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Chem Empirical formula question!!?
Consider a chemical known to contain only C, H, S and O. The chemical is then subjected to two combustion analysis experiments in the presence of excess oxygen. One experiment measures the mass of CO2 and H20 produced, while the other experiment measures the mass of SO2 produced. In the 1st experiment, the combustion of 32.149 mg of the sample yielded 57.271 mg of CO2 and 23.444 mg of H20. In the second experiment, the combustion of 33.565 mg of the sample yieled 14.507 mg of SO2. What is the empirical formula?
2 Answers
- FernLv 74 years ago
57.271 mg CO2 x 12.0 mg C/44.0 mg CO2 = 15.6 mg C
23.444 mg H2O x 2 mg H/18.0 mg H2O = 2.60 mg H
14.507 mg SO2 x 32.0 mg S/64.0 mg SO2 = 7.25 mg S
Percent S in compound: 7.25 mg S/33.565 mg Compound * 100 = 21.6% sulfur
Sulfur present in 32.149 mg of compound = 0.216 * 32.149 mg = 6.94 mg Sulfur
Weight of oxygen in 32.149 g =
32.149 – 15.6 – 2.6 – 7.3 – 6.94 = 7.05 mg
Millimoles of each element in 32.149
15.6 mg C x 1 millimole/ 12.0 mg = 1.3
2.60 mg H x 1 millimole / 1.00 mg = 2.6
7.05 mg S x 1 millimole/32.0 mg = 0.22
7.01 mg O x 1millimole/16.0 mg = 0.44
C1.3; H2.6; S 0.22; O 0.44 divide the moles by the smallest, namely 0.23
C5.9H11.8S1.0O2 : empirical formula = C6H12SO2
You might want to determine the molar masses to 5 significant figures.