A chemist vaporizes a volatile liquid hydrocarbon (carbon, hydrogen and oxygen) compound to a gas and determines the density (6v g/L) of the gas to be 3.277 g/L at a temperature of 98.4°C and a pressure of 743.2 mmHg. 1. Use the Ideal Gas Law (PV = nRT ) to calculate the molar mass (MM~ g/mol) of the volatile hydrocarbon compound. 2. If the volatile hydrocarbon contains (om/m): 7o.531% carbon (C) and 13.811% hydrogen (H), what is the empirical formula (EF) and the molecular formula (MF) of the hydrocarbon compound (CxHy02)?

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Equations and Constants:
PIV1
P:V2
---- =
----
T1
T2
K = °C + 273.15
Vmalar = 22.41 L/mol
PV = nRT
R = 0.082057 (L-Atm)/(mol-K);
1 atm = 760 mmHg
n= m/MM
n = mols of gas, m = mass of gas, MM = molar mass of gas
density (d) = m/V
V = volume of gas (L)
A chemist vaporizes a volatile liquid hydrocarbon (carbon, hydrogen and oxygen) compound to a gas and determines the density (6~ g/L) of the gas to be 3.277 g/L at a temperature of 98.4°C and a pressure of 743.2 mmHg.
1. Use the Ideal Gas Law (PV = nRT ) to calculate the molar mass (MM~ g/mol) of the volatile hydrocarbon compound.
2. If the volatile hydrocarbon contains (%m/m): 70.531% carbon (C) and 13.811% hydrogen (H), what is the empirical formula (EF) and the molecular formula (MF) of the hydrocarbon compound (CxHy02)?
Data:
& g/L) =
Patm =
Tk =
% C =
% H =
% 0 =
1. The Pressure in Atmospheres: Select
Atm
2. The Temperature in kelvin: Seloct)
v k
3. Molar Mass (MM v g/ mol) of the Hydrocarbon Compound:
( Sekct |
g/ mol
4. The %(m/m) of Oxygen in the Compound:
I Sckct |
v %(m/m)
5. Mols of Oxygen in 100.000 grams of the compound:
(Sekect
|
mol O
6. Mols of Carbon in 100.000 grams of the compound:
(Select)
mol C
7. Mols of Hydrogen in 100.000 grams of the compound:
( Sclect )
v mol H
8. The Ratio of Carbon : Hydrogen : Oxygen atoms:
( Sckct
v C:H:0
9. The Empirical Formula of the Hydrocarbon Compound:
Formula A: CH002
Formula B: C3HO2
Formula C: CH140
Formula D: CSH1201
Formula: (Select |
10. The Molecular Formula of the Hydrocarbon Compound:
Formula A: Cı2H2002
Formula B: CH1601
Formula C: CsH;02
Formula D: CoH1101
Formula: (Select)
Transcribed Image Text:Equations and Constants: PIV1 P:V2 ---- = ---- T1 T2 K = °C + 273.15 Vmalar = 22.41 L/mol PV = nRT R = 0.082057 (L-Atm)/(mol-K); 1 atm = 760 mmHg n= m/MM n = mols of gas, m = mass of gas, MM = molar mass of gas density (d) = m/V V = volume of gas (L) A chemist vaporizes a volatile liquid hydrocarbon (carbon, hydrogen and oxygen) compound to a gas and determines the density (6~ g/L) of the gas to be 3.277 g/L at a temperature of 98.4°C and a pressure of 743.2 mmHg. 1. Use the Ideal Gas Law (PV = nRT ) to calculate the molar mass (MM~ g/mol) of the volatile hydrocarbon compound. 2. If the volatile hydrocarbon contains (%m/m): 70.531% carbon (C) and 13.811% hydrogen (H), what is the empirical formula (EF) and the molecular formula (MF) of the hydrocarbon compound (CxHy02)? Data: & g/L) = Patm = Tk = % C = % H = % 0 = 1. The Pressure in Atmospheres: Select Atm 2. The Temperature in kelvin: Seloct) v k 3. Molar Mass (MM v g/ mol) of the Hydrocarbon Compound: ( Sekct | g/ mol 4. The %(m/m) of Oxygen in the Compound: I Sckct | v %(m/m) 5. Mols of Oxygen in 100.000 grams of the compound: (Sekect | mol O 6. Mols of Carbon in 100.000 grams of the compound: (Select) mol C 7. Mols of Hydrogen in 100.000 grams of the compound: ( Sclect ) v mol H 8. The Ratio of Carbon : Hydrogen : Oxygen atoms: ( Sckct v C:H:0 9. The Empirical Formula of the Hydrocarbon Compound: Formula A: CH002 Formula B: C3HO2 Formula C: CH140 Formula D: CSH1201 Formula: (Select | 10. The Molecular Formula of the Hydrocarbon Compound: Formula A: Cı2H2002 Formula B: CH1601 Formula C: CsH;02 Formula D: CoH1101 Formula: (Select)
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