Acompound is composed of only c h and o. combustion of a 519 gram sample of the compound yields 1.24 grams of co2 and?

C3H3O

Explanation:

Question incomplete needs to be rewritten:

A compound is composed of only C, H, and O. The combustion of a .519-g sample of the compound yields 1.24g of CO_2 and 0.255 g of H_2 O. What is the empirical formula of the compound

We can get the answer through calculations as follows.

From the mass of carbon iv oxide produced, we can get the number of moles of carbon produced. We first divide the mass by the molar mass of carbon iv oxide. The molar mass of carbon iv oxide is 44g/mol

The number of moles of carbon iv oxide is 1.24/44= 0.0282

Since there is only one carbon atom in CO2, the number of moles of carbon is same as above

The mass of carbon in the compound is simply the number of moles multiplied by the atomic mass unit. The atomic mass unit of carbon is 12. The mass of carbon in the compound is thus 12 * 0.0282= 0.338

From the number of moles of water, we can get the number of moles of hydrogen. To get the number of moles of water, we need to divide the mass of water by its molar mass. Its molar mass is 18g/mol. The number of moles here is thus 0.255/18= 0.0142 moles

But there are 2 atoms of hydrogen in 1 mole of water and thus, the number of moles of hydrogen is 2 * 0.0142= 0.0283

The mass of hydrogen is thus 0.0283 * 1 = 0.0283g

The mass of oxygen equals the mass of the compound minus that of hydrogen and that of carbon.

= 0.519 - 0.338 - 0.0283= 0.1527g

The number of moles of oxygen is the mass of oxygen divided by its atomic mass unit.

That equals 0.1527/16= 0.00954375 moles

The empirical formula can be obtained by dividing the number of moles of each by the smallest which is that oxygen 0.00954375 moles

H = 0.0284/0.00954375 = 2.97 = 3

O = 0.00954375/0.00954375= 1

C = 0.0282/0.00954375 = 2.95 = 3

The empirical formula is thus C3H3O