An organic compound was found to contain only C, H, and Cl. When a 1.30 - g sample of the compound was completely combusted in air, 3.05 g of CO2 was formed. In a separate experiment the chlorine in a 1.00 - g sample of the compound was converted to 1.27 g of AgCl.

See explanation

Explanation:

First, I'm assuming you need the empirical formula of this compound.

In this case, we need to calculate the mass of carbon in the CO2.

In order to do that, we use the molar mass of CO2, molar mass of C and the mass of CO2 obtained in the experiment

MMCO2 = 44 g/mol

MMC = 12 g/mol

so, to get the mass of Carbon:

moles of CO2 = 3.05 / 44 = 0.0693 moles

In CO2 we have 1 mole of each element so:

mass of C = 0.0693 * 12 = 0.83 g of C

Now, we will do the same with the chlorine in AgCl to get the mass of Cl:

MMAgCl = 143.25 g/mol

MMCl = 35.45 g/mol

mass of Cl = 1.27 / 143.25 * 35.45 = 0.3143 g of Cl

Now, by simple difference we can get the mass of Hydrogen:

mass of H = 1.30 - 0.83 + 0.31 = 0.16 g of H

Now that we have the masses of each element, we can calculate the empirical formula of the compound.

First step, let's calculate the percentage of each element in the compound and then, it's mole:

C: 0.83/1.3 * 100 = 63.85% / 12 = 5.32

Cl: 0.3143/1 * 100 = 31.43% / 35.45 = 0.89

H: 100 - 63.85 - 31.43 = 4.72% / 1 = 4.72

The second step, is get the ratio of these elements between them, and to find that, we just divide the moles between the lowest of them. In this case, Cl has the lower number so:

C: 5.32/0.89 = 5.97 rounded to 6.

Cl: 0.89/0.89 = 1

H: 4.72/0.89 = 5.3 rounded to 5.

Hence, the empirical formula is: C6H5Cl