The empirical formula is CH2, and the molecular formulas are C2H4 and C4H8 for the compounds with molecular weights 28.03 amu and 56.06 amu, respectively.
a. Empirical Formula:
To obtain the empirical formula, we start by assuming a 100g sample. From the given percentages, we have 85.62g of carbon (C) and 14.38g of hydrogen (H). Next, we convert these masses to moles by dividing by their respective molar masses (12.01 g/mol for C and 1.01 g/mol for H). The resulting mole ratios are then simplified to the smallest whole numbers, giving us the simplest ratio of atoms in the empirical formula.
Assuming a 100g sample:
Moles of C: 85.62 g / 12.01 g/mol ≈ 7.13 mol C
Moles of H: 14.38 g / 1.01 g/mol ≈ 14.24 mol H
The simplest ratio is approximately 1:2 (C:H), leading to the empirical formula CH2.
b. Molecular Formulas:
Now, we use the molecular weights to determine the molecular formulas. Given that one compound has a molecular weight of 28.03 amu and the other 56.06 amu, we divide each molecular weight by the molar mass of the empirical formula (14.03 g/mol) to find the respective ratios.
For the compound with a molecular weight of 28.03 amu:
28.03 amu / 14.03 g/mol ≈ 2
For the compound with a molecular weight of 56.06 amu:
56.06 amu / 14.03 g/mol ≈ 4
Thus, the molecular formulas are:
For the compound with a molecular weight of 28.03 amu: (CH2)2 or C2H4
For the compound with a molecular weight of 56.06 amu: (CH2)4 or C4H8
The question probable may be:
Two compounds have the same composition: 85.62% C and 14.38% H.
a. Obtain the empirical formula corresponding to this composition.
b. One of the compounds has a molecular weight of 28.03 amu; the other, of 56.06 amu. Obtain the molecular formulas of both compounds