Question

Q.7. A pure sample of compound is found to contain 2.04 g of sodium,2.65 x 1022 atoms of carbon and 0.132 mole of oxygen atoms. Determine the empirical formula of the compound.

Answer 1

Moles of Na in sample = Given mass/Molar mass of Na= 2.04 g/23 g mol-1​ = 0. 088Mole of carbon1 mole C = 6. 022 x 1023 atoms of C​Therefore , 2.65 x 1022 atoms = 2.65 ​x 1022 / 6. 022 x 1023= 0. 044Moles of oxygen = 0.132Molar ratio of Na , C and O = 0.088 : 0.044 : 0.132The simple whole number ration = 2 : 1 : 3Empirical formula of compound = Na2CO3

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Answer 2

Hey mate!

Here's your answer!!

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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