The following data were obtained from the analysis of five volatile phosphorous compounds. The percentages of phosphorous in these compounds having the molecular weights 222, 154, 140, 300 and 126 are respectively 27.9%, 20-2%, 22.5%, 43.7% and 24-6%. Calculate the atomic weight of phosphorous (to the nearest whole number).

Explanation

The smallest weight of an element present in 1 mole of its various compounds gives either the atomic weight of the element or a simple multiple of the atomic weight.
(i) Weight of P in 1 mole (222 g) `=(27.9)/(100) xx 222 = 61.94.`
(ii) Weight of P in 1 mole (154 g) `= (20.2)/(100) xx 154 = 31.08.`
(iii) Weight of P in 1 mole (140) `=(22.5)/(100) xx140 = 31.50`
(iv) Weight of P in 1 mole (300g) = `(43.7)/(100) xx300= 131.10`
(iv) Weight of P in 1 mole (126 g) `= (24.6)/(100) xx126 = 30.99`
The smallest weight of P being 31 (to the nearest whole number) may be its atomic weight.