Answer:
C3H8
Explanation:
Step 1:
Data obtained from the question. This includes the following:
Mass of alkane = 1.17g
Volume (V) = 674 mL
Temperature (T) = 28°C
Pressure (P) = 741 mmHg.
Gas constant (R) = 0.08206 atm.L/Kmol
Step 2:
Conversion to appropriate unit.
For Volume:
1000mL = 1L
Therefore, 674mL = 674/1000 = 0.674L
For Temperature:
Temperature (Kelvin) = Temperature (celsius) + 273
Temperature (celsius) = 28°C
Temperature (Kelvin) = 28°C + 273 = 301K
For Pressure:
760mmHg = 1atm
Therefore, 741 mmHg = 741/760 = 0.975atm
Step 3:
Determination of the number of mole of the alkane..
The number of mole of the alkane can be obtained by using the ideal gas equation. This is illustrated below:
Volume (V) = 0.674L
Temperature (T) = 301K
Pressure (P) = 0.975atm
Gas constant (R) = 0.08206 atm.L/Kmol
Number of mole (n) =?
PV = nRT
n = PV /RT
n = (0.975 x 0.674)/(0.08206x301)
n = 0.0266 mole
Step 4:
Determination of the molar mass of the alkane.
Mass of alkane = 1.17g
Number of mole of the alkane = 0.0266mole
Molar Mass of the alkane =?
Number of mole = Mass/Molar Mass
Molar Mass = Mass/number of mole
Molar Mass of the alkane = 1.17/0.0266 = 44g/mol
Step 5:
Determination of the molecular formula of the alkane.
This is illustrated below:
The general formula for the alkane is CnH2n+2
To obtain the molecular formula for the alkane we shall assume n = 1, 2, 3 or more till we arrive at molar Mass of 44.
When n = 1
CnH2n+2 = CH4 = 12 + (4x1) = 16g/mol
When n = 2
CnH2n+2 = C2H6 = (12x2) + (6x1) = 30g/mol
When n = 3
CnH2n+2 = C3H8 = (12x3) + (8x1) = 44g/mol
We can see that when n is 3, the molar mass is 44g/mol.
Therefore, the molecular formula for the alkane is C3H8.
