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:

P b C l_{2}

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P b_{2} C l_{2}

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P b C l

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P b C l_{4}

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Solution

The correct option is C $P b C l_{4}$
Element Mass Atomic weight Relative no. of atoms Simplest ratio
Lead 6.21 g 207 6.21/207 = 0.03 0.03/0.03 = 1
Chlorine 4.26 g 35.5 4.26/35.5 = 0.12 0.12/0.03 = 4

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

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Similar questions

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]

(a) A flask contains 3.2 g of sulphur dioxide. Calculate the following :
(i) The moles of sulphur dioxide present in the flask.
(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.
(S = 32, O = 16)
(b) A n experiment showed that in a lead chloride solution 6.21 g of lead is combined with 4.26 g of chlorine. What is the empirical formula of this chloride?
(Pb = 2107; Cl = 35.5).

F o u r g r a m s

of copper chloride on analysis was found to contain

1.890 g

of copper (

C u

) and

2.110 g

of chlorine (

C l

). What is the empirical formula of copper chloride?

$P b (N O_{3})_{2} (a q) + 2 K C l (a q) \to 2 K N O_{3} (a q) + P b C l_{2} (s)$

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 _________________.

Q. Four gram of copper chloride on analysis was found to contain 1.890 g of copper (

C u

) and 2.110 g of chlorine (

C l

). What is the empirical formula of copper chloride?

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Element	Mass	Atomic weight	Relative no. of atoms	Simplest ratio
Lead	6.21 g	207	6.21/207 = 0.03	0.03/0.03 = 1
Chlorine	4.26 g	35.5	4.26/35.5 = 0.12	0.12/0.03 = 4

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:

The correct option is C PbCl4ElementMass Atomic weight Relative no. of atoms Simplest ratioLead 6.21 g 207 6.21/207 = 0.03 0.03/0.03 = 1Chlorine4.26 g 35.5 4.26/35.5 = 0.12 0.12/0.03 = 4Hence, empirical formula is PbCl4.

The correct option is C $P b C l_{4}$
Element Mass Atomic weight Relative no. of atoms Simplest ratio
Lead 6.21 g 207 6.21/207 = 0.03 0.03/0.03 = 1
Chlorine 4.26 g 35.5 4.26/35.5 = 0.12 0.12/0.03 = 4

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