3.10g of an organic compound that contains carbon hydrogen and oxygen was completely burned in oxygen .in the process 4.4g of carbon 4 oxide and 2.7g of water was produced determine the empirical formula of the compound(c=12,h=1,o=16)
First, we need to determine the number of moles of carbon dioxide and water produced:
moles of CO2 = 4.4g / 44.01 g/mol = 0.1 mol
moles of H2O = 2.7g / 18.02 g/mol = 0.15 mol
Next, we need to determine the number of moles of carbon, hydrogen, and oxygen in the compound before it was burned:
moles of C = 3.10g / 12.01 g/mol = 0.258 mol
moles of H = (2 * 0.15 mol) = 0.3 mol
moles of O = [(0.1 mol CO2) + (0.15 mol H2O)] - moles of C - moles of H
= (0.1 + 0.15) - 0.258 - 0.3
= 0.292 mol
Now, we need to find the smallest whole number ratio of moles of C, H, and O. To do this, we divide each by the smallest number of moles:
moles of C = 0.258 / 0.258 = 1
moles of H = 0.3 / 0.258 = 1.16 (approximately)
moles of O = 0.292 / 0.258 = 1.13 (approximately)
The empirical formula of the compound is CH1.16O1.13, which can be simplified to CHO.