Question

1. An inorganic salt on analysis gave the following percentage composition :
Pb = 62-6, N = 84,0 = 29
What is empirical formula of the compound ? Also name the compound. (At. mass Pb = 207, N = 14,
0 = 16).
c a n be the mercentage composition : N = 25.94 and 0 = 74.06
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Answer 1

Na = 43.4%

C = 11.3

O = 43.3%

relative number of moles

of Na = 43.4/23 = 1.88

of C = 11.3/12 = 0.94

of O = 43.3/16 = 2.71

simple ratio of moles 

of Na = 1.88/0.94 = 2

of C = 0.94/0.94 = 1

of O = 2.71/0.94 = 2.87 ~ 3

empirical formula = Na2CO3

Answer 2

Explanation:

Given: Pb = 62-6, N = 84,0 = 29

To find: What is empirical formula of the compound ? Also name the compound.

Solution:  (At. mass Pb = 207, N = 14,

0 = 16).

Na = 43.4%

C = 11.3

O = 43.3%

relative number of moles

of Na = 43.4/23 = 1.88

of C = 11.3/12 = 0.94

of O = 43.3/16 = 2.71

simple ratio of moles

of Na = 1.88/0.94 = 2

of C = 0.94/0.94 = 1

of O = 2.71/0.94 = 2.87 ~ 3

empirical formula = Na2CO3

Convert the mass of each element to moles using the periodic table's molar mass. By the least determined number of moles, multiply each mole amount. Round to the closest whole number. The empirical formula uses subscriptions to represent this mole ratio of the components.

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