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A compound containing carbon, hydrogen and oxygen. Combustion analysis of a $ 4.30-g $ sample of butyric acid produced $ 8.59g\text{ }C{{O}_{2}} $ and $ 3.52\text{ }{{H}_{2}}O. $ Find the empirical formula.

Last updated date: 24th Jul 2024
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Hint :We know that the empirical formula provides the idea about the lowest whole number ratio in which atoms are exiting in the compound. It does not give any idea about the arrangement of the atoms in the compound. It is the relative ratio of elements in a compound.

Complete Step By Step Answer:
The empirical formula is the simplest integer ratio in which atoms combine. The relative number of atoms in the compound is explained by considering the empirical formula. The following are the step used for the determination of empirical formula,
First write down the number of grams each element provides to us.
We will consider that, the total mass of the compound formed after the combination of elements is equal to $ 100 $ grams and masses are provided in the percentage considered.
In this step, find out the number of moles of the element. Divide the moles of the element by the lowest value of the mole computed.
Round up the obtained values to the closest whole number and denote this as the subscript concerning each element.
If subscripts can be further simplified, do so and we have the empirical formula for the compound.
Mass of $ C{{O}_{2}}\text{ }=\text{ }8.59\text{ }g~ $ .
Molar mass of $ C{{O}_{2}}\text{ }=\text{ }44\text{ }g/mol $
Moles of $ C{{O}_{2}}=\dfrac{Mass\text{ }of\text{ }C{{O}_{2}}}{Molar\text{ }mass\text{ }of\text{ }C{{O}_{2}}}=\dfrac{\text{ }8.59\text{ }g}{44g/mol}~~=\text{ }0.1952\text{ }moles $
Now there is one mole of $ C $ in one mole of $ C{{O}_{2}} $
Therefore, moles of carbon $ = $ moles of $ C{{O}_{2}}=0.1952moles $
Mass of carbon $ ~= $ moles of $ C\text{ }\times $ the atomic weight of $ C $
 $ =\text{ }0.1952\text{ }moles\text{ }\times \text{ }12\text{ }g/mol=\text{ }2.3427\text{ }g $
Mass of $ {{H}_{2}}O\text{ }=\text{ }3.52g~ $
Molar mass of $ {{H}_{2}}O\text{ }=\text{ }18g/mol $
Moles of $ {{H}_{2}}O=~\dfrac{3.52g}{18g/mol}=0.1956\text{ }moles $
Now there is two moles of Hydrogen in one mole of $ {{H}_{2}}O $
Moles of $ H=0.1956\text{ }moles\text{ }\times \text{ }2~=\text{ }0.3912\text{ }moles $
Mass of $ H\text{ }=\left( 0.3912\text{ }moles \right)\times \left( atomic\text{ }mass\text{ }of\text{ }H \right)~=\left( 0.3912\text{ }moles \right)~\times \left( 1\text{ }g/mol \right)=0.3912\text{ }g $
Mass of butyric acid $ =\text{ }4.30g~ $
Mass of $ O\text{ }=\text{ }mass\text{ }of\text{ }butyric\text{ }acid\text{ }-\text{ }\left( mass\text{ }of\text{ }C\text{ }+\text{ }mass\text{ }of\text{ }H \right) $
 $ =\text{ }4.30\text{ }g\text{ }-\text{ }2.3427\text{ }g\text{ }-\text{ }0.3912\text{ }g\text{ }=\text{ }1.5661\text{ }g $
Moles of $ O\text{ }=\text{ }mass\text{ }of\text{ }O\text{ }/\text{ }atomic\text{ }mass\text{ }of\text{ }O\text{ }=\text{ }1.5661\text{ }g\text{ }/\text{ }16\text{ }g/mol=\text{ }0.0978\text{ }moles~ $
Divide by smallest mole to get the molar ratio
 $ C\text{ }=\text{ }\dfrac{0.1952\text{ }moles}{0.0978\text{ }moles}=\text{ }2 $
 $ H\text{ }=~\dfrac{0.3912\text{ }moles}{0.0978\text{ }moles~}=\text{ }4 $
 $ O\text{ }=\text{ }\dfrac{0.0978\text{ }moles}{0.0978\text{ }moles}~=\text{ }1 $
Empirical formula of butyric acid is $ {{C}_{2}}{{H}_{4}}O~ $
Molecular formula of butyric acid is $ {{C}_{4}}{{H}_{8}}{{O}_{2}} $

Note :
Remember that the molecular formula expresses the arrangement of the toms in the compound but the empirical formula does not. Different molecular formulas but have the same empirical formula. So do not get confused with the empirical formula and molecular formula.