CHM
2045C Module Five Sample
Name:_______________
Module
Four Part P Determining the Formula of a Compound
by Combustion 5 points
A 0.523 g sample of the unknown compound CxHy,
is burned in air to give 1.612 g of CO2 and 0.7425g of H2O.
A separate experiment gave a molar mass for CxHy of 114
g/mol. Determine the empirical and
molecular formula for the hydrocarbon.
CxHy + O2
----> CO2 + H2O
0.523g 1.612g 0.7425g
All the Carbon from the hydrocarbon becomes
Carbon Dioxide, so the grams of Carbon in the hydrocarbon can be determined
from the mass of Carbon Dioxide formed:
All the Hydrogen from the hydrocarbon
becomes water, so the grams of Hydrogen in the hydrocarbon can be determined
from the mass of Water formed:
0.4396 g Carbon + 0.0831 g Hydrogen = 0.5227 g Hydrocarbon
This agrees with the original mass of
hydrocarbon 0.523 g
The moles of Carbon in the unknown
hydrocarbon:
The moles of Hydrogen in the unknown
hydrocarbon:
Therefore the
formula of the Unknown Hydrocarbon:
C0.0366 H0.0824
Divide both fractional
subscripts by 0.0366 to convert decimals to whole numbers:
C1H2.25
Now Multiple Both
subscripts by 4 to convert the 2.25 to a whole number:
Empirical formula of
unknown hydrocarbon: C4H9
The empirical Formula Mass is:
Molar Mass CxHy: 114 g/mol
114 g/mole ÷ 57 g/mol = 2 empirical formulas to
1 chemical formula
The chemical
formula is:
C4x2H9x2 or C8H18 which is octane (or gasoline)
Additional
Combustion Problems:
Read Section 3.12
pages 93-96 McMurry 5th
See worked
example 3.18 page 95 and try problems 3.27 page 95.
End of chapter
3.98, 3.99 page 104.
Q# 36-43 p 142
Kotz 5th Edition