Question

When 2.86 g of a mixture of 1-butene, ${{C}_{4}}{{H}_{8}}$ and butane, ${{C}_{4}}{{H}_{10}}$ was burned in excess of oxygen, 8.80 g of $C{{O}_{2}}$ and 4.14 g of ${{H}_{2}}O$ were obtained. What is percentage by mass of butane in the mixture? (Write your answer by dividing it by 10 and round up to nearest integer)

Answer 1

Percentage by mass of any substance is the amount of that substance present in the compound upon total amount of substance multiplied by 100. The amount when in form of mass is called mass percent, while in the form of volume is called percent by volume.

Formula used: percent by mass = $\dfrac{mass}{total\,mass}\times 100$

Complete step-by-step answer: We have been given a mixture of 1-butene, ${{C}_{4}}{{H}_{8}}$and butane, ${{C}_{4}}{{H}_{10}}$, whose amount is 2.86 g. This mixture of combustion produces 8.80 g of $C{{O}_{2}}$and 4.14 g of ${{H}_{2}}O$. We have to find out the percent by mass of butane in the mixture.
For this let’s assume the value (amount) of butane to be x g. Therefore, the mass of 1-butene is 2.86- x g. The reaction that happens is,

& {{C}_{4}}{{H}_{8}}+6{{O}_{2}}\to 4C{{O}_{2}}+4{{H}_{2}}O \\\ & {{C}_{4}}{{H}_{10}}+\dfrac{13}{2}{{O}_{2}}\to 4C{{O}_{2}}+5{{H}_{2}}O \\\ \end{aligned}$$ The molar mass of 1-butene is 56 g/mol, and that of butane is 58 g/mol. So, the number of moles of both of them will be, $\dfrac{2.86-x}{56}$ for 1-butene, and $\dfrac{x}{58}$ for butane. The amount of carbon dioxide, these moles will produce will be given from the stoichiometry of the above equation, as $\dfrac{4(2.86-x)}{56}$from 1-butene, and$\dfrac{4x}{58}$from butane. So, by calculating total number of moles of carbon dioxide from these values, and solving for x, we will get the mass of butane, as, $\dfrac{4(2.86-x)}{56}+\dfrac{4x}{58}=\dfrac{8.80}{44}$, Therefore, x = 1.738 g. Now, using the mass percent formula, we have, Mass percent of butane = $\dfrac{1.738}{2.86}\times 100$ Mass percent of butane = 60.8 Hence, by dividing this by 10 and rounding it we get the answer as, 6. **Note:** As calculated for the carbon dioxide, we can also do the same way with the number of moles of water, and then solve for x. But for that we have to make the number of moles of water for both reactions to be the same.