Analysis of vitamin C shows that it contains \[40.92%\] of C by \[4.58%\] of H and \[54.5%\] of O.
Determine the empirical formula of vitamin C
(\[C = 12,\text{ }O = 16,H = 1\] )
Answer
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Hint: In order to find out the empirical formula, first we have to find out the moles of the atoms using their molar masses and then by that you can easily predict the molecular formula of that very compound.
Complete step by step answer:
- First of all, we should know what an empirical formula is. By the empirical formula we mean the simple whole number ratio of each atom in a comp[found. It can be calculated as; suppose that you have the \[100g\] of the compound and the percentage of each atom is considered to be taken as; \[x\text{ }g\] of the atom in \[100g\] of the compound.
- Now, the next step is to convert the given masses into the moles using the molar mass of the given atoms and then there, molar ratio gives the empirical formula of the compound.
Now, let’s calculate the empirical formula of the given compound according to above given points:
\[40.92%\] of C means \[40.92\text{ }g\] of the C in \[100g\] of the compound. if;
\[12\text{ }g\] of carbon atom contains = \[1\text{ }mole\] of C
\[1\text{ }g\] of the carbon atom contains = $\dfrac{1}{12}mole$ of C
Similarly,
\[40.92\text{ }g\] of the carbon atom contains = $\dfrac{1}{12}\times 40.92\text{ =}3.4\text{ }moles$ of C
Thus, for H atom,
\[4.58%\] of H means \[4.58g\]of the H in \[100g\] of the compound. if;
\[1\text{ }g\] of hydrogen atom contains = \[1\text{ }mole\] of H
4.58 g of the hydrogen atom contains = $1\times 4.58=\text{ }4.58\text{ } moles$ of H
- Similarly, for O atom
\[54.5%\] of O means \[54.5g\] of the O in \[100g\] of the compound. if;
\[16\text{ }g\] of oxygen atom contains = \[1\text{ }mole\] of C
\[1\text{ }g\] of the oxygen atom contains = $\dfrac{1}{16} mole$ of O
\[54.5g\] of the oxygen atom contains = $\dfrac{1}{16}\times 54.5=~3.40\text{ }moles$ of O
thus, the mole ratio of carbon, hydrogen and oxygen is as;
\[3.4:4.58:\text{ }3.40 =\text{ }3:4:3\]
So, the empirical formula = \[{{C}_{3}}{{H}_{4}}{{O}_{3}}\]
Hence, the empirical formula of vitamin C having (\[C = 12,\text{ }O = 16,H = 1\])= \[{{C}_{3}}{{H}_{4}}{{O}_{3}}\].
Note: Don’t get confused in the molecular and the empirical formulas . they both are different terms. Molecular formula shows the number of each type of atom which is present in a molecule whereas the empirical formula is the simplest whole ratio of the atoms in the compound.
Complete step by step answer:
- First of all, we should know what an empirical formula is. By the empirical formula we mean the simple whole number ratio of each atom in a comp[found. It can be calculated as; suppose that you have the \[100g\] of the compound and the percentage of each atom is considered to be taken as; \[x\text{ }g\] of the atom in \[100g\] of the compound.
- Now, the next step is to convert the given masses into the moles using the molar mass of the given atoms and then there, molar ratio gives the empirical formula of the compound.
Now, let’s calculate the empirical formula of the given compound according to above given points:
\[40.92%\] of C means \[40.92\text{ }g\] of the C in \[100g\] of the compound. if;
\[12\text{ }g\] of carbon atom contains = \[1\text{ }mole\] of C
\[1\text{ }g\] of the carbon atom contains = $\dfrac{1}{12}mole$ of C
Similarly,
\[40.92\text{ }g\] of the carbon atom contains = $\dfrac{1}{12}\times 40.92\text{ =}3.4\text{ }moles$ of C
Thus, for H atom,
\[4.58%\] of H means \[4.58g\]of the H in \[100g\] of the compound. if;
\[1\text{ }g\] of hydrogen atom contains = \[1\text{ }mole\] of H
4.58 g of the hydrogen atom contains = $1\times 4.58=\text{ }4.58\text{ } moles$ of H
- Similarly, for O atom
\[54.5%\] of O means \[54.5g\] of the O in \[100g\] of the compound. if;
\[16\text{ }g\] of oxygen atom contains = \[1\text{ }mole\] of C
\[1\text{ }g\] of the oxygen atom contains = $\dfrac{1}{16} mole$ of O
\[54.5g\] of the oxygen atom contains = $\dfrac{1}{16}\times 54.5=~3.40\text{ }moles$ of O
thus, the mole ratio of carbon, hydrogen and oxygen is as;
\[3.4:4.58:\text{ }3.40 =\text{ }3:4:3\]
So, the empirical formula = \[{{C}_{3}}{{H}_{4}}{{O}_{3}}\]
Hence, the empirical formula of vitamin C having (\[C = 12,\text{ }O = 16,H = 1\])= \[{{C}_{3}}{{H}_{4}}{{O}_{3}}\].
Note: Don’t get confused in the molecular and the empirical formulas . they both are different terms. Molecular formula shows the number of each type of atom which is present in a molecule whereas the empirical formula is the simplest whole ratio of the atoms in the compound.
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