Final answer:
The compound with a molar mass of about 70 g/mol and a composition of 19.7% nitrogen and 80.3% fluorine is nitryl fluoride (NF3). The Lewis structure shows nitrogen atom with sp3 hybridization and a trigonal pyramidal geometry. The formal charges on both nitrogen and fluorine in this compound are zero.
Step-by-step explanation:
Formation of Nitryl Fluoride
To determine the chemical formula and Lewis structure of a compound with a molar mass of about 70 g/mol that contains 19.7% nitrogen and 80.3% fluorine by mass, we start with the percent composition to deduce the empirical formula. Given the percentages, we can assume a total mass of 100 grams, so we would have 19.7 grams of nitrogen and 80.3 grams of fluorine. Using their molar masses (N: 14.01 g/mol, F: 19 g/mol), we convert these masses to moles:
- N: 19.7 g / 14.01 g/mol ≈ 1.41 mol
- F: 80.3 g / 19 g/mol ≈ 4.23 mol
We divide by the smallest number of moles to get the simplest whole number ratio:
- N: 1.41 mol / 1.41 = 1
- F: 4.23 mol / 1.41 ≈ 3
Thus, the empirical formula is NF3, which has a molar mass of approximately 71 g/mol (14.01 for N + 3*19 for F), close enough to the target molar mass of 70 g/mol to suggest this is our compound. The structure of NF3 involves sp3 hybridization of the nitrogen atom with a pyramidal geometry.
The formal charge of each atom can be calculated by subtracting the number of non-bonding electrons and half the number of bonding electrons from the number of valence electrons for each atom.
In the case of NF3:
- Nitrogen (N) has 5 valence electrons and is bonded to three fluorine atoms with one lone pair, giving it a formal charge of 0.
- Each Fluorine (F) has 7 valence electrons, is single-bonded to the nitrogen, and has three lone pairs, also giving each a formal charge of 0.