A sample containing only carbon, calcium, and nitrogen is analyzed in a combustion train, producing solely carbon dioxide, calcium oxide, and nitrogen dioxide. The reaction of a 426.74 mg sample of the unknown produces 298.75 mg of calcium oxide, and 490.13 mg of nitrogen dioxide. What is the empirical formula of the compound?

CaCN2

Explanation:

For convenience sake, we would be working with g instead of mg in the question. Note to get g, simply divide mg by 1000.

We should know that we can obtain the individual masses from their respective compounds number of moles. For example, we can obtain the mass of calcium from that of calcium oxide.

First, we need to know the number of moles of calcium oxide produced. This is the mass of calcium oxide divided by the molar mass of calcium oxide. Molar mass of calcium oxide is 56g/mol

The mass here is 298.75mg = 0.29875g.

Thus, the number of moles is 0.29875/56 = 0.005334821428571 mole

Hence the mass of calcium is thus thus number of moles multiplied by the atomic mass. The atomic mass is 40g/mol

The mass is thus 0.005334821428571 * 40 = 0.2134g

We can proceed from here to get the mass of nitrogen produced. We can get this from the number of moles of nitrogen dioxide NO2. The molar mass of NO2 is 46g/mol.

The number of moles is thus 0.49013/46 = 0.010655

Since we have just one atom of nitrogen, the number of moles are equal. The mass of nitrogen is thus 0.010655 * 14 = 0.14917g

The mass of carbon is thus the total mass minus the masses of nitrogen and calcium.

0.42674 - 0.14917 - 0.2134 = 0.06417g

The number of moles of carbon is thus 0.06417/12 = 0.0053475 mole

Now, we divide through by the smallest number of moles which is that of calcium:

Ca = 0.00533/0.00533 = 1

N = 0.010655/0.0053 = 2

C = 0.0054/0.0053 = 1

Hence, the empirical formula of the compound is : CaCN2