Question

What is the empirical formula of a compound consisting of 29.6% oxygen and 70.4% fluorine by mass?

Answer 1

Answer : The empirical formula of the compound is, $OF_2$

Solution : Given,

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of O = 29.6 g

Mass of F = 70.4 g

Molar mass of F = 19 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{29.6g}{16g/mole}=1.85moles$

Moles of F = $\frac{\text{ given mass of F}}{\text{ molar mass of F}}= \frac{70.4g}{19g/mole}=3.70moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For O = $\frac{1.85}{1.85}=1$

For F = $\frac{3.70}{1.85}=2$

The ratio of O : F = 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $O_1F_2=OF_2$

Therefore, the empirical formula of the compound is, $OF_2$

Answer 2

O1Fl2

1. Assume an 100g sample, so the percentage will stay the same

2. Covert each element into their molar mass
29.6/16.00=1.8 mols of O
70.4/19.00=3.7 mols of Fl

3. Divide both by the smallest value of mol
1.8/1.8=1 O
3.7/1.8=2 Fl

4. Write the empirical formula:
O1Fl2