Question

A 0.08541 g sample of gas occupies 10.0-ml at 288.5 k and 1.10 atm. upon further analysis, the compound is found to be 13.068% c and 86.932% br. what is the molecular formula of the compound? (order: c, x)

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

C2Br2 First, we need to determine how many moles of the gas we have. For that, we'll use the Ideal Gas Law which is PV = nRT where P = pressure (1.10 atm = 111458 Pa) V = volume (10.0 ml = 0.0000100 m^3) n = number of moles R = Ideal gas constant (8.3144598 (m^3 Pa)/(K mol) ) T = Absolute temperature Solving for n, we get PV/(RT) = n Now substituting our known values into the formula. (111458 Pa * 0.0000100 m^3) / (288.5 K * 8.3144598 (m^3 Pa)/(K mol)) = (1.11458/2398.721652) mol = 0.000464656 mol Now let's calculate the empirical formula for this compound. Atomic weight carbon = 12.0107 Atomic weight bromine = 79.904 Relative moles carbon = 13.068 / 12.0107 = 1.08802984 Relative moles bromine = 86.932 / 79.904 = 1.087955547 So the relative number of atoms of the two elements is 1.08802984 : 1.087955547 After dividing all numbers by the smallest, the ratio becomes 1.000068287 : 1 Which is close enough to 1:1 for me to consider the empirical formula to be CBr Now calculate the molar mass of CBr 12.0107 + 79.904 = 91.9147 Finally, let's determine if the compound is actually CBr, or something like C2Br2, or some other multiple. Using the molar mass of CBr, multiply by the number of moles and see if the result matches the mass of the gas. So 91.9147 g/mol * 0.000464656 mol = 0.042708701 g 0.0427087 g is a lot smaller than 0.08541 g. So the compound isn't exactly CBr. Let's divide them to see what the factor is. 0.08541 / 0.0427087 = 1.99982673 1.99982673 is close enough to 2 to within the number of significant digits we have for me to claim that the formula for the unknown gas isn't CBr, but instead is C2Br2.