Question

When 0.25 g of Na combines with elemental oxygen under certain conditions, it forms 0.42 grams of a binary compound Na_xO_y - made of Na and O atoms only. Find its empirical formula:

Find its molecular formula, given its molar mass of 78g/mol:

Answer 1

we have mass of each elements as:

Na: 0.25 g

O: 0.42 - 0.25 = 0.17 g

Divide by molar mass to get number of moles of each:

Na: 0.25/22.99 = 0.0109

O: 0.17/16.0 = 0.0106

Divide by smallest to get simplest whole number ratio:

Na: 0.0109/0.0106 = 1

O: 0.0106/0.0106 = 1

So empirical formula is:

NaO

Molar mass of NaO,

MM = 1*MM(Na) + 1*MM(O)

= 1*22.99 + 1*16.0

= 38.99 g/mol

Now we have:

Molar mass = 78.0 g/mol

Empirical formula mass = 38.99 g/mol

Multiplying factor = molar mass / empirical formula mass

= 78.0/38.99

= 2

Hence the molecular formula is : Na2O2

Answer:

Empirical formula : NaO

Molecular formula : Na2O2