Question

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A pure sample of compound is found to contain 2.04g of Sodium, 2.65×10²² atoms of carbon and 1.32 mole of oxygen atoms. Determine the empirical formula of the compound.
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Answer 1

Answer:-

Empirical formula of the compound

is Na₂CO₃ .

Explanation:-

• Molar mass of Sodium (Na) = 23g/mol

• Molar mass of Carbon (C) = 12g/mol

• Molar mass of Oxygen (O) = 16g/mol

• Avogadro Number = 6.022×10²³

Number of moles in 2.04g of Na :-

= Given Mass/Molar mass

= 2.04/23

= 0.887 mole

Number of moles in 2.65×10²² atoms

of C :-

= No of atoms/Avogadro Number

= 2.65×10²²/6.022×10²³

= 0.044 mole

Number of moles of O atoms = 0.132

Thus, the atomic ratio Na : C : O is :-

= 0.0887 : 0.044 : 0.132

• Least value = 0.044

Now, let's divide each value by the least value:-

0.0887/0.044 : 0.044/0.044 : 0.132/0.044

= 2 : 1 : 3

Hence, empirical formula is Na₂CO₃ .

Answer 2

Answer:

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Given,

2.04g of Na present in a sample.

Number of moles of sodium present in 2.04 g of Na = 2.04/23= 0.088

2.65 × 10^22 atoms of C present.

So, number of moles of C present in the compound

= (2.65 × 10^22 )/ (6.023 × 10^23 )

= 0.044

And 0.132 mole of oxygen

So, Carbon: Sodium: Oxygen= 1:2:3

Thus, the compound is C1N2O3 (and the molar mass of the compound= 12+46+48=106 as given).

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