To check the answers for the questions, let's go through each one and explain how to determine the empirical formula:
1) The empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen:
To find the empirical formula, we first need to assume we have 100 grams of the compound. This means we have 25.9 grams of nitrogen and 74.1 grams of oxygen. Next, we convert these masses to moles by dividing by their respective atomic masses (nitrogen: 14 g/mol, oxygen: 16 g/mol).
Moles of nitrogen = 25.9 g / 14 g/mol = 1.85 mol
Moles of oxygen = 74.1 g / 16 g/mol = 4.63 mol
Now we find the simplest whole number ratio between the elements.
Nitrogen:Oxygen ≈ 1:2
Therefore, the empirical formula is N2O.
2) The empirical formula of a compound that consists of 4.80 grams of carbon, 1.20 grams of hydrogen, and 2.80 grams of nitrogen:
Again, assuming we have 100 grams of the compound, we convert the masses to moles.
Moles of carbon = 4.80 g / 12 g/mol = 0.40 mol
Moles of hydrogen = 1.20 g / 1 g/mol = 1.20 mol
Moles of nitrogen = 2.80 g / 14 g/mol = 0.20 mol
Next, we find the simplest whole number ratio.
Carbon:Hydrogen:Nitrogen ≈ 2:6:1
Therefore, the empirical formula is C2H6N, which matches your answer.
3) The empirical formula of a substance composed of 3.05% carbon, 0.26% hydrogen, and 96.69% iodine:
Once again, assuming we have 100 grams of the substance, we convert the percentages to grams.
Grams of carbon = 3.05 g
Grams of hydrogen = 0.26 g
Grams of iodine = 96.69 g
Now we convert the masses to moles.
Moles of carbon = 3.05 g / 12 g/mol = 0.254 mol
Moles of hydrogen = 0.26 g / 1 g/mol = 0.26 mol
Moles of iodine = 96.69 g / 127 g/mol = 0.76 mol
Next, we find the simplest whole number ratio.
Carbon:Hydrogen:Iodine ≈ 1:1:3
Therefore, the empirical formula is CHI3, which matches your answer.
4) Symclosene has an empirical formula of CClNO and a molecular mass of 232.42 u. What is the molecular formula of symclosene?
To find the molecular formula, we need to find the ratio between the empirical formula mass and the molecular formula mass.
Empirical formula mass = 1(C) + 1(Cl) + 1(N) + 1(O) = 12.01 g/mol + 35.45 g/mol + 14.01 g/mol + 16.00 g/mol = 77.47 g/mol
Next, we calculate the ratio:
Molecular formula mass / Empirical formula mass = 232.42 g/mol / 77.47 g/mol ≈ 3
Therefore, the molecular formula is C3Cl3N3O3, which matches your answer.
5) The empirical formula of a compound is C2H3O. The gram formula mass of the compound is 172 g. What is the molecular formula?
To find the molecular formula, we need to compare the empirical formula mass with the gram formula mass.
Empirical formula mass = 2(C) + 3(H) + 1(O) = 2(12.01 g/mol) + 3(1.01 g/mol) + 16.00 g/mol = 42.05 g/mol
Next, we find the ratio:
Gram formula mass / Empirical formula mass = 172 g/mol / 42.05 g/mol ≈ 4
Therefore, the molecular formula is C8H12O4, which matches your answer.
6) The molecular formula of a given compound is C4H10. The empirical formula of the same compound would be:
The empirical formula represents the simplest whole number ratio of atoms in a compound. To find this, we divide the subscripts in the molecular formula by their greatest common divisor (GCD).
C4H10 (Divide by GCD of 4 and 10, which is 2) = C2H5
Therefore, the empirical formula is C2H5, which matches your answer.
Regarding question 1, the empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen is N2O5, not N2O.