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# The empirical formula of a compound containing 47.9% potassium, 5.5% beryllium and 46.6% fluorine by mass is-[Atomic weight of \$Be\$ =9; \$F\$=19; \$K\$=39]A. \$K_2BeF_2\$ B. \$KBeF_4\$C. \$K_2Be_2F_4\$D. \$K_2BeF_4\$

Last updated date: 30th May 2024
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Hint: The empirical formula of a compound represents the ratio of elements present in that compound and doesn’t take into consideration the actual numbers of atoms found in the molecule. These ratios are represented by subscripts next to the elemental symbols.

Step-by-step solution: The empirical formula and molecular formula are two different concepts. So, first let’s try to differentiate between these two.
An empirical formula is the simplest method of expressing the elemental composition of a compound and can be derived by using the mass or weight percentage data of the elements present in the compound.
On the other hand, the molecular formula represents the actual composition of elements contained in a given compound. It can be derived by making use of empirical formula.
In order to calculate the empirical formula of the compound from the data given in the question, we need to follow the following steps-

Step1- By making use of atomic weights and percentage composition, we will first calculate the moles composition. For determining the relative numbers of moles of each element present in the compound, we will use the relation,
Relative no. of moles= (given mass composition of element)/ (atomic mass of element)
We will calculate the relative number of moles for beryllium, potassium and fluorine by using the above relation and then we will write the respective values in column IV of the table.

Step2- By making use of moles composition that we have calculated in the above step now we will calculate the smallest whole number ratio of atoms. For this, first we will find out the smallest number from column IV, and then divide every data of column IV with that smallest number.
Here, 0.6 is the smallest number out of 1.2, 0.6 and 2.4. So, we will divide 1.2/0.6, 0.6/0.6 and 2.4/0.6
The respective answers have been then entered into our last column, i.e., 2, 1, 4, respectively.

 Name of the Elements Percentage composition (%) Atomic weights of elements(g) Relative number of moles Simplest ratio Potassium 47.9 39 47.9/39= 1.2 1.2/0.6= 2 Beryllium 5.5 9 5.5/9= 0.6 0.6/0.6= 1 Fluorine 46.6 19 46.6/19= 2.4 2.4/0.6= 4

By using this simplest ratios of potassium (\$K\$ ), beryllium (\$Be\$ ) and fluorine (\$F\$ ), as the subscripts, we can write our empirical formula as \$K_2BeF_4\$
So, the correct option is D.

Note:
● Sometimes determining the whole number ratio can be tricky and we will need to use trial and error to get the correct value. For example, if we are getting 1.5 for a solution, then we need to multiply each number in the problem by 2 in order to make the 1.5 into 3.
● By making use of the empirical formula, we can easily calculate the molecular formula if we know the molar mass of the compound, by using the relation,
Ratio= (molar mass of the compound) / (empirical formula mass)
By multiplying all of the subscripts present in empirical formula by the above calculated ratio, we can obtain the molecular formula.