Advanced Chem Lecture #6
Percent Composition
Suppose we need to verify the purity of a substance
First determine what percentage of mass should come from each element
Then determine what percentages are actually contributed
Use percent composition to determine mass of elements in molecule
Mass % element X = moles X in formula x molar mass of x (g/mol)
mass (g) of 1 mol compound
X 100
Note that the individual components ALWAYS add up to 100%
Glucose (C6H12O6). Determine mass percent.
How many grams carbon in 16.55g?
Determination of Empirical Formula
Can determine empirical formula of unknown compound from masses of elements
An analytical chemist investigating a compound decomposes it into simpler substances, finds the mass of
each component element, converts these masses to numbers of moles, and then mathematically converts the
moles to whole-number subscripts. This procedure yields the empirical formula, the simplest wholenumber ratio of moles of each element in the compound.
Analysis shows that an unknown compound contains 0.21mol Zn, 0.14mol P, and
0.56mol O. Determine empirical formula. First, divide by smallest subscript. If
any subscripts are still not integers, multiply through by the smallest integer that
will make them all integers.
Zn0.21P0.14O0.56 / 0.14  Zn1.5P1.0O4.0 * 2 = Zn3P2O8
Note that since we multiplied all the subscripts by two, we did not change the
relative amounts of moles in each
Different substances can have the same empirical formula but different molecular
Name
Formaldehyde
Acetic Acid
Lactic Acid
Erythrose
Ribose
Glucose
Molecular
Formula
CH2O
C2H4O2
C3H6O3
C4H8O4
C5H10O5
C6H12O6
Multiple
1
2
3
4
5
6
g/mol
30.03
60.05
90.08
120.1
150.13
180.16
Use
biologoical preservative
vinegar 5% solution
forms in muscle during exercise
forms during sugar metabolism
nucleic acid component
cellular nutrient
Balancing Chemical Equations
Vitamin C (176.12g/mol) is a compound of C,H,O found in citrus fruits. When a
1.000g sample is placed in a combustion chamber and burned, following data obtained:
Mass CO2 after combustion: 85.35g
Mass CO2 before combustion: 83.85g
Mass H2O absorber after combustion: 37.96g
Mass H2O absorber before combustion: 37.55g
What is the molecular formula for Vitamin C? Find mass of CO2 added, from this, can
use mass fraction of C to find carbon content in vitamin C. Similarly, find mass of H
from mass fraction of H2O added. Mass of vitamin C (1.000g) minus C and H gives
mass of O. Knowing grams of each, calculate moles of each, construct empirical
formula, determine whole-number multiple from given molar mass, then give molecular
formula
Mass CO2 = 85.35-83.85 = 1.50g CO2
Mass H2O = 37.96-37.55 = 0.41g H2O
Mass C = 1.50g CO2 * (12.01gC/44.01gCO2) = 0.409g
C
Mass H = 0.41g H20 * (2.016g H/18.02g H2O) = 0.046g
H
Mass vitamin - mass C - mass H = mass O = 0.545g
Divide grams by molar mass to give mols
0.0341 mol C, 0.046mol H, 0.0341mol O
Preliminary formula: C0.0341H0.046O0.0341
Divide by smallest subscript: C1H1.3O1
Multiply through by 3: C3,H4O3
Molar mass vitamin C / molar mass emprical = 2
(Empirical * 2) = C6H8O6
Check
Stoichiometry
Can use example from combustion reaction above to balance.