Question

When $2.86g$ of a mixture of $1 -$butene, ${C_4}{H_8}$ and butane, ${C_4}{H_{10}}$ was burned in excess of oxygen, $8.80g$ of $C{O_2}$ and $4.14g$ of ${H_2}O$ were obtained. What is the percentage by mass of butane in the mixture? (Write your answer by dividing it with $10$ and round up to nearest integer)

Answer 1

To solve this question first we have to write the reactions involved. After writing the reaction we have to compare the number of moles of reactant to produce the product. By solving the equations thus formed we can get the mass of butane in the mixture.

Formula used: Number of moles$= \dfrac{{{\text{given mass}}}}{{{\text{molecular mass}}}}$
Percentage by mass$= \dfrac{{{\text{mass of substance}}}}{{{\text{mass of mixture}}}} \times 100$

Complete step by step answer:
Percentage by mass can be calculated by dividing mass of the substance with total mass of the mixture and then multiplying it by $100$. In this question we have given a mixture of butane and butane. This mixture is burnt in excess of oxygen and the mass of the mixture and the products formed is given. We have to find the percentage by mass of butane in this mixture. Reactions involved when butane and butane will be heated are:
${C_4}{H_8} + 6{O_2}\xrightarrow[{}]{}4C{O_2} + 4{H_2}O$ and ${C_4}{H_{10}} + 6{O_2}\xrightarrow[{}]{}4C{O_2} + 4{H_2}O$
Let the mass of butane be $x$. Total mass of mixture is $2.86g$ therefore mass of butene will be $2.86 - x$ . Molecular mass is the mass of one molecule of a substance. It is calculated by adding atomic mass of all the components present in the molecule. Molecular mass of butane is $58g$ and butene is $56g$ . With the help of this we can calculate number of moles of butane and butene with the formula, Number of moles$= \dfrac{{{\text{given mass}}}}{{{\text{molecular mass}}}}$.
Number of moles of butane$= \dfrac{x}{{58}}$
Number of moles of butene$= \dfrac{{2.86 - x}}{{56}}$
From the reaction we can see that one mole of butane and butene react to form four moles of carbon dioxide. Therefore,
$\dfrac{x}{{58}}$ Moles of butane$= \dfrac{{4x}}{{58}}$ moles of carbon dioxide (as one mole of butane is equal to four moles of carbon dioxide)
$\dfrac{{2.86 - x}}{{56}}$ Moles of butene$= \dfrac{{4\left( {2.86 - x} \right)}}{{56}}$ mole of carbon dioxide
A total mole of carbon dioxide formed from the mixture is equal to the sum of moles of carbon dioxide calculated above from butane and butene.
Total moles of carbon dioxide$= \dfrac{{4\left( {2.86 - x} \right)}}{{56}} + \dfrac{{4x}}{{58}}$
Molecular mass of carbon dioxide is $44g$. Therefore mass of carbon dioxide formed can be written as:
Mass of carbon dioxide formed$= \left( {\dfrac{{4\left( {2.86 - x} \right)}}{{56}} + \dfrac{{4x}}{{58}}} \right) \times 44$
Mass of carbon dioxide formed is given and is equal to $8.80g$. Therefore,
$\left( {\dfrac{{4\left( {2.86 - x} \right)}}{{56}} + \dfrac{{4x}}{{58}}} \right) \times 44 = 8.80$
$\left( {\dfrac{{4\left( {2.86 - x} \right)}}{{56}} + \dfrac{{4x}}{{58}}} \right) = \dfrac{{8.80}}{{44}} = 0.2$
Solving this equation we get,
$x = 1.74g$
Therefore, mass of butane is $1.74g$ (we assumed the mass of butane equal to $x$)
Now, we have both mass of butane and mass of mixture. With the help of this data we can calculate the percentage by mass of butane using the formula,
Percentage by mass$= \dfrac{{1.74}}{{2.86}} \times 100 = 60.8\% \approx 61\%$
Therefore the percentage by mass of butane in this mixture is $61\%$.

Note:
Percentage by mass can be calculated by dividing the mass of substance with the total mass of mixture and then multiplying it by $100$. Atomic percent of a particular atom in a molecule can be calculated by dividing the mass of that atom with the total mass of the molecule and then multiplying it by $100$.