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)
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
To determine the empirical formula of the compound, we need to find the simplest whole-number ratio of the elements present.
First, let's find the number of moles of water (H2O) produced:
Mass of water (H2O) = 7.749 g
Molar mass of water (H2O) = 2(1 g/mol of H) + 16 g/mol of O = 18 g/mol
Number of moles of water (H2O) = Mass of water / Molar mass of water
Number of moles of water (H2O) = 7.749 g / 18 g/mol = 0.4305 mol
Next, let's find the number of moles of carbon dioxide (CO2) produced:
Mass of carbon dioxide (CO2) = 18.942 g
Molar mass of carbon dioxide (CO2) = 12 g/mol of C + 2(16 g/mol of O) = 44 g/mol
Number of moles of carbon dioxide (CO2) = Mass of carbon dioxide / Molar mass of carbon dioxide
Number of moles of carbon dioxide (CO2) = 18.942 g / 44 g/mol = 0.4305 mol
Now, let's find the number of moles of carbon (C) in the compound:
Number of moles of carbon (C) = Number of moles of carbon dioxide (CO2) = 0.4305 mol
Lastly, let's find the number of moles of hydrogen (H) in the compound:
Number of moles of hydrogen (H) = Number of moles of water (H2O) x 2 (since there are 2 moles of hydrogen per mole of water)
Number of moles of hydrogen (H) = 0.4305 mol x 2 = 0.861 mol
Now, we can find the empirical formula by dividing the number of moles of each element by the smallest number of moles (in this case, 0.4305).
Empirical formula of the compound: C0.4305H2O0.861
To simplify this formula, let's multiply all the subscripts by 2 to get a whole number ratio:
C0.861H1.722O1.722
Rounding to the nearest whole number ratio, we have:
C1H2O2
Therefore, the empirical formula of the compound is CH2O.