Calculating molecular formulas

Question

1. Analysis of a compound reveals 64.9% carbon, 13.5% hydrogen, and 21.6% oxygen. If it’s molecular mass is 148.0 g/mol, compute its molecular      formula.

 

 

2. Data from a lab yields these values: mass of calcium is 2.39 g; mass of chlorine is 4.23 g; mass of oxygen is 5.72 g. If the molecular mass is found    to be 207.1 g/mol. Determine the molecular formula of the compound.

Answer 1

1). Assume 100 g of compound and then % is the same as grams, which can then be converted to moles.

moles C = 64.9 g x 1 mol / 12 g = 5.41 moles C

moles H = 13.5 g x 1 mol / 1 g = 13.5 moles H

moles O = 21.6 g x 1 mol / 16 g = 1.35 moles O

Divide all by the lowest value to try to get all whole numbers...

5.41 / 1.35 = 4.0 moles C

13.5 / 1/35 = 10 moles H

1.35 / 1.35 = 1 mol O

Empirical formula = C₄H₁₀O

molar mass of this is 48 + 10 + 16 = 74 g/mol

If molecular mass is 148, then formula would have to be 2x that of the empirical (148 / 74 = 2).

Molecular formula = C₈H₂₀O₂

2) Find moles of each element and then convert to whole numbers to find empirical formula.

moles Ca = 2.39 g x 1 mol / 40 g = 0.0598 mols Ca

moles Cl = 4.23 g x 1 mol / 35.5 g = 0.119 moles Cl

moles O = 5.72 g x 1 mol / 16 g = 0.358 moles O

Dividing all by 0.0598 to try to get whole numbers we have...

0.0598/0.0598 = 1 mole Ca

0.119/0.0598 = 1.99 = 2 moles Cl

0.358/0.0598 = 5.99 = 6 moles O

Empirical formula = CaCl₂O₆

molar mass of empirical formula = 40 + 71 + 96 = 207

Since this is also the molecular mass of the molecular formula, the molecular formula is the same as the empirical formula

Molecular formula = CaCl₂O₆