An experiment showed that in a lead chloride Solution, \[6.21{ }g\] of lead combined with \[4.26{ }g\] of chlorine. The empirical formula of this chloride is:
A.$PbC{{l}_{2}}$
B.$P{{b}_{2}}C{{l}_{2}}$
C.$PbCl$
D.$PbC{{l}_{4}}$
Answer
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Hint: Using the given data we need to first calculate the percentage of lead and chlorine in lead chloride. Using that percentage of lead and chlorine we will find an atomic ratio which further will help us in knowing and calculating the simplest whole number ratio or empirical formula required.
Complete step by step answer:
In lead chloride
Mass of lead = \[6.21g\]
Mass of chlorine = \[4.26g\]
Total mass of lead chloride $= 6.21+4.16 = 10.47g$
Percentage of lead in lead chloride $= \dfrac{6.21}{10.47}\times 100 = 59.31%$
Percentage of chlorine in lead chloride $ = \dfrac{4.26}{10.47}\times 100 = 40.69%$
To calculate empirical formula of lead chloride:
After finding the simplest whole number ratio we can directly put these simple whole numbers with their respective elements as these simple whole numbers are nothing but just telling us the number of atoms of respective elements which are combining to form the desired compound. So the empirical formula of lead chloride is $PbC{{l}_{4}}$
So, option (D) i.e. $PbC{{l}_{4}}$ is correct.
Note:
Don’t get confused between chemical formula and empirical formula. We will be discussing a few differences between chemical and empirical formulas in order to solve your confusion among the two.
Complete step by step answer:
In lead chloride
Mass of lead = \[6.21g\]
Mass of chlorine = \[4.26g\]
Total mass of lead chloride $= 6.21+4.16 = 10.47g$
Percentage of lead in lead chloride $= \dfrac{6.21}{10.47}\times 100 = 59.31%$
Percentage of chlorine in lead chloride $ = \dfrac{4.26}{10.47}\times 100 = 40.69%$
To calculate empirical formula of lead chloride:
| Element | Percentage | Atomic mass | Atomic ratio | Simplest atomic ratio | Simplest whole number ratio |
| Lead | $59.31$ | $207$ | $\dfrac{59.31}{207}=0.287$ | $\dfrac{0.287}{0.287}=1$ | $1$ |
| Chlorine | $40.69$ | $35.5$ | $\dfrac{40.69}{35.5}=1.146$ | $\dfrac{1.146}{0.287}=3.99$ | $4$ |
After finding the simplest whole number ratio we can directly put these simple whole numbers with their respective elements as these simple whole numbers are nothing but just telling us the number of atoms of respective elements which are combining to form the desired compound. So the empirical formula of lead chloride is $PbC{{l}_{4}}$
So, option (D) i.e. $PbC{{l}_{4}}$ is correct.
Note:
Don’t get confused between chemical formula and empirical formula. We will be discussing a few differences between chemical and empirical formulas in order to solve your confusion among the two.
| Empirical formula | Chemical formula or molecular formula |
| Empirical formula is the simplest form of expressing the elemental composition of a compound. | Chemical or molecular formula is the actual representation of the elemental composition of the compound. |
| The empirical formula is derived first from the weight percentages of the element present in the compound. | The chemical formula is related to the total weight of the compound in question and often is derived after obtaining the empirical formula. |
| The empirical formula contains the most simplified ratio of the moles of elements in the compound. | The molecular formula needs to be a multiple of the empirical formula. |
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