Question

For the gas propane (C₃H₈), (a) how many grams are in 1.50 moles of C₃H₈, (b) how many moles are in 25.0 grams of C₃H₈, (c) how many molecules of C₃H₈ are in 25.0 moles of C₃H₈ and finally (d) what is the percent carbon in C₃H₈?

Answer 1

a)
Molar mass of C3H8,
MM = 3*MM(C) + 8*MM(H)
= 3*12.01 + 8*1.008
= 44.094 g/mol

use:
mass of C3H8,
m = number of mol * molar mass
= 1.5 mol * 44.09 g/mol
= 66.14 g
Answer: 66.1 g

b)
Molar mass of C3H8,
MM = 3*MM(C) + 8*MM(H)
= 3*12.01 + 8*1.008
= 44.094 g/mol


mass(C3H8)= 25.0 g

use:
number of mol of C3H8,
n = mass of C3H8/molar mass of C3H8
=(25 g)/(44.09 g/mol)
= 0.567 mol
Answer: 0.567 mol

c)
use:
number of molecules = number of mol * Avogadro’s number
number of molecules = 25 * 6.022*10^23 molecules
number of molecules = 1.506*10^25 molecules
Answer: 1.51*10^25 molecules

d)

Molar mass of C3H8,
MM = 3*MM(C) + 8*MM(H)
= 3*12.01 + 8*1.008
= 44.094 g/mol


MM of C3H8 = 44.094 g/mol
Lets calculate the mass of element

mass of C = 3*MM(C)
= 36.03 g

Mass % of C = (36.03*100)/44.094
= 81.7118 %
Answer: 81.7 %