A compound contains 64.27% carbon, 7.19% hydrogen, and 28.54% oxygen with a molar mass of 168.19 g/mol. What is its molecular formula?

To determine the molecular formula of the compound, we can follow these steps:

Step 1: Assume a sample size of 100 grams to make the percentages easier to work with. This means in the 100 g sample, we have 64.27 g of carbon (C), 7.19 g of hydrogen (H), and 28.54 g of oxygen (O).

Step 2: Convert grams of each element to moles by using their atomic masses: - Carbon: 12.01 g/mol - Hydrogen: 1.008 g/mol - Oxygen: 16.00 g/mol

For carbon: 64.27 g C * (1 mol C / 12.01 g C) = 5.351 mol C

For hydrogen: 7.19 g H * (1 mol H / 1.008 g H) = 7.134 mol H

For oxygen: 28.54 g O * (1 mol O / 16 g O) = 1.784 mol O

Step 3: Determine the mole ratio of elements by dividing the number of moles by the smallest value obtained in the previous step.

In this case, the smallest number of moles is that of oxygen, so we divide all amounts of moles by 1.784 mol: C: 5.351 mol C / 1.784 mol O = 3 H: 7.134 mol H / 1.784 mol O = 4 O: 1.784 mol O / 1.784 mol O = 1

Step 4: These mole ratios can be interpreted directly as the empirical formula, which gives the simplest whole-number ratio of elements in a compound. In our case, it will be C3H4O.

Step 5: To determine the molecular formula, we have to compare the molar mass of the empirical formula to the given molar mass of the compound. First, we have to calculate the molar mass of the empirical formula (C3H4O):

C3H4O molar mass = (3 * 12.01 g/mol) + (4 * 1.008 g/mol) + (16 g/mol) = 36.03 g/mol + 4.032 g/mol + 16 g/mol = 56.062 g/mol

Step 6: Divide the given molecular mass (168.19 g/mol) by the molar mass of the empirical formula (56.062 g/mol) to find the multiple:

168.19 g/mol / 56.062 g/mol ≈ 3

Step 7: Multiply the subscripts in the empirical formula by this multiple to find the molecular formula:

C3H4O * 3 = C9H12O3

Therefore, the molecular formula of the compound is C9H12O3.

Extra: Molecular formulas provide the actual number of atoms of each element in a molecule, in contrast to empirical formulas which give only the simplest whole-number ratio. The molar mass is used to scale up from the empirical formula to the molecular formula.

Sometimes, the empirical and molecular formulas are the same, which means the compound's molar mass is the same as the mass of the empirical unit. However, if the compound's molar mass is a multiple of the empirical formula's mass, the molecular formula will be that multiple of the empirical formula, as demonstrated in the example above.

Understanding the relationship between percent composition, empirical formulas, and molecular formulas is an essential part of chemical stoichiometry, which is the quantitative relationship between reactants and products in a chemical reaction.