Question

An analysis of a 4.2-gram sample of a compound is found to contain 54.76 percent sodium (Na) and 45.24 percent fluorine (F). Determine the empirical formula of the compound.



Na2F


NaF2


NaF


Na2F2

Answer 1

If you think in terms of the hypothetical 100g sample, then you can calculate it like this:

Number of moles (n) = mass (m) / molecular mass (Mr). You can find the molecular mass formula by looking in the periodic table.

So;

n(Na) = 54.76/22.99 = 2.381905... mol
n(F) = 45.24/19.00 = 2.381052... mol

Na has the smallest number of moles calculated and we will therefore divide each mole value on this smallest value, like this:

Na: 2.381905... mol/2.381905... mol = 1
F: 2.381052... mol/2.381905... mol \[\approx 1\]

So the empirical formula will then be: NaF