Q. 23 - An Introduction to General, Organic, and Biological Chemistry

Question

Acetone (propanone), a clear liquid solvent with an acrid odor, is used to remove nail polish, paints, and resins. It has a low boiling point and is highly flammable. Acetone has a density of $0.786 g/mL$. Acetone is a solvent used in nail polish remover.

1.  What is the molecular formula of acetone? 
2. What is the molar mass of acetone? 
3.  Identify the bonds $\mathrm{C}-\mathrm{C} , \mathrm{C}-\mathrm{H}, \mathrm{C}-\mathrm{O}$ in a molecule of acetone as polar covalent or nonpolar covalent.
4.  Write the balanced chemical equation for the combustion of acetone.
5.  How many grams of oxygen gas are needed to react with $15 mL$of acetone? 

Step-by-Step Solution

Answer

Our required answers are:

1. ${\mathrm{C}}_3{\mathrm{H}}_6\mathrm{O}$ is molecular formula of acetone
2. $58 g/mol$ is molar mass of acetone
3. $\mathrm{C}-\mathrm{C}$ is non-polar covalent bond , $\mathrm{C}-\mathrm{H}, \mathrm{C}-\mathrm{O}$ are polar covalent bond
4. ${\mathrm{C}}_3{\mathrm{H}}_6\mathrm{O} + 4{\mathrm{O}}_2\to 3{\mathrm{H}}_2\mathrm{O}+3{\mathrm{CO}}_2\left(\mathrm{g}\right)+ \mathrm{heat}$  is required equation on combustion of acetone
5. $26 g$oxygen is required for complete reaction with acetone.

Part(a) Step 1: Given Information

We need to write molecular formula of acetone.

Part(a) Step 2: Explanation

Acetone is a organic compound with three carbon atoms and having  ketonic group as functional group.

So, molecular formula of acetone is ${\mathrm{C}}_3{\mathrm{H}}_6\mathrm{O}$, having structure as 

Part(b) Step 1: Given Information

We need to find molar mass of acetone.

Part(b) Step 2: Explanation

Molar mass of acetone = $6\times$(molar mass of data-custom-editor="chemistry" $\mathrm{H}$) + $3\times$(molar mass of data-custom-editor="chemistry" $\mathrm{C}$)+ molar mass of data-custom-editor="chemistry" $\mathrm{O}$

$$\begin{gathered} =6\times 1 + 3\times 12 + 16 \\ =58 g/mol \end{gathered}$$

Hence, molar mass of acetone is $58 g/mol$.

Part(c) Step 1: Given Information

Identify the bonds $\mathrm{C}-\mathrm{C}, \mathrm{C}-\mathrm{H} , \mathrm{C}-\mathrm{O}$ in a molecule of acetone as polar covalent or nonpolar covalent. 

Part(c) Step 2: Explanation

Non polar covalent bonds does not have dipole moment between them. So, $\mathrm{C}-\mathrm{C}$ bond is non polar covalent bond as dipole moment is not present between them as difference in electronegativity is zero.

$\mathrm{C}-\mathrm{H}, \mathrm{C}-\mathrm{O}$ have dipole between them. So, they are polar covalent bond.

Part(d) Step 1: Given Information

We need to write the balanced chemical equation for the combustion of acetone.

Part(d) Step 2: Explanation

When a chemical combines with oxygen to release energy, it is called a combustion reaction.

Required balanced equation is ${\mathrm{C}}_3{\mathrm{H}}_6\mathrm{O}\left(\mathrm{g}\right)+ 4{\mathrm{O}}_2\left(\mathrm{g}\right) \to 3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)+3{\mathrm{CO}}_2\left(\mathrm{g}\right)+ \mathrm{heat}$

Part(e) Step 1: Given Information

We need to find grams of oxygen gas  needed to react with $15.00 mL$of acetone

Part(e) Step 2: Explanation

From reaction shown in part(d) Step 2, we analyze that for combustion of one mole acetone we need 4 mole of oxygen gas.

Mass of $15 mL$acetone=$15\times 0.786=11.79g$

Moles of acetone=$\frac{11.79}{58}=0.203$

Moles of oxygen required= $4\times 0.203=0.81$

Grams of oxygen required=$0.81\times 32\simeq 26 g$

Hence, $26 g$ oxygen is required to react with $15 mL$ of acetone.