A 1.50 gram sample of cadaverine gives 3.23 g of CO_{2}, 1.58 g of N_{2}O_{5}, and 1.865 g of H_{2}O. Its molar mass is 102.2 g/mol. Determine the empirical and molecular formulas.

**Step One:** determine the mass of each element present.

Carbon: 3.23 g x (12.011 / 44.0098) = 0.8815 g

Hydrogen: 1.865 g x (2.016 / 18.0152) = 0.2087 g

Nitrogen: 1.58 g x (28.014 / 108.009) = 0.4098 g

Oxygen: 1.50 g minus (0.8815 + 0.2087 + 0.4098) = zero

Although the problem does not specifically mention oxygen, there may be some in the compound. Consequently, make sure you check by adding the masses of the other elements and subtracting the answer from the total mass of the starting compound. Any difference would be the oxygen (or a mistake in the calculation to this point!!). If the masses add up exactly to the starting total, as they do here, that's a nice indicator you've done the calculation correctly (or committed two mistakes that canceled each other out!!).

**Step Two:** Convert mass of each element to moles.

Carbon: 0.8815 g ÷ 12.011 g/mol = 0.0734 mol

Hydrogen: 0.2087 g ÷ 1.008 g/mol = 0.207 mol

Nitrogen: 0.4098 g ÷ 14.007 g/mol = 0.02926 mol

**Step Three:** find the ratio of molar amounts, expressed in smallest, whole numbers.

Carbon: 0.0734 mol ÷ 0.02926 mol = 2.51

Hydrogen: 0.207 mol ÷ 0.02926 mol = 7.07

Nitrogen: 0.02926 mol ÷ 0.02926 mol = 1

Doubling each value gives C = 5, H = 14.14, N = 2, so the empirical formula is C_{5}H_{14}N_{2}. Since the "EFW" of this is about 102, we know that it is also the molecular formula.

Notice that the hydrogen value is off by about 10%, but I went ahead and rounded off to the nearest whole number. This raises the question "How far off can a value be and be rounded off?" Another way to express this would be "When can I round off, rather than continuing to search for a set of smallest whole-number values?"

Frankly, this can be a difficult question to answer, since the borderline cases require a bit of a judgement call, based on experience. However this is one fact in the example above that justifies our decision. The problem gave the molar mass as about 102 and we got 102 with our formula of C_{5}H_{14}N_{2}. This validates our decision to round off.

You might think that having the molar mass so easily available is just a bit too easy, too magical. However, I assure you that determining molar masses in this modern area is a perfectly straightforward process, with very exact values easily obtainable. This was not the case in the 1800's, but that is a different story.

In the empirical formula section, there was a practice problem, numbered 6. It shows a problem where you might be tempted to round off too early.