**Title: Determining the Molecular Formula and Lewis Structure of a Compound****Objective:**To determine the molecular formula of a given compound based on its elemental composition and gas properties, and to illustrate its Lewis structure.**Task Description:**A 0.02847 g sample of gas occupies 10.0 mL at 292.0 K and 1.10 atm. Upon further analysis, the compound is found to be composed of 38.734% Carbon (C) and 61.266% Fluorine (F).**Question:**What is the molecular formula of the compound?**Input Section:**- Molecular Formula: [__________]**Instructions for Drawing the Lewis Structure:****Tool Options:**- Select: Choose your tool from the menu.- Draw: Use single, double, or triple bond lines for drawing. - Single bond: - Double bond: - Triple bond: - Elements available for selection: - C (Carbon) - Br (Bromine) - F (Fluorine) - Cl (Chlorine)- Additional Options: - Undo: - Redo: - Zoom In/Out: **Explanation:**The task involves calculating the empirical formula based on the percentage composition and then determining the molecular formula given the gas properties. The Lewis structure should depict the atom connectivity and any lone pairs necessary for the complete description of the molecule. Use the tools to draw and modify the Lewis structure as needed.

Recall ideal gas equation PV=nRTAs moles of gas(n)=mass of gasmmolar mass of gas(M) PV=mRTM M=mRTPVGiven that, mass of gas(m)=0.02847 g Molar gas constant(R)=0.08206 L.atm.mol-1.K-1 Temperature(T)=292.0 K Pressure of gas(P)=1.10 atm Volume of gasV=10.0 mL =10.0 mL×1 L1000 mL =0.01 LSubstitute above expression with given values M=0.02847 g×0.08206 L.atm.mol-1.K-1×292.0 K1.10 atm×0.01 LM=62.02 g/molFind empirical formula of the compound Ratio of number of atoms of C:F=percentage of Catomic weight of C:percentage of Fatomic weight of F =38.734%12.01:61.266%19 =3.225:3.225 =1:1Empirical formula of the compound is CFMolecular mormula of the compound is (CF)n where, n is atomicityNow molecular weight=(12.01 + 19)×n=31.01n Compare this expression of molecular weight with actual molecular weight of compound 31.01n=62.02n=62.0231.01n=2 Molecular formula=CF2=C2F2 Therefore molecular formula of the compound is C2F2

Answered: A 0.02847 g sample of gas occupies 10.0…

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A 0.02847 g sample of gas occupies 10.0 mL at 292.0 K and 1.10 atm. Upon further analysis, the compound is found to be 38.734% C and 61.266% F. What is the molecular formula of the compound?

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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