An orgnaic compound contains C, H, No, and O. Combustion of 0.1023 g of the compound in excess oxygen yielded 0.2766 g of CO2 and 0.991g of H2O. A asample of 0.4831 g of the compound was analyzed for nitrogen by the Dumas method. At STP, 27.6 mL of dry N2was obtained. In a third experiment, t he density of the compound as a gas was found to be 4.02 g/ L at 127 degrees Celsius and 256 torr. What are the empirical and molecular formulas of the compound?
I believe you should check the numbers in your problem, particularly the number for grams H2O. I don't think it can be that much. Check the other numbers while you are at it.
please someone answer!!!!
Anyone figure this one out yet?
To determine the empirical and molecular formulas of the compound, we need to analyze the given information step by step.
Step 1: Calculate the moles of CO2 and H2O produced from the combustion reaction.
We are given the mass of CO2 produced, which is 0.2766 g. The molar mass of CO2 is 44.01 g/mol. We can use this information to calculate the moles of CO2:
moles of CO2 = mass of CO2 / molar mass of CO2
moles of CO2 = 0.2766 g / 44.01 g/mol
moles of CO2 = 0.00628 mol
Next, we calculate the moles of H2O produced. The given mass of H2O is 0.991 g, and the molar mass of H2O is 18.02 g/mol:
moles of H2O = mass of H2O / molar mass of H2O
moles of H2O = 0.991 g / 18.02 g/mol
moles of H2O = 0.055 mol
Step 2: Determine the empirical formula from the moles of elements.
Since the combustion reaction produced CO2 and H2O, we can determine the ratio of carbon to hydrogen in the compound.
From the moles of CO2, we know that there are 0.00628 moles of carbon. From the moles of H2O, we know that there are 0.055 moles of hydrogen.
To simplify the ratio, we divide both moles by the lowest value, which is 0.00628:
carbon:hydrogen = 0.00628 mol / 0.00628 mol : 0.055 mol / 0.00628 mol
carbon:hydrogen = 1 : 8.77
Since empirical formulas represent the simplest whole-number ratio of elements, we need to round these values to the nearest whole number:
carbon:hydrogen = 1 : 9
The empirical formula of the compound is CH9.
Step 3: Calculate the moles of nitrogen (N2) from the Dumas method.
We are given the volume of dry N2 obtained, which is 27.6 mL. At STP, 1 mole of any gas occupies 22.4 L. We can use this information to calculate the moles of N2:
moles of N2 = volume of N2 / 22.4 L/mol
moles of N2 = 27.6 mL * (1 L / 1000 mL) / 22.4 L/mol
moles of N2 = 0.00123 mol
Step 4: Determine the empirical formula from the moles of elements.
Since the Dumas method yielded nitrogen gas (N2), we know that there are 0.00123 moles of nitrogen.
The empirical formula only includes whole numbers, so we need to determine the simplest ratio. In this case, the ratio is already in its simplest form, so the empirical formula remains as N2.
Step 5: Calculate the molar mass of the empirical formula.
To determine the molecular formula, we need to find the molar mass of the empirical formula. The molar mass of CH9 is:
(1 * atomic mass of carbon) + (9 * atomic mass of hydrogen)
Using the atomic masses from the periodic table,
(1 * 12.01 g/mol) + (9 * 1.01 g/mol) = 21.1 g/mol
Step 6: Calculate the empirical formula mass.
The empirical formula mass is the sum of the atomic masses in the empirical formula. For N2, the empirical formula mass is:
(2 * atomic mass of nitrogen) = (2 * 14.01 g/mol) = 28.02 g/mol
Step 7: Determine the molecular formula.
The molecular formula represents the actual number of atoms of each element present in a molecule. To find the molecular formula, we divide the molar mass of the empirical formula by the empirical formula mass and multiply by the subscript ratio:
molar mass of empirical formula / empirical formula mass = molecular ratio
molar mass of empirical formula / empirical formula mass = (21.1 g/mol) / (28.02 g/mol)
(molar mass of empirical formula / empirical formula mass) * empirical formula ratio
(21.1 g/mol / 28.02 g/mol) * (2 N atoms / 1 N atom) = 1.5
Therefore, the molecular formula of the compound is N3.