A compound of carbon, hydrogen, and oxygen was burned in oxygen, and 2.00 g of the compound produced 2.868 g CO2 and 1.567 g H2O. In another experiment 0.1107g of the compound was dissolved in 25.0 g of water. This solution had a freezing point of -0.0894 degrees C. What is the molecular formula of the compound?
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To find the molecular formula of the compound, we need to determine the empirical formula first, and then use the molar mass to find the molecular formula.
1. Find the moles of carbon, hydrogen, and oxygen in the reactants:
From the given information, we can calculate the moles of CO2 and H2O produced.
Molar mass of CO2 = 12.01 g/mol (C) + 2 * 16.00 g/mol (O) = 44.01 g/mol
Mole of CO2 = 2.868 g / 44.01 g/mol = 0.0652 mol
Molar mass of H2O = 2 * 1.01 g/mol (H) + 16.00 g/mol (O) = 18.02 g/mol
Mole of H2O = 1.567 g / 18.02 g/mol = 0.0869 mol
2. Determine the ratio of moles of carbon, hydrogen, and oxygen:
From the balanced chemical equation for the combustion of hydrocarbons, we know that the ratio of moles of carbon and hydrogen is 1:2 and the ratio of moles of carbon to oxygen in CO2 is 1:2.
Moles of carbon = 0.0652 mol
Moles of hydrogen = 0.0869 mol * (2/1) = 0.1738 mol
Moles of oxygen = 0.0652 mol * (2/1) = 0.1304 mol
3. Find the empirical formula:
Divide the moles of each element by the smallest number of moles to get the simplest whole number ratio.
Empirical formula: CH2O (1 carbon, 2 hydrogen, 1 oxygen)
4. Find the molar mass of the empirical formula:
Molar mass of CH2O = 12.01 g/mol (C) + 2 * 1.01 g/mol (H) + 16.00 g/mol (O) = 30.03 g/mol
5. Calculate the molar mass factor:
Molar mass factor = molar mass of compound / molar mass of empirical formula
Molar mass factor = (2.00 g / (2.868 g/mol)) / (0.1107 g / 30.03 g/mol)
Molar mass factor = 20.907
6. Calculate the molecular formula:
Multiply the subscripts in the empirical formula by the molar mass factor.
Molecular formula: C6H12O6
Therefore, the molecular formula of the compound is C6H12O6.