Answer:
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
