A hydrocarbon gas burned completely in O2 to give CO2 and H2O. It was found that 1volume of a hydrocarbon gas at STP produced 2volumes of CO2 and 3 volumes of steam at STP. What is the emperical formula of this gas? Please help me on how to work out the formula of the gas.

1CxHy + O2 ==> 2CO2 + 3H2O

If all are in the gaseous state, volumes are the same as mols; therefore, x must be 2 and y must be 6.

To determine the empirical formula of the gas, we need to analyze the ratios of the atoms in the given compounds.

Start by considering the ratio of CO2 to the hydrocarbon gas by volume. It states that 1 volume of the hydrocarbon gas produced 2 volumes of CO2. This suggests that the ratio of CO2 to the hydrocarbon gas is 2:1.

Next, consider the ratio of H2O to the hydrocarbon gas by volume. It states that 1 volume of the hydrocarbon gas produced 3 volumes of steam (H2O). This indicates that the ratio of H2O to the hydrocarbon gas is 3:1.

Now, let's assume the hydrocarbon gas has an empirical formula of C⟨a⟩H⟨b⟩, where ⟨a⟩ represents the number of carbon atoms, and ⟨b⟩ represents the number of hydrogen atoms.

From the first ratio, we know that the carbon in CO2 comes from the carbon in the hydrocarbon gas. Therefore, the number of carbon atoms in the hydrocarbon gas would be equal to the number of carbon atoms in 2CO2, which is 2.

Next, let's calculate the number of hydrogen atoms. From the second ratio, we know that the hydrogen in H2O comes from the hydrogen in the hydrocarbon gas. Therefore, the number of hydrogen atoms in the hydrocarbon gas would be equal to the number of hydrogen atoms in 3H2O, which is 6.

Hence, the empirical formula of the gas is CH2.

To confirm this, you can calculate the molecular mass of the empirical formula CH2, which is:

(1 x 12.01) + (2 x 1.008) = 14.026 g/mol

Then, divide the molecular mass of the hydrocarbon gas given by the experiment by the molecular mass of CH2:

Molecular mass of hydrocarbon gas = (2 x 12.01) + (6 x 1.008) = 30.068 g/mol

So, 30.068 g/mol ÷ 14.026 g/mol = approximately 2.143.

Since the ratio is close to 2, it confirms our assumption that the empirical formula of the gas is CH2.

Therefore, the empirical formula of the gas is CH2.