Combustion analysis - sample. How many grams of C is this?...

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Combustion analysis When a compound containing carbon and hydrogen is subject to combustion with oxygen in a special combustion apparatus all the carbon is converted to CO 2 and the hydrogen to H 2 O. The amount of carbon produced can be determined by measuring the amount of CO 2 produced. This is trapped by the sodium hydroxide, and thus we can monitor the mass of CO 2 produced by determining the increase in mass of the CO 2 trap. Likewise, we can determine the amount of H produced by the amount of H 2 O trapped by the magnesium perchlorate. Consider the combustion of isopropyl alcohol. The sample is known to contain only C, H and O. Combustion of 0.255 grams of isopropyl alcohol produces 0.561 grams of CO 2 and 0.306 grams of H 2 O. From this information we can quantitate the amount of C and H in the sample: Since one mole of CO 2 is made up of one mole of C and two moles of O, if we have 0.0128 moles of CO 2 in our sample, then we know we have 0.0128 moles of C in the
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Unformatted text preview: sample. How many grams of C is this? How about the hydrogen? Since one mole of H 2 O is made up of one mole of oxygen and two moles of hydrogen, if we have 0.017 moles of H 2 O, then we have 2*(0.017) = 0.034 moles of hydrogen. Since hydrogen is about 1 gram/mole, we must have 0.034 grams of hydrogen in our original sample. When we add our carbon and hydrogen together we get: 0.154 grams (C) + 0.034 grams (H) = 0.188 grams But we know we combusted 0.255 grams of isopropyl alcohol. The 'missing' mass must be from the oxygen atoms in the isopropyl alcohol: 0.255 grams - 0.188 grams = 0.067 grams oxygen This much oxygen is how many moles? Overall therfore, we have: 0.0128 moles Carbon 0.0340 moles Hydrogen 0.0042 moles Oxygen Divide by the smallest molar amount to normalize: C = 3.05 atoms H = 8.1 atoms O = 1 atom Within experimental error, the most likely empirical formula for propanol would be: C 3 H 8 O...
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Combustion analysis - sample. How many grams of C is this?...

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