Question

How do you calculate the mass percent composition of carbon in each carbon containing compound?
a) $C{{H}_{4}}$
b) ${{C}_{2}}{{H}_{6}}$
c) ${{C}_{2}}{{H}_{2}}$
d) ${{C}_{2}}{{H}_{5}}Cl$

Answer 1

Hint: Mass percentage of an element in a compound or a molecule is the percentage composition, or the content of that specific element in that molecule. It can be calculated by the use of mass of that individual element and the total mass of the molecule.

Complete step-by-step answer:
First we need to define the mass percent composition of an element in a molecule, in order to solve this question. It is a way of expressing the concentration of a component or a part of a material present in a certain type of mixture.

The mass percentage composition of an element can be calculated by the division of the mass of that substance and the total mass of the mixture. In case of the mass percent of an element in a molecule we divide the mass of that individual element and the total mass of that molecule in which the element is present. This equation can be mathematically expressed as,
$mass%=\dfrac{{{W}_{(E)}}}{{{W}_{(M)}}}\times 100$
Where, ${{W}_{(E)}}$ is the weight of the element whole mass percent composition is to be determined and ${{W}_{(M)}}$ is the total mass of the molecule.

a) In the case of $C{{H}_{4}}$ , we know that the mass of carbon is $12.01u$ and that of hydrogen is $1.008u$. So the total mass of this compound would be, $12.01+(4\times 1.008)=16.032u$. So the mass percentage would become, \[mass%(C)=\dfrac{12.01u}{16.032u}\times 100%=74.91%\]

b) In the next case ${{C}_{2}}{{H}_{6}}$, the total mass of the compound would be, \[(2\times 12.01{ }u)+(6\times 1.008u)=30.068{ }u\]
And the mass of carbon would be $24.02u$. Now, in the same way, we will calculate the mass percentage of carbon, which is, \[mass%(C)=\dfrac{24.02u}{30.068u}\times 100%=79.89{ }%\]

c) In the third case ${{C}_{2}}{{H}_{2}}$, we would have total mass of the compound as \[\left( 2\times 12.01u \right)+\left( 2\times 1.008u \right)=24.036u\]
The mass of carbon would be $24.02u$. Now, we will calculate the mass percentage of carbon which is,
\[\dfrac{24.02u}{26.036u}\times 100%=92.26{ }%\]

d) Finally in case of ${{C}_{2}}{{H}_{5}}Cl$, the total mass of the compound would be,
\[\left( 2\times 12.01u \right)+\left( 5\times 1.008u \right)+35.45u=64.51u\]
And the mass of carbon is again $24.02u$.
By doing the same calculation, we get,
\[mass%\left( C \right)=\dfrac{24.02u}{64.51u}\times 100%=37.23{ }%\]

Note: The mass percent composition of an element in a compound decreases, as the number of different elements in the compound increases. This is because every element would contribute to the total mass of the compound hence lowering the value of ratio of mass of element by the mass of compound.