Question

calculate the (a) empirical formula, and (b) molecular formula of a compound having the following percentage composition sodium=42.1%, phosphorus= 18.9% and oxygen= 39%. given that the molecular mass of the compound is 164.[na=23, p=31 and 0=16]

Answer 1

Answer:

(a) Empirical Formula is Na3PO4.

(b) Molecular Formula is Na3PO4.

Explanation:

In 100gm of compound, 42.1 gm Na, 18.9gm P and 39gm O are present.

Now

Number of moles of each element

n(Na) = 42.1÷23 = 1.83

n(P) = 18.9 ÷ 31 = 0.61

n(O) = 39 ÷16 = 2.44

Now, Number of moles of each element is divided by 0.61, then

N(Na) = 1.83 ÷ 0.61 = 3

N(P) = 0.61 ÷ 0.61 = 1

N(O) = 2.44 ÷ 0.61 = 4

Now,

Empirical formula of compound is Na3PO4

Molecular formula of compound is Na3PO4

Answer 2

Answer:

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