Question

A 0.050 02-g sample of impure piperazine contained 71.29 wt% piperazine (FM 86.136). How many grams of product (FM 206.240) will be formed when this sample is analyzed by Reaction 27-6?

Short answer

The mass of the product is 0.08538 g.

Step-by-step solution

calculate the mass of piperazine:

In this task we have 0.05002 g of impure piperazine which contains 71.29 wt% of piperazine (M=86.136 g/mol). Here we calculate the mass of the product (M = 206.240 g/mol) formed when this sample is analyses by reaction 27-6.

Consider the following:

$1\mathrm{mol}\left(\mathrm{product}\right)=1\mathrm{mol}\left(\mathrm{piperazine}\right)$

Calculate the mass of piperazine in a sample:

$$\begin{gathered} \mathrm{m}(\mathrm{piperazine})=\mathrm{m}(\mathrm{piperazine})/1\mathrm{g}\mathrm{of}\mathrm{sample}\times \mathrm{m}(\mathrm{sample}) \\ \mathrm{m}(\mathrm{piperazine})=0.7129\mathrm{g}/1\mathrm{g}\times 0.05002\mathrm{g} \\ \mathrm{m}(\mathrm{piperazine})=0.03566\mathrm{g} \end{gathered}$$

Calculate the mass of product:

$$\begin{gathered} \mathrm{m}\left(\mathrm{product}\right)=\frac{\mathrm{M}\left(\mathrm{product}\right)}{\mathrm{M}\left(\mathrm{piperazine}\right)}\times \mathrm{m}\left(\mathrm{piperazine}\right) \\ \mathrm{m}\left(\mathrm{product}\right)=\frac{206.240\mathrm{g}/\mathrm{mol}}{86.136\mathrm{g}/\mathrm{mol}}\times 0.03566\mathrm{g} \\ \mathrm{m}(\mathrm{product})=0.08538\mathrm{g} \end{gathered}$$