A liquid hydrocarbon (Cx Hy) is found to be 16.37% H by mass. A 1.158-g vaporized sample of the hydrocarbon has a 358 mL volume at 71.0 C and 749 mmHg. what is the molecular formula for this hydrocarbon?
Calculate the Empirical Formula given 16.37% H and (100-16.37)% C. Refer to Example #2.
http://www.chem.tamu.edu/class/majors/tutorialnotefiles/empirical.htm
Use PV=nRT.
n=mass/molar mass
units: P=atm, V=L, T=K
R=0.08206 atm*L/mol*k
Solve for molar mass and calculate the Molecular Formula.
http://chemistry.about.com/od/workedchemistryproblems/a/Molecular-Formula-Simplest-Formula-Example-Problem-1.htm
To determine the molecular formula of the hydrocarbon, we need to use the given information about the percent composition of hydrogen (H) and the experimental data of the vaporized sample's mass, volume, temperature, and pressure.
Let's start by calculating the molar mass of the hydrocarbon:
1. Calculate the number of moles of the vaporized sample:
- Convert the volume of the vaporized sample from mL to L:
358 mL ÷ 1000 mL/L = 0.358 L
- Convert the pressure from mmHg to atm:
749 mmHg ÷ 760 mmHg/atm = 0.985 atm
- Use the ideal gas law (PV = nRT) to calculate the moles:
n = PV / RT
where P is the pressure in atm, V is the volume in L, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin (K).
n = (0.985 atm) * (0.358 L) / [(0.0821 L·atm/mol·K) * (71.0°C + 273.15)]
n = 0.01575 mol
2. Calculate the mass of hydrogen (H) in the vaporized sample:
- Given that the hydrocarbon is 16.37% hydrogen by mass, we can calculate the mass of hydrogen in the sample:
Mass of H = 0.1637 * 1.158 g = 0.1894 g
3. Calculate the moles of hydrogen (H) in the sample:
- Use the molar mass of hydrogen (1.008 g/mol) to convert the mass to moles:
Moles of H = Mass of H / Molar mass of H
= 0.1894 g / 1.008 g/mol
≈ 0.188 mol
4. Determine the moles of carbon (C) in the sample:
- Calculate the moles of carbon in the hydrocarbon by subtracting the moles of hydrogen from the total moles of the sample:
Moles of C = Total moles of sample - Moles of H
= 0.01575 mol - 0.188 mol
≈ -0.173 mol
Notice that we obtained a negative value for the moles of carbon, which indicates an error in the calculation.
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