Question

A compound is made up of 28 g N, 24 g C, 48 g O, and 8 g H .What is the empirical formula?

Answer 1

Answer:

$\rm C_2H_8N_2O_3$.

Explanation:

Step One: calculate the coefficients.

Look up the relative atomic mass of these four elements on a modern periodic table:

• $\rm C$: approximately $12$.
• $\rm H$: approximately $1$.
• $\rm N$: approximately $14$.
• $\rm O$: approximately $16$.

The relative atomic mass of an element is numerically equal to the mass (in grams, $\rm g$,) of one mole of atoms of this element.

For example, the relative atomic mass of $\rm C$ is approximately $12$. Therefore, each mole of $\rm C\!$ atoms would have a mass of $12\; \rm g$.

This sample contains $24\; \rm g$ of carbon. That would correspond to approximately $\displaystyle \left(\frac{24}{12}\right)\; \rm mol = 2\; \rm mol$ of $\rm C$ atoms.

Similarly, for the other three elements:

$\displaystyle n(\mathrm{H}) \approx \frac{8\; \rm g}{1\; \rm g \cdot mol^{-1}} = 8\; \rm mol$.

$\displaystyle n(\mathrm{N}) \approx \frac{28\; \rm g}{14\; \rm g \cdot mol^{-1}} = 2\; \rm mol$.

$\displaystyle n(\mathrm{O}) \approx \frac{48\; \rm g}{16\; \rm g \cdot mol^{-1}} = 3\; \rm mol$.

Hence, the ratio between these elements in this compound would be:

$n(\mathrm{C}): n(\mathrm{H}): n(\mathrm{N}):n(\mathrm{O}) = 2: 8 : 2 : 3$.

In the empirical formula of a compound, the coefficients should represent the smallest possible integer ratio between the number of atoms of these elements.

$n(\mathrm{C}): n(\mathrm{H}): n(\mathrm{N}):n(\mathrm{O}) = 2: 8 : 2 : 3$ is indeed the smallest possible integer ratio between the number of atoms of these elements.

Step Two: arrange the elements in an appropriate order

Apply the Hill System to arrange these four elements in the empirical formula. In the Hill System:

If carbon, $\rm C$, is present in this compound, then:

• $\rm C$ (carbon) and then $\rm H$ (hydrogen) will be the first two elements listed in the formula (ignore the hydrogen if it is not in the compound.)
• The other elements in this compound will be listed in alphabetical order.

If there is no carbon $\rm C$ in this compound, then list all the elements in this compound in alphabetical order.

Both $\rm C$ (carbon) and $\rm H$ (hydrogen) are found in this compound. Therefore, the first element in the list would be $\rm C\!$. The second would be $\rm H\!$, followed by $\rm N\!$ and then $\rm O\!$.

Hence, the empirical formula of this compound would be $\rm C_2H_8N_2O_3$.

Answer 2

The empirical formula of the compound is C₂H₈N₂O₃  

From the question given above, the following data were:

Carbon (C) = 24 g

Hydrogen (H) = 8 g

Nitrogen (N) = 28 g

Oxygen (O) = 48 g

Empirical formula =?

The empirical formula of the compound can be obtained as follow:

C = 24 g

H = 8 g

N = 28 g

O = 48 g

Divide by their molar mass

C = 24 /12 = 2

H = 8 /1 = 8

N = 28 /14 = 2

O = 48 / 16 = 3

We can see from the above calculations that the elements are in simple whole number ratio.

Thus, the empirical formula of the compound is C₂H₈N₂O₃

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