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 .

sodium (na), for example, has one electron in its valence shell. this is an unstable state because that valence shell needs eight electrons to be full. in order to fill its valence shell, find a way to add seven electrons to its valence shell