Problem 113 - Chemistry: The Central Science

Question

A compound composed of \(2.1 \% \mathrm{H}, 29.8 \% \mathrm{~N}\), and \(68.1 \% \mathrm{O}\) has a molar mass of approximately \(50 \mathrm{~g} / \mathrm{mol}\). (a) What is the molecular formula of the compound? (b) What is its Lewis structure if \(\mathrm{H}\) is bonded to \(\mathrm{O}\) ? (c) What is the geometry of the molecule? (d) What is the hybridization of the orbitals around the \(\mathrm{N}\) atom? (e) How many \(\sigma\) and how many \(\pi\) bonds are there in the molecule?

Step-by-Step Solution

Answer

a) The molecular formula is HNO2. b) The Lewis structure is O = N - O with H bonded to N ( O double bonded to N, O single bonded to N, H single bonded to N). c) The geometry of the molecule is trigonal planar. d) The hybridization of the orbitals around the N atom is sp2. e) There are 3 sigma bonds and 1 pi bond in the molecule.

Step 1: Find the empirical formula

To find the empirical formula, we need to assume 100 grams of the compound. Then we can convert the percentages given into grams and divide by the molar mass of each element to find the mole ratio. Assuming a 100g sample: - 2.1g of Hydrogen (H) - 29.8g of Nitrogen (N) - 68.1g of Oxygen (O) Now, we will convert grams to moles for the three elements. Moles of H = (2.1 g) / (1.01 g/mol) ≈ 2.08 mol Moles of N = (29.8 g) / (14.01 g/mol) ≈ 2.13 mol Moles of O = (68.1 g) / (16.00 g/mol) ≈ 4.26 mol To find the empirical formula, we divide all the mole values by the smallest one (2.08) and round to the nearest whole number. H: 2.08 / 2.08 ≈ 1 N: 2.13 / 2.08 ≈ 1 O: 4.26 / 2.08 ≈ 2 So, the empirical formula is HNO2.

Step 2: Find the molecular formula

The molecular formula is obtained by multiplying the empirical formula with a whole number (n). The molar mass of the molecular formula should be approximately 50 g/mol, as given in the problem. Empirical formula mass of HNO2 = 1.01g (H) + 14.01g (N) + 2 * 16.00g (O) = 47.02 g/mol Now, we need to find the value of n: n = (Molar Mass of molecular formula) / (Empirical formula mass) = 50 / 47.02 ≈ 1 Since n ≈ 1, the molecular formula is the same as the empirical formula: Molecular Formula = HNO2

Step 3: Draw the Lewis structure

To draw the Lewis structure, we first need to count the total number of valence electrons. - H has 1 valence electron. - N has 5 valence electrons. - O has 6 valence electrons. Total valence electrons = 1(H) + 5(N) + 2 * 6(O) = 18 valence electrons The Lewis structure should have the least electronegative atom in the center, with the other atoms surrounding it. In this case, N is in the center, and O and H are bonded to it: O - N - O | H To complete the octet rule, we add lone pairs to the remaining valences: O = N - O || || H (6 lone pairs + 3 sigma bonds = 18 valence electrons)

Step 4: Determine the geometry and hybridization

The geometry of the molecule can be determined by the VSEPR theory, considering the number of electron domains. N has 3 domains (2 single bonds and 1 double bond). So it will have a trigonal planar geometry. The hybridization can be determined using the formula: n + l = (number of orbitals) n = number of sigma bonds l = number of lone pairs This molecule has: n = 3 (2 single bonds and 1 double bond) l = 0 (N has no lone pairs) So, 3 + 0 = 3 would indicate that the hybridization of the orbitals around the N atom is sp2.

Step 5: Count sigma and pi bonds

Sigma bonds are single bonds, and pi bonds are additional bonds in double and triple bonds. Sigma bonds in HNO2: - 1 bond between H and N - 1 bond between N and O (single bond) - 1 bond between N and O (double bond) Total sigma bonds = 3 Pi bonds in HNO2: - 1 bond between N and O (double bond) Total pi bonds = 1 In summary: a) Molecular formula: HNO2 b) Lewis structure: O = N - O / H (O double bonded to N, O single bonded to N, H single bonded to N) c) Geometry: Trigonal planar d) Hybridization of N: sp2 e) Sigma bonds: 3, Pi bonds: 1