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Determine the empirical formula of the following compound:

A 0.4987-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.9267 g of carbon dioxide and 0.1897 g of water.

Respuesta :

meerkat18
the combustion reaction is  CHO + O2 = H20 + CO2 
The amount of carbon  can be determined by the amount of carbon dioxide.
moles C = 0.9267 g CO2 / 44 g/mol CO2 = 0.0211 moles C = 0.2327 g C
The amount of H can be derived from the amount of water
mole H = 0.1897 g H2O / 18 g / mol H2O = 0.0105 g H

grams O is 0.4987 - 0.2327 g C - 0.0105 g H = 0.2555 g O. 

transforming the determined masses to moles and dividing each to the least among the three, the emprical formula is  C2HO

Answer: The empirical formula for the given organic compound is [tex]C_3H_3O_2[/tex]

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

[tex]C_xH_yO_z+O_2\rightarrow CO_2+H_2O[/tex]

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of [tex]CO_2=0.9267g[/tex]

Mass of [tex]H_2O=0.1897g[/tex]

We know that:

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 0.9267 g of carbon dioxide, [tex]\frac{12}{44}\times 0.9267=0.253g[/tex] of carbon will be contained.

  • For calculating the mass of hydrogen:

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

So, in 0.1897 g of water, [tex]\frac{2}{18}\times 0.1897=0.021g[/tex] of hydrogen will be contained.

  • Mass of oxygen in the compound = (0.4987) - (0.253 + 0.021) = 0.2247 g

To formulate the empirical formula, we need to follow some steps:

  • Step 1: Converting the given masses into moles:

Moles of Carbon =[tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.253g}{12g/mole}=0.021moles[/tex]

Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.021g}{1g/mole}=0.021moles[/tex]

Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.2247g}{16g/mole}=0.014moles[/tex]

  • Step 2: 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 0.014 moles.

For Carbon = [tex]\frac{0.021}{0.014}=1.5[/tex]

For Hydrogen  = [tex]\frac{0.021}{0.014}=1.5[/tex]

For Oxygen  = [tex]\frac{0.014}{0.014}=1[/tex]

Converting the mole ratios into whole numbers by multiplying it by 2.

For Carbon = [tex]1.5\times 2=3[/tex]

For Hydrogen = [tex]1.5\times 2=3[/tex]

For Oxygen = [tex]1\times 2=2[/tex]

  • Step 3: Taking the mole ratio as their subscripts.

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

Hence, the empirical formula for the given compound is [tex]C_3H_{3}O_2[/tex]