C1H4O2
Step-by-step explanation:
First, we need to find the number of moles of carbon dioxide and water produced from the combustion of 6.38 mg of ethylene glycol. Using the molar masses of CO2 (44.01 g/mol) and H2O (18.02 g/mol), we get:
moles of CO2 = 9.06 mg / 44.01 g/mol = 0.000206 mol
moles of H2O = 5.58 mg / 18.02 g/mol = 0.000309 mol
Next, we can find the number of moles of carbon, hydrogen, and oxygen in the sample by multiplying the number of moles of CO2 and H2O by the corresponding atom ratios in each compound. From the balanced chemical equation for the combustion of ethylene glycol, we know that:
C2H6O2 + 3O2 → 2CO2 + 3H2O
So, the atom ratios for carbon, hydrogen, and oxygen in ethylene glycol are 2:6:2. Thus, we have:
moles of C = 2 × 0.000206 mol = 0.000412 mol
moles of H = 6 × 0.000309 mol = 0.001854 mol
moles of O = 2 × (0.000206 mol + 0.000309 mol) = 0.001030 mol
Finally, we need to convert these mole ratios into the simplest whole-number ratio. We can do this by dividing each of the mole ratios by the smallest value, which in this case is 0.000412 mol. This gives us:
C: 1.000
H: 4.500 (approximately)
O: 2.500 (approximately)
Rounding these ratios to the nearest whole number, we get the empirical formula for ethylene glycol: C1H4O2.
Therefore, the empirical formula of ethylene glycol is First, we need to find the number of moles of carbon dioxide and water produced from the combustion of 6.38 mg of ethylene glycol. Using the molar masses of CO2 (44.01 g/mol) and H2O (18.02 g/mol), we get:
moles of CO2 = 9.06 mg / 44.01 g/mol = 0.000206 mol
moles of H2O = 5.58 mg / 18.02 g/mol = 0.000309 mol
Next, we can find the number of moles of carbon, hydrogen, and oxygen in the sample by multiplying the number of moles of CO2 and H2O by the corresponding atom ratios in each compound. From the balanced chemical equation for the combustion of ethylene glycol, we know that:
C2H6O2 + 3O2 → 2CO2 + 3H2O
So, the atom ratios for carbon, hydrogen, and oxygen in ethylene glycol are 2:6:2. Thus, we have:
moles of C = 2 × 0.000206 mol = 0.000412 mol
moles of H = 6 × 0.000309 mol = 0.001854 mol
moles of O = 2 × (0.000206 mol + 0.000309 mol) = 0.001030 mol
Finally, we need to convert these mole ratios into the simplest whole-number ratio. We can do this by dividing each of the mole ratios by the smallest value, which in this case is 0.000412 mol. This gives us:
C: 1.000
H: 4.500 (approximately)
O: 2.500 (approximately)
Rounding these ratios to the nearest whole number, we get the empirical formula for ethylene glycol: C1H4O2.
Therefore, the empirical formula of ethylene glycol is C1H4O2.