A 7.005 gram sample of an organic compound containing C, H, and O is analyzed by combustion analysis?
2017-09-30 4:10 am
A 7.005 gram sample of an organic compound containing C, H, and O is analyzed by combustion analysis and 14.00 grams of CO2 and 5.730 grams of H2O are produced. In a separate experiment, the molar mass is found to be 88.11g/mol. Determine the empirical formula and the molecular formula of the organic compound.
回答 (1)
2017-09-30 4:37 am
(14.00 g CO2) / (44.00964 g CO2/mol) × (1 mol C / 1 mol CO2) × (12.01078 g C/mol) = 3.82077 g C
(5.730 g H2O) / (18.01532 g H2O/mol) × (2 mol H / 1 mol H2O) × (1.007947 g H/mol) = 0.6411805 g H
(7.005 g total) - (3.82077 g C) - (0.6411805 g H) = 2.5430495 g O
(88.11g/mol) x (3.82077 g C) / (7.005 g total) / (12.01078 g C/mol) = 4.001
(88.11g/mol) x (0.6411805 g H) / (7.005 g total) / (1.007947 g H/mol) = 8.001
(88.11g/mol) x (2.5430495 g O) / (7.005 g total) / (15.99943 g O/mol) = 1.999
Round to the nearest whole numbers to find the molecular formula:
C4H8O2
All the coefficients in the molecular formula have the common factor 2, so divide all the coefficients by 2 to find the empirical formula:
C2H4O
(5.730 g H2O) / (18.01532 g H2O/mol) × (2 mol H / 1 mol H2O) × (1.007947 g H/mol) = 0.6411805 g H
(7.005 g total) - (3.82077 g C) - (0.6411805 g H) = 2.5430495 g O
(88.11g/mol) x (3.82077 g C) / (7.005 g total) / (12.01078 g C/mol) = 4.001
(88.11g/mol) x (0.6411805 g H) / (7.005 g total) / (1.007947 g H/mol) = 8.001
(88.11g/mol) x (2.5430495 g O) / (7.005 g total) / (15.99943 g O/mol) = 1.999
Round to the nearest whole numbers to find the molecular formula:
C4H8O2
All the coefficients in the molecular formula have the common factor 2, so divide all the coefficients by 2 to find the empirical formula:
C2H4O
收錄日期: 2021-05-01 21:52:59
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