Question

The mass percentage of carbon in acetone is:
A.$41.36\mathrm{\%}$
B.$$31.034\mathrm{\%}$$
C.$$20.68\mathrm{\%}$$
D.$$62.068\mathrm{\%}$$

Answer 1

Hint: Acetone is a universal colourless, volatile, flammable organic solvent with the chemical formula $$(C{H}_3{)}_{2}CO$$. To get the mass percentage of carbon in acetone, we will divide the molar mass of carbon times the number of atoms in the compound by the molar mass of the whole compound and finally multiply it by $$100$$ to get the percentage.

Complete answer: The molecular formula of acetone is $$(C{H}_3{)}_{2}CO$$ where we have three carbon atoms, six hydrogen atoms and one oxygen atom.
The molar mass of a carbon atom is $$12.011g/mol$$. Similarly, the molar mass of hydrogen is $$1.008g/mol$$ and molar mass of oxygen is $$16g/mol$$. Hence, the molar mass of whole acetone molecule can be calculated as follows:
Molar mass of acetone $$=12.011\times 3+1.008\times 6+16$$
$$=36.033+6.048+16$$
$$=58.081g/mol$$
Percentage of mass of carbon $$={\displaystyle \frac{\text{mass of carbon}}{\text{mass of acetone}}}\times 100\mathrm{\%}$$
$$={\displaystyle \frac{3\times 12.011g/mol}{58.081g/mol}}\times 100\mathrm{\%}$$
Here, we have multiplied the molar mass of carbon by $$3$$ because there are three carbon atoms in the molecule of acetone.
On further solving,
$$={\displaystyle \frac{36.033g/mol}{58.081g/mol}}\times 100\mathrm{\%}$$
$$\cong 62.039\mathrm{\%}$$
Hence, the percentage mass of carbon in acetone is approximately equal to $$62.039\mathrm{\%}$$. This is in line to the value $$62.068\mathrm{\%}$$.
Therefore, the correct answer is option D.

Note:
The value in the solution is not written directly as equal to and an approximation is used. This is done because the molar mass of elements is written in more than three decimal places and can be very large. In high school calculation, it is not recommended to write such large values and the values used are only equal to two or three decimal places. That is why there is a bit of inaccuracy and hence approximation is used. If students use the only whole numbers for their calculation, the inaccuracy will further increase as the solution will be around $$60\mathrm{\%}$$.