Question

A sample of a compound is made up of 57.53 g c, 3.45 g h, and 39.01 g f. Determine the empirical formula of this compound

Answer 1

Answer: The empirical formula is $C_2H_2F$

Explanation:

We are given:

Mass of hydrogen = 3.45 g

Mass of carbon = 57.53 g

Mass of fluorine = 39.01 g

To find the empirical formula of the compound, we must follow some steps:

• Step 1: Converting the given masses into moles.

Moles of Carbon =$\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{57.53g}{12g/mole}=4.79moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{3.45g}{1g/mole}=3.45moles$

Moles of Fluorine = $\frac{\text{Given mass of fluorine}}{\text{Molar mass of Fluorine}}=\frac{39.01g}{19g/mole}=2.05moles$

• Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 2.05 moles.

For Carbon = $\frac{4.79}{2.05}=2.34\approx 2$

For Hydrogen = $\frac{3.45}{2.05}=1.68\approx 2$

For Fluorine = $\frac{2.05}{2.05}=1$

• Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : F = 2 : 2 : 1

Hence, the empirical formula is $C_2H_2F$

Answer 2

Answer:    C7H5F3

Explanation:

First, use the grams to calculate the number of moles of each atom present.

mol C=57.53gC×1molC12.011gC=4.790molC

mol H=3.45gH×1molH1.008gH=3.423molH

mol F=39.01gF×1molF18.998gF=2.053molF

Next, divide by the smallest number of moles.

subscript C=4.790molC2.053molF=2.33≈73

 

subscript H=3.423molH2.053molF=1.66≈53

 

subscript F=2.053molF2.053molF=1

Notice that the values obtained for carbon and hydrogen are approximately equal to 73 and 53 respectively; therefore, multiplying each of the values above by 3 will yield the whole number subscripts to be used in the empirical formula.

Subscript C:73×3=7Subscript H:53×3=5Subscript F:1×3=3

Therefore, the empirical formula is C7H5F3.