Question

How many milliliters of 2.15 wt% dimethylglyoxime (DMG, 116.12 g/mol) solution should be used to provide a 50.0% excess for analyzing the nickel content of 0.9984g of steel with an expected 2.07 wt% nickel? Please show work and explain! Thank you.

Answer 1

Mass of steel = 0.9984 g.

Nickel content in steel = 2.07 wt%.

Therefore, mass of nickel in 0.9984 g steel = (2.07 wt%)*(0.9984 g)

= (2.07/100)*(0.9984 g)

= 0.020667 g.

Atomic mass of nickel = 58.6934 g/mol.

Mol(s) nickel corresponding to 0.020667 g = (0.020667 g)/(58.6934 g/mol)

= 3.52118*10^-4 mol.

Nickel and dimethylglyoxime react on a 1:1 molar ratio to give nickel-dmg complex.

Hence, mol(s) dimethylglyoxime required to react with the nickel in steel = 3.52118*10^-4 mol.

Molar mass of dimethylglyoxime = 116.12 g/mol.

Mass of dimethylglyoxime reacted with nickel = (3.52118*10^-4 mol)*(116.12 g/mol)

= 0.040888 g.

Since the sample of dimethylglyoxime contains 50% excess dimethylglyoxime, hence, the mass of dimethylglyoxime present in the sample =

(0.040888 g) + (50%)*(0.040888 g)

= (0.040888 g) + (50/100)*(0.040888 g)

= (0.040888 g) + (0.020444 g)

= 0.061332 g.

The given solution of dimethylglyoxime is 2.15 wt%, i.e., 100 mL solution contains 2.15 g dimethylglyoxime.

Therefore, 0.061332 g dimethylglyoxime is contained in

(100 mL)*(0.061332 g)/(2.15 g)

= 2.85265 mL

≈ 2.85 mL (ans, correct to 3 sig. figs).

Volume of dimethylglyoxime solution required = 2.85 mL (ans).