Question 3.EP.7: A demolition engineer is planning to use the explosive ethyl...

A demolition engineer is planning to use the explosive ethylenedinitramine (C_{2}H_{6}N_{4}O_{4}, also known as halite) to bring down an abandoned building. Calculations show that 315 moles of the compound will provide the necessary explosive force. How many pounds of C_{2}H_{6}N_{4}O_{4} should be used?

Strategy
We are asked to convert from a number of moles to a mass. We know that the link between these two quantities is the molar mass of the compound, so we can start by calculating that. Once we know the molar mass, we can use it to find the required mass of explosive. Finally, we can easily convert that result from grams to pounds.

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We will need to calculate the molar mass of the compound.

\ \begin{matrix} C_{2}H_{6}N_{4}O_{4}: & 2 \ moles \ C: \ 2\times 12.011 \ g/mol=24.022 \ g/mol \\ & 6 \ moles \ H: \ 6\times 1.0079 \ g/mol=6.0474 \ g/mol \\ & 4 \ moles \ N: \ 4\times 14.0067 \ g/mol=56.0268 \ g/mol \\ & 4 \ moles \ O: \ 4\times 15.9994 \ g/mol=\underline{63.9976 \ g/mol} \\ & \quad\quad\quad\quad Molar \ mass \quad\quad = 150.094 \ g/mol \end{matrix}

Now we can use the molar mass and the required number of moles to find the appropriate mass.

315 \ moles \ C_{2}H_{6}N_{4}O_{4}\times \frac{150.094 \ g \ C_{2}H_{6}N_{4}O_{4}}{1 \ mol \ C_{2}H_{6}N_{4}O_{4}}=4.73 × 10^{4}g \ C_{2}H_{6}N_{4}O_{4}

Finally, we can convert this from grams to pounds:

4.73 × 10^{4} \ g \ C_{2}H_{6}N_{4}O_{4}\times \frac{1 \ pound}{454 \ g} = 104 \ pounds \ C_{2}H_{6}N_{4}O_{4} \ needed

Analyze Your Answer
We can assess the reasonableness of this answer by comparing it to Example Problem 3.6. For these two problems, the molar masses of the compounds are roughly comparable. In this problem, we have about 300 moles, whereas in the previous one, we had about 3 moles, so it’s not surprising that our answer is close to 100 times larger than that in our last exercise.

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