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?

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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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 .

Hence, this is the required solution.