116 amu = 2
empirical mass
58.0 amu
The subscripts in the empirical formula are x by 2 to give the molecular formula
C2 x 3 H2 x 6 O 2 x 1 = C6H12O2
PROBLEM
Coniine, a toxic substance isolated from poison hemlock, contains only carbon,
hydrogen and nitrogen. Combustion analysis of a 5.024 mg sample yields
13.90 mg CO2, 6.048 mg of H2O. What is the formula for coniine?
Ans: C8 H17 N
3
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