
A molecular compound is composed of 60.4% $Xe$ ,21.1% $O$ and 17.5% $F$ , by mass. If the molecular weight is $217.3u$, then what is the molecular formula?
A. $Xe{O_3}{F_2}$
B. $Xe{O_3}F$
C. $Xe{O_2}{F_3}$
D. None of these
Answer
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Hint: Empirical formula gives the simplest whole – number ratio of atoms in the chemical compound, while the molecular formula gives the kind and number of atoms of each element present in the chemical compound. Sometimes both empirical formulas and molecular formulas can be the same.
Complete Step by step answer: First we will start with determining empirical formula from the given percentage and molecular mass of atoms.
-Here the percentage is given, so we will assume the total mass is $100g$ so that the mass of each element will be equal to the percent given. So we have $60.4gXe$ and $21.1gO$ and $17.5gF$ .
-Now we will convert the mass of each element to moles using the molar mass of each atom.
So moles of $Xe$ = $\dfrac{{60.4g}}{{131gmo{l^{ - 1}}}} = 0.461$mole
Moles of $O$ = $\dfrac{{21.1g}}{{16gmo{l^{ - 1}}}} = 1.318mol$
Moles of $F$ = $\dfrac{{17.5g}}{{19gmo{l^{ - 1}}}} = 0.921mol$
-Now we will divide each mole value by the smallest number of moles calculated that is 0.461 and round of to the nearest whole number. We get the number of $Xe$ comes out as 1. The number of atoms of oxygen $O$ comes out as 3. At last the number of fluorine $F$ comes out as 2. Thus these are the mole ratio of the elements and are represented by subscripts in the empirical formula.
-So the empirical formula is $Xe{O_3}{F_2}$ $Xe{O_3}{F_2}$
now the chemical formula will always be some integer multiple of the empirical formula
-Molecular formula = n (empirical formula). Here the given formula weight is $217.3u$, and given formula weight is almost identical with the empirical formula that is $Xe{O_3}{F_2}$ .
Therefore here the value of n is 1
So the correct option will be A.
Note: There are three main categories of chemical formula analytical, molecular and structural. If no subscription exists in the formula, this means that one atom is present in the compound. Empirical formulas show the simplest total number ratio of atoms in that compound, while molecular formula shows the number of each atom in a molecule that is bonded together.
Complete Step by step answer: First we will start with determining empirical formula from the given percentage and molecular mass of atoms.
-Here the percentage is given, so we will assume the total mass is $100g$ so that the mass of each element will be equal to the percent given. So we have $60.4gXe$ and $21.1gO$ and $17.5gF$ .
-Now we will convert the mass of each element to moles using the molar mass of each atom.
So moles of $Xe$ = $\dfrac{{60.4g}}{{131gmo{l^{ - 1}}}} = 0.461$mole
Moles of $O$ = $\dfrac{{21.1g}}{{16gmo{l^{ - 1}}}} = 1.318mol$
Moles of $F$ = $\dfrac{{17.5g}}{{19gmo{l^{ - 1}}}} = 0.921mol$
-Now we will divide each mole value by the smallest number of moles calculated that is 0.461 and round of to the nearest whole number. We get the number of $Xe$ comes out as 1. The number of atoms of oxygen $O$ comes out as 3. At last the number of fluorine $F$ comes out as 2. Thus these are the mole ratio of the elements and are represented by subscripts in the empirical formula.
-So the empirical formula is $Xe{O_3}{F_2}$ $Xe{O_3}{F_2}$
now the chemical formula will always be some integer multiple of the empirical formula
-Molecular formula = n (empirical formula). Here the given formula weight is $217.3u$, and given formula weight is almost identical with the empirical formula that is $Xe{O_3}{F_2}$ .
Therefore here the value of n is 1
So the correct option will be A.
Note: There are three main categories of chemical formula analytical, molecular and structural. If no subscription exists in the formula, this means that one atom is present in the compound. Empirical formulas show the simplest total number ratio of atoms in that compound, while molecular formula shows the number of each atom in a molecule that is bonded together.
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