Question

What is the empirical formula of a substance that contains 2.64 g of c, 0.444 g of h, and 7.04 g of o?

Answer 1

Answer: The empirical formula for the given compound is $CH_2O_2$

Explanation:

We are given:

Mass of C = 2.64 g

Mass of H = 0.444 g

Mass of O = 7.04 g

To formulate the empirical formula, we need to follow some steps:

• Step 1: Converting the given masses into moles.

Moles of Carbon =$\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{2.64g}{12g/mole}=0.22moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.444g}{1g/mole}=0.444moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{7.04g}{16g/mole}=0.44moles$

• 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 0.22 moles.

For Carbon = $\frac{0.22}{0.22}=1$

For Hydrogen  = $\frac{0.444}{0.22}=2.02\approx 2$

For Oxygen  = $\frac{0.44}{0.22}=2$

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

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

Hence, the empirical formula for the given compound is $CH_2O_2$

Answer 2

To calculate empirical formula, number of moles of different atoms calculated first:

Number of moles of Carbon , n=  

=$\frac{Given mass of carbon}{Molar mass of carbon}$

=$\frac{2.64 g}{12.011 g mol^{-1}}$

= 0.220 mol  

Number of moles H =$\frac{Given mass of hydrogen}{Molar mass of hydrogen}$

=$\frac{0.444 g}{1.008 g mol^{-1}}$

= 0.440 mol

Number of moles of oxygen =$\frac{Given mass of oxygen}{Molar mass of oxygen}$

=$\frac{7.04 g}{15.999 g mol^{-1}}$

= 0.440 mol

Ratio of carbon

=$\frac{0.220}{0.220}$

= 1

Ratio of hydrogen

=$\frac{0.440}{0.220}$

= 2  

Ratio of oxygen

=$\frac{0.440}{0.220}$

= 2  

So, the empirical formula is CH₂O₂