Stoichiometry

Unit 3.6-3.10 Chem 121

3.6 Determining the formula of a compound

• Combustion analysis

  • when a substance containing carbon and hydrogen is burned it is converted to co2 and h2o (just like normal shit)

• Determining the formula of a compound

  • ex

    • a 15g sample of a hydrocarbon is burned in oxygen to give 50.769g of co2 and 10.386g
      • $50.769gC{O}_2\ast \frac{1molC{O}_2}{44.01gC{O}_2}\ast \frac{1molC}{1molC{O}_2}=1.154mol$
    • A compound containing only phosphorus and oxygen is analyzed to contain 43.64% P by mass what is the empirical formula
      • Assuming 100g $43.64gP\ast \frac{1molP}{30.97g}=1.409106molP$
        • $100-43.64g=56.36gO\ast \frac{1molO}{16.00g}=3.5225molO$
      • $\frac{P_{1.409106}{0}_{3.5225}}{1.409106}$ dividing through the smallest%
        • $P_2{O}_5$
    • A sample of a compound that is known to contain only C=40.92%, H=4.58% and O the molar mass is 176.1
      • %C = mass of C in 1 mol/mass of 1 mol of the substance
        • $\frac{40.92}{100}\ast 176.1g=72.060gC\ast \frac{1molC}{12.01}=6molC$
      • %H
        • $\frac{4.58}{100}\ast 176.1g=8.06538gH=\frac{1molH}{1.008g}=8.00molH$
      • %O
        • $\frac{54.50}{100}\ast 176.1g=95.9745g\ast \frac{1molO}{16.00g}=5.998molO$

      • $C_6{H}_8{O}_6$

• Molecular formula

  • Empirical formula will not represent all the actual numbers of atoms

3.8 Balancing chemical Equations

3.9 Stoichiometric calculations

3.10 Calculations involving a limiting reagent