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)

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