Final answer:
First, the empirical formula based on the given weight ratio and the atomic masses is found to be CH28O10. Upon adjusting this for the given molecular weight of 88, the ratio suggests the molecular formula is identical to the empirical formula: C2H4O2.
Step-by-step explanation:
To determine the molecular formula of a compound with carbon (C), hydrogen (H), and oxygen (O) present in 6:14 by weight and a molecular weight of 88, we start by deducing the empirical formula. By assuming a 100g sample, we would have 6g of C, 14g of H, and the remainder being O. To convert the masses to moles, we use atomic masses (C=12, H=1, O=16): C moles = 6g/12g/mol = 0.5 mol, H moles = 14g/1g/mol = 14 mol.
Since oxygen is not given, we subtract the weight of carbon and hydrogen from the total mass (100g) to get oxygen's mass: O mass = 100g - (6g + 14g) = 80g. O moles = 80g/16g/mol = 5 mol. Now, we get the simplest whole number ratio by dividing by the smallest number of moles (0.5): C=1, H=28, O=10.
To find the molecular formula, we need to adjust the empirical formula to the given molecular mass, 88 g/mol. Dividing the molecular mass by the empirical formula mass (CH28O10), we get approximately 1. Therefore, the molecular formula is the same as the empirical formula: C2H4O2.