An oxide of nitrogen is 25.9% N by mass, has a molar mass of 108 g/mol, and contains no nitrogen-nitrogen or oxygen-oxygen bonds
. Draw its Lewis structure (including all lone pair electrons) and name it. Include any nonzero formal charges in your structure.
1 answer:
Answer:
The compound is N2O4
Explanation:
We have certain important pieces of information about the compound;
1) it is an oxide (a binary compound of nitrogen and oxygen)
2) there are no N-N bonds present
3) there are no O-O bonds present
Since it contains only nitrogen and oxygen then nitrogen accounts for 25.9% of the molecule by mass then oxygen should account for (100-25.9) = 74.1% oxygen
Relative atomic mass of oxygen = 16
Relative atomic mass of nitrogen = 14
We now deduce the empirical formula
Nitrogen. Oxygen
25.9/14. 74.1/16
1.85/1.85. 4.6/1.85 (divide through by the lowest ratio)
1 2
Empirical formula is NO2
To find the molecular formula
(NO2)n = 108
(14+2(16))n= 108
46n=108
n= 108/46
n= 2
Therefore molecular formula= N2O4
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The enthalpy change : -196.2 kJ/mol
<h3>Further explanation </h3>The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation
The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)
(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)
The value of ° H ° can be calculated from the change in enthalpy of standard formation:
∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)
Reaction
2 H₂O₂(l)-→ 2 H₂O(l) + O₂(g)
∆H ° rxn = 2. ∆Hf ° H₂O - 2. ∆Hf °H₂O₂