First, we need to determine the moles of water and carbon dioxide produced:
moles of H2O = 7.749 g / 18.015 g/mol = 0.4306 mol
moles of CO2 = 18.942 g / 44.01 g/mol = 0.4301 mol
Next, we need to determine the number of moles of each element present in the compound. We can do this by dividing the number of moles of each product by their respective coefficients in the balanced chemical equation for the combustion reaction:
C + O2 → CO2
2H2 + O2 → 2H2O
For CO2, 0.4301 mol / 1 mol CO2 = 0.4301 mol C
For H2O, 0.4306 mol / 2 mol H2O = 0.2153 mol H
We can then convert the moles to grams and find the mass of oxygen present in the compound (by subtracting the mass of carbon and hydrogen from the total):
mass of C = 0.4301 mol C * 12 g/mol = 5.1612 g
mass of H = 0.2153 mol H * 1.008 g/mol = 0.2168 g
mass of O = 12.915 g - 5.1612 g - 0.2168 g = 7.537 g
Finally, we can divide each element's mass by its molar mass to get the number of moles of each element:
moles of C = 5.1612 g / 12 g/mol = 0.4301 mol
moles of H = 0.2168 g / 1.008 g/mol = 0.2153 mol
moles of O = 7.537 g / 16 g/mol = 0.4711 mol
Dividing each number of moles by the smallest number (0.2153 mol) and rounding to the nearest whole number gives us the empirical formula:
C2H4O2
12.915 g of a compound was combusted in excess oxygen in which 7.749 g of H2O and 18.942 g of CO2 were produced. Determine the empirical formula of the compound. The compound contains C, H and O.
(Molar mass: N=14 g/mol; C=12 g/mol; O= 16 g/mol)
1 answer