Question

A 1.000-g sample of unknown gave 2.500 g of bis (dimethylglyoximate)nickel (II) (FM 288.91) when analyzed by Reaction 27-7. Find the weight percent of Ni in the unknown.

Short answer

The weight percent of Ni is 0.5079.

Step-by-step solution

calculate the moles of bis (dimethylglyoximate)nickel (II):4

In this task we have a 1 g of sample which gave 2.5 g of bis (dimethylglyoximate)nickel(II) (M=288.91 g/mol) when analyzed by Reaction 27-7. Here we will find the weight percent of Ni in the unknown sample.

Calculate the moles of bis (dimethylglyoximate)nickel(II):

$$\begin{gathered} \mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}} \\ \mathrm{n}=\frac{2.5\mathrm{g}}{288.91\mathrm{g}/\mathrm{mol}} \\ \mathrm{n}=8.6532\times {10}^{-8}\mathrm{mol} \end{gathered}$$

consider that n (bis (dimethylglyoximate)nickel(II)) =n (Ni)

calculate mass and weight percent of Ni:

The mass of Ni is,

$$\begin{gathered} \mathrm{m}\left(\mathrm{Ni}\right)=\mathrm{n}\times \mathrm{M} \\ \mathrm{m}\left(\mathrm{Ni}\right)=(8.6532\times {10}^{-8}\mathrm{mol})\times 58.6934\mathrm{g}/\mathrm{mol} \end{gathered}$$

The weight percent of Ni is,

$$\begin{gathered} \mathrm{wt}\%=\frac{\mathrm{m}\left(\mathrm{Ni}\right)}{\mathrm{m}\left(\mathrm{sample}\right)} \\ \mathrm{wt}\%=\frac{0.50788\mathrm{g}}{1\mathrm{g}} \\ \mathrm{wt}\%=0.5079 \end{gathered}$$