The number of moles of CuO, that will be utilized in Dumas method for estimating nitrogen in a sample of 57.5 g of N, N-dimethylaminopentane is _________ $$\times 10^{-2}$$. (Nearest integer)

We are asked to calculate how many moles of copper(II) oxide, CuO, take part in the Dumas combustion of a given mass of the amine “N, N-dimethylaminopentane”.

First we write the molecular formula of the compound. A pentyl group contributes $$\mathrm{C_5H_{11}}$$ and two additional methyl groups are attached to the nitrogen, giving two more carbon atoms and six more hydrogens. Hence we have

$$\mathrm{C_{5+2}H_{11+6}N}=\mathrm{C_7H_{17}N}.$$

Its molar mass is obtained term by term:

$$M=\underbrace{7\times 12}_{\text{carbon}}+\underbrace{17\times 1}_{\text{hydrogen}}+\underbrace{1\times 14}_{\text{nitrogen}} =84+17+14 =115\ \text{g mol}^{-1}.$$

The sample mass is $$57.5\ \text{g}$$, so the number of moles of the amine present is

$$n_{\text{amine}}=\frac{57.5\ \text{g}}{115\ \text{g mol}^{-1}}=0.500\ \text{mol}.$$

Now we recall the stoichiometry of the Dumas combustion. For a general molecule $$\mathrm{C_xH_yN_z},$$ complete oxidation with CuO satisfies the balanced relation

$$\mathrm{C_xH_yN_z} \;+\;(2x+\tfrac{y}{2})\,\mathrm{CuO} \;\longrightarrow\; x\,\mathrm{CO_2}+ \tfrac{y}{2}\,\mathrm{H_2O}+ \tfrac{z}{2}\,\mathrm{N_2} + (2x+\tfrac{y}{2})\,\mathrm{Cu}.$$

This equality is obtained as follows. Each carbon atom requires two oxygen atoms to become $$\mathrm{CO_2}$$, and each pair of hydrogen atoms requires one oxygen atom to become $$\mathrm{H_2O}$$. Therefore, per mole of the organic compound, the total number of oxygen atoms demanded is $$2x+\dfrac{y}{2}$$. Since every mole of CuO supplies exactly one oxygen atom, the same numerical amount, $$2x+\dfrac{y}{2},$$ gives the moles of CuO used.

Substituting $$x=7$$ and $$y=17$$ for our specific amine, we get

$$n_{\text{CuO per mol amine}}=2(7)+\frac{17}{2}=14+\frac{17}{2}=14+8.5=22.5\ \text{mol}.$$

We actually have only half a mole of the amine, so the total moles of CuO consumed become

$$n_{\text{CuO total}} =0.500\ \text{mol amine}\;\times\;22.5\ \frac{\text{mol CuO}}{\text{mol amine}} =11.25\ \text{mol}.$$

It is convenient to express $$11.25$$ as a product with $$10^{-2}$$:

$$11.25\ \text{mol}=1125\times10^{-2}\ \text{mol}.$$

The problem asks for the nearest integer appearing as the coefficient in “$$\times10^{-2}$$”, which is clearly $$1125$$.

So, the answer is $$1125$$.