Question

One of the chemical controversies of the 19th century concerned the element Beryllium (Be). The debate Centered around whether Beryllium naturally reacted with two ions (divalent) or three ions trivalent). In 1894, an experiment was done to test this. In the experiment Beryllium was reacted with the anion C5 H₂O₂- and measured the density of the gaseous produc Two experiments were performed & the following was collected. MASS: 2022g Volume: 22.6 cm3 Temp: 13°C Pressure: 765.2 mmity I .22249 26.0 cms 17°C 764.6 mmila If beryllium is a divalent metal, the molecular formula of the product would be Belcs H₂O. If trivalent the formula would be Be Cs H₂O₂). Show how this data helps to confirm that Berillium is a divalent metal,

Answer 1

Answer:

Here, we will use the density expression from ideal gas law.

Density = Mass/Volume = (Pressure x Molecular mass)/(R x Temperature)

R = Gas constant = 0.08206 L.atm/K.mol

For Data Set 1:

T = 13 C = 286 K; P = 765.2 torr (mm Hg) = 1.006 atm

V = 22.6 cm³ or 22.6 x 10^-3 L

Molecular mass = (0.2022 g x 0.08206 L.atm/K.mol x 286 K)/(1.006 atm x 22.6 x 10^-3L) = 208.7 g/mol

For Data Set 2:

T = 13 C = 290 K; P = 765.2 torr (mm Hg) = 1.006 atm

V = 26 cm³ or 26 x 10^-3 L

Molecular mass = (0.2224 g x 0.08206 L.atm/K.mol x 290 K)/(1.006 atm x 26 x 10^-3L) = 202.3 g/mol

From above calculation; (from there atomic weight)

Be(C₅H₇O₂)₂ gives = 207 g/mol

Be(C₅H₇O₂)₂ gives = 306 g/mol

Hence, the above dataset represents the Be²⁺ to be active ion.

Thanks in advance