Question

When 0.273 g of Mg is heated strongly in a nitrogen (${N_2}$) atmosphere, a chemical reaction occurs. The product of the reaction weighs 0.378 g. Empirical formula of the compound is:
A. $MgN$
B. $M{g_3}{N_2}$
C. $Mg{N_2}$
D. $M{g_2}N$

Answer 1

An empirical formula provides information about the ratio of atoms of different elements present in a molecule of the compound. It is the simplest form of notation. Empirical formula also helps in calculating the percent composition of the compound. Molecular formula can be equal or can be a multiple of its empirical formula.

Formula used: To calculate empirical formula, we will need to calculate the number of moles of each element present in that compound.
The number of moles can be calculated as:
$n = \dfrac{m}{M}$
$n$ = Moles of the element
$m =$ Mass of the element in g
$M =$ Molar mass of the element

Complete step by step answer:
Here, we have to determine the empirical formula of the compound containing magnesium (Mg) and nitrogen (N).
Let us first understand the given values.
Mass of Mg = 0.273g
Mass of the product formed after magnesium reacted with nitrogen = 0.378g
Step 1: To calculate the empirical formula of the compound, we need to calculate the moles of magnesium and nitrogen. For calculating it, we require the mass of both magnesium and nitrogen in grams.
The mass of magnesium is given as 0.273g.
We need to calculate the mass of nitrogen. This can be found by calculating the difference between the amount of product formed and amount of magnesium reacted.
Therefore, mass of nitrogen = $0.378 - 0.273 = 0.105g$
The mass of nitrogen is 0.105g.
Step 2: To calculate the number of moles of magnesium and nitrogen, we need the value of molar mass for each. The molar mass of an element is equal to its atomic weight and it is expressed in g/mol.
The molar mass of magnesium 24g/mol and the molar mass of nitrogen element is 14 g/mol.
Let us calculate the number of moles for magnesium and nitrogen. By putting the respective values in the formula for calculating number of moles, we get:
$n = \dfrac{m}{M}$
Number of moles of magnesium are:
$n = \dfrac{{0.273g}}{{24g/mol}}$
$\therefore n = 0.0113moles$
Number of moles of nitrogen gas are:
$n = \dfrac{{0.105g}}{{14g/mol}}$
$\therefore n = 0.0075moles$
Step 3: As the values of n obtained for both magnesium and nitrogen are not whole numbers, we need to convert them into whole numbers. For that, we need to divide both values of n with the smallest value of n amongst magnesium and nitrogen.
For magnesium,
$\dfrac{{0.0113}}{{0.0113}} = 1$
For nitrogen,
$\dfrac{{0.0075}}{{0.0113}} = 0.66$
Now, multiply both values by 2 as moles of nitrogen is not a whole number. On multiplying both values by 2 and then taking the ratio, we get,
$Mg:N = 3:2$
Therefore, the empirical formula of the compound will be $M{g_3}{N_2}$ and the name of the compound is magnesium nitride.

So, the correct answer is Option B.

Additional information:
Empirical formula does not provide any information about the structural aspects of the compound. It does not describe the different structures or isomers with different physical properties.
Magnesium nitride is an inorganic compound at room temperature.

Note: The empirical formula for the given compound can be obtained in an alternate way by using the following method.
When 0.273 g of Mg is heated strongly with ${N_2}$ gas 0.378g of compound is formed.
On calculating the amount of nitrogen in the compound, we get,
Amount of nitrogen in the compound = $0.378 - 0.273 = 0.105g$
Let us calculate the mass % of both elements in the compound.
The mass % of nitrogen in compound = $\dfrac{{0.105}}{{0.378}} \times 100 = 27.77\%$
The mass % of Mg in compound = $\dfrac{{0.273}}{{0.378}} \times 100 = 72.23\%$
Assume the mass of chemical compound formed by combining magnesium and nitrogen gas is 100g
Therefore, mass of Mg $= 72.23\%$ and mass of N = 27.77g $= 27.77\%$
The moles of both magnesium and nitrogen can be calculated as follows:
Number of moles of Mg in compound = $\dfrac{{72.23}}{{24}} \approx 3$
Number of moles of nitrogen gas in the compound = $\dfrac{{27.73}}{{14}} \approx 2$
From the above calculations, the 3 moles of magnesium and 2 moles of nitrogen gas combine to form the product and the empirical formula of $M{g_3}N_2$.