In: Chemistry
A compound used in glue has a composition of 54.5% C, 9.1% H, and 36.4% O
At 100 Degrees C, and 1.00 atm, the compound is a gas and 0.345 g occupies a volume of 120.0mL
What is the molecular formula of the compound?
step 1: find empirical formula
we have mass of each elements as:
C: 54.5 g
H: 9.1 g
O: 36.4 g
Divide by molar mass to get number of moles of each:
C: 54.5/12.01 = 4.5379
H: 9.1/1.008 = 9.0278
O: 36.4/16.0 = 2.275
Divide by smallest to get simplest whole number ratio:
C: 4.5379/2.275 = 2
H: 9.0278/2.275 = 4
O: 2.275/2.275 = 1
So empirical formula is:C₂H₄O
step 2:
find molar mass
we have:
P = 1.0 atm
V = 120.0 mL
= (120.0/1000) L
= 0.12 L
T = 100.0 oC
= (100.0+273) K
= 373 K
find number of moles using:
P * V = n*R*T
1 atm * 0.12 L = n * 0.08206 atm.L/mol.K * 373 K
n = 3.919*10^-3 mol
mass(solute)= 0.345 g
we have below equation to be used:
number of mol = mass / molar mass
3.919*10^-3 mol = (0.345 g)/molar mass
molar mass = 88.04 g/mol
step 3: find molecular formula
Molar mass of C2H4O = 2*MM(C) + 4*MM(H) + 1*MM(O)
= 2*12.01 + 4*1.008 + 1*16.0
= 44.052 g/mol
Now we have:
Molar mass = 88.04 g/mol
Empirical formula mass = 44.052 g/mol
Multiplying factor = molar mass / empirical formula mass
= 88.04/44.052
= 2
So molecular formula is:C₄H₈O₂