Question

Calculating the molecular formula of caffeine? Caffeine contains $49.48\%$ carbon, $5.15\%$ Hydrogen, $28.87\%$ nitrogen, $16.49\%$ oxygen. It has a molar mass of $194.2gmo{l^{ - 1}}$ .

Answer 1

The chemical formula of the molecule is of two types, that is molecular formula and empirical formula. The molecular formula is the actual number of different atoms present in the molecule, whereas the empirical formula of the compound is the simplest ratio of the number of different atoms present in the molecule.
Formula used:
The general relation between the molecular and empirical formula is:
$Molecular\;Formula = n \times Empirical\;Formula$
Where, $n$ can be calculated by simply dividing the molar mass and the empirical formula of the compound.
To make use of this formula, we should know about molecular formulas or empirical formulas.

Complete answer:
To solve this question we should know about empirical formulas as well as molecular formulas.
Empirical formula is the simplest positive integer ratio of atoms present in the molecule whereas molecular formula is the formula which tells us the actual number of atoms in the compound.
Hence by using empirical formula table we will find the empirical formula of the given compound:

Element	Percentage	Atomic mass	Moles of atom	Atomic ratio	Simplest ratio
C	$49.48\%$	$12$	$\dfrac{{49.48}}{{12}} = 4.123$	$\dfrac{{4.123}}{{1.03}} = 4.003$	$4$
H	$5.15\%$	$1$	$\dfrac{{5.15}}{1} = 5.15$	$\dfrac{{5.15}}{{1.03}} = 5$	$5$
N	$28.87\%$	$14$	$\dfrac{{28.87}}{{14}} = 2.062$	$\dfrac{{2.062}}{{1.03}} = 2.002$	$2$
O	$16.49\%$	$16$	$\dfrac{{16.49}}{{16}} = 1.03$	$\dfrac{{1.03}}{{1.03}} = 1$	$1$

Hence the empirical formula of caffeine is:
${C_4}{H_5}{N_2}O$
Now, we will check the molar mass of the empirical formula we obtained by:
$Molar\;Mass = (12 \times 4 + 1 \times 5 + 14 \times 2 + 16 \times 1)$
$\Rightarrow Molar\;Mass = 97$
Which is not equal to the given molar mass, i.e. $194.2gmo{l^{ - 1}}$. So we will divide this molar mass by the molar mass we get from empirical formula:
$\Rightarrow \dfrac{{Given\;Molar\;Mass}}{{empirical\;Molar\;Mass}}(n) = \dfrac{{194.2}}{{97}} \approx 2$
Now we will put the values in the given formula:
$Molecular\;Formula = n \times Empirical\;Formula$
$MolecularFormula = \,2 \times ({C_4}{H_5}{N_2}O)$
And we will get:
${C_8}{H_{10}}{N_4}{O_2}$
Thus, this is the required molecular formula of caffeine.

Note:
Remember that the empirical formula is used to simply show the elements present in the molecule. This is used when one wants to know at a glance what elements they are dealing with. Molecular formula is more useful for the determination of elements in compounds.