First, we determine the mass of each element from the data collected. We can get the mass of molybdenum Mo from the difference between the mass of crucible and molybdenum and the mass of crucible:
Mass of molybdenum = 39.52 – 38.26 = 1.26 g Mo
We can calculate for the mass of molybdenum oxide from the difference between the mass of crucible and molybdenum oxide and the mass of crucible:
Mass of molybdenum oxide = 39.84 – 38.26 = 1.58g
We can now compute for the mass of oxygen O by subtracting the mass of molybdenum from the mass of molybdenum oxide:
Mass of oxygen in molybdenum oxide = 1.58 – 1.26 = 0.32g O
To convert mass to moles, we use the molar mass of each element.
1.26 g Mo * 1 mol Mo / 95.94 g Mo = 0.0131 mol Mo
0.32 g O * 1 mol O / 15.999 g O = 0.0200 mol O
0.0131 mol is the smallest number of moles. We divide each mole value by this number:
0.0131 mol Mo / 0.0131 = 1
0.0200 mol O / 0.0131 = 1.53
Multiplying these results by 2 to get the lowest whole number ratio,
0.0131 mol Mo / 0.0131 = 1 * 2 = 2
0.0200 mol O / 0.0131 = 1.5 * 2 = 3
Thus, we can write the empirical formula as Mo2O3.
Answer:
ΔH rxn = -1010 kJ/molC₂H₂
Explanation:
To obtain the enthalpy change for a reaction from bond energies what we do is to make an inventory of the bonds broken and formed for the balanced chemical reaction:
C₂H₂ + 5/2O₂ ⇒ 2CO₂ + H₂O
Bond Broken Bonds Formed
2 C-H + 1 C≡C + 5/2 O=O 4C=O + 2 H-O
Enthalpy bonds broken:
2 mol (456 kJ/mol)+ 1 mol (962 kJ/mol) + 5/2 mol (499 kJ/mol) = 3121.5 kJ
Enthalpy bond formed:
4 mol (802 kJ/mol) + 2 mol (462 kJ/mol) = 4132.0 kJ
ΔH rxn = H broken - H formed = 3121.5 kJ - 4132.0 kJ = - 1010 kJ (per mol C₂H₂ )
