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 C_xH_y, is burned in air to give 1.612 g of CO₂ and 0.7425g of H₂O. A separate experiment gave a molar mass for C_xH_y of 114 g/mol. Determine the empirical and molecular formula for the hydrocarbon.

C_xH_y + 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:

C_0.0366 H_0.0824

Divide both fractional subscripts by 0.0366 to convert decimals to whole numbers:

C₁H_2.25

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 C_xH_y: 114 g/mol

114 g/mole ÷ 57 g/mol = 2 empirical formulas to 1 chemical formula

The chemical formula is:

C_4x2H_9x2 or C₈H₁₈ which is octane (or gasoline)

Additional Combustion Problems:

Read Section 3.12 pages 93-96 McMurry 5^th

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 5^th Edition