Question

A 0.04358 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 25.305% c and 74.695% cl. what is the molecular formula of the compound?

Answer 1

PV = mRT: gas law.

M = $\frac{n}{m}$ : molecular mass

Use: M = $\frac{mRT}{PV}$

Where

m = 0.04343 g

P = 1.10 atm

V = 0.0100 L

R = 0.08205746 (14) L atm K  mole  

T = 293.0 K

M: the molecular mass of the gas.

Then take the mass of this molecule, the molecular mass of the atom (C & Cl), and also the proportion of mass given to search out the amount of moles per mole of gas.

Further explanation

The chemical formula is a component of a formula, that has associate understanding of a substance that determines the kind and a relative range of atoms that area unit in this substance. Or in different languages, a formula are often given that contains data concerning the atoms that form up a selected chemical composition.

Chemical formula is split into 2 specifically chemical formula and formula. The formula announces that the formula mentioned will prove the molecules of the mix atoms. whereas the formula is that the original formula of a composition. The chemical formula are often determined if the relative molecular mass is thought.

Learn more

Gas molecular calculations brainly.com/question/6250579

Molecular Mass brainly.com/question/7233727

Details

Class: highschool

Subject: Chemistry

Keywords: Mole, Formula, Mass

Answer 2

Answer: A 0.04403 g sample of gas occupies 10.0-mL at 289.0 K and 1.10 atm. Upon further analysis, the compound is found to be 25.305% C and 74.695% Cl. What is the molecular formula of the compound? --------------------------------------... Seems like I did a problem very similar to this--this must be the "B" test. But the halogen was different. 25.305% C/12 = 2.108 74.695% Cl/35.5 = 2.104 So the empirical formula would be CH. However, there are many compounds which fit this bill, so we have to use the gas data. (And I made, in the previous problem, the simplifying assumption that 289C and 1.10 atm would offset each other, so I'll do that, too.) 0.044 grams/10 ml = x/22.4 liters 0.044g/0.010 liters = x/22.4 liters 22.4 liters/0.010 liters = 2240 (ratio) 2240 x .044 = 98.56 (actual atomic weight) CCl = 35.5+12 or 47.5, so two of those is 95 grams/mole. This is sufficiient to distinguish C2CL2, (dichloroacetylene) from C6CL6 (hexachlorobenzene) which would mass 3 times as much.