When 6.114 grams of a hydrocarbon, cxhy, were burned in a combustion analysis apparatus, 20.67 grams of co2 and 4.231 grams of h2o were produced. in a separate experiment, the molar mass of the compound was found to be 26.04 g/mol. determine the empirical formula and the molecular formula of the hydrocarbon.

Molecular formula : C₂H₂

Empirical formula H-C≡C-H

Explanation:

First of all, think the equation

CₓHy  + O₂  →  CO₂  +  H₂O

We have produced grams, of products, so we can find those moles.

Mass / molar weight = 20.67 g/ 44 g/m = 0.469 mol CO₂

Mass / molar weight   = 4.231 g/ 18g/m = 0.235 mol  H₂O

6.411 of a hydrocargon / 26.04 g/m = 0.246 mol

Now that we have the moles, we can work by thinking the relation 1:1

0.246 mol CxHy ___ produce __ 0.469 mol

1 mol CxHy __ produce ___ (0.469 /0.246) = 1.90 mol (almost 2)

0.246 mol CxHy___produce __ 0.235 mol

1 mol CxHy __ produce ___ (0.235/0.246) = 0.95 (almost 1)

Molar weight C = 12g/m

Molar weight H = 1g/m

Now we can write the equation as this:

CₓHy  + O₂  →  2CO₂  + H₂O

We have 2 carbons on product side, 5 O and 2 H. So in reactant side, we can add 2C to the hydrocarbon, so if the molar mass is 26.04 g/m and we have 2 carbons (24 g/m), H must be 2. The molecular formula for the hydrocarbon must be C2H2. Hydrocarbon's name is acetylene

Finally we have to ballance the oxigen.

C₂H₂ + 5/2 O₂ →  2CO₂  + H₂O

sodium (na), for example, has one electron in its valence shell. this is an unstable state because that valence shell needs eight electrons to be full. in order to fill its valence shell, find a way to add seven electrons to its valence shell