The atomic number of Li is 3
Electron configuration of Li : 1s² 2s¹
The atomic number of Na is 11
Electron configuration of Na : 1s²2s²2p⁶3s¹
Thus there is one electron in the valence shell of Li (2s¹) and that of Na (3s¹). However, the valence electron in Na is in a shell that is farther away from the nucleus compared to that of Li. As a result, the Na valence electron will be held less tightly by the nucleus i.e. it will experience a reduced nuclear attraction and can be removed easily than the Li 2s electron.
Answer;
C7H14O2
Solution;
Isobutyl contains , oxygen, carbon and hydrogen (total mass is 1.152 g)
Mass of carbon = 12/44 × 2.726 g
= 0.743455 g
Mass of Hydrogen = 2/18 × 1.116 g
= 0.124 g
Mass of oxygen = 1.152 - (0.7435 + 0.124)
= 0.2845 g
Moles of carbon ; 0.7435/12 = 0.06196 moles
Moles of hydrogen; 0.124/1 = 0.124 moles
Moles of oxygen; 0.2845/16 = 0.01778 moles
Ratios ; 0.06196/0.01778 ; 0.124/0.01778 : 0.01778/0.01778
= 3.5 : 7.0 : 1
To make them whole numbers ; we multiply the ratios by 2 to get;
(3.5 : 7.0 : 1 )2 = 7 : 14 : 2
Thus, the empirical formula of Isobutyl propionate is C7H14O2
Limiting reactant in this experiment would be Magnesium since it will run out first
