Question

A compound is found to contain 47.25% copper and 52.75% chlorine. Find the
empirical formula for this compound.
​

Answer 1

Answer:

CuCl2

Explanation:

Step I. Percentage of the element

Percentage of Cu=1.894.0×100=47.25,

Percentage of Cl=2.114.0×100=52.75

Step II. Empirical formula of copper chloride

ElementCuClPercentage47.2552.75Atomic mass63.535.5Gram atoms (Moles)47.2563.5=0.7452.7535.5=1.48Atomic ratio (Molar ratio)0.740.74=11.480.74=2Simplest whole no. ratio12

Element

Percentage

Atomic mass

Gram atoms (Moles)

Atomic ratio (Molar ratio)

Simplest whole no. ratio

Cu

47.25 63.5

47.25

63.5

=0.74

0.74

0.74

=1 1

Cl

52.75 35.5

52.75

35.5

=1.48

1.48

0.74

=2 2

Empirical formula of the compound =CuCl2

Answer 2

Answer:

$CuCl_{2}$.

Given:

Percent composition of compound: 47.25 % Copper and 52.75 % Chlorine.

To find: Empirical formula

Explanation:

Molecular formula of compound is a way to tell how many atoms of each type are present in the compound.

Whereas empirical formula is the smallest positive integer ratio of the atoms present in the compound.

Now, the composition of the compound is 47.25 % Copper (Cu) , 52.75 % Chlorine (Cl).

Step 1: Now, assume that the mass of compound is 100 g.

Thus, as per the percent composition,

Mass of copper = 47.25 % of 100 g = 47.25 g

Mass of Chlorine = 52.75 % of 100 g = 52.75 g

Step 2: Now, from the above mass, determine the moles of Cu and Cl:

Molar mass of Cu = 63.55 g/mol

Thus, moles of Cu are-

$Moles\ of\ Cu\ = \frac{Mass\ of\ Cu}{Molar\ mass\ of\ Cu} =\frac{47.25\ g}{63.55\ g/mol} = 0.74\ moles$

Molar mass of Cl = 35.45 g/mol.

Thus, moles of Cl are-$Moles\ of\ Cl\ = \frac{Mass\ of\ Cl}{Molar\ mass\ of\ Cl} =\frac{52.75\ g}{35.45\ g/mol} = 1.49\ moles$

step 3: Now divide the moles by the smallest number of moles-

Moles of Cu = 0.74 moles

Moles of Cl = 1.49 moles.

The smallest number of moles is 0.74 moles.

Thus,

$Cu =\frac{0.74}{0.74} = 1\\ \\Cl =\frac{0.149}{0.74} = 2$

Therefore, the empirical formula of the given compound is $CuCl_{2}$.

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