A sample weighing 3.00 g of a compound that contains only C, H, and O is completely burned in air to form 4.30g of CO2 and 2.35g of H2O. What is the empirical formula of this compound?

CH3O

Explanation:

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 4.30/44 = 0.0977 mole

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 * 0.0977 = 1.173g

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 2.35/18 = 0.131 mole

But there are 2 atoms of hydrogen in 1 mole of water and thus, the number of moles of hydrogen is 2 * 0.131 = 0.262 mole

The mass of hydrogen is thus 0.262 * 1 = 0.262g

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

= 3 - 0.262 - 1.173 = 1.565g

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

That equals 1.565/16 = 0.0978moles

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

H = 0.262/0.0977 = 2.7 = 3

O = 0.0978/0.0977 = 1

C = 0.0978/0.0977 = 1

We multiply through by 3 to give CH3O