A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.8635 g of CO2 and 0.1767 g of H2O. What is the empirical formula of the compound?

The empirical formula is .

Explanation:

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:

Mass of = 0.8635

Mass of = 0.1767 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.8635 g of carbon dioxide, = of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.1767 g of water, = of hydrogen will be contained.

Mass of oxygen in the compound = (0.4647) - (0.2355+0.0196) = 0.2096 g

Mass of C = 0.2355 g

Mass of H = 0.0196 g

Step 1 : convert given masses into moles.

Moles of C =

Moles of H=

Moles of O=

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =

For H =

For O =

The ratio of C : H: O = 1.5 : 1 .5 : 1

Converting to simple whole number ratio:

The ratio of C : H: O = 3 : 3: 2

Hence the empirical formula is .

  the   balanced   equation   for   reaction of   ca metal   with   o2 to   produce   cao   is as below

2 ca+ o2→ 2 cao2

  explanation

the reaction is   balanced     since the   number of atoms   in reactant side is   equal to   number of atoms   in   product side.   for   example   they   are 2   atom of   ca in   reactant   side   and   2   atoms of ca   in   the product side.

  the   process   is a   meet the   requirement   of   redox   reaction because :     calcium   is   oxidized   from   oxidation state 0   to 2+   while   oxygen   is reduced from   oxidation   state   0   to 2-