Answered

A compound containing Na, C, and O is found to have 1.06 mol Na, 0.528 mol C, and 1.59 mol O. What is the empirical formula of the compound?

Answer :

Given :

Moles of Na : 1.06

Moles of C : 0.528

Moles of O : 1.59

To Find :

The empirical formula of the compound.

Solution :

Dividing moles of each atom with the smallest one i.e 0.528 .

So,

Na : 1.06/0.528 = 2.007 ≈ 2

C : 0.528/0.528 = 1

O : 1.59/0.528 = 3.011 ≈ 3

Rounding all them to nearest integer, we will get the number of each atom in the empirical formula.

So, empirical formula is [tex]Na_2CO_3[/tex] .

Hence, this is the required solution.

pstnonsonjoku

The empirical formula of the compound is Na2CO3.

We have the following information from the question;

Number of moles Na = 1.06 mol

Number of moles C = 0.528

Number of moles O = 1.59 mol

Divide through by the lowest ratio;

Na - 1.06/0.528          C- 0.528/0.528           O - 1.59/0.528

Na - 2                          C- 1                                O - 3

The empirical formula of the compound is Na2CO3

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