Question

Calculate the molecular formula of a compound which contains $20\% $ $Ca$ and $80\% $ $Br$ (by weight) if the molecular weight of the compound is $200$ . (Atomic weight $Ca = 40,Br = 80$)
A.$C{a_{1/2}}Br$
B.$CaB{r_2}$
C.$CaBr$
D.$C{a_2}Br$

Answer 1

Hint: Molecular weight is the weight of one mole of a compound . Since the molecular weight of compound is given in the question so we may assume the molecular formula of compound in suitable form and calculate the weight of one mole of compound .

Complete step by step answer:
Let us assume the formula of the compound is $C{a_x}B{r_y}$ .
Molecular weight of this compound $ = x + y$
Also , molecular weight of given compound $ = 200$ .
So , $x + y$ $ = $ $200$ $\left( 1 \right)$
Suppose we have one mole of the above compound that is $200$ grams .
According to the question percentage mass of $Ca = 20\% $
Now mass of $Ca$ present in one mole of compound $ = 200 \times \dfrac{{20}}{{100}}$$ = 40$
Therefore $40gm$ of $Ca$ is present in one mole of compound .
Number of moles of $Ca$ $ = 1$ i.e. $x = 1$ (Atomic weight of calcium is $40$)
Percentage mass of $Br$ present is compound $ = 80$
Mass of $Br$ present in one mole of compound $ = 200 \times \dfrac{{80}}{{100}}$$ = 160$ .
Therefore $160gm$ of $Br$ is present in one mole of the compound .
Number of moles of $Br$ $ = $$\dfrac{{160}}{{80}}$$ = 2$ (Atomic weight of bromine is $80$)
i.e. $y = 2$
Therefore the formula of the compound is $CaB{r_2}$ .
Therefore the correct answer is option B.


Note:
A molecular formula represents chemical symbols for the constituent elements followed by numeric subscripts describing the number of atoms of each element present in the molecule . While empirical formula represents the simplest whole integer ratio of atoms in a compound .For example empirical formula of glucose is $C{H_2}O$ while its molecular formula is ${C_6}{H_{12}}{O_6}$ . Structural formula indicates not only the number of atoms but it also represents their arrangement in space . To convert between empirical and molecular formulas we multiply empirical formulas by a whole number .