<u>Answer:</u> The empirical formula for the given compound is
<u>Explanation:</u>
The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:
where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.
We are given:
Conversion factor: 1 g = 1000 mg
Mass of
Mass of
Mass of compound = 2.78 mg = 0.00278 g
We know that:
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
- <u>For calculating the mass of carbon:</u>
In 44g of carbon dioxide, 12 g of carbon is contained.
So, in 0.00632 g of carbon dioxide, of carbon will be contained.
- <u>For calculating the mass of hydrogen:</u>
In 18g of water, 2 g of hydrogen is contained.
So, in 0.00258 g of water, of hydrogen will be contained.
- Mass of oxygen in the compound = (0.00278) - (0.00172 + 0.000286) = 0.000774 g
To formulate the empirical formula, we need to follow some steps:
- <u>Step 1:</u> Converting the given masses into moles.
Moles of Carbon =
Moles of Hydrogen =
Moles of Oxygen =
- <u>Step 2:</u> Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is
For Carbon =
For Hydrogen =
For Oxygen =
- <u>Step 3:</u> Taking the mole ratio as their subscripts.
The ratio of C : H : O = 3 : 6 : 1
Hence, the empirical formula for the given compound is
