An experiment showed that in a lead chloride solution, 6.21 g of lead combined with 4.26 g of chlorine.

What is the empirical formula of this chloride? [Pb = 207, Cl = 35.5]

PbCl₂

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PbCl₃

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PbCl₄

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PbCl₆

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Solution

The correct option is C
PbCl₄

$\begin{matrix} E l e m e n t & % c o m p o s i t i o n & A t . W t . & A t o m i c r a t i o & S i m p l e s t A t . r a t i o P b & \frac{6.21}{10.47} \times 100 = 59.31 % & 207 & \frac{59.31}{207} = 0.286 & \frac{0.286}{0.286} = 1 C l & \frac{4.26}{10.47} \times 100 = 40.69 % & 35.5 & \frac{40.69}{35.5} = 1.146 & \frac{1.146}{0.286} = 4 \end{matrix}$

Hence, the empirical formula of the compound is $P b C l_{4}$ .

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Similar questions

Q. An experiment showed that a lead chloride solution is formed when 6.21 g of lead combines with 4.26 g of chlorine. What is the empirical formula of this chloride?

[Pb = 207; Cl = 35.5]

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(ii) the number of molecules of sulphur dioxide present in the flask.
(iii) the volume occupied by 3.2 g of sulphur dioxide at STP.
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of copper chloride on analysis was found to contain

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C u

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When lead nitrate solution is mixed with potassium chloride solution, a precipitate of lead chloride and solution of potassium nitrate are formed. This type of reaction is called _________________.

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An experiment showed that in a lead chloride solution, 6.21 g of lead combined with 4.26 g of chlorine. What is the empirical formula of this chloride? [Pb = 207, Cl = 35.5]

The correct option is C PbCl4 Element% compositionAt. Wt.Atomic ratioSimplest At. ratioPb6.2110.47×100=59.31 %20759.31207=0.2860.2860.286=1Cl4.2610.47×100=40.69 %35.540.6935.5=1.1461.1460.286=4 Hence, the empirical formula of the compound is PbCl4.

An experiment showed that in a lead chloride solution, 6.21 g of lead combined with 4.26 g of chlorine.

What is the empirical formula of this chloride? [Pb = 207, Cl = 35.5]