8750 g of a compound which contains only carbon, hydrogen, and nitrogen were burned in oxygen to produce, 2.1930 g of N2O4 , and 2.2900 g of water. What is the empirical formula of the compound? (Hint: do carbon by difference)
To find the empirical formula of the compound, we need to determine the ratio of the elements present in the compound. In this case, we have the masses of nitrogen and hydrogen in the products of the combustion reaction.
First, let's find the amount of nitrogen in N2O4:
The molar mass of N2O4 is 92 g/mol (2 x 14 g/mol for N and 4 x 16 g/mol for O).
Using the molar mass, we can calculate the number of moles of N2O4:
moles of N2O4 = mass of N2O4 / molar mass of N2O4 = 2.1930 g / 92 g/mol ≈ 0.0238 mol.
Since there are two moles of nitrogen atoms in one mole of N2O4, the moles of nitrogen in N2O4 is also 0.0238 mol.
Next, let's find the amount of hydrogen in water (H2O):
The molar mass of H2O is 18 g/mol (2 x 1 g/mol for H and 1 x 16 g/mol for O).
Using the molar mass, we can calculate the number of moles of water:
moles of water = mass of water / molar mass of water = 2.2900 g / 18 g/mol ≈ 0.1272 mol.
Since there are two moles of hydrogen atoms in one mole of water, the moles of hydrogen in water is also 0.1272 mol.
To find the amount of carbon, we can use the fact that the total mass is 8750 g, and carbon is the only remaining element in the compound.
mass of carbon = total mass - (mass of N2O4 + mass of water)
mass of carbon = 8750 g - (2.1930 g + 2.2900 g)
mass of carbon = 8750 g - 4.4830 g ≈ 8745.5170 g
Now we have the masses of nitrogen, hydrogen, and carbon:
mass of nitrogen = 0.0238 mol x 14 g/mol ≈ 0.3332 g
mass of hydrogen = 0.1272 mol x 1 g/mol ≈ 0.1272 g
mass of carbon = 8745.5170 g
Finally, we can calculate the empirical formula by dividing each element's mass by its atomic mass and simplifying the ratio. Let's divide each element's mass by the smallest mass (carbon):
ratio of nitrogen = mass of nitrogen / mass of carbon = 0.3332 g / 8745.5170 g ≈ 0.000038 mol
ratio of hydrogen = mass of hydrogen / mass of carbon = 0.1272 g / 8745.5170 g ≈ 0.000015 mol
ratio of carbon = mass of carbon / mass of carbon = 8745.5170 g / 8745.5170 g = 1 mol
Now we can express the empirical formula using these ratios. Since the ratios are very small, we can multiply them by a certain factor to make them whole numbers. In this case, let's multiply by 26,315 (which is the reciprocal of the smallest ratio, 0.000038):
ratio of nitrogen = 0.000038 mol x 26,315 ≈ 1 mol
ratio of hydrogen = 0.000015 mol x 26,315 ≈ 1 mol
ratio of carbon = 1 mol x 26,315 = 26,315 mol
Hence, the empirical formula of the compound is NH₃C₂₆H₃₁₅ or simplified as NH₃C₂₆H₁₅.