Question

A compound is found to contain 13.65 % carbon and 86.35 % fluorine by mass. To answer the question, enter the elements in the order presented above. QUESTION 1: The empirical formula for this compound is . QUESTION 2: The molar mass for this compound is 88.01 g/mol. The molecular formula for this compound is __________.

Answer 1

Answer: The molecular formula will be $CF_4$

Explanation:

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 C= 13.65 g

Mass of F = 86.35 g

Step 1 : convert given masses into moles.

Moles of C =$\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{13.65g}{12g/mole}=1.138moles$

Moles of F=$\frac{\text{ given mass of F}}{\text{ molar mass of F}}= \frac{86.35g}{19g/mole}=4.545moles$

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

For C = $\frac{1.138}{1.138}=1$

For F = $\frac{4.545}{1.138}=4$

The ratio of C : F= 1: 4

Hence the empirical formula is $CF_4$

The empirical weight of $CF$ = 1(12)+4(19)= 88g.

The molecular weight = 88.01 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{88.01}{88}=$

The molecular formula will be=$1\times CF_4=CF_4$