HELP on how to solve a Gas Chemistry Problem?

Question

A gas consisting of only carbon and hydrogen has an empirical forumla of CH2. The gas has a density of 1.65g/L at 27C and 734mmHg. Determine the molar mass and molecular formula of the gas.

Answer 1

I don't know about Your Numbers, but I have Built a Device that Safely Converts Regular Pump Gasoline, set up to run Vehicle Engines at 14.7 Parts of Air to 1 Part of Fuel, into what is, for all Intents and Purposes, Propane. 100 Parts of Air to 1 Part of Fuel.The Proof is Reflected in My Avatar. In Fact, I've Lit it on Fire to Prove it Works. Compare the Color Spread of the Flame to that of a Butane Lighter Flame. Go to My 360 for even more info.

Answer 2

Start from the ideal gas state equation:

p*V = n*R*T

Now n = m/Mr (n: number of moles, m: mass, Mr: molar mass) and m = d*V (d: density). Substituting we get:

p = d*R*T/Mr => Mr = d*R*T/p

p = 734 mmHg = 734/760 atm, R = 0.082 L*atm/mol*K, T = 27 + 273 = 300 K

Mr = 1.65*0.082*300/(734/760) = 42

The empirical formula of the hydrocarbon is (CH2)x and it's molar mass:

Mr = (12 + 2*1)x = 14x (where 12 the atomic mass of C and 1 the atomic mass of H). So, we have the equation:

14x = 42 => x = 3.

So, the molecular formula of the substance is: C3H6 (propene or cyclopropane).

Answer 3

MUCH simpler than you may first think:

1.65 g/l X 22.4 l/mol = about 37 g/mol

MW of CH2 is 14 g/mol

therefore, you must multiply by 2.5 to get approximately 37 g/mol.

thus you get C2.5 H5

Since you cannot use fractional parts as coefficients, multiple everything by 2, and get

C5 H10.

The slight difference between the 35 g/mol and the value of 37 g/mol is due to the pressure difference (the 734 mm Hg is NOT at STP, but it won't make much difference.)

Therefore I bet the answer is C5H10

Answer 4

By PV = nRT
and n = Number of mole = g/MW
where g = gram of gas
MW = molecular weight of gas
so PV = gRT/MW or MW = gRT/PV
let D is density of gas = g/V
thus MW = DRT/P
Convert P to atm P = 734/760 = 0.96 atm
MW = 1.65 g/L * 300 K * 0.082 L.atm/mole.K / 0.96 atm
= 42.6 g/mole
Empirical formula is (CH2)n = 14n = 42.6
So n = 3
Answer: Molar mass is 42.6 g/mole
Molecular formula is C3H6

Answer 5

the quantity of a gas relies upon on the temperature and the rigidity by way of fact the rigidity drops the quantity will improve. Temperature this is not changing, so which you will use: P1V1 = P2V2 subsequently: V1 x (P1/P2) = V2; 2.50L x (a million.15/a million.00) = 2.875L